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Binary Calculator | Online Tool for Binary Arithmetic | SEO Tools Audit
Binary Calculator
Use our Binary Calculator for quick and precise binary arithmetic. Whether you're a professional or new to binary math, our user-friendly tool simplifies addition, subtraction, multiplication,
division, and logic operations. Enhance accuracy, save time, and enjoy privacy—our calculator doesn't store any data. Simplify your binary math tasks with us today!
There is a need to understand the binary numeral system because everything from text to pictures is processed and stored as a series of 0s and 1s. It serves as the framework for all computers and
other digital devices.
Binary numbers can be used for arithmetic operations, however, they can be time-consuming. The Binary Calculator can help in this situation by providing a seamless and effective method of managing
binary computations.
What is a Binary Calculator?
A binary calculator is an online tool that makes simple mathematical operations on binary numbers. When working with binary numbers, the binary calculator tool is essential, especially for performing
complex operations like binary multiplication.
Even if you have no previous knowledge of dealing with binary calculations, you can easily do it with the binary calculator because it has a beginner-friendly interface. And you can be assured you
will get quick and precise results, saving time and effort in the process.
Benefits and Features of Binary Calculator
Here are reasons to consider using the binary calculator
Various Binary Calculations
The binary calculator supports different operations including addition, subtraction, multiplication, division, AND, OR, NOT, XOR, left shift, right shift, and zero fill right shift. These tools are
adaptable for many jobs since they can handle both fundamental and sophisticated binary operations.
It is well known that binary multiplication is difficult, especially for long binary strings. This process is made simpler by binary calculators, which provide prompt and precise results. They also
make binary division easier, which is a crucial process in computer architecture.
Don’t be worried about getting ‘errors’ instead of results as the binary calculators can handle binary values of various lengths and still give accurate results. This comes in handy when working with
digital circuits or data manipulation jobs.
These calculators are used for binary logic, such as Boolean algebra or bitwise computations. They make it simple to carry out logical operations such as AND (conjunction), OR (disjunction), NOT
(negation), and XOR (exclusive OR).
It also helps when dealing with binary-encoded data, such as flags or conditions in programming.
Binary calculators enable bit shifting, a fundamental operation in computer science.
Binary digits are shifted left or right in binary code, while zero-fill right shift adds zeros during right shifts. The effective handling of binary data and code optimization depend on these
Saves Time
One of the main benefits of binary calculators is their ability to speed up and improve accuracy in binary computations.
Manually performing binary calculations, particularly multiplication and division, can be laborious. Binary calculators quickly complete these processes, allowing you to concentrate on solving
problems rather than laborious calculations.
Manual binary calculations are prone to mistakes, especially when dealing with long binary sequences. A single error can produce inaccurate outcomes. By automating the procedure, binary calculators
reduce this risk and produce accurate results every time.
Binary calculators' convenience is valued by users. They guarantee easy access and privacy protection because they don't require registration or personal information.
These resources are easily accessible online without the need to register or provide personal information. Users are encouraged to utilize the calculators whenever necessary, whether for academic,
professional, or personal reasons, thanks to their accessibility.
Privacy and Security
Tools for binary calculation do not save or maintain any user data. Input values are immediately removed after the calculations are finished. This dedication to privacy gives users confidence in the
tool by assuring them that their data is secure.
Conversion Capabilities
Many binary calculators provide flexible conversion settings that let users convert between various number systems
Many binary calculators can convert binary values to decimals and vice versa in addition to performing binary operations. When you need to express binary results in a more traditional decimal style,
this functionality is helpful.
Some binary calculators, including SEOToolsaudit’s binary calculator, also make it simple to convert binary integers to hexadecimal. This is especially helpful in hexadecimal notation-heavy fields
like computing and digital design.
Cross-Platform Accessibility
Typically, binary calculators are accessible online and on a variety of platforms and devices.
Regardless of the device's operating system, users can use web browsers on PCs, tablets, and smartphones to access binary calculators. Due to its adaptability, binary operations can be carried out
anywhere and on any device.
How to use the Binary Calculator Tool?
Using a Binary Calculator is straightforward:
1. Go to the Binary Calculator tool page on the SEOToolsaudit site.
2. Type the first binary number in the first field.
3. Enter the second number in the second field.
4. Choose the desired operation (e.g., addition, subtraction, etc.).
5. Click on the "Calculate" button.
The results will be in binary, decimal, and hexadecimal formats. To start new calculations, click on "reset" to clear the input values.
The binary calculator proves to be a useful tool. It maintains user privacy while streamlining complicated binary processes, saving time, and ensuring accuracy. This tool is essential for anyone
working with binary data, whether they are professionals or students doing binary math issues.
Anyone working with binary numbers in today's digital environment should use it because of its simplicity, versatility, and accessibility.
Frequently Asked Questions
Why use a binary calculator?
It can be very stressful to do mathematical operations on long binary data. Binary calculators make things much simpler and relieve you of the headache.
Can I use the binary calculator in programming?
Yes. The binary calculator supports some operations that are useful in programming and it also works with binary-encoded data.
How does a binary calculator work?
The results for the numbers you enter into the binary calculator are given using robust calculation algorithms for fundamental arithmetic operations.
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Want to calculate the PDF (Probability Density Function) of a distribution? No problem, just use:
pdf(my_dist, x); // Returns PDF (density) at point x of distribution my_dist.
Or how about the CDF (Cumulative Distribution Function):
cdf(my_dist, x); // Returns CDF (integral from -infinity to point x)
// of distribution my_dist.
And quantiles are just the same:
quantile(my_dist, p); // Returns the value of the random variable x
// such that cdf(my_dist, x) == p.
If you're wondering why these aren't member functions, it's to make the library more easily extensible: if you want to add additional generic operations - let's say the n'th moment - then all you
have to do is add the appropriate non-member functions, overloaded for each implemented distribution type.
Random numbers that approximate Quantiles of Distributions
If you want random numbers that are distributed in a specific way, for example in a uniform, normal or triangular, see Boost.Random.
Whilst in principal there's nothing to prevent you from using the quantile function to convert a uniformly distributed random number to another distribution, in practice there are much more
efficient algorithms available that are specific to random number generation.
For example, the binomial distribution has two parameters: n (the number of trials) and p (the probability of success on any one trial).
The binomial_distribution constructor therefore has two parameters:
binomial_distribution(RealType n, RealType p);
For this distribution the random variate is k: the number of successes observed. The probability density/mass function (pdf) is therefore written as f(k; n, p).
Random Variates and Distribution Parameters
The concept of a random variable is closely linked to the term random variate: a random variate is a particular value (outcome) of a random variable. and distribution parameters are conventionally
distinguished (for example in Wikipedia and Wolfram MathWorld) by placing a semi-colon or vertical bar) after the random variable (whose value you 'choose'), to separate the variate from the
parameter(s) that defines the shape of the distribution.
For example, the binomial distribution probability distribution function (PDF) is written as f(k| n, p) = Pr(K = k|n, p) = probability of observing k successes out of n trials. K is the random
variable, k is the random variate, the parameters are n (trials) and p (probability).
By convention, random variate are lower case, usually k is integral, x if real, and random variable are upper case, K if integral, X if real. But this implementation treats all as floating point
values RealType, so if you really want an integral result, you must round: see note on Discrete Probability Distributions below for details.
As noted above the non-member function pdf has one parameter for the distribution object, and a second for the random variate. So taking our binomial distribution example, we would write:
pdf(binomial_distribution<RealType>(n, p), k);
The ranges of random variate values that are permitted and are supported can be tested by using two functions range and support.
The distribution (effectively the random variate) is said to be 'supported' over a range that is "the smallest closed set whose complement has probability zero". MathWorld uses the word 'defined' for
this range. Non-mathematicians might say it means the 'interesting' smallest range of random variate x that has the cdf going from zero to unity. Outside are uninteresting zones where the pdf is
zero, and the cdf zero or unity.
For most distributions, with probability distribution functions one might describe as 'well-behaved', we have decided that it is most useful for the supported range to exclude random variate values
like exact zero if the end point is discontinuous. For example, the Weibull (scale 1, shape 1) distribution smoothly heads for unity as the random variate x declines towards zero. But at x = zero,
the value of the pdf is suddenly exactly zero, by definition. If you are plotting the PDF, or otherwise calculating, zero is not the most useful value for the lower limit of supported, as we
discovered. So for this, and similar distributions, we have decided it is most numerically useful to use the closest value to zero, min_value, for the limit of the supported range. (The range remains
from zero, so you will still get pdf(weibull, 0) == 0). (Exponential and gamma distributions have similarly discontinuous functions).
Mathematically, the functions may make sense with an (+ or -) infinite value, but except for a few special cases (in the Normal and Cauchy distributions) this implementation limits random variates to
finite values from the max to min for the RealType. (See Handling of Floating-Point Infinity for rationale).
Discrete Probability Distributions
Note that the discrete distributions, including the binomial, negative binomial, Poisson & Bernoulli, are all mathematically defined as discrete functions: that is to say the functions cdf and pdf
are only defined for integral values of the random variate.
However, because the method of calculation often uses continuous functions it is convenient to treat them as if they were continuous functions, and permit non-integral values of their parameters.
Users wanting to enforce a strict mathematical model may use floor or ceil functions on the random variate prior to calling the distribution function.
The quantile functions for these distributions are hard to specify in a manner that will satisfy everyone all of the time. The default behaviour is to return an integer result, that has been rounded
outwards: that is to say, lower quantiles - where the probability is less than 0.5 are rounded down, while upper quantiles - where the probability is greater than 0.5 - are rounded up. This
behaviour ensures that if an X% quantile is requested, then at least the requested coverage will be present in the central region, and no more than the requested coverage will be present in the
This behaviour can be changed so that the quantile functions are rounded differently, or return a real-valued result using Policies. It is strongly recommended that you read the tutorial
Understanding Quantiles of Discrete Distributions before using the quantile function on a discrete distribution. The reference docs describe how to change the rounding policy for these
For similar reasons continuous distributions with parameters like "degrees of freedom" that might appear to be integral, are treated as real values (and are promoted from integer to floating-point
if necessary). In this case however, there are a small number of situations where non-integral degrees of freedom do have a genuine meaning. | {"url":"https://live.boost.org/doc/libs/1_86_0/libs/math/doc/html/math_toolkit/stat_tut/overview/generic.html","timestamp":"2024-11-10T09:48:05Z","content_type":"text/html","content_length":"19160","record_id":"<urn:uuid:0f43c50d-21c6-4c8f-bdeb-ad514d6d2df5>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00550.warc.gz"} |
Neural networks: Nodes and hidden layers | Machine Learning | Google for Developers
To build a neural network that learns nonlinearities, begin with the following familiar model structure: a linear model of the form $y' = b + w_1x_1 + w_2x_2 + w_3x_3$.
We can visualize this equation as shown below, where $x_1$, $x_2$, and $x_3$ are our three input nodes (in blue), and $y'$ is our output node (in green).
Exercise 1
In the model above, the weight and bias values have been randomly initialized. Perform the following tasks to familiarize yourself with the interface and explore the linear model. You can ignore the
Activation Function dropdown for now; we'll discuss this topic later on in the module.
1. Click the Play (▶️) button above the network to calculate the value of the output node for the input values $x_1 = 1.00$, $x_2 = 2.00$, and $x_3 = 3.00$.
2. Click the second node in the input layer, and increase the value from 2.00 to 2.50. Note that the value of the output node changes. Select the output nodes (in green) and review the Calculations
panel to see how the output value was calculated.
3. Click the output node (in green) to see the weight ($w_1$, $w_2$, $w_3$) and bias ($b$) parameter values. Decrease the weight value for $w_3$ (again, note that the value of the output node and
the calculations below have changed). Then, increase the bias value. Review how these changes have affected the model output.
Adding layers to the network
Note that when you adjusted the weight and bias values of the network in Exercise 1, that didn't change the overall mathematical relationship between input and output. Our model is still a linear
But what if we add another layer to the network, in between the input layer and the output layer? In neural network terminology, additional layers between the input layer and the output layer are
called hidden layers, and the nodes in these layers are called neurons.
The value of each neuron in the hidden layer is calculated the same way as the output of a linear model: take the sum of the product of each of its inputs (the neurons in the previous network layer)
and a unique weight parameter, plus the bias. Similarly, the neurons in the next layer (here, the output layer) are calculated using the hidden layer's neuron values as inputs.
This new hidden layer allows our model to recombine the input data using another set of parameters. Can this help our model learn nonlinear relationships?
Exercise 2
We've added a hidden layer containing four neurons to the model.
Click the Play (▶️) button above the network to calculate the value of the four hidden-layer nodes and the output node for the input values $x_1 = 1.00$, $x_2 = 2.00$, and $x_3 = 3.00$.
Then explore the model, and use it to answer the following questions.
How many
(weights and biases) does this neural network model have?
Our original model in
Exercise 1
had four parameters: w
, w
, w
, and b. Because this model contains a hidden layer, there are more parameters.
Note that the total number of parameters includes both the parameters used to calculate the node values in the hidden layer from the input values, and the parameters used to calculate the output
value from the node values in the hidden layer.
Note that the total number of parameters includes both the weight parameters and the bias parameters.
There are 4 parameters used to calculate each of the 4 node values in the hidden layer—3 weights (one for each input value) and a bias—which sums to 16 parameters. Then there are 5 parameters used to
calculate the output value: 4 weights (one for each node in the hidden layer) and a bias. In total, this neural network has 21 parameters.
Try modifying the model parameters, and observe the effect on the hidden-layer node values and the output value (you can review the Calculations panel below to see how these values were calculated).
Can this model learn nonlinearities?
Click on each of the nodes in the hidden layer and the output node, and review the calculations below. What do you notice about all these calculations?
If you click on each of the nodes in the hidden layer and review the calculations below, you'll see that all of them are linear (comprising multiplication and addition operations).
If you then click on the output node and review the calculation below, you'll see that this calculation is also linear. Linear calculations performed on the output of linear calculations are also
linear, which means this model cannot learn nonlinearities. | {"url":"https://developers.google.cn/machine-learning/crash-course/neural-networks/nodes-hidden-layers?authuser=2","timestamp":"2024-11-06T18:32:51Z","content_type":"text/html","content_length":"176511","record_id":"<urn:uuid:9f504425-d343-4ccd-801e-5fd37a135914>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00725.warc.gz"} |
In her short life, mathematician Emmy Noether changed the face of physics
Noether linked two important concepts in physics: conservation laws and symmetries
Sam Falconer
On a warm summer evening, a visitor to 1920s Göttingen, Germany, might have heard the hubbub of a party from an apartment on Friedländer Way. A glimpse through the window would reveal a gathering of
scholars. The wine would be flowing and the air buzzing with conversations centered on mathematical problems of the day. The eavesdropper might eventually pick up a woman’s laugh cutting through the
din: the hostess, Emmy Noether, a creative genius of mathematics.
At a time when women were considered intellectually inferior to men, Noether (pronounced NUR-ter) won the admiration of her male colleagues. She resolved a nagging puzzle in Albert Einstein’s
newfound theory of gravity, the general theory of relativity. And in the process, she proved a revolutionary mathematical theorem that changed the way physicists study the universe.
It’s been a century since the July 23, 1918, unveiling of Noether’s famous theorem. Yet its importance persists today. “That theorem has been a guiding star to 20th and 21st century physics,” says
theoretical physicist Frank Wilczek of MIT.
Noether was a leading mathematician of her day. In addition to her theorem, now simply called “Noether’s theorem,” she kick-started an entire discipline of mathematics called abstract algebra.
E. Otwell
But in her career, Noether couldn’t catch a break. She labored unpaid for years after earning her Ph.D. Although she started working at the University of Göttingen in 1915, she was at first permitted
to lecture only as an “assistant” under a male colleague’s name. She didn’t receive a salary until 1923. Ten years later, Noether was forced out of the job by the Nazi-led government: She was Jewish
and was suspected of holding leftist political beliefs. Noether’s joyful mathematical soirees were extinguished.
She left for the United States to work at Bryn Mawr College in Pennsylvania. Less than two years later, she died of complications from surgery — before the importance of her theorem was fully
recognized. She was 53.
Although most people have never heard of Noether, physicists sing her theorem’s praises. The theorem is “pervasive in everything we do,” says theoretical physicist Ruth Gregory of Durham University
in England. Gregory, who has lectured on the importance of Noether’s work, studies gravity, a field in which Noether’s legacy looms large.
Making connections
Noether divined a link between two important concepts in physics: conservation laws and symmetries. A conservation law — conservation of energy, for example — states that a particular quantity must
remain constant. No matter how hard we try, energy can’t be created or destroyed. The certainty of energy conservation helps physicists solve many problems, from calculating the speed of a ball
rolling down a hill to understanding the processes of nuclear fusion.
Symmetries describe changes that can be made without altering how an object looks or acts. A sphere is perfectly symmetric: Rotate it any direction and it appears the same. Likewise, symmetries
pervade the laws of physics: Equations don’t change in different places in time or space.
Photo Archives/Special Collections Department/Bryn Mawr College Library
Noether’s theorem proclaims that every such symmetry has an associated conservation law, and vice versa — for every conservation law, there’s an associated symmetry.
Conservation of energy is tied to the fact that physics is the same today as it was yesterday. Likewise, conservation of momentum, the theorem says, is associated with the fact that physics is the
same here as it is anywhere else in the universe. These connections reveal a rhyme and reason behind properties of the universe that seemed arbitrary before that relationship was known.
During the second half of the 20th century, Noether’s theorem became a foundation of the standard model of particle physics, which describes nature on tiny scales and predicted the existence of the
Higgs boson, a particle discovered to much fanfare in 2012 (SN: 7/28/12, p. 5). Today, physicists are still formulating new theories that rely on Noether’s work.
When Noether died, Einstein wrote in the New York Times: “Noether was the most significant creative mathematical genius thus far produced since the higher education of women began.” It’s a hearty
compliment. But Einstein’s praise alluded to Noether’s gender instead of recognizing that she also stood out among her male colleagues. Likewise, several mathematicians who eulogized her remarked on
her “heavy build,” and one even commented on her sex life. Even those who admired Noether judged her by different standards than they judged men.
Symmetry leads the way
There’s something inherently appealing about symmetry (SN Online: 4/12/07). Some studies report that humans find symmetrical faces more beautiful than asymmetrical ones. The two halves of a face are
nearly mirror images of each other, a property known as reflection symmetry. Art often exhibits symmetry, especially mosaics, textiles and stained-glass windows. Nature does, too: A typical
snowflake, when rotated by 60 degrees, looks the same. Similar rotational symmetries appear in flowers, spider webs and sea urchins, to name a few.
But Noether’s theorem doesn’t directly apply to these familiar examples. That’s because the symmetries we see and admire around us are discrete; they hold only for certain values, for example,
rotation by exactly 60 degrees for a snowflake. The symmetries relevant for Noether’s theorem, on the other hand, are continuous: They hold no matter how far you move in space or time.
One kind of continuous symmetry, known as translation symmetry, means that the laws of physics remain the same as we move about the cosmos.
The conservation laws that relate to each continuous symmetry are basic tools of physics. In physics classes, students are taught that energy is always conserved. When a billiard ball thwacks
another, the energy of that first ball’s motion is divvied up. Some goes into the second ball’s motion, some generates sound or heat, and some energy remains with the first ball. But the total amount
of energy remains the same — no matter what. Same goes for momentum.
These rules are taught as rote facts, but there’s a mathematical reason behind their existence. Energy conservation, according to Noether, comes from translation symmetry in time. Similarly, momentum
conservation is due to translation symmetry in space. And conservation of angular momentum, the property that allows ice skaters to speed up their spins by hugging their arms close to their bodies,
emerges from rotational symmetry, the idea that physics stays the same as we spin around in space.
In Einstein’s general theory of relativity, there is no absolute sense of time or space, and conservation laws become more difficult to comprehend. It’s that complexity that brought Noether to the
topic in the first place.
Gravity gets Noether’d
In 1915, general relativity was a fascinating new theory. German mathematicians David Hilbert and Felix Klein, both at the University of Göttingen, were immersed in the new theory’s quirks. Hilbert
had been competing with Einstein to develop the mathematically complex theory, which describes gravity as the result of matter curving spacetime (SN: 10/17/15, p. 16).
But Hilbert and Klein stumbled on a puzzle. Attempts to use the framework of general relativity to write an equation for conservation of energy resulted in a tautology: Like writing “0 equals 0,” the
equation had no physical significance. This situation was a surprise to the pair; no previously accepted theories had energy conservation laws like this. The duo wanted to understand why general
relativity had this peculiar feature.
The two recruited Noether, who had expertise in relevant areas of mathematics, to join them in Göttingen and help them solve the riddle.
Noether showed that the seemingly strange type of conservation law was inherent to a certain class of theories known as “generally covariant.” In such theories, the equations associated with the
theory hold whether you’re moving steadily or accelerating wildly, because both sides of the theory’s equations change in sync. The result is that generally covariant theories — including general
relativity — will always have these nontraditional conservation laws. This discovery is known as Noether’s second theorem.
This is what Noether did best: fitting specific concepts into their broader mathematical context. “She was just able to see what’s right at the heart of what’s going on and to generalize it,” says
philosopher of science Katherine Brading of Duke University, who has studied Noether’s theorems.
On her way to proving the second theorem, Noether proved her first theorem, about the connection between symmetries and conservation laws. She presented both results in a July 23, 1918, lecture to
the Göttingen Mathematical Society, and in a paper published in Göttinger Nachrichten.
It’s not easy to find quotes of Noether reflecting on the significance of her work. Once she made a discovery, she seemed to move on to the next thing. She referred to her own Ph.D. thesis as “crap,”
or “Mist” in her native German. But Noether recognized that she changed mathematics: “My methods are really methods of working and thinking; this is why they have crept in everywhere anonymously,”
she wrote to a colleague in 1931.
“Warm like a loaf of bread”
Born in 1882, Noether (her full name was Amalie Emmy Noether) was the daughter of mathematician Max Noether and Ida Amalia Noether. Growing up with three brothers in Erlangen, Germany, young Emmy’s
mathematical talent was not obvious. However, she was known to solve puzzles that stumped other children.
At the University of Erlangen, where her father taught, women weren’t officially allowed as students, though they could audit classes with the permission of the professor. When the rule changed in
1904, Emmy Noether was quick to take advantage. She enrolled and earned her Ph.D. in 1907.
As a woman, Noether struggled to find a paid academic position, even after being recruited to the University of Göttingen. Her supporters there argued that her sex was irrelevant. “After all, we are
a university and not a bathing establishment,” Hilbert reportedly quipped. But that wasn’t enough to get her a salary.
Although Göttingen finally began paying Noether in 1923, she never became a full-fledged professor. Hermann Weyl, a prominent mathematician at the university, said, “I was ashamed to occupy such a
preferred position beside her whom I knew to be my superior as a mathematician in many respects.”
Noether took these knocks in stride. She was beloved for her buoyant personality. Weyl described her demeanor as “warm like a loaf of bread.”
She made a habit of taking long walks in the countryside with her students and colleagues, holding lengthy, math-fueled debates. When legs began to ache, Noether and company would plop down in a
meadow and continue chatting. Sometimes she’d take students to her apartment for homemade “pudding à la Noether,” conversing until remnants of the dessert had dried on the dishes, according to a 1970
biography, Emmy Noether 1882–1935, by mathematical historian Auguste Dick.
When she landed at Bryn Mawr, Noether continued her research and taught classes of women — a change of pace from her previous students, who were known as “the Noether boys.” She also lectured at the
Institute for Advanced Study in Princeton, N.J. Her death, less than two years after her 1935 arrival, left the academic community grieving.
Russian mathematician Pavel Aleksandrov called Noether “one of the most captivating human beings I have ever known,” and lamented the unfortunate circumstances of her employment. “Emmy Noether’s
career was full of paradoxes, and will always stand as an example of shocking stagnancy and inability to overcome prejudice,” he said in 1935 at a meeting of the Moscow Mathematical Society.
Elusive partners
But Noether’s theorems remained relevant, particularly within particle physics. In the minute, enigmatic world of fundamental particles, teasing out what’s going on is difficult. “We have to rely on
theoretical insight and concepts of beauty and aesthetics and symmetry to make guesses about how things might work,” Wilczek says. Noether’s theorems are a big help.
In particle physics, the relevant symmetries are hidden kinds known as gauge symmetries. One such symmetry is found in electromagnetism and results in the conservation of electric charge.
Gauge symmetry appears in the definition of electric voltage. A voltage — between two ends of a battery, for example — is the result of a difference in electric potential. The actual value of the
electric potential itself doesn’t matter, only the difference.
This creates a symmetry in electric potential: Its overall value can be changed without affecting the voltage. This property explains why a bird can sit on a single power line without getting
electrocuted, but if it simultaneously touches two wires at different electric potentials — bye-bye, birdie.
In the 1960s and ’70s, physicists extended this idea, finding other hidden symmetries associated with conservation laws to develop the standard model of particle physics.
“There’s this conceptual link that — once you realize it — you have a hammer and you go in search of nails to use it on,” Wilczek says. Anywhere they found a conservation law, physicists looked for a
symmetry, and vice versa. The standard model, which Wilczek shared a 2004 Nobel Prize for his role in developing, explains a plethora of particles and their interactions. It is now considered by many
physicists to be one of the most successful scientific theories ever, in terms of its ability to precisely predict the results of experiments.
At the Large Hadron Collider, at CERN in Geneva, physicists are still searching for new particles predicted using Noether’s insights. A hypothetical hidden symmetry, dubbed supersymmetry because it
proposes another level of symmetry in particle physics, posits that each known particle has an elusive heavier partner.
So far, no such particles have been found, despite high hopes for their detection (SN: 10/1/16, p. 12). Some physicists are beginning to ask if supersymmetry is correct. Perhaps symmetry can only
take physicists so far.
That notion is leaving some physicists in a bit of a lurch: “If that’s not going to be your guiding motto all the time — that more symmetry is better — then what will be your guiding motto?” asks
mathematical physicist John Baez of the University of California, Riverside.
Holograms get symmetric
Despite such disappointments, symmetry maintains its luster in physics at large. Noether’s theorems are essential tools for developing potential theories of quantum gravity, which would unite two
disparate theories: general relativity and quantum mechanics. Noether’s work helps scientists understand what kinds of symmetries can appear in such a unified theory.
One candidate relies on a proposed connection between two types of complementary theories: A quantum theory of particles on a two-dimensional surface without gravity can act as a hologram for a
three-dimensional theory of quantum gravity in curved spacetime. That means the information contained in the 3-D universe can be imprinted on a surrounding 2-D surface (SN: 10/17/15, p. 28).
Picture a soda can with a label that describes the size and location of each bubble inside. The label catalogs how those bubbles merge and pop. A curious researcher could use the behavior of the
can’s surface to understand goings-on inside the can, for example, calculating what might happen upon shaking it. For physicists, understanding a simpler, 2-D theory can help them comprehend a more
complicated mess — namely, quantum gravity — going on inside. (The theory of quantum gravity for which this holographic principle holds is string theory, in which particles are described by wiggling
“Noether’s theorem is a very important part of that story,” says theoretical physicist Daniel Harlow of MIT. Symmetries in the 2-D quantum theory show up in the 3-D quantum gravity theory in a
different context. In a satisfying twist, Noether’s first and second theorems become linked: Noether’s first theorem in the 2-D picture makes the same statement as Noether’s second theorem in 3-D.
It’s like taking two sentences, one in Japanese and one in English, and realizing upon translating them that both say the same thing in different ways.
New directions for Noether
Everyday physics relies on Noether’s theorem as well. The conservation laws it implies help to explain waves on the surface of the ocean and air flowing over an airplane wing.
Simulating such systems helps scientists make predictions — about weather patterns, vibrations of bridges or the effects of a nuclear blast, for example. Noether’s theorem doesn’t automatically apply
in computer simulations, which simplify the world by slicing it up into small chunks of space and time. So programmers have to manually add in conservation laws for energy and momentum.
“They throw away all of the physics, and then they have to try and force it all back in somehow,” says mathematician Elizabeth Mansfield of the University of Kent in England. But Mansfield has found
new ways to make Noether’s theorem apply in simulations. She and colleagues have simulated a person beating a drum inside a simplified Stonehenge, determining how sound waves would wrap around the
stone — while automatically conserving energy. Mansfield says her method, which she will present in September in London at a Noether celebration, could eventually be used to create simulations that
behave more like the real world.
In addition to Noether’s importance in physics, in mathematics her ideas are so prominent that her name has become an adjective. References to Noetherian rings, Noetherian groups and Noetherian
modules are sprinkled throughout current mathematical literature.
Noether’s work “should have been a wake-up call to society that women could do mathematics,” Gregory says. Eventually, society did awaken. In a 2015 lecture she gave about Noether at the Perimeter
Institute for Theoretical Physics in Waterloo, Canada, Gregory showed a slide of herself with five female colleagues, then at the center for particle theory at Durham University. While women in
science still face challenges, no one in the group had to struggle to get paid for her work. “That is Noether’s legacy, and I honestly think she would have been really jazzed,” Gregory says. “I think
this would have been her real … vindication.” | {"url":"https://www.sciencenews.org/article/emmy-noether-theorem-legacy-physics-math?fbclid=IwAR27EdT0MArupfodoDnZSrBEJbNKH4X1mG4NKN38XgamT2u-EcjUXTmZFrE","timestamp":"2024-11-14T08:18:49Z","content_type":"text/html","content_length":"323275","record_id":"<urn:uuid:2b3b1e49-c90c-46ac-ad09-74011c37e167>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00480.warc.gz"} |
Statistical significance
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In statistics, a result is called statistically significant if it is unlikely to have occurred by chance. "A statistically significant difference" simply means there is statistical evidence that
there is a difference; it does not mean the difference is necessarily large, important, or significant in the common meaning of the word.
The significance level of a test is a traditional frequentist statistical hypothesis testing concept. In simple cases, it is defined as the probability of making a decision to reject the null
hypothesis when the null hypothesis is actually true (a decision known as a Type I error, or "false positive determination"). The decision is often made using the p-value: if the p-value is less than
the significance level, then the null hypothesis is rejected. The smaller the p-value, the more significant the result is said to be.
In more complicated, but practically important cases, the significance level of a test is a probability such that the probablility of making a decision to reject the null hypothesis when the null
hypothesis is actually true is no more than the stated probability. This allows for those applications where the probability of deciding to reject may be much smaller than the significance level for
some sets of assumptions encompassed within the null hypothesis.
Use in practice[]
The significance level is usually represented by the Greek symbol, α (alpha). Popular levels of significance are 5%, 1% and 0.1%. If a test of significance gives a p-value lower than the α-level, the
null hypothesis is rejected. Such results are informally referred to as 'statistically significant'. For example, if someone argues that "there's only one chance in a thousand this could have
happened by coincidence," a 0.1% level of statistical significance is being implied. The lower the significance level, the stronger the evidence.
In some situations it is convenient to express the statistical significance as 1 − α. In general, when interpreting a stated significance, one must be careful to note what, precisely, is being tested
Different α-levels have different advantages and disadvantages. Smaller α-levels give greater confidence in the determination of significance, but run greater risks of failing to reject a false null
hypothesis (a Type II error, or "false negative determination"), and so have less statistical power. The selection of an α-level inevitably involves a compromise between significance and power, and
consequently between the Type I error and the Type II error.
In some fields, for example nuclear and particle physics, it is common to express statistical significance in units of "σ" (sigma), the standard deviation of a Gaussian distribution. A statistical
significance of "${\displaystyle n\sigma}$" can be converted into a value of α via use of the error function:
${\displaystyle \alpha = 1 - \operatorname{erf}(n/\sqrt{2})}$
The use of σ is motivated by the ubiquitous emergence of the Gaussian distribution in measurement uncertainties. For example, if a theory predicts a parameter to have a value of, say, 100, and one
measures the parameter to be 109 ± 3, then one might report the measurement as a "3σ deviation" from the theoretical prediction. In terms of α, this statement is equivalent to saying that "assuming
the theory is true, the likelihood of obtaining the experimental result by coincidence is 0.27%" (since 1 − erf(3/√2) = 0.0027).
Fixed significance levels such as those mentioned above may be regarded as useful in exploratory data analyses. However, modern statistical advice is that, where the outcome of a test is essentially
the final outcome of an experiment or other study, the p-value should be quoted explicitly. And, importantly, it should be quoted whether or not the p-value is judged to be significant. This is to
allow maximum information to be transferred from a summary of the study into meta-analyses.
A common misconception is that a statistically significant result is always of practical significance, or demonstrates a large effect in the population. Unfortunately, this problem is commonly
encountered in scientific writing.^[1] Given a sufficiently large sample, extremely small and non-notable differences can be found to be statistically significant, and statistical significance says
nothing about the practical significance of a difference.
One of the more common problems in significance testing is the tendency for multiple comparisons to yield spurious significant differences even where the null hypothesis is true. For instance, in a
study of twenty comparisons, using an α-level of 5%, one comparison will likely yield a significant result despite the null hypothesis being true. In these cases p-values are adjusted in order to
control either the false discovery rate or the familywise error rate.
An additional problem is that frequentist analyses of p-values are considered by some to overstate "statistical significance".^[2]^[3] See Bayes factor for details.
Yet another common pitfall often happens when a researcher writes the ambiguous statement "we found no statistically significant difference," which is then misquoted by others as "they found that
there was no difference." Actually, statistics cannot be used to prove that there is exactly zero difference between two populations. Failing to find evidence that there is a difference does not
constitute evidence that there is no difference. This principle is sometimes described by the maxim "Absence of evidence is not evidence of absence."
According to J. Scott Armstrong, attempts to educate researchers on how to avoid pitfalls of using statistical significance have had little success. In the papers "Significance Tests Harm Progress in
Forecasting,"^[4] and "Statistical Significance Tests are Unnecessary Even When Properly Done,"^[5] Armstrong makes the case that even when done properly, statistical significance tests are of no
value. A number of attempts failed to find empirical evidence supporting the use of significance tests. Tests of statistical significance are harmful to the development of scientific knowledge
because they distract researchers from the use of proper methods. Armstrong suggests authors should avoid tests of statistical significance; instead, they should report on effect sizes, confidence
intervals, replications/extensions, and meta-analyses.
Use of the statistical significance test has been called seriously flawed and unscientific by authors Deirdre McCloskey and Stephen Ziliak. They point out that "insignificance" does not mean
unimportant, and propose that the scientific community should abandon usage of the test altogether, as it can cause false hypotheses to be accepted and true hypotheses to be rejected.^[6]^[1]
Signal–noise ratio conceptualisation of significance[]
Statistical significance can be considered to be the confidence one has in a given result. In a comparison study, it is dependent on the relative difference between the groups compared, the amount of
measurement and the noise associated with the measurement. In other words, the confidence one has in a given result being non-random (i.e. it is not a consequence of chance) depends on the
signal-to-noise ratio (SNR) and the sample size.
Expressed mathematically, the confidence that a result is not by random chance is given by the following formula by Sackett:^[7]
${\displaystyle \mathrm{confidence} = \frac{\mathrm{signal}}{\mathrm{noise}} \times \sqrt{\mathrm{sample\ size}}.}$
For clarity, the above formula is presented in tabular form below.
Dependence of confidence with noise, signal and sample size (tabular form)
Parameter Parameter increases Parameter decreases
Noise Confidence decreases Confidence increases
Signal Confidence increases Confidence decreases
Sample size Confidence increases Confidence decreases
In words, the dependence of confidence is high if the noise is low and/or the sample size is large and/or the effect size (signal) is large. The confidence of a result (and its associated confidence
interval) is not dependent on effect size alone. If the sample size is large and the noise is low a small effect size can be measured with great confidence. Whether a small effect size is considered
important is dependent on the context of the events compared.
In medicine, small effect sizes (reflected by small increases of risk) are often considered clinically relevant and are frequently used to guide treatment decisions (if there is great confidence in
them). Whether a given treatment is considered a worthy endeavour is dependent on the risks, benefits and costs.
Tests of statistical significance[]
See also[]
1. ↑ ^1.0 ^1.1 Ziliak, Stephen T. and Deirde N. McCloskey. "Size Matters: The Standard Error of Regressions in the American Economic Review" (August 2004). [1]
2. ↑ Goodman S (1999). Toward evidence-based medical statistics. 1: The P value fallacy.. Ann Intern Med 130 (12): 995–1004.
3. ↑ Goodman S (1999). Toward evidence-based medical statistics. 2: The Bayes factor.. Ann Intern Med 130 (12): 1005–13.
4. ↑ Armstrong, J. Scott (2007). Significance tests harm progress in forecasting. International Journal of Forecasting 23: 321–327.
5. ↑ Armstrong, J. Scott (2007). Statistical Significance Tests are Unnecessary Even When Properly Done. International Journal of Forecasting 23: 335–336.
6. ↑ McCloskey, Deirdre N.; Stephen T. Ziliak (2008). The Cult of Statistical Significance: How the Standard Error Costs Us Jobs, Justice, and Lives (Economics, Cognition, and Society), The
University of Michigan Press.
7. ↑ Sackett DL (October 2001). Why randomized controlled trials fail but needn't: 2. Failure to employ physiological statistics, or the only formula a clinician-trialist is ever likely to need (or
understand!). CMAJ 165 (9): 1226–37.
External links[]
• Raymond Hubbard, M.J. Bayarri, P Values are not Error Probabilities. A working paper that explains the difference between Fisher's evidential p-value and the Neyman-Pearson Type I error rate ${\
displaystyle \alpha}$.
• The Concept of Statistical Significance Testing - Article by Bruce Thompon of the ERIC Clearinghouse on Assessment and Evaluation, Washington, D.C. | {"url":"https://psychology.fandom.com/wiki/Statistical_significance","timestamp":"2024-11-06T15:24:33Z","content_type":"text/html","content_length":"186414","record_id":"<urn:uuid:552d24dd-dfd4-4b8d-94df-ca5f134e8d85>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00662.warc.gz"} |
How I created a simple cross-multiplication for entrepreneurs
Hello, first post on DEV for this professional account, it will be written in English and French to facilitate reading.
Why I created that?
I'm a freelancer since april 2023, I invoice by the hour according to my tasks. So I used a classic website or even a calculator to get my result for each task. Except it was taking me a bit longer,
and I was thinking it could be an even simpler and faster process.
So, as a web developer, I designed this web interface to meet my needs. It's accessible to everyone, and free. The magic of the Internet.
What is a cross-multiplication?
In mathematics, specifically in elementary arithmetic and elementary algebra, given an equation between two fractions or rational expressions, one can cross-multiply to simplify the equation or
determine the value of a variable.
That's all. Thanks Wikipedia. ๐
In mathematics, specifically in elementary arithmetic and elementary algebra, given an equation between two fractions or rational expressions, one can cross-multiply to simplify the equation or
determine the value of a variable.
How I created this tool?
The website is under Nuxt, I use SASS for styling.
The script is quite simple, written in two parts, one to calculate the total, and the second to transform "4h23" into "263m".
How I store datas?
All data is stored via localStorage, based on your client. This makes it easier to have both values (time and price) saved when you return to the page.
watch([defaultMinutes, defaultHourlyRate, defaultTimePerMinutes, result], () => {
localStorage.setItem('defaultMinutes', defaultMinutes.value);
localStorage.setItem('defaultHourlyRate', defaultHourlyRate.value);
onMounted(() => {
if (localStorage.getItem('defaultMinutes')) {
defaultMinutes.value = localStorage.getItem('defaultMinutes');
if (localStorage.getItem('defaultHourlyRate')) {
defaultHourlyRate.value = localStorage.getItem('defaultHourlyRate');
When the page is loaded (onMounted), we retrieve the localstorage if available to set them to the correct values. Nothing mysterious! ๐ ง โ โ ๏ธ
Converting hours into minutes
Okay, this part takes a bit longer to explain, but if you understand a bit of code logic, you'll notice quite easily how I do it.
First I simply check if defaultTimePerMinutes is not undefined, then other conditions to avoid errors, and then we check if there's the letter "h" to split the two parts.
We calculate the hours value, i.e. the value to the left of "h" in minutes, and add the two parts to obtain a total in minutes.
watch([defaultMinutes, defaultHourlyRate, defaultTimePerMinutes], () => {
if (!defaultTimePerMinutes.value) {
result.value = null;
if (isNaN(defaultTimePerMinutes.value) && !defaultTimePerMinutes.value.includes('h')) {
result.value = null;
totalMinutes.value = null;
// if defaultTimePerMinutes contains a value of type "1h30".
if (defaultTimePerMinutes.value.includes('h')) {
let [hours, minutes] = defaultTimePerMinutes.value.split('h');
minutes = minutes ? parseInt(minutes) : 0;
let totalHours = hours * 60;
totalMinutes.value = totalHours + minutes;
result.value = (totalMinutes.value * defaultHourlyRate.value) / defaultMinutes.value;
result.value = result.value.toFixed(2);
} else {
totalMinutes.value = defaultTimePerMinutes.value;
result.value = (totalMinutes.value * defaultHourlyRate.value) / defaultMinutes.value;
result.value = result.value.toFixed(2);
if (totalMinutes.value === 0 || !totalMinutes) {
totalMinutes.value = null;
Don't hesitate to give me feedback, or even a tool like this one, which can be great to set up for calculations in a professional environment. It's always interesting to know what's missing on the
Top comments (0)
For further actions, you may consider blocking this person and/or reporting abuse | {"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/conceptweb/how-i-created-a-simple-cross-multiplication-for-entrepreneurs-435l","timestamp":"2024-11-10T07:41:46Z","content_type":"text/html","content_length":"83969","record_id":"<urn:uuid:9bf4bf46-8985-4997-87f2-74bfb5a03e25>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00797.warc.gz"} |
Logarithms and pH
n this essay we explain the mathematical concept of exponential behavior and logarithms and apply them to a few properties of water. Let’s take a common and important property of water (see the
Harebrain topic Water). The vapor pressure of water increases as the temperature of the water rises (Table 1). As the temperature increases the water molecules gain kinetic energy and more and more
molecules have sufficient energy to break away from the collective interactions of bulk liquid water. At the normal boiling point, 100 °C and 760 mm Hg pressure, many molecules have enough energy to
break away and form gaseous water bubbles within the liquid—at that point the water begins to boil.
Table 1. Vapor Pressure (mm Hg) of water from 0 to 100 °C
Temperature, ^oC mm Hg
0 4.6
10 9.2
20 17.5
25 23.8
30 31.8
40 55.3
50 92.5
60 149.4
70 233.7
80 355.1
90 525.8
95 633.9
100 760.0
If we plot the vapor pressure against the temperature we obtain what you see in Figure 1.
Figure 1. A plot of the vapor pressure of water in mm Hg vs the temperature of the water in °C.
This is not a straight-line, or linear, relationship. If you look at a small range, say, from 20-50 °C, it appears to be approximately linear as shown in Figure 2, where a straight line is drawn
through four points. But even in this small range it is clear that there is curvature in the plot (notice that the two middle points lie below the line and the outer two lie above the line).
Figure 2. A plot of the vapor pressure of water in mm Hg vs the temperature of the water in °C over the range 10 to 40 °C.
Before we discuss this behavior in more depth, let’s take a look at some made-up numbers for vapor pressure. We constructed this made-up trend by taking the real vapor pressure of water at 10 °C and
then adding 10 to that value to get the vapor pressure at 20 °C, then adding 10 to that value, and so on. This data (Figure 3) is clearly linear: the points fall on a straight line and the line has a
slope of 1 (this is the number in front of the x in the equation y = x – 0.8) and an intercept of -0.8 (this is where the line meets the y axis, that is, at 0 °C the vapor pressure is -0.8 mm Hg).
Figure 3. A plot of some artificial numbers for the vapor pressure of water vs temperature.
Now let’s go back to the real vapor pressure data. If you take 10, 20, 30, and 40 °C, does the vapor pressure vary by a constant amount as you go from 10 to 20, from 20 to 30, and so on? No, the
vapor pressures at the first two temperatures are 9.2 and 17.5, which is a difference of 8.3. Is the difference between the vapor pressures at 20 and 30 °C (notice that we are considering
temperatures that have the same increment (10 °C)) 8.3? No, in fact the difference is 6.3 °C. This tells us that if we increment the temperature by 10 °C, the vapor pressure does not change by the
same value no matter whether we go from 10 to 20 °C or from 70 to 80 °C. [Look at the made-up data and notice that the vapor pressure increases by 10 mm Hg for any 10 °C jump in temperature.]
The real vapor pressures do not vary linearly with temperature; they vary exponentially with temperature. So let’s see what Excel will give us if we insist that the trendline is an exponential one
(Figure 4). Figure 4 contains both our previous linear fit to these four points, as well as an exponential fit. The exponential curve goes through each of the four points, whereas the linear fit
provides a much less satisfactory description of the points. The equation obtained for the exponential curve is y = 5.17e ^0.0598x. This equation contains e, the base of natural logarithms with an
approximate value of 2.718. The equation could also be written using base 10: y = 5.17(10^0.0598x/2.303), where 2.303 is an approximate conversion factor to convert from base 10 to base e. The real
power (pun intended) of this equation lies in the fact that x is part of the exponent. Let’s see what happens to y when we use a value for x of 10, 20, and 30 in the exponent.
Figure 4. The vapor pressure of water at four temperatures with an exponential fit.
For x = 10, y = 5.1754e^0.0598(10) = 9.41
For x = 20, y = 5.1754e^0.0598(20) = 17.1
For x = 30, y = 5.1754e^0.0598(30) = 31.1
Because these exponential equations and very large or very small numbers like Avogadro’s number (6.023 x 10^23) are a bit unwieldy it is often convenient to use logarithms. Logarithms can be used
with any base number, but those for base e and base 10 are most frequently used. For base e the logarithm is designated ln, and for base 10 the designation is log (though many writers use log for
both bases). We will use mainly base 10.
Let’s take a number like 100. We could express this number as 10^2. Or, let’s take 500 and express it as an exponent, which would be 10^2.70. The log of 100 is the number to which 10 (the base) must
be raised to obtain 100. So,
Log(100) = 2, that is, 10^2 = 100.
Likewise, the log of 500 is 2.70. When the log has numbers to the right of the decimal point, this part of the logarithm is called the mantissa. In fact we could write the log of 500 as
log(100) + log(5) = 2 + 0.70 = 2.70.
Notice that while we must multiply 5 x 100 to get 500, we can add log(100) to log(5) to obtain the log of 500. It is important also to be aware of significant figures. The antilog (sometimes called
inverse log) of 1.0 is 10; that is, log(10) = 1 The antilog of 0.70 is 5.0119; the antilog of 0.71 is 5.1286 and the antilog of 0.69 is 4.8978. So, the use of just two significant figures in the
mantissa (0.70) of 2.70 gives us a number somewhere between 490 and 513. If we want to express the number more precisely, we need more significant figures. If we use 2.700, we imply that the mantissa
is known to ± 0.001. The antilog of 2.701 is 502 and for 2.699 is 500. Thus, three significant figures in the mantissa provide an antilog that also has three significant figures.
Logarithms and powers of 10 are used frequently in chemistry. Logs have a reputation of being difficult to use but they are much easier to understand since modern scientific calculators have log[10]x
and 10^x functions. Logs are used in measurements with exponential behavior and to compress large and small numbers to a more manageable form. Because logs are based on powers they are dimensionless
numbers; that is, they have no units.
Avogadro’s number is a very large, uncountable, number 6.023 X 10^23 but its log is 23.78. The reciprocal of Avogadro’s number, 1/(6.023 X 10^23) = 1.67 X 10^-24, is the mass of one atomic mass unit
and its log is -23.78. Oh yes, -23.78 and 23.78 have only two significant numbers, the log mantissa 0.78. The 23 is only the power of 10 and sets the decimal point.
Addition is a simpler operation than multiplication and for many years tables of logarithms were used for accurate multiplication by adding the logs of the numbers. Also slide rules (the precursor of
digital calculators) had log scales, which allowed rapid addition (multiplication) or subtraction (division) of the logs of numbers to obtain answers to three significant figures.
The dissociation of water into hydrogen and hydroxide ions makes good use of logs. The pH of an acid or basic solution is measured with a volt meter. In order to keep the scale positive and readable
the pH was defined as –log [H^+] in which the square brackets mean molar concentration. The p in pH stands for power; that is, 10^-pH and the powers are usually negative which makes 10^-pHa positive
For the auto dissociation equilibrium of water
H[2]O + H[2]O --> H^+ + OH^-,
The dissociation constant at room temperature is K[w] = 10^-14 and is written as:
[ H^+ ] [ OH^- ] = K[w] = 10^-14.
Taking the negative log of both sides of this equation gives:
-log { [ H^+ ] [ OH^- ] = K[w] = 10^-14}.
Because -log[H^+] = pH and -log[OH^-] = pOH,
pH + pOH = pK[w] =14,
which is very useful in working acid-base equilibrium problems. | {"url":"https://wiredchemist.com/chemistry/instructional/harebrain-corner/logarithms-and-ph","timestamp":"2024-11-07T05:36:52Z","content_type":"text/html","content_length":"19124","record_id":"<urn:uuid:b284dbf2-6fce-4b58-9e71-a11f82b6e959>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00625.warc.gz"} |
Fractions – Definition and Actual-Life Examples
Fractions are used to point the variety of elements of a complete. Suppose you go to Domino’s pizza and order a medium pizza. In case your pizza has 8 slices and also you didn’t eat the entire pizza,
which means that elements of the entire pizza shall be left on the pizza field.
The pizza on the left reveals six-eighths (6/8) and the pizza on the proper reveals four-eighths (4/8).
6/8 and 4/8 are referred to as fractions.
The highest quantity is the numerator of the fraction and it tells us what number of slices there are.
The underside quantity is the denominator of the fraction and it tells us what number of equal slices there are in complete.
Equally, suppose a rectangle has 4 equal elements and we shade 1 a part of the rectangle. The fraction is 1/4
Discover {that a} ahead slash (/) or a horizontal line could also be used to separate the numerator from the denominator.
As a result of
is extra interesting, we choose to make use of it as an alternative of two/8.
Equal fractions
Now, check out the determine under. You’ll be able to clearly see that six-eighths (6/8) of a complete pizza is identical as three-fourths (3/4) of a complete pizza.
6/8 and three/4 are equal fractions. Though 6/8 and three/4 have completely different numerators and completely different denominators, they’re each equal.
Actually, 6/8 = 3/4 = 0.75
Extra examples of equal fractions
Greater phrases and decrease phrases of fractions
Discover that to get from 1/2 to 4/8, all we have to do is to multiply each the numerator and the denominator of 1/2 by the identical quantity or 4.
In the identical approach, to get from 1/3 to five/15, we will multiply each the numerator and the denominator of 1/3 by 5.
4/8 and 5/15 are referred to as larger phrases of 1/2 and 1/3 respectively.
Watch out! 4/8 just isn’t larger than 1/2 and 5/15 just isn’t larger than 1/3.
4/8 is a better time period of 1/2 solely as a result of it has a much bigger numerator and a much bigger denominator. Nonetheless, we noticed earlier than that 1/2 and 4/8 are equal fractions since
1/2 = 4/8 = 0.5
Discover too which you can go from a better time period to a decrease time period. As an example, to carry 5/15 to a decrease time period, all it is advisable to do is to divide each numerator and
denominator by 5. You’re going to get 1/3.
Simplifying a fraction is the method of going from larger phrases to decrease phrases. To get the best type of a fraction, divide the numerator and the denominator by the best widespread issue (gcf)
or the most important quantity that divides into each numerator and denominator evenly.
As an example, to carry 10/40 to its easiest from, divide each 10 and 40 by 10. You’re going to get 1/4.
1/2 and 1/3 are the bottom phrases of 4/8 and 5/15 respectively.
Including fractions
When the fractions have a typical denominator (identical denominator), add the numerators and maintain the identical denominator.
4/8 + 2/8 = (4 + 2)/8 = 6/8
Discover that 1/2 = 4/8 and 1/4 = 2/8. Subsequently, 1/2 + 1/4 = 4/8 + 2/8 = 6/8.
If the denominators will not be the identical, you have to to search for the least widespread a number of (LCM) of the denominators. See including fractions for extra explanations.
Utilizing fractions to point out ratios
You should use a fraction to point out a ratio. In a ratio, the numerator reveals the a part of a bunch you’re contemplating and the denominator present the remainder of the group or the entire
Suppose a category has 6 boys and 10 ladies. What’s the ratio of boys to ladies. On this instance, the numerator is the variety of boys and the denominator is the remainder of the group or the
variety of ladies.
The ratio of boys to ladies is
What’s the ratio of ladies to the full variety of college students? On this case, the numerator is the variety of ladies and the denominator is the entire group or the full variety of college
The ratio of ladies to complete variety of college students is
Utilizing fractions to point out division
A fraction will also be used to characterize division.
The expression
may imply 4 divided by 5 or 4 ÷ 5.
You’ll be able to then write the fraction as a decimal. For instance, 4/5 = 4 ÷ 5 = 80/100 = 0.8
Benchmark fractions
Benchmark fractions are fractions which are used lots in fundamental math and they’re additionally useful in picturing different fractions.
With benchmark fractions, you are able to do the followings:
• Rapidly evaluate and order fractions. For instance, 2/3 is larger than 1/4.
• Spherical fractions and blended numbers. For instance, 3/4 rounds as much as 1 since it’s nearer to 1 than to 0.
• Estimate sums and variations of fractions and blended numbers. For instance, 3/4 + 1/3 is larger than 1 since 1/3 is larger than 1/4.
A few examples exhibiting how fractions are utilized in on a regular basis life
1. If you happen to cook dinner lots by following a recipe, then you need to have used fractions lots earlier than.
Suppose a recipe says to make use of 12 tablespoons of sugar to make a cake weighing 3 kilos. How a lot sugar will you employ for a one-pound cake?
Since you make 1/3 of the entire cake, chances are you’ll cause that it is advisable to use 1/3 of 12 tablespoons.
2. Nurses and pharmacists should completely be good at fractions to be able to give affected person the right dosage of a drugs. Suppose the dosage energy of a drugs is 100 mg. If the physician
orders 25 mg, what number of tablets will you give? Nurses and pharmacists may have the next system to unravel this downside,
Variety of tablets to present =
Treatment energy ordered by the physician
Dosage energy of 1 pill
Variety of tablets to present =
25 mg
100 mg
Nurses and pharmacists should know that 25/100 and 1/4 are equal fractions to allow them to give 1/4 of the pill to the affected person.
How properly do you perceive this lesson? Take this quiz about fractions!
Nonetheless scuffling with fractions? Eliminate your fears and frustrations as soon as and for all!
If you do not know fractions very properly, you’ll in all probability wrestle to do properly on most math checks. Construct a robust basis in math at this time earlier than it’s too late!
Purchase my fractions e book now. It affords a radical protection of fractions! | {"url":"https://keiseronlineuniversity.com/fractions-definition-and-actual-life-examples/","timestamp":"2024-11-12T13:25:57Z","content_type":"text/html","content_length":"43442","record_id":"<urn:uuid:2241280f-2060-46c1-a977-9d14f83b5b62>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00090.warc.gz"} |
weitrix: Tools for matrices with precision weights, test and explore weighted or sparse data version 1.2.0 from Bioconductor
Data type and tools for working with matrices having precision weights and missing data. This package provides a common representation and tools that can be used with many types of high-throughput
data. The meaning of the weights is compatible with usage in the base R function "lm" and the package "limma". Calibrate weights to account for known predictors of precision. Find rows with excess
variability. Perform differential testing and find rows with the largest confident differences. Find PCA-like components of variation even with many missing values, rotated so that individual
components may be meaningfully interpreted. DelayedArray matrices and BiocParallel are supported.
Author Paul Harrison [aut, cre] (<https://orcid.org/0000-0002-3980-268X>)
Bioconductor views DataRepresentation DimensionReduction GeneExpression RNASeq Regression SingleCell Software Transcriptomics
Maintainer Paul Harrison <paul.harrison@monash.edu>
License LGPL-2.1 | file LICENSE
Version 1.2.0
Package repository View on Bioconductor
Install the latest version of this package by entering the following in R:
Installation if (!requireNamespace("BiocManager", quietly = TRUE))
Any scripts or data that you put into this service are public.
weitrix documentation
built on Nov. 8, 2020, 8:10 p.m. | {"url":"https://rdrr.io/bioc/weitrix/","timestamp":"2024-11-09T19:09:19Z","content_type":"text/html","content_length":"28499","record_id":"<urn:uuid:99fabd4f-d757-4e51-b686-740657554f52>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00638.warc.gz"} |
Machine-learning-aided cognitive reconfiguration for flexible-bandwidth HPC and data center networks [Invited] (Journal Article) | NSF PAGES
This paper presents a cognitive flexible-bandwidth optical interconnect architecture for datacom networks. The proposed architecture leverages silicon photonic reconfigurable all-to-all switch
fabrics interconnecting top-of-rack switches arranged in a Hyper-X-like topology with a cognitive control plane for optical reconfiguration by self-supervised learning. The proposed approach makes
use of a clustering algorithm to learn the traffic patterns from historical traces. We developed a heuristic algorithm for optimizing the intra-pod connectivity graph for each identified traffic
pattern. Further, to mitigate the scalability issue induced by frequent clustering operations, we parameterized the learned traffic patterns by a support vector machine classifier. The classifier is
trained offline by self-labeled data to enable the classification of traffic matrices during online operations, thereby facilitating cognitive reconfiguration decision making. The simulation results
show that compared with a static all-to-all interconnection, the proposed approach can improve the throughput by up to$1.62×<#comment/>$while reducing the end-to-end packet latency and flow
completion time by up to$3.84×<#comment/>$and$20×<#comment/>$, respectively.
more » « less | {"url":"https://par.nsf.gov/biblio/10210534-machine-learning-aided-cognitive-reconfiguration-flexible-bandwidth-hpc-data-center-networks-invited","timestamp":"2024-11-15T00:28:09Z","content_type":"text/html","content_length":"260550","record_id":"<urn:uuid:01e514c9-fe2d-49aa-86b0-e6729c271f16>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00024.warc.gz"} |
Project Report - P.PDFKUL.COM
Compressed Sensing for Background Subtraction Digital Image Processing(ECE 278A) Project Report Karthikeyan Shanmuga Vadivel March 16, 2009
Compressed Sensing is a new exciting eld which challenges the celebrated Nyquist Theorem. It proposes that signals which are sparse in some transform domain can be reconstructed accurately with KlogN
samples where K is the sparsity of the signal in the Transform Domain and N is the length of the original signal. By utilizing the compressed sensing notion of obtaining signals, the complexity of
the design of the sensor (camera) is reduced. This could practically materialize the eld of sensor networks. Whenever a scene is captured using a still video camera, background subtracted images
latch on to the relevant new information in each frame. Hence compressed sensing applied to background subtracted images is an important issue to be handled. Hence, background subtracted images are
compressed sensed and reconstructed using three dierent algorithms L1 norm minimization, Weighted L1 norm minimization and Orthogonal Pursuit Matching and the eciency and the complexities involved in
these processes are analysed in this project.
Every scene obtained by a still video camera (single pixel camera) consists of moving and stationary objects. The stationary objects form the background and the moving objects comprise the
foreground. In every new frame obtained, the additional information carried by the frame is some sense captured by the foreground or the background subtracted image. Hence, instead of transmitting
the entire image, it would be almost sucient if the background subtracted image is transmitted assuming the receiver knows the background information. Hence, compressed sensing background subtracted
images is an important issue as they can be reconstructed using much fewer samples of the original signal. This is intuitively expected as background subtracted images are more sparse and can be
compressed to a much greater extent and this compressability can be transferred to the sensor by sampling the signal at a much lower rate. The primary reference paper in this project where background
subtracted images are reconstructed is [5].
Compressed Sensing
Compressive sensing is a non adaptive sampling technique which requires the signal to be compressible (to be sparse in some domain, transform domain) to be taken advantage of. It also requires the
measurement functions(sampling functions) to be incoherent to the transform domain. When these conditons are satised a sparse signal can be recovered using much smaller sampling rate than one
suggested by Nyquist Theorem. Compressed sensing cameras have several applications in astronomy, MRI imaging, sensor networks where conventional cameras have inherent limitations. Hence, the
reconstruction of compressed sensed images is an important problem to be handled. In this project three dierent reconstruction algorithms are analysed - L1 norm minimization, Weighted L1 norm
minimization and Orthogonal Pursuit Matching to recover the original signal from compressed sensed images and the performance of these techniques are compared. A owchart of compressed sensing is
shown in gure below
Background subtracted images are inherently sparse. Hence, the transform Ψ can be considered to be the identity matrix I. So, the reconstruction algorithms fall into the simple category of sparse
signal recovery.
Implementation Details
A vehicle monitoring video http://i2lwww.ira.uka.de/image_sequences/ using a still video camera was the prototype on which compressed sensing and reconstruction was performed. First the background
was estimated using simple median operation and median ltering over a small segment of the video. Then, the background subtracted image was obtained for a frame selected randomly and reconstruction
of this frame was the problem under analysis. The measurement functions used for compressed sensing are noiselets i.e randomly generated 1s and +1s. The measurement matrix is similar to an incomplete
Walsh-Hadamard Transform matrix. After sampling(compressed sensing) the background subtracted image, three reconstruction algorithms were used to reconstruct the original image. Two softwares, SEDUMI
and CVX were installed in MATLAB to perform convex optimization. The performances of the reconstruction algorithms were analysed in the following sections
L1 norm minimization
L1 norm minimization is a very standard technique for sparse signal recovery suggested in the preliminary papers [2, 3, 4] on compressed sensing. The selective advantage of L1 norm optimization over
L2 norm optimization is that it equinorm surface i.e recovery signal search direction is more oriented towards more sparser directions though we pay in terms of computational complexity. The
following convex minimization is implemented minimize
subject to ϕf = yk
This technique produces a sparse solution and the simulation results on a 50Ö50 resized image is shown below
Better reconstruction was obtained when the number of samples to reconstruct the image is increased. Compressed sensing sampling theorem suggests a rate of around 1300 samples to almost exactly
reconstruct the image. But, suboptimal reconstructions using 400 - 1200 samples was analysed and for 1200 samples nearly perfect reconstruction was obtained.
Weighted L1 norm reconstruction
This method tries to emulate L0 norm reconstruction using L1 norm reconstruction. In this technique instead of minimizing ||x||1 , ||W x||1 should be minimized, subject to the same constraints as L1
norm in 1. The weight matrix is intelligently chosen to get better performance than L1- norm minimization. Here W was chosen to be a diagonal 1 matrix with the diagonal entries wii to be xo,i where
x0 is the original signal. As the original signal is M-sparse only M elements in W matrix are non-zero and the rest of the values are innity in the main diagonal. Intuitively, minimizing ||W x||1
makes sure that the rest of the values other than the M non-zero values in x go to zero and hence a M sparse reconstruction is obtained. But, this technique has a glaring assumption about the prior
knowledge of the original signal. Hence to converge to this technique, x is estimated using L1 norm minimization and then W matrix is computed using this estimate and the process is repeated. A good
analysis of this techique is presented in [7].
In this simulation, the original image (resized 50×50 image) is reconstructed by compressed sensing the background subtracted image using 1200 samples using Weighted L1 norm minimization(5
iterations). It was observed that Weighted L1 norm minimization gave a lower Mean square error compared to L1 norm minimization, but not a signicantly astonishing improvement. The Weighted L1 norm
minimization only gives a very good reconstruction after a lot of iterations. A comparision of the MSE of L1 and Weighted L1 techniques is shown below (MSE is dened as the square of the frobenius
norm of the dierence between the original and the nal reconstructed image).
Orthogonal Pursuit Matching
This is a greedy algorithm which is less computationally complex compared to L1 norm minimization because it makes use of L2 norm minimization instead. The basic idea is to split the measurement
matrix into columns φ = [φ1 φ2 ... φN ]. Now φf = yk is solved. Here, only M values in φcontribute to the reconstruction of f . Next, the best M φi which contribute yk is estimated. The key idea here
is we consider yk as a weighted linear combination M φi . We estimate the best φi by correlating it with the output yk and reconstruct the signal fˆ at each stage using L2 norm reconstruction until
we nd the best M φi . At each stage we also estimate the the correspinding fˆi The complete algorithm is presented in [6].
In this simulation the background subtracted image (size - 50×50) was reconstructed using Orthogonal Matching Pursuit algorithm assuming the sparsity of the image to be 100,300 and 500 for a sampling
rate of 1200 samples. The reconstruction accuracy increased as higher sparsity was assumed. It should be noted that the actual sparsity of the background subtracted image was around 500.
Comparision of Reconstruction Algorithms
L1 norm and Weighted L1 norm minimization reconstruction techniques are computationally more complicated as they try to minimize L1 norm which has a computational complexity of O(N 3 ) using linear
programming. But, the critical parameter that aected the project was storage complexity. When images of size greater than 50×50 were compressed sensed, Matlab was out of memory. The problem is that
the reconstruction algorithms try to reduce computational complexity by increasing storage complexity when they use techniques like dynamic programming. One plausible reason for this issue in this
project was that Sedumi and CVX use dynamic programming to solve L1 norm minimization. But Orthogonal Matching pursuit uses L2 norm minimization at each stage. This reduces storage complexity issues,
but computational complexity was still an issue because of repeated L2 norm minimization. But one pays for gain in storage complexity of Orthogonal Matching Pursit for the accuarcy of reconstruction
as it is a pretty suboptimal method and an estimate of the sparsity of the original signal is required. In terms of accuarcy of reconstruction, Weighted L1 performed better than L1 as expected. The
relation below summarizes this discussion. With respect to Computational complexity : Weighted L1 > L1 ≈ Orthogonal Pursuit Matching With respect to Storage complexity : Weighted L1 ≈ L1 > Orthogonal
Pursuit Matching With respect to Reconstruction accuracy : Weighted L1 > L1 > Orthogonal Pursuit Matching
Conclusion In this project the reconstruction of compressed sensed background subtracted images was acheived using three different reconstruction algorithms - L1 norm minimization, Weighted L1 norm
minimization and Orthogonal Pursuit 6
Matching. A detailed analysis and comparision of all the three algorithms was performed. The primary issue in the project was the storage complexity of the reconstruction algorithms and hence
compressed sensing could only be applied to images of very small sizes. Finally, can be concluded that these reconstruction algorithms work reliably for recovering background subtracted images which
are inherently sparse.
References [1] Emmanuel Candès, Compressive Sampling. (Int. Congress of Mathematics, 3, pp. 1433-1452, Madrid, Spain,2006 ) [2] Richard Baraniuk, Compressive sensing. (IEEE Signal Processing
Magazine, 24(4), pp. 118-121, July 2007) [3] Emmanuel Candès and Michael Wakin, An introduction to compressive sampling. (IEEE Signal Processing Magazine, 25(2), pp. 21 - 30, March 2008)
[High-resolution version] [4] Justin Romberg, Imaging via compressive sampling. (IEEE Signal Processing Magazine, 25(2), pp. 14 - 20, March 2008) [5] Volkan Cevher, Aswin Sankaranarayanan, Marco
Duarte, Dikpal Reddy, Richard Baraniuk, and Rama Chellappa, Compressive sensing for background subtraction. (European Conf. on Computer Vision (ECCV), Marseille, France, October 2008) [6] Joel Tropp
and Anna Gilbert, Signal recovery from random measurements via orthogonal matching pursuit. (IEEE Trans. on Information Theory, 53(12) pp. 4655-4666, December 2007) [7] Emmanuel Candès, Michael
Wakin, and Stephen Boyd, Enhancing sparsity by reweighted ell-1 minimization. (Preprint, 2008) | {"url":"https://p.pdfkul.com/project-report_59c0cfe41723dd99436882b5.html","timestamp":"2024-11-08T01:31:12Z","content_type":"text/html","content_length":"62419","record_id":"<urn:uuid:da2eadfd-0775-48a4-8f06-ebcf37f2378c>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00792.warc.gz"} |
Question: If a particle has a defined position at every time, must it necessarily also have a defined velocity? Consider a particle moving along a line, so its position along the line at time t is x
(t). Suppose we define x(t) as follows:
{ 1 if t > 0
x(t) = {
{ 0 if t <= 0
If I remember my calculus correctly, x'(0) is undefined, while x(t) = 0. Does it therefore follow that at time t=0, the particle has a position and is moving but has no velocity? Would it be
physically possible (i.e. compatible with the laws of physics as we currently understand them) for a particle with that behaviour to actually exist? -- SJK
In classical (non-quantum) mechanics a particle with mass cannot make such an instantaneous jump in position. It implies infinite acceleration which implies infinite force. So this case is not
physically possible in classical mechanics (assuming zero-mass particles are not physically possible). -- Eob
I am in browse mode tonight... but at some point mention will have to made of tangent spaces and tie the discussion back to differential geometry.
Eob: What about in quantum mechanics? IIRC, quantum mechanics predicts instantaneous jumps in position (consider e.g. the Bohr model of the atom). And it has a zero-mass particle, the photon... --
I was not sure about quantum mechanics which was why I explicitly restricted my comments to classical mechanics. But now that I consider it more I would hazard that the question that was posed is
not meaningful in quantum mechanics because you can never know x(t) exactly. As for the The Bohr Model, it has been superceeded by a model of the atom surrounded by orbitals which are standing
waves of the wave function, so I am not sure it is relevant. --Eob | {"url":"https://chita.us/wikipedia/nost/index.pl?action=browse&diff=1&id=Velocity/Talk","timestamp":"2024-11-04T13:49:36Z","content_type":"text/html","content_length":"5453","record_id":"<urn:uuid:23770bb8-2f76-4d8e-a4c2-33cc725f661e>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00706.warc.gz"} |
An R package for space-time anomaly detection using scan statistics.
Installing the package
To install the latest (CRAN) release of this package, type the following:
To install the development version of this package, type this instead:
What are scan statistics?
Scan statistics are used to detect anomalous clusters in spatial or space-time data. The gist of the methodology, at least in this package, is this:
1. Monitor one or more data streams at multiple locations over intervals of time.
2. Form a set of space-time clusters, each consisting of (1) a collection of locations, and (2) an interval of time stretching from the present to some number of time periods in the past.
3. For each cluster, compute a statistic based on both the observed and the expected responses. Report the clusters with the largest statistics.
Main functions
Scan statistics
• scan_eb_poisson: computes the expectation-based Poisson scan statistic (Neill 2005).
• scan_pb_poisson: computes the (population-based) space-time scan statistic (Kulldorff 2001).
• scan_eb_negbin: computes the expectation-based negative binomial scan statistic (Tango et al. 2011).
• scan_eb_zip: computes the expectation-based zero-inflated Poisson scan statistic (Allévius & Höhle 2017).
• scan_permutation: computes the space-time permutation scan statistic (Kulldorff et al. 2005).
• scan_bayes_negbin: computes the Bayesian Spatial scan statistic (Neill 2006), extended to a space-time setting.
Zone creation
• knn_zones: Creates a set of spatial zones (groups of locations) to scan for anomalies. Input is a matrix in which rows are the enumerated locations, and columns the k nearest neighbors. To create
such a matrix, the following two functions are useful:
□ coords_to_knn: use stats::dist to get the k nearest neighbors of each location into a format usable by knn_zones.
□ dist_to_knn: use an already computed distance matrix to get the k nearest neighbors of each location into a format usable by knn_zones.
• flexible_zones: An alternative to knn_zones that uses the adjacency structure of locations to create a richer set of zones. The additional input is an adjacency matrix, but otherwise works as
• score_locations: Score each location by how likely it is to have an ongoing anomaly in it. This score is heuristically motivated.
• top_clusters: Get the top k space-time clusters, either overlapping or non-overlapping in the spatial dimension.
• df_to_matrix: Convert a data frame with data for each location and time point to a matrix with locations along the column dimension and time along the row dimension, with the selected data as
Example: Brain cancer in New Mexico
To demonstrate the scan statistics in this package, we will use a dataset of the annual number of brain cancer cases in the counties of New Mexico, for the years 1973-1991. This data was studied by
Kulldorff (1998), who detected a cluster of cancer cases in the counties Los Alamos and Santa Fe during the years 1986-1989, though the excess of brain cancer in this cluster was not deemed
statistically significant. The data originally comes from the package rsatscan, which provides an interface to the program SaTScan, but it has been aggregated and extended for the scanstatistics
To get familiar with the counties of New Mexico, we begin by plotting them on a map using the data frames NM_map and NM_geo supplied by the scanstatistics package:
# Load map data
# Plot map with labels at centroids
ggplot() +
geom_polygon(data = NM_map,
mapping = aes(x = long, y = lat, group = group),
color = "grey", fill = "white") +
geom_text(data = NM_geo,
mapping = aes(x = center_long, y = center_lat, label = county)) +
ggtitle("Counties of New Mexico")
We can further obtain the yearly number of cases and the population for each country for the years 1973-1991 from the data table NM_popcas provided by the package:
#> year county population count
#> 1 1973 bernalillo 353813 16
#> 2 1974 bernalillo 357520 16
#> 3 1975 bernalillo 368166 16
#> 4 1976 bernalillo 378483 16
#> 5 1977 bernalillo 388471 15
#> 6 1978 bernalillo 398130 18
It should be noted that Cibola county was split from Valencia county in 1981, and cases in Cibola have been counted to Valencia in the data.
A scan statistic for Poisson data
The Poisson distribution is a natural first option when dealing with count data. The scanstatistics package provides the two functions scan_eb_poisson and scan_pb_poisson with this distributional
assumption. The first is an expectation-based scan statistic for univariate Poisson-distributed data proposed by Neill et al. (2005), and we focus on this one in the example below. The second scan
statistic is the population-based scan statistic proposed by Kulldorff (2001).
Using the Poisson scan statistic
The first argument to any of the scan statistics in this package should be a matrix (or array) of observed counts, whether they be integer counts or real-valued “counts”. In such a matrix, the
columns should represent locations and the rows the time intervals, ordered chronologically from the earliest interval in the first row to the most recent in the last. In this example we would like
to detect a potential cluster of brain cancer in the counties of New Mexico during the years 1986-1989, so we begin by retrieving the count and population data from that period and reshaping them to
a matrix using the helper function df_to_matrix:
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> filter, lag
#> The following objects are masked from 'package:base':
#> intersect, setdiff, setequal, union
counts <- NM_popcas %>%
filter(year >= 1986 & year < 1990) %>%
df_to_matrix(time_col = "year", location_col = "county", value_col = "count")
#> Warning: `spread_()` was deprecated in tidyr 1.2.0.
#> Please use `spread()` instead.
#> This warning is displayed once every 8 hours.
#> Call `lifecycle::last_lifecycle_warnings()` to see where this warning was generated.
Spatial zones
The second argument to scan_eb_poisson should be a list of integer vectors, each such vector being a zone, which is the name for the spatial component of a potential outbreak cluster. Such a zone
consists of one or more locations grouped together according to their similarity across features, and each location is numbered as the corresponding column index of the counts matrix above (indexing
starts at 1).
In this example, the locations are the counties of New Mexico and the features are the coordinates of the county seats. These are made available in the data table NM_geo. Similarity will be measured
using the geographical distance between the seats of the counties, taking into account the curvature of the earth. A distance matrix is calculated using the spDists function from the sp package,
which is then passed to dist_to_knn and on to knn_zones:
# Remove Cibola since cases have been counted towards Valencia. Ideally, this
# should be accounted for when creating the zones.
zones <- NM_geo %>%
filter(county != "cibola") %>%
select(seat_long, seat_lat) %>%
as.matrix %>%
spDists(x = ., y = ., longlat = TRUE) %>%
dist_to_knn(k = 15) %>%
The advantage of expectation-based scan statistics is that parameters such as the expected value can be modelled and estimated from past data e.g. by some form of regression. For the
expectation-based Poisson scan statistic, we can use a (very simple) Poisson GLM to estimate the expected value of the count in each county and year, accounting for the different populations in each
region. Similar to the counts argument, the expected values should be passed as a matrix to the scan_eb_poisson function:
mod <- glm(count ~ offset(log(population)) + 1 + I(year - 1985),
family = poisson(link = "log"),
data = NM_popcas %>% filter(year < 1986))
ebp_baselines <- NM_popcas %>%
filter(year >= 1986 & year < 1990) %>%
mutate(mu = predict(mod, newdata = ., type = "response")) %>%
df_to_matrix(value_col = "mu")
Note that the population numbers are (perhaps poorly) interpolated from the censuses conducted in 1973, 1982, and 1991.
We can now calculate the Poisson scan statistic. To give us more confidence in our detection results, we will perform 999 Monte Carlo replications, by which data is generated using the parameters
from the null hypothesis and a new scan statistic calculated. This is then summarized in a P-value, calculated as the proportion of times the replicated scan statistics exceeded the observed one. The
output of scan_poisson is an object of class “scanstatistic”, which comes with the print method seen below.
poisson_result <- scan_eb_poisson(counts = counts,
zones = zones,
baselines = ebp_baselines,
n_mcsim = 999)
#> Data distribution: Poisson
#> Type of scan statistic: expectation-based
#> Setting: univariate
#> Number of locations considered: 32
#> Maximum duration considered: 4
#> Number of spatial zones: 415
#> Number of Monte Carlo replicates: 999
#> Monte Carlo P-value: 0.005
#> Gumbel P-value: NULL
#> Most likely event duration: 4
#> ID of locations in MLC: 15, 26
As we can see, the most likely cluster for an anomaly stretches from 1986-1989 and involves the locations numbered 15 and 26, which correspond to the counties
These are the same counties detected by Kulldorff (1998), though their analysis was retrospective rather than prospective as ours was. Ours was also data dredging as we used the same study period
with hopes of detecting the same cluster.
A heuristic score for locations
We can score each county according to how likely it is to be part of a cluster in a heuristic fashion using the function score_locations, and visualize the results on a heatmap as follows:
# Calculate scores and add column with county names
county_scores <- score_locations(poisson_result, zones)
county_scores %<>% mutate(county = factor(counties[-length(counties)],
levels = levels(NM_geo$county)))
# Create a table for plotting
score_map_df <- merge(NM_map, county_scores, by = "county", all.x = TRUE) %>%
arrange(group, order)
# As noted before, Cibola county counts have been attributed to Valencia county
score_map_df[score_map_df$subregion == "cibola", ] %<>%
mutate(relative_score = score_map_df %>%
filter(subregion == "valencia") %>%
select(relative_score) %>%
.[[1]] %>% .[1])
ggplot() +
geom_polygon(data = score_map_df,
mapping = aes(x = long, y = lat, group = group,
fill = relative_score),
color = "grey") +
scale_fill_gradient(low = "#e5f5f9", high = "darkgreen",
guide = guide_colorbar(title = "Relative\nScore")) +
geom_text(data = NM_geo,
mapping = aes(x = center_long, y = center_lat, label = county),
alpha = 0.5) +
ggtitle("County scores")
A warning though: the score_locations function can be quite slow for large data sets. This might change in future versions of the package.
Finding the top-scoring clusters
Finally, if we want to know not just the most likely cluster, but say the five top-scoring space-time clusters, we can use the function top_clusters. The clusters returned can either be overlapping
or non-overlapping in the spatial dimension, according to our liking.
top5 <- top_clusters(poisson_result, zones, k = 5, overlapping = FALSE)
# Find the counties corresponding to the spatial zones of the 5 clusters.
top5_counties <- top5$zone %>%
purrr::map(get_zone, zones = zones) %>%
purrr::map(function(x) counties[x])
# Add the counties corresponding to the zones as a column
top5 %<>% mutate(counties = top5_counties)
The top_clusters function includes Monte Carlo and Gumbel P-values for each cluster. These P-values are conservative, since secondary clusters from the original data are compared to the most likely
clusters from the replicate data sets.
Other univariate scan statistics can be calculated practically in the same way as above, though the distribution parameters need to be adapted for each scan statistic.
If you think this package lacks some functionality, or that something needs better documentation, please open an issue here.
Allévius, B., M. Höhle (2017): An expectation-based space-time scan statistic for ZIP-distributed data, (under review).
Kleinman, K. (2015): Rsatscan: Tools, Classes, and Methods for Interfacing with SaTScan Stand-Alone Software, https://CRAN.R-project.org/package=rsatscan.
Kulldorff, M., Athas, W. F., Feuer, E. J., Miller, B. A., Key, C. R. (1998): Evaluating Cluster Alarms: A Space-Time Scan Statistic and Brain Cancer in Los Alamos, American Journal of Public Health
88 (9), 1377–80.
Kulldorff, M. (2001), Prospective time periodic geographical disease surveillance using a scan statistic, Journal of the Royal Statistical Society, Series A (Statistics in Society), 164, 61–72.
Kulldorff, M., Heffernan, R., Hartman, J., Assunção, R. M., Mostashari, F. (2005): A space-time permutation scan statistic for disease outbreak detection, PLoS Medicine, 2 (3), 0216-0224.
Neill, D. B., Moore, A. W., Sabhnani, M., Daniel, K. (2005): Detection of Emerging Space-Time Clusters, In Proceedings of the Eleventh ACM SIGKDD International Conference on Knowledge Discovery in
Data Mining, 218–27. ACM.
Neill, D. B., Moore, A. W., Cooper, G. F. (2006): A Bayesian Spatial Scan Statistic, Advances in Neural Information Processing Systems 18: Proceedings of the 2005 Conference.
Tango, T., Takahashi, K. Kohriyama, K. (2011), A Space-Time Scan Statistic for Detecting Emerging Outbreaks, Biometrics 67 (1), 106–15. | {"url":"https://cran-r.c3sl.ufpr.br/web/packages/scanstatistics/readme/README.html","timestamp":"2024-11-10T13:55:31Z","content_type":"application/xhtml+xml","content_length":"40361","record_id":"<urn:uuid:159e1eb9-85f2-4372-ae2f-622bfb91a619>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00026.warc.gz"} |
The observed point pattern, from which an estimate of \(K(r)\) will be computed. An object of class "ppp", or data in any format acceptable to as.ppp().
Numeric values giving the range of angles inside which points will be counted. Angles are measured in degrees (if units="degrees", the default) or radians (if units="radians") anti-clockwise from
the positive \(x\)-axis.
Units in which the angles begin and end are expressed.
Optional. Vector of values for the argument \(r\) at which \(K(r)\) should be evaluated. Users are advised not to specify this argument; there is a sensible default.
This argument is for internal use only.
Optional. A character vector containing any selection of the options "none", "border", "bord.modif", "isotropic", "Ripley", "translate", "translation", "none", "good" or "best". It specifies the
edge correction(s) to be applied. Alternatively correction="all" selects all options.
Optional window. The first point \(x_i\) of each pair of points will be constrained to lie in domain.
Logical. If TRUE, the numerator and denominator of each edge-corrected estimate will also be saved, for use in analysing replicated point patterns.
Logical value indicating whether to print progress reports and warnings. | {"url":"https://www.rdocumentation.org/packages/spatstat.explore/versions/3.0-6/topics/Ksector","timestamp":"2024-11-12T17:07:53Z","content_type":"text/html","content_length":"72457","record_id":"<urn:uuid:7541be58-df9c-4eaa-8638-b96832461890>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00165.warc.gz"} |
Use the Fig. 9.57 showing the layout of a farm house: Each banana tree required 1.25 m² of ground space. How many banana trees can there be in the orchard
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
Use the Fig. 9.57 showing the layout of a farm house:
Each banana tree required 1.25 m² of ground space. How many banana trees can there be in the orchard?
From the figure,
Area covered by banana orchard = 20 m x 15.7 m
= 314 m²
Given that 1.25 m² area is required for each banana tree
The number of trees in 314 m² area = 314/1.25
= 251.2
≈ 251 trees
Therefore, 251 banana trees can be in the orchard.
✦ Try This: How much distance, in metres, a wheel of 45 cm radius will cover if it rotates 550 times?
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 11
NCERT Exemplar Class 7 Maths Chapter 9 Problem 112 (d)
Use the Fig. 9.57 showing the layout of a farm house: Each banana tree required 1.25 m² of ground space. How many banana trees can there be in the orchard?
Using the Fig. 9.57 showing the layout of a farm house, each banana tree required 1.25 m² of ground space. 251 banana trees can be there in the orchard.
☛ Related Questions:
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/ncert-solutions/use-the-fig-9-57-showing-the-layout-of-a-farm-house-each-banana-tree-required-1-25-m2-of-ground-space-how-many-banana-trees-can-there-be-in-the-orchard/","timestamp":"2024-11-02T17:39:56Z","content_type":"text/html","content_length":"226214","record_id":"<urn:uuid:7537d90c-a5c7-49c8-ba1e-b0dbc21acb82>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00864.warc.gz"} |
A typical machine will use a sequence of repetitive steps that can be clearly identified. Ladder logic can be written that follows this sequence. The steps for this design method are;
2. Write the steps of operation in sequence and give each step a number.
3. For each step assign a bit.
4. Write the ladder logic to turn the bits on/off as the process moves through its states.
5. Write the ladder logic to perform machine functions for each step.
6. If the process is repetitive, have the last step go back to the first.
Consider the example of a flag raising controller in Figure 9.1 A Process Sequence Bit Design Example and Figure 9.1 A Process Sequence Bit Design Example (continued). The problem begins with a
written description of the process. This is then turned into a set of numbered steps. Each of the numbered steps is then converted to ladder logic.
Figure 9.1 A Process Sequence Bit Design Example
Figure 9.1 A Process Sequence Bit Design Example (continued)
The previous method uses latched bits, but the use of latches is sometimes discouraged. A more common method of implementation, without latches, is shown in Figure 9.1 Process Sequence Bits Without
Figure 9.1 Process Sequence Bits Without Latches
Similar methods are explored in further detail in the book Cascading Logic (Kirckof, 2003). | {"url":"https://engineeronadisk.com/V2/book_PLC/engineeronadisk-105.html","timestamp":"2024-11-13T17:28:13Z","content_type":"text/html","content_length":"4052","record_id":"<urn:uuid:47bbb56d-fd15-456c-897a-1a449160b774>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00011.warc.gz"} |
Understanding the Magnetic Insulation Boundary Condition
When using the AC/DC Module, the most common interface used for solving electromagnetic fields is the Magnetic Fields interface. This interface solves for the magnetic vector potential field, and
from that computes the electric field. This guide addresses the default boundary condition within the Magnetic Fields interface, the Magnetic Insulation (MI) boundary condition.
The Mathematical Form of the Equation and Its Implications
Within the Magnetic Fields interface, the Magnetic Insulation boundary condition enforces the equation:
This equation means that the cross product of the normal vector to the surface,
There are three quantities that are defined from the
Since the
where the
This means that, for a steady-state model, the What is Gauge Fixing? A Theoretical Introduction".
It is useful to visualize these vector fields on a surface, as shown below. The fields are excited via a sinusoidally time-varying excitation from within the model, and this visualization considers
only a small patch of the boundary. The
An animation of the magnetic vector potential (gray arrows) over a surface (transparent blue) with the Magnetic Insulation boundary condition surrounding a sinusoidally time-varying source. The
colored surface is a visualization of the potential field on the surface.
An animat ion of the magnetic vector potential (gray arrow) as well as the magnetic field (red arrow), the surface current (green arrow), and the electric field (blue arrow).
There might appear to be a contradiction in the above equations, as they mean that current will flow tangentially to the surface, even though the electric field has no component tangent to the
surface. This appears to be in contradiction with the constitutive relation:
The Magnetic Insulation Boundary Condition as a Type of Symmetry Condition
A common example of a material with very high conductivity is the layer of metal on a mirror, a plane that produces a symmetric image. If the Magnetic Insulation boundary condition is applied at a
planar boundary of the modeling domain, and if no portion of the modeling domain extends past that boundary, then it can be interpreted as a symmetry condition. All geometry of the modeled structure
is mirrored about this plane, but the polarity of sources is switched.
As an example, consider a loop of conductive wire connected to a DC source (a battery) leading to an electric field and current flow through the material, and a magnetic field in the surrounding
space. A Magnetic Insulation boundary condition on one side of this loop implies that all of the structure exists on the other side of the plane, but with the terminals of the battery switched. This
leads to currents flowing along the mirrored path but in the opposite direction.
A battery connected to a wire and placed next to a Magnetic Insulation boundary condition. The current (green arrows) leads to a magnetic field (red arrows) The current flows through the battery and
wires. Note the polarity of the battery.
The implication of Magnetic Insulation as a symmetry boundary condition is that a second, mirrored, structure exists on the other side with the polarity of sources reversed, leading to currents
flowing in the opposite direction along the mirrored paths. The current flowing through the battery is not pictured.
This symmetry condition also holds if the conductors are going into the boundary. The figures below show how the fields are reflected when the wire intersects the boundary. Note how the polarity of
the source is reversed. Since only half of the source (the battery) exists on one side of the boundary, the magnitude of the potential due to that source (the battery voltage) is halved but excites
the same magnitude current through the system.
The Magnetic Insulation boundary condition can cut through a source and imposes symmetry such that the polarity of the source is reversed.
It is possible to combine multiple Magnetic Insulation boundary conditions on different sets of planar surfaces and interpret them as a symmetry condition as long as the effect of the reflection is
to fill the entirety of space. Both Cartesian symmetry and symmetry about an axis are allowed, as illustrated in the two figures below. For the case of symmetry about an axis, the angle of the
modeling space must be 180°/N, where N is an integer.
Three orthogonal Magnetic Insulation boundary conditions and their symmetry implication.
Symmetry about an axis is imposed via the Magnetic Insulation boundary conditions when the angle between the faces divides the unit circle into an even number of sectors.
If the object is placed at the center of a cubical domain with all boundaries set to Magnetic Insulation, this implies that the entirety of an infinite space is filled with a 3D, periodically
repeating pattern of the object, with spacing equal to the cube size.
An alternative type of symmetry can be imposed via the Perfect Magnetic Conductor boundary condition. It implies that a second, mirrored, structure exists on the other side with the same polarity of
sources, leading to currents flowing in the same direction along the mirrored paths.
There are two additional boundary conditions that can be used to model other types of symmetry. First, the Perfect Magnetic Conductor boundary condition similarly imposes mirror symmetry on the
structure but does not imply a switch of the polarity of sources. Second, the Periodic Condition boundary condition imposes that the fields on two boundaries of identical shape have identical fields.
For additional details on these conditions, see: "Exploiting Symmetry to Simplify Magnetic Field Modeling".
The Magnetic Insulation Boundary Condition as an Approximation of Infinite Free Space
The previous section showed that a model of a system sitting at the center of a cubical domain with Magnetic Insulation boundaries is equivalent to an infinite 3D periodic pattern of that same
system. There is a second interpretation of the Magnetic Insulation boundary conditions that entirely surround the system of interest, and that is as an approximation of infinite free space. Under
this interpretation it is not necessary to restrict the Magnetic Insulation boundaries to be planar, but they do have to define a convex domain entirely enclosing the system. A set of spherical
surfaces enclosing the system represents a good way to understand this interpretation.
Enclosing the system within a convex set of Magnetic Insulation boundary conditions represents an approximation of free space. Current flows both through the battery and wire, and the resultant
magnetic field intensity drops off with distance.
Assuming that the system sits within an infinite space, empty of any other structure, the magnetic fields due to the current flow will extend to infinity but drop off exponentially with increasing
radial distance. Consider two points, one sitting just within the sphere and the other just outside. We place ourselves at the point outside of the sphere and, if we temporarily ignore the Magnetic
Insulation boundary conditions, know that there must be some nonzero magnetic field due to the currents flowing through the system.
To get a zero magnetic field at this point, one needs to introduce a canceling current. One could introduce a canceling current that is exactly opposite the current flow already in the device, but
this would simply lead to a system with no current flow and no magnetic fields. It is also possible to introduce currents on the surface of the sphere. An appropriate current distribution over this
surface will lead to zero fields outside of the sphere. The Magnetic Insulation boundary condition, which allows currents to flow along the surface, can be thought of as solving for this canceling
surface current.
The surface currents on the Magnetic Insulation boundary condition are the currents that will lead to zero magnetic fields outside of the domain.
As the radius of the domain is increased, the surface currents will decrease in magnitude and have less and less of an effect on the fields just within the boundaries. However, increasing the radius
of the surrounding sphere will increase the domain size and overall computational cost, so it is desirable to keep this domain as small as possible. The quantities of interest, such as the system
inductance and impedance, can be studied with increasing domain size; these will converge monotonically. The Perfect Magnetic Conductor boundary condition can alternatively be used to truncate the
modeling domain. With increasing domain size, this approach will converge toward the identical solution. For additional details on this approach, as well as using infinite elements, and the tradeoffs
involved, see: "How to Choose Between Boundary Conditions for Coil Modeling".
The Magnetic Insulation Boundary Condition as an Electrically Unconnected Perfect Electric Conductor
The previous section showed that there are currents flowing along the surface of a Magnetic Insulation boundary condition, and that these exist even in the case of a DC analysis, implying no
variation with respect to time. This leads to results that require some careful interpretation. Consider the system shown below, of a sphere placed within the center of a loop of the wire connected
to a DC source, with free space between the wire and sphere. The currents on the surface of this sphere are plotted, along with the currents through the wire. This might appear counterintuitive — how
can a DC current flowing through a wire induce current to flow in an electrically unconnected object?
A spherical set of Magnetic Insulation boundary condition placed inside of a wire with DC current. There are surface currents induced on these boundaries.
An appropriate way to think about this is to view a stationary model as a frequency-domain model being excited at a very low, albeit nonzero, frequency and to introduce the concept of skin depth.
This is the distance into a conductor over which the majority of the induced current will flow. A simple expression for the skin depth within a good conductor is:
The Magnetic Insulation boundary condition means that Magnetic Insulation boundary condition is an idealization that should typically not be used to approximate the boundaries of a highly conductive
domain that is electrically unconnected.
On the other hand, when solving a time-domain problem, it can be valid to interpret these currents as an approximation of the currents flowing near the surface of a good conductor and the resultant
shielding. This does depend on how the system is excited and the objectives of the model. Verifying the validity of the approximation will require comparing against a model that solves for the fields
within the volume of the good conductor.
When solving a frequency-domain model, the Magnetic Insulation boundary condition has a more clear interpretation if modeling a real conductor where the skin depth is much smaller than the part
dimensions. In this case, it is a reasonable approximation, as long as the losses on the conductive material are not of interest. Alternatively, the Impedance Boundary Condition models the boundary
of a domain with finite conductivity.
A set of Magnetic Insulation boundary conditions on interior boundaries. The magnetic vector potential is solved for on both sides, so fields and surface currents will be different on either side.
For an AC excitation, this will reasonably approximate the shielding due to a thin conductor.
It can also be reasonable to apply a Magnetic Insulation boundary condition to an interior boundary, where the magnetic vector potential is being solved for on both sides, and there will be a
different magnetic field, electric field, and surface current on either side of the boundary. This can be appropriate for modeling a relatively thin-walled structure. In the frequency domain, this
implies that the skin depth is much smaller than the wall thickness and thus represents a perfect shielding condition. Alternatively, the Transition Boundary Condition models interior boundaries of
thin-walled lossy materials.
The Magnetic Insulation Boundary Condition as an Electrical Ground Condition
Here, an electrical ground is meant to be any domain that is relatively large compared to the system in question, or has a relatively high electric conductivity, or both, that can act as an infinite
source and sink of current. This includes both earth ground (such as bedrock, soil, or a body of water) as well as chassis ground (such as an airplane fuselage or car frame). Any part of the
electrical circuit that is nearby or touching this boundary will lead to induced currents flowing along the surfaces.
A Magnetic Insulation boundary condition of any shape can be interpreted as this ground condition. It is important to keep in mind that the currents flowing on the surface are induced due to the
structure of the nearby electrical system; they do not follow the shortest path. For time-domain and frequency-domain analyses, this is reasonable. On the other hand, for stationary analysis, the
currents through the medium of the domain should be modeled if accuracy of the nearby fields is of interest.
A Magnetic Insulation boundary condition of any shape can be interpreted as a ground condition. Currents are induced on the surface due to the time variation of currents flowing within the modeling
A structure within the modeling domain with the Magnetic Insulation boundary condition applied to the boundaries can be interpreted as a chassis ground. Connecting the electrical circuit to this
structure will also lead to currents flowing along the surface, and these should similarly be interpreted as induced currents and hence will not necessarily take the shortest return path through the
When modeling earth or chassis ground in frequency-domain models, the Impedance Boundary Condition and the Transition boundary conditions are alternatives that will additionally compute losses. In
time-domain models, the entire volume of the material might need to be modeled. In DC models, it is often sufficient to only consider current flow through part of the domain.
The Magnetic Insulation boundary condition on the boundaries of a structure sitting within the modeling domain can be interpreted as a chassis ground when electrically connected to a source. If this
was instead a DC model, only the part of the frame between the wires would need to be modeled.
The Magnetic Insulation Boundary Condition as the Boundaries of a Source
There are many ways to excite a system. Currents can be imposed to flow along edges, boundaries, or through volumes. These currents, at any one instant in time, are flowing into one terminal of the
source and out of the other terminal, thus maintaining conservation of the current. Although the details of the current flow through the interior of source are assumed to not be relevant, there must
still always be a way for the current to flow in a solenoidal path. The Magnetic Insulation boundary condition can be applied at the boundaries of the source, and in this context it represents a
current closure that shields the interior details of the source. This interpretation is valid in DC, AC, and transient modeling regimes. The geometry of the source can be simplified, es pec ially if
the source is far away from the fields of interest.
Current flowing around the loop of the wire are driven by a source. The currents flowing on the surface of the Magnetic Insulation boundary condition represent the interior of the source.
The Magnetic Insulation Boundary Condition as a Waveguiding Condition
When solving in the time or frequency domain, there will be a nonzero electric field, and as a consequence there will be a nonzero Poynting vector, or power flow, which is:
At a Magnetic Insulation boundary condition, this vector has to be parallel to the surface current and does not vary in time. The implication of this is that power can be flowing through the adjacent
modeled medium and will be flowing parallel to the Magnetic Insulation boundary condition. This interpretation is more applicable at higher frequencies, when the size of the model is comparable to
the wavelength within the material. When solving in the frequency domain, the Impedance Boundary Condition and the Transition Boundary Condition can also be used to model a waveguiding structure.
A model of the interior of a coaxial cable. The power flow (magenta arrows) within the modeling domain is parallel to the current flow on the surfaces.
The Magnetic Insulation boundary condition has several different interpretations. You are free to choose whichever interpretation of this boundary condition that you want for the modeling purposes at
hand. Currents are always free to flow along any set of adjacent Magnetic Insulation boundaries. These induced surface currents will often require different interpretations when evaluating the model,
so it is important to be familiar with all of the implications of using this boundary condition. | {"url":"https://www.comsol.de/support/learning-center/article/86381","timestamp":"2024-11-12T10:38:53Z","content_type":"text/html","content_length":"88324","record_id":"<urn:uuid:e603e21d-c727-4320-8932-386af3b22ac3>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00608.warc.gz"} |
Technical Analysis from A to Z
Technical Analysis from A to Z
by Steven B. Achelis
The New Highs/Lows Ratio ("NH/NL Ratio") displays the daily ratio between the number of stocks reaching new 52-week highs and the number of stocks reaching new 52-week lows.
The NH/NL Ratio is another useful method to visualize the relationship of stocks that are making new highs and new lows. High readings indicate that a large number of stocks are making new highs
(compared to the number of stocks making new lows). Readings less than one indicate that the number of stocks making new highs are less than the number of stocks making lows.
Refer to the New Highs-New Lows indicator for more information on interpreting the NH/NL Ratio.
The following chart shows the S&P 500 and the NH/NL Ratio.
The Ratio increased dramatically when the S&P 500 began making new highs in 1990. However, as the S&P has continued to move on to new highs, the Ratio has failed to reach new highs. This implies that
the S&P 500 is weaker than it appears.
This online edition of Technical Analysis from A to Z is reproduced here with permission from the author and publisher. | {"url":"https://www.metastock.com/customer/resources/taaz/?p=78","timestamp":"2024-11-08T04:29:24Z","content_type":"text/html","content_length":"147816","record_id":"<urn:uuid:f839c62a-ea6f-4682-8985-55f6faa34dbd>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00589.warc.gz"} |
Greater Than Tree
Solution to LeetCode Problem 538
Example of a Binary Search Tree
I recently interviewed for a junior developer position and wanted to share the solution to one of the interview problems I had. The problem is called Convert BST to Greater Tree, and it happens to be
on LeetCode.
The basis of the problem is to take a binary search tree and convert it to another tree where the node values are equal to the sum of itself and all of the node values greater than itself.
Greater than tree conversion
The solution to this problem is actually pretty straightforward. We want to traverse our tree to the right, then add the value of the node to a sum variable, then return to the root node, and finally
traverse left. It’s actually just reversed inorder traversal. Because we need to visit each node in the tree, the time complexity is O(N). I used a recursive approach to solve the problem, which is
And that’s it. If you are familiar with binary search trees, this should be a fairly easy problem to solve. Thanks for reading! | {"url":"https://julianrosenthal.medium.com/greater-than-tree-65e94ddd5338","timestamp":"2024-11-14T11:51:29Z","content_type":"text/html","content_length":"93696","record_id":"<urn:uuid:cab5199f-61ba-40ca-b2df-81a9d00e097b>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00346.warc.gz"} |
Understanding Mathematical Functions: How To Find Horizontal Intercept
Understanding Mathematical Functions and Horizontal Intercepts
Mathematical functions play a fundamental role in various fields such as engineering, economics, and statistics. They are used to model relationships between different variables and are crucial for
making predictions, analyzing data, and solving real-world problems. One important aspect of mathematical functions is finding their horizontal intercepts, which hold significant value in graphing
functions and solving equations.
A Definition of mathematical functions and their role in various fields
Mathematical functions can be defined as a rule that assigns to each input value exactly one output value. In other words, it takes an input, performs certain operations on it, and produces an
output. Functions are used in a wide range of fields such as engineering, economics, physics, and computer science to model relationships between different quantities and to make predictions about
real-world phenomena.
An overview of what horizontal intercepts represent and their significance in graphing functions
A horizontal intercept of a function is a point on the graph where the function intersects the x-axis. This means that the value of the function at the horizontal intercept is zero. In graphing
functions, horizontal intercepts provide crucial information about where the function crosses the x-axis, which helps in understanding the behavior of the function and its relationship with the input
variable. Horizontal intercepts also provide valuable insights into the roots or solutions of the function.
The importance of understanding horizontal intercepts for solving real-world problems
Understanding horizontal intercepts is crucial for solving real-world problems that involve finding the roots of functions or analyzing the behavior of a system. For example, in economics, finding
the horizontal intercepts of a demand or supply function can help in determining the equilibrium price or quantity of a product. In engineering, horizontal intercepts of a system model can provide
insights into the stability and performance of the system. Therefore, having a thorough understanding of horizontal intercepts is essential for making informed decisions and solving practical
Key Takeaways
• Horizontal intercept is where function crosses x-axis.
• Set y = 0 and solve for x.
• Use algebra to isolate x in the equation.
• Graphically, horizontal intercept is the x-coordinate of the point.
• Understanding horizontal intercept helps analyze function's behavior.
The Concept of Horizontal Intercepts
Horizontal intercepts are the points at which a function crosses the x-axis on a graph. These points are also known as x-intercepts or roots of the function. Understanding how to find horizontal
intercepts is essential in analyzing the behavior of a function and solving equations.
Explanation of horizontal intercepts as the points where the function crosses the x-axis
When graphing a function, the horizontal intercepts are the points where the graph intersects the x-axis. At these points, the value of y (or the function's output) is zero. In other words, the
x-values at the horizontal intercepts are the solutions to the equation f(x) = 0, where f(x) represents the function.
The relationship between horizontal intercepts and the roots or zeros of a function
The horizontal intercepts of a function are directly related to the roots or zeros of the function. The roots of a function are the values of x for which the function equals zero. Therefore, the
horizontal intercepts represent the x-values of the roots of the function. Finding the horizontal intercepts is equivalent to solving the equation f(x) = 0 to determine the roots of the function.
Understanding that a function may have multiple, one, or no horizontal intercepts
It's important to note that a function may have multiple, one, or no horizontal intercepts. If a function has multiple horizontal intercepts, it means that the graph of the function crosses the
x-axis at more than one point. If a function has only one horizontal intercept, the graph intersects the x-axis at a single point. On the other hand, if a function has no horizontal intercepts, the
graph does not intersect the x-axis at any point.
Finding Horizontal Intercepts Algebraically
One of the fundamental concepts in understanding mathematical functions is finding their horizontal intercepts. This process involves determining the points at which a function crosses the x-axis. By
setting the function equal to zero, we can solve for the x-values where the function intersects the x-axis.
A Step-by-step method to find horizontal intercepts by setting the function equal to zero
To find the horizontal intercepts of a function algebraically, we can follow a step-by-step method:
• Step 1: Set the function equal to zero: f(x) = 0
• Step 2: Solve for x by using algebraic techniques such as factoring, the quadratic formula, or other methods depending on the type of function
• Step 3: The solutions for x represent the x-coordinates of the horizontal intercepts
Illustration of this process with various types of functions, such as linear, quadratic, and polynomial functions
Let's illustrate the process of finding horizontal intercepts with different types of functions:
Linear Function: For a linear function f(x) = mx + b, setting f(x) = 0 gives us mx + b = 0. Solving for x, we get x = -b/m, which represents the x-coordinate of the horizontal intercept.
Quadratic Function: For a quadratic function f(x) = ax^2 + bx + c, setting f(x) = 0 gives us ax^2 + bx + c = 0. We can use the quadratic formula or factorization to solve for x and find the
horizontal intercepts.
Polynomial Function: For a polynomial function of higher degree, the process involves setting the function equal to zero and using algebraic techniques such as factoring or synthetic division to find
the horizontal intercepts.
Techniques on simplifying equations to make the process of finding horizontal intercepts more manageable
When dealing with complex functions, simplifying the equations can make the process of finding horizontal intercepts more manageable. Techniques such as factoring, grouping like terms, and using the
rational root theorem for polynomial functions can help simplify the equations and make it easier to solve for the horizontal intercepts.
Graphical Interpretation and Analysis
Understanding mathematical functions involves analyzing their graphical representations. One important aspect of this analysis is identifying the horizontal intercepts of a function, which are the
points where the function crosses the x-axis. This chapter will discuss how to use graphs to visually identify horizontal intercepts, provide tips on accurately sketching functions to locate
intercepts, and highlight the importance of graphing calculators and software in finding intercepts.
A. Using graphs to visually identify horizontal intercepts
Graphs provide a visual representation of functions, making it easier to identify their key features, including horizontal intercepts. When graphing a function, the horizontal intercepts are the
points where the graph crosses the x-axis. These points are crucial in understanding the behavior of the function and its relationship with the x-axis.
By examining the graph of a function, you can visually identify the x-coordinates of the horizontal intercepts. This visual approach allows you to quickly grasp the behavior of the function and
locate the points where it intersects with the x-axis.
B. Tips on how to accurately sketch functions to locate intercepts
Sketching functions accurately is essential for locating intercepts. When sketching a function, it's important to consider the key characteristics of the function, such as its shape, direction, and
points of intersection with the axes.
Tip 1: Start by identifying the key points of the function, such as the intercepts, maximum and minimum points, and points of inflection.
Tip 2: Pay attention to the behavior of the function as it approaches the x-axis, as this will help you accurately locate the horizontal intercepts.
Tip 3: Use a ruler or graphing software to ensure that your sketch is as accurate as possible, allowing you to pinpoint the exact location of the intercepts.
C. Discuss the importance of graphing calculators and software in finding intercepts
Graphing calculators and software play a crucial role in finding intercepts, especially for complex functions that are difficult to sketch by hand. These tools provide a more accurate and efficient
way to visualize functions and identify their horizontal intercepts.
With the use of graphing calculators and software, you can input the function and quickly generate its graph, allowing you to visually identify the horizontal intercepts with precision. This not only
saves time but also reduces the margin of error in locating intercepts.
Furthermore, graphing calculators and software offer advanced features such as zooming, tracing, and analyzing functions, which make it easier to explore the behavior of functions and locate their
intercepts in a more detailed manner.
Role of Horizontal Intercepts in Function Analysis
Horizontal intercepts play a crucial role in the analysis of mathematical functions. They provide valuable insights into the behavior and characteristics of a function, aiding in its understanding
and interpretation.
A The way horizontal intercepts aid in understanding the behavior of a function
The horizontal intercepts of a function, also known as the x-intercepts, are the points at which the function intersects the x-axis. These points are significant as they indicate the values of x for
which the function equals zero. By identifying these intercepts, we can gain a better understanding of the behavior of the function, particularly in relation to its roots and the points at which it
crosses the x-axis.
Understanding the horizontal intercepts allows us to determine the critical points of the function and analyze its behavior in different regions of the coordinate plane. This information is essential
for comprehending the overall nature of the function and its relationship with the x-axis.
B Horizontal intercepts in context of function's increasing and decreasing intervals and overall shape
The presence and location of horizontal intercepts are closely linked to the increasing and decreasing intervals of a function. By examining the x-intercepts, we can identify the intervals over which
the function is either increasing or decreasing. This insight is valuable for understanding the overall shape and behavior of the function, as well as its concavity and turning points.
Furthermore, horizontal intercepts contribute to the visualization of the function's graph, providing key points that help in sketching its shape and understanding its overall trajectory. They serve
as reference points for determining the behavior of the function as it extends across the coordinate plane.
C Examples demonstrating the application of horizontal intercepts in optimizing functions within real-world scenarios
The application of horizontal intercepts extends beyond theoretical analysis and finds practical relevance in real-world scenarios. For instance, in the field of economics, the horizontal intercepts
of a cost function can be used to optimize production levels and minimize costs. By identifying the points at which the cost function intersects the x-axis, businesses can make informed decisions
about production and pricing strategies.
Similarly, in engineering and physics, the horizontal intercepts of a function representing a physical phenomenon can provide insights into the behavior of the system and aid in optimizing its
performance. Understanding the x-intercepts allows for the identification of critical points and the determination of optimal conditions for various applications.
Overall, horizontal intercepts play a fundamental role in the analysis and interpretation of mathematical functions, offering valuable insights into their behavior, shape, and practical implications.
Troubleshooting Common Problems
When it comes to finding the horizontal intercept of a mathematical function, there are several common problems that individuals may encounter. Understanding these issues and knowing how to
troubleshoot them is essential for accurate calculations.
A Common mistakes made when attempting to find horizontal intercepts
One of the most common mistakes when attempting to find horizontal intercepts is incorrectly setting the function equal to zero. This can lead to inaccurate results and frustration. Additionally,
misinterpreting the x-intercept as the horizontal intercept can also lead to errors in calculations.
Another mistake is failing to consider the domain of the function. Some functions may have restrictions on the values of x for which they are defined, and overlooking this can result in incorrect
horizontal intercepts.
B How to check and verify the accuracy of calculated intercepts
After calculating the horizontal intercept of a function, it is important to check and verify the accuracy of the result. One way to do this is by graphing the function and visually inspecting the
point where it intersects the x-axis. This can help confirm the calculated intercept.
Another method is to substitute the calculated x-value back into the original function and ensure that the resulting y-value is indeed zero. If the y-value is not zero, then there may have been an
error in the calculation.
C Solutions to typical challenges encountered with complex functions, including higher-degree polynomials or rational functions
Complex functions, such as higher-degree polynomials or rational functions, can present unique challenges when trying to find horizontal intercepts. One common solution is to factor the function and
use the zero-product property to identify the x-intercepts. This method can be particularly useful for higher-degree polynomials.
For rational functions, it is important to identify any vertical asymptotes and holes in the graph, as these can affect the existence of horizontal intercepts. Understanding the behavior of the
function as x approaches infinity or negative infinity can also provide insight into the location of horizontal intercepts.
Overall, by being aware of these common mistakes, verifying the accuracy of calculated intercepts, and employing appropriate solutions for complex functions, individuals can effectively troubleshoot
and find the horizontal intercepts of mathematical functions.
Conclusion & Best Practices
Understanding how to find the horizontal intercept of a function is an essential skill in mathematics. It allows us to determine the points at which a function crosses the x-axis, providing valuable
information about the behavior and properties of the function.
A Recap of the importance of horizontal intercepts and their role in mathematical functions
Horizontal intercepts play a crucial role in understanding the behavior of a function. They provide insight into the roots or solutions of the function, helping us to identify where the function
equals zero. This information is valuable in various mathematical and real-world applications, such as determining break-even points in business or analyzing the motion of objects in physics.
Summary of best practices for finding and verifying horizontal intercepts, including cross-checking with graphical and algebraic methods
• Identify the function: Begin by clearly identifying the function for which you want to find the horizontal intercept. This may involve rearranging the function into standard form if necessary.
• Set y = 0: To find the horizontal intercept, set the function equal to zero and solve for the value of x. This will give you the x-coordinate of the intercept.
• Verify with graphical methods: Plot the function on a graph and visually identify the points where the function crosses the x-axis. This can serve as a helpful visual confirmation of the
horizontal intercepts.
• Use algebraic methods: If necessary, use algebraic techniques such as factoring or the quadratic formula to solve for the x-intercepts of the function.
• Cross-check your results: Always cross-check your calculated intercepts using both graphical and algebraic methods to ensure accuracy.
Encouragement to continue practicing with various functions to gain a solid understanding and proficiency in identifying horizontal intercepts
Like any mathematical skill, the ability to find horizontal intercepts improves with practice. I encourage you to continue working with various functions, including linear, quadratic, and
higher-order functions, to gain a solid understanding and proficiency in identifying horizontal intercepts. As you become more familiar with different types of functions, you will develop a keen
intuition for recognizing and analyzing horizontal intercepts in mathematical contexts. | {"url":"https://dashboardsexcel.com/blogs/blog/understanding-mathematical-functions-how-to-find-horizontal-intercept","timestamp":"2024-11-13T02:32:36Z","content_type":"text/html","content_length":"228087","record_id":"<urn:uuid:718dac38-f711-4afa-abd1-dd37a4ffa62b>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00279.warc.gz"} |
[Solved] The radius of the first permitted Bohr orbit, for the electr
The radius of the first permitted Bohr orbit, for the electron, in a hydrogen atom equals 0.51 Å and its ground state energy equals −13.6 eV. If the electron in the hydrogen atom is replaced by muon
(μ−) [charge same as electron and mass 207 me], the first Bohr radius and ground state energy will be,
Answer (Detailed Solution Below)
Option 1 : 2.56 × 10−13 m, −2.8 keV
NEET CT 1: Zoology (Structural Organisation in Animals (Animal Tissue))
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25 Questions 100 Marks 20 Mins
→The reduced mass is written as;
\(μ = \frac{mM}{m+M}\)
Here we have m is the mass of the first particle and M is the mass of the second particle.
→The Bohr radius is written as
\(r = \frac{n^2h^2}{4 \pi^2mkze^2}\)
Here we have m as the mass, h as Planck's constant, m is the mass, z is the atomic number, and e is the electron mass.
Given: m[μ] = 207m[e]
q[μ] = qe^-
and M[nucleus ]= 1836m[e]
radius, r = 0.51 Ȧ
The reduced mass is written as;
\(μ = \frac{m_eM_{nucleus}}{m_e+M_{nucleus}}\)
\(\Rightarrow μ = \frac{207m_e× 1836 m_e}{207m_e+ 1836 m_e}\)
\(\Rightarrow μ = \frac{380052 1836 m^2_e}{2043 m_e}\)
⇒μ = 186 m[e]
The radius of the first orbit is written as;
\(r_1 = \frac {m_e}{186 m_e} × 0.51\)
m⇒r[1] = 2.56×10^-13 m
and energy of the first Bohr is written as;
\(E_1 = \frac{μ}{m}E\)
\(\Rightarrow E_1 = \frac{186 m_e}{m_e}× (-13.6)\)
⇒E[1] = -2.8 keV
Hence, option 1) is the correct answer.
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Mode Superposition Quiz | Ansys Innovation Courses
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What are angles of elevation?
but i think
If a person stands and looks up at an object, the angle of elevation is the angle between the horizontal line of sight and the object. If a person stands and looks down at an object, the angle of
depression is the angle between the horizontal line of sight and the object.
3 Answers
angle of elevation is the angle from an horizontal line facing upward towards an object
maybe use a calculator of tan-1
angle of elevation is the angle between the horizontal line of sight and the object
Step 1 The two sides we know are Opposite (300) and Adjacent (400).
Step 2 SOHCAHTOA tells us we must use Tangent.
Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.
Step 4 Find the angle from your calculator using tan-1
The angle of elevation is angle formed on the horizontal ground from a point of looking upwards | {"url":"https://classhall.com/topics/what-are-angles-of-elevation/","timestamp":"2024-11-09T17:48:23Z","content_type":"text/html","content_length":"187010","record_id":"<urn:uuid:65c32b8e-64f2-41ff-bafb-1668edb0f8f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00209.warc.gz"} |
Determine 8183 - math word problem (8183)
Determine 8183
Determine the length of the rectangle if the width is 28cm in a ratio of 7:4 (d: s).
Correct answer:
Did you find an error or inaccuracy? Feel free to
write us
. Thank you!
Tips for related online calculators
You need to know the following knowledge to solve this word math problem:
Units of physical quantities:
Grade of the word problem:
We encourage you to watch this tutorial video on this math problem:
Related math problems and questions: | {"url":"https://www.hackmath.net/en/math-problem/8183","timestamp":"2024-11-03T23:40:22Z","content_type":"text/html","content_length":"47899","record_id":"<urn:uuid:22a5f1b0-767b-4b8f-b249-cdb479b89df8>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00331.warc.gz"} |
Eileen Wharmby
In memory of Eileen Wharmby, who passed away on December 15th, 2004, and to whom this Lounge is dedicated.
Despite the odds, we're still here, Eileen!
Re: Eileen Wharmby
Indeed Leif. I wonder how many current loungers are aware that she made all the Smillies used here, but to think it's already that it's already 8 years ago, quite amazing, hopefully, we'll be going
on for a long time. And hopefully more professional than the last place here
Cheers, Claude.
Re: Eileen Wharmby
It's sad that we don't have Eileen among us, but I'm happy that we have this place named after her, so that her name lives on.
Best wishes,
Re: Eileen Wharmby
It is a pity that more Internet communities didn't have Eileen's guiding hand. This place is unique.
Re: Eileen Wharmby
Merry Christmas, "Mum."
Re: Eileen Wharmby
Thank you for the reminder. It's good to remember people who made a difference.
Re: Eileen Wharmby
She will always be missed.
I am so far behind, I think I am First
Genealogy....confusing the dead and annoying the living
Re: Eileen Wharmby
As Dave said, she will forever be missed and remembered here.
Bob's yer Uncle
Dell Intel Core i5 Laptop, 3570K,1.60 GHz, 8 GB RAM, Windows 11 64-bit, LibreOffice,and other bits and bobs
Re: Eileen Wharmby
She created a very special place, indeed, and the reason we are here.
Thank you Leif and all admins!
Byelingual When you speak two languages but start losing vocabulary in both of them.
Re: Eileen Wharmby
This is a wonderful place she helped create, and I truly am thankful for her and everyone here!
Re: Eileen Wharmby
Eileen's Lounge is a wonderful place to come and I think it reflects what Eileen was all about. Helpfulness and friendliness!
I was lucky enough to have a little bit of interaction with her, which I thought was really special. And my regret was that I hadn't found the Lounge sooner than I had!
I've enjoyed my time spent here reading, interacting and learning! And I appreciate that this current lounge was created in her memory! Thank you for those responsible for keeping the rest of us
tucked into a safe and knowledgeable environment on the web!
A cup of coffee shared with a friend is happiness tasted and time well spent.
Re: Eileen Wharmby
Eileen was not only a leader by example, and had an ability to recruit and inspire a dedicated group of torchbearers, but she set the tone for a supportive, always considerate and civilised group of
experts and amateurs in "her" Lounge. A quick glance at so many other online computer groups reveals how exceptional this "family" is- the nastiness and ill-humour which pervades so many other groups
makes one's toes curl. Thank you Eileen and all who keep the good ship Eileen's Lounge afloat and thriving.
Re: Eileen Wharmby
Eileen sounds like a lovely lady, and I am SOOOO glad she established Eileen's Lounge, and the rules, the way she did. I've been on other discussion groups and bulletin boards, and sometimes the
sniping and mean-ness is just awful. This is a great place to learn and enjoy other folks. Thanks, all you Administrators and members, for making this a great place.
Re: Eileen Wharmby
Well, in fact Eileen passed away more than five years before this board was created. She was one of the founders of another board, Woody's Lounge, where Charlotte, Claude, Leif, Mike, Stuart and I
were administrators. When Woody's Lounge was taken over and became the Windows Secrets Lounge, we wanted to recreate the cosy atmosphere of the "old" Lounge, so we started a small new board, and to
honor Eileen's memory, we named it after her.
Best wishes,
Re: Eileen Wharmby
This is a really friendly board. I can't remember the last time the dreaded wart gun was used in anger!
Re: Eileen Wharmby
Hans, thank you!!!!
Re: Eileen Wharmby
I never knew Eileen.
But her legacy (clearly) still lives on. | {"url":"https://eileenslounge.com/viewtopic.php?style=33&p=87487","timestamp":"2024-11-14T21:55:40Z","content_type":"text/html","content_length":"52596","record_id":"<urn:uuid:b2253ea7-9f08-4e97-a25e-b3240a833f96>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00293.warc.gz"} |
Geometric Tools for Computer Graphics-MIRI
Master in Innovation and Research in Informatics (MIRI)
Departament de Matemàtica Aplicada II
Facultat d’Informàtica de Barcelona
Universitat Politècnica de Catalunya
This course has been designed to provide students with the mathematical geometric tools most ubiquitously used in computer graphics.
By the end of the course, students should be able to:
• Use and manipulate affine coordinates (homogeneous or not).
• Compute distance and angular measures.
• Describe and control linear objects.
• Describe and control parametrized curves and surfaces.
• Locate a given geometric object in the desired position in space, using different techniques.
1. Basics of affine and metric geometry
□ Vectorial spaces.
□ Affine spaces. Coordinate systems. Affine varieties in dimensions 2 and 3.
□ Euclidean spaces. Distances and angles. Projections. Cartesian coordinate systems.
□ Changing coordinates.
2. Geometric objects
□ Linear objects.
□ Curves in dimensions 2 and 3. Parametrizations. Rudiments of differential geometry of curves.
□ Surfaces in dimension 3. Parametrizations. Rudiments of differential geometry of surfaces. Intersections.
3. Affine transforms
□ Rigid motions, similarities and affinities.
□ Using quaternions in rotations.
The contents of the course being of mathematical nature, the activities associated with it are oriented accordingly:
• Theory classes: aimed at presenting and discussing the geometric techniques included in the syllabus, they will be mainly conducted by the instructor.
• Problems solving classes: students will present and discuss their solutions to the problems that will have been posed in advance.
• Laboratory classes: for the students to implement the above mentioned solutions to geometric problems. Emphasis will be put into implementing the geometric components of the solutions, while
essentially avoiding graphics issues.
• Exam: in addition to the homework (problems solving and implementing), there will be a final written exam, mainly devoted to problems solving.
The length of the course, together with its level, entail the need of some non irrelevant amount of student’s personal work. This includes exploring course topics beyond the material presented in
class, proposing solutions to new problems and proving the correctness of the proposed solutions, and designing and carrying out their implementations.
Along the course, students will get assigned some problems solving and implementing. This homework will be presented in class by the students, and revised by the instructor, giving as a result the
homework component of the final grade (H). There will also be a final written exam, mainly devoted to problems solving, which will give the exam component of the final grade (E). The final grade (F)
will be obtained by the following formula: F = max (E, (H+E)/2).
General information
The general information related to a specific semester of the course (calendar, schedule, class rooms, evaluation, etc.) can be found at the FIB web page. Announcements along the semester will be
done at the FIB corner (Racó). In this web page (which is always under construction) I will post some material for the course.
Material for the course
• Linear algebra. Need to refresh it?
• Conics:
• Quaternions:
□ Comparing Foley – Van Dam’s and the quaternions methods: An example.
□ Comparing Foley – Van Dam’s and the quaternions methods: General case.
• Euler angles:
□ Computing Euler’s angles: An example
□ Tait-Bryan angles
□ What is a gimbal lock and why it may be annoying? Take a look of this video.
• Algorithms
Are you looking for a textbook containing most of what we will see in class?
Are you interested in reading what computer graphics people consider to be the mathematical tools useful for their work?
Do you want to learn more about differential geometry on curves and surfaces?
Do you want to learn more about geometric algorithms?
Do you need help with SAGE?
Further resources
Mathematical edition is almost always and everywhere done using LaTeX. Not only it is used in universities (LaTeX has been used to write all the documents of this course, and probably all the
problems lists, exams, and other mathematical texts that you had in your hands during your previous studies) but it is the most extended editor of scientific texts (LaTeX is used to write all the
textbooks of scientific publishers as important as Springer, and also most of the mathematics and computer science conferences around the world).
It’s a free source software that has versions for all operating systems (Linux, Mac, Windoxs, etc.), it helps writing all sort of scientific texts, such as articles, books and presentations, while it
allows to incorporate figures previously produced in PDF by any drawing program.
You can download LaTeX from http://www.tug.org/
For those of you who work on windows, WinEdt is a very convenient editor for writing LaTeX code. You can download it from http://www.winedt.com/
Each person likes preparing his/her figures with his/her favorite drawing program. I use IPE (an evolution of xfig designed by a computational geometer), because it allows me to draw the geometric
figures that I need, because it integrates text in LaTeX, and because it allows me to also prepare my presentations in a very easy what-you-see-is-what-you-get way. It’s a free source software that
has versions for all the operating systems (Linux, Windoxs, Unix, etc.).
You can download IPE from http://ipe.otfried.org/
In addition to producing your own drawings, you may wish to create and experiment with geometric constructions. If so, I recommend GeoGebra: http://www.geogebra.org/
Thesis, grants, and projects | {"url":"https://dccg.upc.edu/people/vera/teaching/courses/geometric-tools-for-computer-graphics/","timestamp":"2024-11-05T00:21:37Z","content_type":"text/html","content_length":"54543","record_id":"<urn:uuid:02b29920-4a84-42fe-b6bc-bb21c59fa295>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00710.warc.gz"} |
The Real Number System
The real number system evolved over time by expanding the notion of what we mean by the word “number.” At first, “number” meant something you could count, like how many sheep a farmer owns. These are
called the natural numbers, or sometimes the counting numbers.
Natural Numbers
or “Counting Numbers”
1, 2, 3, 4, 5, . . .
• The use of three dots at the end of the list is a common mathematical notation to indicate that the list keeps going forever.
At some point, the idea of “zero” came to be considered as a number. If the farmer does not have any sheep, then the number of sheep that the farmer owns is zero. We call the set of natural numbers
plus the number zero the whole numbers.
Whole Numbers
Natural Numbers together with “zero”
0, 1, 2, 3, 4, 5, . . .
About the Number Zero
What is zero? Is it a number? How can the number of nothing be a number? Is zero nothing, or is it something?
Well, before this starts to sound like a Zen koan, let’s look at how we use the numeral “0.” Arab and Indian scholars were the first to use zero to develop the place-value number system that we use
today. When we write a number, we use only the ten numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. These numerals can stand for ones, tens, hundreds, or whatever depending on their position in the
number. In order for this to work, we have to have a way to mark an empty place in a number, or the place values won’t come out right. This is what the numeral “0” does. Think of it as an empty
container, signifying that that place is empty. For example, the number 302 has 3 hundreds, no tens, and 2 ones.
So is zero a number? Well, that is a matter of definition, but in mathematics we tend to call it a duck if it acts like a duck, or at least if it’s behavior is mostly duck-like. The number zero
obeys most of the same rules of arithmetic that ordinary numbers do, so we call it a number. It is a rather special number, though, because it doesn’t quite obey all the same laws as other
numbers—you can’t divide by zero, for example.
Note for math purists: In the strict axiomatic field development of the real numbers, both 0 and 1 are singled out for special treatment. Zero is the additive identity, because adding zero to a
number does not change the number. Similarly, 1 is the multiplicative identity because multiplying a number by 1 does not change it.
Even more abstract than zero is the idea of negative numbers. If, in addition to not having any sheep, the farmer owes someone 3 sheep, you could say that the number of sheep that the farmer owns is
negative 3. It took longer for the idea of negative numbers to be accepted, but eventually they came to be seen as something we could call “numbers.” The expanded set of numbers that we get by
including negative versions of the counting numbers is called the integers.
Whole numbers plus negatives
. . . –4, –3, –2, –1, 0, 1, 2, 3, 4, . . .
About Negative Numbers
How can you have less than zero? Well, do you have a checking account? Having less than zero means that you have to add some to it just to get it up to zero. And if you take more out of it, it will
be even further less than zero, meaning that you will have to add even more just to get it up to zero.
The strict mathematical definition goes something like this:
For every real number n, there exists its opposite, denoted – n, such that the sum of n and – n is zero, or
n + (– n) = 0
Note that the negative sign in front of a number is part of the symbol for that number: The symbol “–3” is one object—it stands for “negative three,” the name of the number that is three units less
than zero.
The number zero is its own opposite, and zero is considered to be neither negative nor positive.
Read the discussion of subtraction for more about the meanings of the symbol “–.”
The next generalization that we can make is to include the idea of fractions. While it is unlikely that a farmer owns a fractional number of sheep, many other things in real life are measured in
fractions, like a half-cup of sugar. If we add fractions to the set of integers, we get the set of rational numbers.
Rational Numbers
All numbers of the form a/b, where a and b are integers (but b cannot be zero)
Rational numbers include what we usually call fractions
• Notice that the word “rational” contains the word “ratio,” which should remind you of fractions.
The bottom of the fraction is called the denominator. Think of it as the denomination—it tells you what size fraction we are talking about: fourths, fifths, etc.
The top of the fraction is called the numerator. It tells you how many fourths, fifths, or whatever.
• RESTRICTION: The denominator cannot be zero! (But the numerator can)
If the numerator is zero, then the whole fraction is just equal to zero. If I have zero thirds or zero fourths, than I don’t have anything. However, it makes no sense at all to talk about a fraction
measured in “zeroths.”
• Fractions can be numbers smaller than 1, like 1/2 or 3/4 (called proper fractions), or they can be numbers bigger than 1 (called improper fractions), like two-and-a-half, which we could also
write as 5/2
All integers can also be thought of as rational numbers, with a denominator of 1:
This means that all the previous sets of numbers (natural numbers, whole numbers, and integers) are subsets of the rational numbers.
Now it might seem as though the set of rational numbers would cover every possible case, but that is not so. There are numbers that cannot be expressed as a fraction, and these numbers are called
irrational because they are not rational.
Irrational Numbers
• Cannot be expressed as a ratio of integers.
• As decimals they never repeat or terminate (rationals always do one or the other)
More on Irrational Numbers
It might seem that the rational numbers would cover any possible number. After all, if I measure a length with a ruler, it is going to come out to some fraction—maybe 2 and 3/4 inches. Suppose I then
measure it with more precision. I will get something like 2 and 5/8 inches, or maybe 2 and 23/32 inches. It seems that however close I look it is going to be some fraction. However, this is not
always the case.
Imagine a line segment exactly one unit long:
Now draw another line one unit long, perpendicular to the first one, like this:
Now draw the diagonal connecting the two ends:
Congratulations! You have just drawn a length that cannot be measured by any rational number. According to the Pythagorean Theorem, the length of this diagonal is the square root of 2; that is, the
number which when multiplied by itself gives 2.
According to my calculator,
But my calculator only stops at eleven decimal places because it can hold no more. This number actually goes on forever past the decimal point, without the pattern ever terminating or repeating.
This is because if the pattern ever stopped or repeated, you could write the number as a fraction—and it can be proven that the square root of 2 can never be written as
for any choice of integers for a and b. The proof of this was considered quite shocking when it was first demonstrated by the followers of Pythagoras 26 centuries ago.
Real Numbers
• Rationals + Irrationals
• All points on the number line
• Or all possible distances on the number line
When we put the irrational numbers together with the rational numbers, we finally have the complete set of real numbers. Any number that represents an amount of something, such as a weight, a volume,
or the distance between two points, will always be a real number. The following diagram illustrates the relationships of the sets that make up the real numbers.
An Ordered Set
The real numbers have the property that they are ordered, which means that given any two different numbers we can always say that one is greater or less than the other. A more formal way of saying
this is:
For any two real numbers a and b, one and only one of the following three statements is true:
1. a is less than b, (expressed as a < b)
2. a is equal to b, (expressed as a = b)
3. a is greater than b, (expressed as a > b)
The Number Line
The ordered nature of the real numbers lets us arrange them along a line (imagine that the line is made up of an infinite number of points all packed so closely together that they form a solid line).
The points are ordered so that
points to the right are greater than points to the left:
• Every real number corresponds to a distance on the number line, starting at the center (zero).
• Negative numbers represent distances to the left of zero, and positive numbers are distances to the right.
• The arrows on the end indicate that it keeps going forever in both directions.
Struggling with The Real Number System in your math class? Get expert assignment help to conquer your challenges today!
Absolute Value
When we want to talk about how “large” a number is without regard as to whether it is positive or negative, we use the absolute value function. The absolute value of a number is the distance from
that number to the origin (zero) on the number line. That distance is always given as a non-negative number.
In short:
• If a number is positive (or zero), the absolute value function does nothing to it:
• If a number is negative, the absolute value function makes it positive:
WARNING: If there is arithmetic to do inside the absolute value sign, you must do it before taking the absolute value—the absolute value function acts on the result of whatever is inside it. For
example, a common error is
The correct result is
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Math 407, Fall 2006
Here are the statements that we collected on Thursday. Arrange them into two columns,
so that the statements in the column on the right can be deduced from the ones on the left.
While you are doing this, you may find that some statements need to be made more
precise and that some terms need to be defined. Make a note of any such situations.
The sum of all angles created by lines intersecting at a point is 360 degrees.
The exterior angles of a polygon add up to 360 degrees.
Angles of a triangle add up to 180 degrees.
All right angles are equal.
Given a line and a point not on it, there a unique line through the point parallel to the line.
Opposite angles are equal.
Pythagoras's theorem.
Theorems about congruent triangles: SAS, SSS, ASA.
A triangle inscribed in a circle with one side on the diameter is a right triangle.
Given two parallel lines and another line transversal to them, corresponding angles are
equal and alternate interior angles are equal.
A straight line is a 180 degree angle.
Triangle inequality: the length of any side of a triangle is less than the sum of the lengths
of the two other sides.
In a triangle, the largest side is opposite the largest angle.
If two sides in a triangle are congruent, then the angles opposite the two congruent sides
are the same.
Triangles tesselate the plane.
If all angles are the same then the sides are proportional.
In a triangle, the three lines from each vertex to the midpoint of the opposite side all
intersect in a point.
The area of a triangle is half its base times its height. | {"url":"https://studylib.net/doc/17690203/math-407--fall-2006","timestamp":"2024-11-14T14:55:30Z","content_type":"text/html","content_length":"55230","record_id":"<urn:uuid:e11d15e3-91f5-4b2c-92e0-e73ed5651bce>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00842.warc.gz"} |
Distribution of shortest path lengths in subcritical Erdos-Rényi networks
Networks that are fragmented into small disconnected components are prevalent in a large variety of systems. These include the secure communication networks of commercial enterprises, government
agencies, and illicit organizations, as well as networks that suffered multiple failures, attacks, or epidemics. The structural and statistical properties of such networks resemble those of
subcritical random networks, which consist of finite components, whose sizes are nonextensive. Surprisingly, such networks do not exhibit the small-world property that is typical in supercritical
random networks, where the mean distance between pairs of nodes scales logarithmically with the network size. Unlike supercritical networks whose structure has been studied extensively, subcritical
networks have attracted relatively little attention. A special feature of these networks is that the statistical and geometric properties vary between different components and depend on their sizes
and topologies. The overall statistics of the network can be obtained by a summation over all the components with suitable weights. We use a topological expansion to perform a systematic analysis of
the degree distribution and the distribution of shortest path lengths (DSPL) on components of given sizes and topologies in subcritical Erdos-Rényi (ER) networks. From this expansion we obtain an
exact analytical expression for the DSPL of the entire subcritical network, in the asymptotic limit. The DSPL, which accounts for all the pairs of nodes that reside on the same finite component (FC),
is found to follow a geometric distribution of the form PFC(L=L<)=(1-c)c-1, where c<1 is the mean degree. Using computer simulations we calculate the DSPL in subcritical ER networks of increasing
sizes and confirm the convergence to this asymptotic result. We also obtain exact asymptotic results for the mean distance, (L)FC, and for the standard deviation of the DSPL, σL,FC, and show that the
simulation results converge to these asymptotic results. Using the duality relations between subcritical and supercritical ER networks, we obtain the DSPL on the nongiant components of ER networks
above the percolation transition.
Bibliographical note
Publisher Copyright:
© 2018 American Physical Society.
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Elliptic point
From Encyclopedia of Mathematics
A point on a regular surface at which the osculating paraboloid is an elliptic paraboloid. In an elliptic point the Dupin indicatrix is an ellipse, the Gaussian curvature of the surface is positive,
the principal curvatures of the surface are of the same sign, and for the coefficients of the second fundamental form of the surface the inequality
holds. In a neighbourhood of an elliptic point the surface is locally convex.
How to Cite This Entry:
Elliptic point. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Elliptic_point&oldid=31652
This article was adapted from an original article by D.D. Sokolov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
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Multiplying Mixed Numbers Worksheets
Multiplying Mixed Numbers Worksheets function as foundational tools in the realm of mathematics, providing an organized yet functional system for learners to check out and master numerical concepts.
These worksheets use an organized approach to comprehending numbers, nurturing a strong foundation whereupon mathematical proficiency thrives. From the simplest checking exercises to the ins and outs
of advanced estimations, Multiplying Mixed Numbers Worksheets accommodate students of varied ages and ability levels.
Unveiling the Essence of Multiplying Mixed Numbers Worksheets
Multiplying Mixed Numbers Worksheets
Multiplying Mixed Numbers Worksheets -
Multiplying Mixed Numbers Turn the two mixed numbers into fractions and proceed to find the product as you would normally do Some mixed numbers have common factors so don t forget to simplify your
product Grab the Worksheet Multiplying Mixed Numbers With Word Problems
Fractions worksheets Multiplying mixed numbers by mixed numbers Below are six versions of our grade 5 math worksheet on multiplying mixed numbers together These worksheets are pdf files Worksheet 1
Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6 5 More
At their core, Multiplying Mixed Numbers Worksheets are vehicles for theoretical understanding. They envelop a myriad of mathematical principles, guiding students via the maze of numbers with a
collection of interesting and purposeful workouts. These worksheets go beyond the limits of typical rote learning, urging energetic interaction and cultivating an instinctive understanding of
numerical partnerships.
Supporting Number Sense and Reasoning
Multiplying Mixed Numbers J Worksheet For 4th 6th Grade Lesson Planet
Multiplying Mixed Numbers J Worksheet For 4th 6th Grade Lesson Planet
Multiplying Mixed Numbers with Whole Numbers Worksheets Recalibrate kids practice with our free multiplying mixed numbers with whole numbers worksheets While multiplying a mixed number with a whole
number can trip up many children it s not as big of a deal as they think it is
These multiplying mixed numbers by mixed numbers worksheets will help to visualize and understand place value and number systems 5th and 6th grade students will learn basic multiplication methods
with mixed numbers and can improve their basic math skills with our free printable multiplying mixed numbers by mixed
The heart of Multiplying Mixed Numbers Worksheets hinges on growing number sense-- a deep understanding of numbers' definitions and interconnections. They urge expedition, welcoming students to
explore arithmetic procedures, figure out patterns, and unlock the secrets of sequences. Through thought-provoking obstacles and logical problems, these worksheets end up being gateways to developing
thinking skills, nurturing the logical minds of budding mathematicians.
From Theory to Real-World Application
Multiply Fractions With Mixed Numbers Worksheet Kidsworksheetfun
Multiply Fractions With Mixed Numbers Worksheet Kidsworksheetfun
These multiply mixed numbers worksheets engage children s cognitive processes and help enhance their creativity and memory retention skills Get started now to make learning interesting for your child
Convert Mixed to Improper Fractions 1 12 22 12 32 2 15 105 15 115 Multiply the fractions multiply the top numbers multiply bottom numbers 32 115 3 112 5 3310 Convert to a mixed number 3310 3 310 If
you are clever you can do it all in one line like this 1 12 2 15 32 115 3310 3 310 One More
Multiplying Mixed Numbers Worksheets act as channels bridging academic abstractions with the palpable truths of daily life. By infusing practical situations right into mathematical workouts, learners
witness the significance of numbers in their surroundings. From budgeting and measurement conversions to recognizing statistical information, these worksheets encourage trainees to possess their
mathematical prowess past the confines of the class.
Diverse Tools and Techniques
Flexibility is inherent in Multiplying Mixed Numbers Worksheets, using an arsenal of instructional devices to accommodate different learning styles. Visual help such as number lines, manipulatives,
and electronic resources work as buddies in imagining abstract principles. This diverse approach makes sure inclusivity, fitting learners with different choices, toughness, and cognitive designs.
Inclusivity and Cultural Relevance
In a progressively diverse globe, Multiplying Mixed Numbers Worksheets embrace inclusivity. They transcend cultural boundaries, integrating instances and problems that resonate with learners from
diverse backgrounds. By including culturally appropriate contexts, these worksheets promote a setting where every student really feels represented and valued, enhancing their connection with
mathematical ideas.
Crafting a Path to Mathematical Mastery
Multiplying Mixed Numbers Worksheets chart a course towards mathematical fluency. They infuse determination, critical thinking, and analytic skills, essential qualities not just in mathematics but in
various aspects of life. These worksheets encourage learners to navigate the complex terrain of numbers, supporting a profound admiration for the elegance and reasoning inherent in mathematics.
Welcoming the Future of Education
In an era noted by technological innovation, Multiplying Mixed Numbers Worksheets perfectly adapt to digital platforms. Interactive user interfaces and digital resources augment typical learning,
supplying immersive experiences that transcend spatial and temporal limits. This combinations of traditional methods with technical innovations proclaims a promising period in education and learning,
cultivating an extra vibrant and appealing understanding atmosphere.
Conclusion: Embracing the Magic of Numbers
Multiplying Mixed Numbers Worksheets represent the magic inherent in mathematics-- a charming trip of expedition, exploration, and mastery. They go beyond conventional rearing, functioning as
stimulants for firing up the flames of curiosity and questions. With Multiplying Mixed Numbers Worksheets, students start an odyssey, unlocking the enigmatic globe of numbers-- one trouble, one
solution, at a time.
Multiply Mixed Numbers Worksheets
Multiplying Mixed Numbers Worksheets
Check more of Multiplying Mixed Numbers Worksheets below
Fractions Worksheets Printable Fractions Worksheets For Teachers
16 Multiplying Mixed Numbers Worksheet Reginalddiepenhorst
Writing Fractions As Mixed Numbers
Grade 5 Fractions Worksheets Multiplying Mixed Numbers K5 Learning Multiplying And Dividing
Multiplying Mixed Numbers By Fractions 5th Grade Maths Worksheets
Multiplying Mixed Numbers Worksheets Teaching Resources
Grade 5 Fractions Worksheets Multiplying Mixed Numbers K5 Learning
Fractions worksheets Multiplying mixed numbers by mixed numbers Below are six versions of our grade 5 math worksheet on multiplying mixed numbers together These worksheets are pdf files Worksheet 1
Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6 5 More
Multiplying Mixed Numbers K5 Learning
Multiplying mixed numbers Grade 5 Fractions Worksheet Find the product 1 2 4 1 5 6 3 2 1 6 1 6 12 2 3 1 2 2 4 5 3 4 1 3 3 4 10 1 5 3 4 3 2 9 3 6 5 6 3 1 2 2 7 1 2 1 1 2 3 8 8 12 1 2 10 3 9 2 6 3 2 3
Fractions worksheets Multiplying mixed numbers by mixed numbers Below are six versions of our grade 5 math worksheet on multiplying mixed numbers together These worksheets are pdf files Worksheet 1
Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6 5 More
Multiplying mixed numbers Grade 5 Fractions Worksheet Find the product 1 2 4 1 5 6 3 2 1 6 1 6 12 2 3 1 2 2 4 5 3 4 1 3 3 4 10 1 5 3 4 3 2 9 3 6 5 6 3 1 2 2 7 1 2 1 1 2 3 8 8 12 1 2 10 3 9 2 6 3 2 3
Grade 5 Fractions Worksheets Multiplying Mixed Numbers K5 Learning Multiplying And Dividing
16 Multiplying Mixed Numbers Worksheet Reginalddiepenhorst
Multiplying Mixed Numbers By Fractions 5th Grade Maths Worksheets
Multiplying Mixed Numbers Worksheets Teaching Resources
Multiplying Mixed Numbers Worksheet Pdf Juvxxi
Multiplying Mixed Numbers Worksheet
Multiplying Mixed Numbers Worksheet
Worksheets For Fraction Multiplication | {"url":"https://szukarka.net/multiplying-mixed-numbers-worksheets","timestamp":"2024-11-08T01:09:27Z","content_type":"text/html","content_length":"26052","record_id":"<urn:uuid:187c2ad0-9356-4353-8f7d-3cdef7cd4175>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00801.warc.gz"} |
Normal Probability Distributions
Abdulla Javeri
30 years: Financial markets trader
Abdulla explains the normal distribution, which is one of the most widely used of all probability distributions. He discusses its basic characteristics and how to determine probability based on the
distance from the average.
Abdulla explains the normal distribution, which is one of the most widely used of all probability distributions. He discusses its basic characteristics and how to determine probability based on the
distance from the average.
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Normal Probability Distributions
Key learning objectives:
• Define and understand what the normal distribution is
• Identify the relationship between mean, standard deviation and the probability of a given outcome
The normal distribution is a popular technique in statistics that shows the probability of different outcomes in an experiment. Terms such as mean and standard deviation are crucial in being able to
interpret the curve.
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What is a normal distribution?
The normal distribution is a representation of the probabilities of outcomes. The curve is symmetrical around a central point, the mean. However, just because the curve is symmetrical, or bell
shaped, it doesn’t necessarily mean that it’s normally distributed. The term bell shaped seems to have been adopted to actually mean a normal distribution, so we’ll stay consistent with that. On the
x axis, the limits are minus infinity and plus infinity, and the curve is asymptotic, which means the curve never touches or crosses the x axis. The x axis represents the outcome of experiments, in
other words, numbers or percentages. And the y axis represents the probability of any given outcome.
Just by looking at it we can draw some conclusions.
• Firstly, the probability of a particular outcome is related to the distance from the mean
• Secondly, the probability of a particular outcome is related to the standard deviation
How is the shape of the curve related to the standard deviation?
We can draw a probability distribution for any level of standard deviation which in turn means that there are an infinite number of normally distributed bell-shaped curves. Some of which are shown in
the graphic. The shape of the curve is dependent on the standard deviation. It could be high and narrow if the standard deviation is low, or low and broad if the standard deviation is high.
How are the mean and standard deviation related to determining the probability of an outcome?
The units on the x axis are likely to be numbers, prices, percentages, or returns. In a normal distribution the probability of finding a level, a certain distance above the mean, is the same as the
probability of finding a level which is the same distance below the mean. For example, if you have a mean of 100, there’s an equal probability of a number being 110 or ten percent above the mean as
it is of it being 90 or ten percent below the mean. The y axis shows the probability of each number given a mean and standard deviation. To reiterate, the actual probability associated with each
outcome will follow the Gaussian function and be dependent on the distance from the mean and the standard deviation.
A characteristic of the normal distribution is that the area under the curve, namely the probability of the outcome being between -1 and plus 1 standard deviations is always approximately 68%,
irrespective of the mean and standard deviation.
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Abdulla Javeri
Abdulla’s career in the financial markets started in 1990 when he entered the trading floor of the London International Financial Futures Exchange, LIFFE, and qualified as a pit trader in equity and
equity index options. In 1996, Abdulla became a trainer for regulatory qualifications and then for non-exam courses, primarily covering all major financial products.
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Geometric random intersection graphs with general connection probabilities
Let (V) and (U) be the point sets of two independent homogeneous Poisson processes on (ℝ). A graph (G)(V) with vertex set (V) is constructed by first connecting pairs of points (v, u) with (V) and
(U) independently with probability g(v-u), where g is a non-increasing radial function, and then connecting two points v1,v2(V) if and only if they have a joint neighbor (U). This gives rise to a
random intersection graph on (R)^d. Local properties of the graph, including the degree distribution, are investigated and quantified in terms of the intensities of the underlying Poisson processes
and the function g. Furthermore, the percolation properties of the graph are characterized and shown to differ depending on whether g has bounded or unbounded support.
• 2024 OA procedure
• Spatial random graphs
• Complex networks
• Degree distribution
• Percolation phase transition
• AB percolation
Dive into the research topics of 'Geometric random intersection graphs with general connection probabilities'. Together they form a unique fingerprint. | {"url":"https://research.utwente.nl/en/publications/geometric-random-intersection-graphs-with-general-connection-prob","timestamp":"2024-11-08T19:16:53Z","content_type":"text/html","content_length":"48308","record_id":"<urn:uuid:0ed559f5-7df9-4c3c-a20f-7df1db2884e0>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00650.warc.gz"} |
past test question
I used ratio test to solve this problem. Since this one has double factorial in its series, so check the limit of $|\frac {a_{n+2}}{a_n}|$ would be helpful. After expanding the limit, you will see
most of the terms will be canceled out. What's left is $\frac{1}{n+2}$, and it equals to 0 as n approaches infinity. It also equals to $\frac {1}{R}$. Thus our radius of convergence is infinity.
Another method is to solve this series by splitting it into its even part and odd part. Then use ratio test to get the radius of convergence of both parts. The smaller radius will be the final | {"url":"https://forum.math.toronto.edu/index.php?PHPSESSID=7ek774377iueclmnm2dm9am3m4&topic=2473.msg7358","timestamp":"2024-11-14T21:04:18Z","content_type":"application/xhtml+xml","content_length":"23143","record_id":"<urn:uuid:ad41de24-1123-4c29-b25e-ed8194f0e3ae>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00648.warc.gz"} |
2-D adaptive noise-removal filtering
The syntax wiener2(I,[m n],[mblock nblock],noise) has been removed. Use the wiener2(I,[m n],noise) syntax instead.
J = wiener2(I,[m n],noise) filters the grayscale image I using a pixel-wise adaptive low-pass Wiener filter. [m n] specifies the size (m-by-n) of the neighborhood used to estimate the local image
mean and standard deviation. The additive noise (Gaussian white noise) power is assumed to be noise.
The input image has been degraded by constant power additive noise. wiener2 uses a pixelwise adaptive Wiener method based on statistics estimated from a local neighborhood of each pixel.
[J,noise_out] = wiener2(I,[m n]) returns the estimates of the additive noise power wiener2 calculates before doing the filtering.
Remove Noise By Adaptive Filtering
This example shows how to use the wiener2 function to apply a Wiener filter (a type of linear filter) to an image adaptively. The Wiener filter tailors itself to the local image variance. Where the
variance is large, wiener2 performs little smoothing. Where the variance is small, wiener2 performs more smoothing.
This approach often produces better results than linear filtering. The adaptive filter is more selective than a comparable linear filter, preserving edges and other high-frequency parts of an image.
In addition, there are no design tasks; the wiener2 function handles all preliminary computations and implements the filter for an input image. wiener2, however, does require more computation time
than linear filtering.
wiener2 works best when the noise is constant-power ("white") additive noise, such as Gaussian noise. The example below applies wiener2 to an image of Saturn with added Gaussian noise.
Read the image into the workspace.
RGB = imread('saturn.png');
Convert the image from truecolor to grayscale.
Add Gaussian noise to the image
J = imnoise(I,'gaussian',0,0.025);
Display the noisy image. Because the image is quite large, display only a portion of the image.
title('Portion of the Image with Added Gaussian Noise');
Remove the noise using the wiener2 function.
Display the processed image. Because the image is quite large, display only a portion of the image.
title('Portion of the Image with Noise Removed by Wiener Filter');
Input Arguments
I — Input image
2-D numeric array
Input image, specified as a 2-D numeric array.
Data Types: single | double | int16 | uint8 | uint16
[m n] — Neighborhood size
[3 3] (default) | 2-element numeric vector
Neighborhood size, specified as a 2-element vector of the form [m n] where m is the number of rows and n is the number of columns.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical
noise — Additive noise
numeric array
Additive noise, specified as a numeric array. If you do not specify noise, then wiener2 uses the mean of the local variance, mean2(localVar).
Data Types: single | double
Output Arguments
J — Filtered image
numeric array
Filtered image, returned as a numeric array of the same size and data type as the input image I.
noise_out — Estimate of additive noise power
numeric array
Estimate of additive noise power, returned as a numeric array.
wiener2 estimates the local mean and variance around each pixel.
$\mu =\frac{1}{NM}\sum _{{n}_{1},{n}_{2}\in \eta }a\left({n}_{1},{n}_{2}\right)$
${\sigma }^{2}=\frac{1}{NM}\sum _{{n}_{1},{n}_{2}\in \eta }{a}^{2}\left({n}_{1},{n}_{2}\right)-{\mu }^{2},$
where $\eta$ is the N-by-M local neighborhood of each pixel in the image A. wiener2 then creates a pixelwise Wiener filter using these estimates,
$b\left({n}_{1},{n}_{2}\right)=\mu +\frac{{\sigma }^{2}-{u }^{2}}{{\sigma }^{2}}\left(a\left({n}_{1},{n}_{2}\right)-\mu \right),$
where ν^2 is the noise variance. If the noise variance is not given, wiener2 uses the average of all the local estimated variances.
[1] Lim, Jae S. Two-Dimensional Signal and Image Processing, Englewood Cliffs, NJ, Prentice Hall, 1990, p. 548, equations 9.44, 9.45, and 9.46.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
wiener2 supports the generation of C code (requires MATLAB^® Coder™). For more information, see Code Generation for Image Processing.
GPU Arrays
Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™.
This function fully supports GPU arrays. For more information, see Image Processing on a GPU.
Version History
Introduced before R2006a
R2022a: Generate C code using MATLAB Coder
wiener2 now supports the generation of C code (requires MATLAB Coder).
R2021a: Support for GPU acceleration
wiener2 now supports GPU acceleration (requires Parallel Computing Toolbox™). | {"url":"https://uk.mathworks.com/help/images/ref/wiener2.html","timestamp":"2024-11-13T15:25:25Z","content_type":"text/html","content_length":"93350","record_id":"<urn:uuid:b681015a-be25-4e07-8e2a-63a1c805f689>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00678.warc.gz"} |
Millimeters to Lightyears c
What is a millimeter?
A millimeter is a thousandth of a meter (1/1000) which is the SI (International System of Units) unit of length. It is normally used to measure small lengths like the thickness of a sheet of paper or
the dimensions of a small object.
One millimeter is approximately equal to 0.03937 inches (about 1/25th). Presicely there are 25.4 millimeters in an inch and it is often used in science and engineering. It is used in countries that
have adopted the metric system.
You may come across millimeters when measuring the size of electronic components, jewelry or even the thickness of a fingernail.
What is a lightyear?
A lightyear is a unit of measurement used in astronomy to describe vast distances in space. It represents the distance that light travels in one year, which is approximately 5.88 trillion miles or
9.46 trillion kilometers. The term "lightyear" is derived from the fact that light, which travels at a speed of about 186,282 miles per second (299,792 kilometers per second), can cover an incredible
distance in the span of a year.
The concept of a lightyear is crucial in understanding the vastness of the universe. Since light travels at a finite speed, it takes time for light to reach us from distant celestial objects.
Therefore, when we observe objects that are millions or billions of lightyears away, we are actually seeing them as they appeared millions or billions of years ago. This allows astronomers to study
the history and evolution of the universe by observing distant galaxies and other cosmic phenomena. | {"url":"https://live.metric-conversions.org/length/millimeters-to-lightyears.htm","timestamp":"2024-11-06T15:10:56Z","content_type":"text/html","content_length":"67930","record_id":"<urn:uuid:a9f58809-193a-4b56-975d-9c51a4780226>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00381.warc.gz"} |
Physics 221 A
Introduction to Quantum Field Theory
This course will provide an introduction to quantum field theory. A knowledge of non-relativistic quantum mechanics and of special relativity is required. For textbook I will be using
"Quantum Theory in a Nutshell" by A. Zee
Astronomy 1
Introduction to Astronomy
This course surveys modern astronomy starting with the solar system and ending with black holes and the universe.
Physics 131
Introduction to General Relativity
This course will provide an introduction to Einstein's theory of gravity. Familiarity with the special theory of relativity is required.
Physics 229 A
Supersymmetric Field Theory
This course will cover some aspects of supersymmetric field theories, starting with an elementary review of the representations of the Lorentz group. In the latter part of the course, I will give an
introduction to Seiberg-Witten theory. If time permits, I may briefly explore other topics, such as the notion of supersymmetry in disordered systems. A working knowledge of relativistic quantum
field theory is assumed.
Physics 221 C
Advanced Quantum Field Theory
Selected topics from the development of quantum field theory over the last three or four decades will be discussed. If time permits, I hope to use examples from condensed matter physics as well as
the more traditional examples from high energy physics. A basic knowledge of relativistic quantum field theory is assumed. | {"url":"https://www.kitp.ucsb.edu/zee/courses","timestamp":"2024-11-01T23:45:53Z","content_type":"text/html","content_length":"81454","record_id":"<urn:uuid:60c99276-d1a1-4550-8be1-5c50bea766e9>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00674.warc.gz"} |
Research shows the best ways to learn math
Students learn math best when they approach the subject as something they enjoy. Speed pressure, timed testing and blind memorization pose high hurdles in the pursuit of math, according to Jo Boaler,
professor of mathematics education at Stanford Graduate School of Education and lead author on a new working paper called "Fluency Without Fear."
"There is a common and damaging misconception in mathematics – the idea that strong math students are fast math students," said Boaler, also cofounder of YouCubed at Stanford, which aims to inspire
and empower math educators by making accessible in the most practical way the latest research on math learning.
Fortunately, said Boaler, the new national curriculum standards known as the Common Core Standards for K-12 schools de-emphasize the rote memorization of math facts. Maths facts are fundamental
assumptions about math, such as the times tables (2 x 2 = 4), for example. Still, the expectation of rote memorization continues in classrooms and households across the United States.
While research shows that knowledge of math facts is important, Boaler said the best way for students to know math facts is by using them regularly and developing understanding of numerical
relations. Memorization, speed and test pressure can be damaging, she added.
Number sense is critical
On the other hand, people with "number sense" are those who can use numbers flexibly, she said. For example, when asked to solve the problem of 7 x 8, someone with number sense may have memorized 56,
but they would also be able to use a strategy such as working out 10 x 7 and subtracting two 7s (70-14).
"They would not have to rely on a distant memory," Boaler wrote in the paper.
In fact, in one research project the investigators found that the high-achieving students actually used number sense, rather than rote memory, and the low-achieving students did not.
The conclusion was that the low achievers are often low achievers not because they know less but because they don't use numbers flexibly.
"They have been set on the wrong path, often from an early age, of trying to memorize methods instead of interacting with numbers flexibly," she wrote. Number sense is the foundation for all
higher-level mathematics, she noted.
Role of the brain
Boaler said that some students will be slower when memorizing, but still possess exceptional mathematics potential.
"Math facts are a very small part of mathematics, but unfortunately students who don't memorize math facts well often come to believe that they can never be successful with math and turn away from
the subject," she said.
Prior research found that students who memorized more easily were not higher achieving – in fact, they did not have what the researchers described as more "math ability" or higher IQ scores. Using an
MRI scanner, the only brain differences the researchers found were in a brain region called the hippocampus, which is the area in the brain responsible for memorizing facts – the working memory
But according to Boaler, when students are stressed – such as when they are solving math questions under time pressure – the working memory becomes blocked and the students cannot as easily recall
the math facts they had previously studied. This particularly occurs among higher achieving students and female students, she said.
Some estimates suggest that at least a third of students experience extreme stress or "math anxiety" when they take a timed test, no matter their level of achievement. "When we put students through
this anxiety-provoking experience, we lose students from mathematics," she said.
Math treated differently
Boaler contrasts the common approach to teaching math with that of teaching English. In English, a student reads and understands novels or poetry, without needing to memorize the meanings of words
through testing. They learn words by using them in many different situations – talking, reading and writing.
"No English student would say or think that learning about English is about the fast memorization and fast recall of words," she added.
Strategies, activities
In the paper, coauthored by Cathy Williams, cofounder of YouCubed, and Amanda Confer, a Stanford graduate student in education, the scholars provide activities for teachers and parents that help
students learn math facts at the same time as developing number sense. These include number talks, addition and multiplication activities, and math cards.
Importantly, Boaler said, these activities include a focus on the visual representation of number facts. When students connect visual and symbolic representations of numbers, they are using different
pathways in the brain, which deepens their learning, as shown by recent brain research.
"Math fluency" is often misinterpreted, with an over-emphasis on speed and memorization, she said. "I work with a lot of mathematicians, and one thing I notice about them is that they are not
particularly fast with numbers; in fact some of them are rather slow. This is not a bad thing; they are slow because they think deeply and carefully about mathematics."
She quotes the famous French mathematician, Laurent Schwartz. He wrote in his autobiography that he often felt stupid in school, as he was one of the slowest math thinkers in class.
Math anxiety and fear play a big role in students dropping out of mathematics, said Boaler.
"When we emphasize memorization and testing in the name of fluency we are harming children, we are risking the future of our ever-quantitative society and we are threatening the discipline of
mathematics," she said. "We have the research knowledge we need to change this and to enable all children to be powerful mathematics learners. Now is the time to use it." | {"url":"https://ed.stanford.edu/news/learning-math-without-fear?utm_source=educator&utm_medium=email&utm_campaign=february-2015","timestamp":"2024-11-10T16:19:34Z","content_type":"text/html","content_length":"49121","record_id":"<urn:uuid:f010e4df-9b86-4066-97a5-505c70483368>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00369.warc.gz"} |
The Paradox of Plato’s Quantum Ball
Generated by DALL-E
Imagine you have a ball at the macro-level. The macro-level ball has a clear identity and a clear set of properties. It also has a definite position (P) and momentum (M) that are identifiable and
independent. Now, imagine that you go down to the quantum-level of the ball. It still exists, but at the quantum-level, we don’t see a ‘ball’ anymore; we just interact with P, M, or ‘particles’ that
have to do with the ball.
But P and M are not fundamentals — they are observables. That is, P and M don’t tell us about any of the properties or the identity of the ball. Furthermore, they are no longer independent. But the
ball must still be there at the macro-level. Its wavefunction still exists. We have, however, begun focusing on another subsidiary wavefunction. This subsidiary wavefunction is of the quantum ball as
it exists in Plato’s Cave.
The Allegory of the Cave appears in Plato’s Republic. It describes reality as experienced by a group of prisoners who have lived their entire lives chained inside a cave. Behind the prisoners is a
fire, and between the fire and the prisoners is a walkway. On the walkway, people pass by carrying objects, which cast shadows on the cave wall in front of the prisoners. The prisoners mistake these
shadows for reality.
Let us refer to this wave function as Plato’s Quantum Ball Wavefunction.
It keeps collapsing when we observe it per the Born rule and as filtered by the dominantly accepted Copenhagen Interpretation of quantum mechanics. Incidentally, that is also why Schrodinger is
reported to have said he wished he had nothing to do with it.
We then have to use statistical analysis to make sense of it. But we can’t understand it because the P and M are just observables. Heisenberg’s insightful idea captured in his uncertainty principle
tells us that the nature of observing observables is that there is always uncertainty in their observed values. We aren’t seeing the quantum ball's properties, identity, or nature.
No matter what our basis states (spin-up/spin-down, horizontal polarization/vertical polarization, ground state of an atom or ion/excited state of an atom or ion, one electron occupancy state/another
electron occupancy state, lower energy state of circuit/higher energy state of circuit) this will always remain true. We must leave the cave to see that we are dealing with a quantum ball.
Then, we will also see that, instead of using aggregation based on P and M to infer other properties of the ball (assuming that is possible), we will be better served by using disaggregation to try
to make sense of P and M, knowing the nature of what we are looking at is a ball. The former seems formidable, given decoherence, lack of scalability, etc. Further, superposition, so long as it
concerns superposed values of P and M, is not telling us anything useful unless we begin to deal with the ball's superposed properties.
That is why Einstein suggested that quantum theory was incomplete.
But we have been approaching it as though it is complete. We have built quantum mechanics using classical mechanics as a basis. We have built the math of Quantum Field Theory (QFT) in terms of the
math of single quantum particles. Everything becomes a nail with a hammer in the hand.
Generated by DALL-E
We have not elevated to “function,” so we can’t leverage the possibilities of disaggregation. Classical physics is focused on precise physical law. When we deal with the quantum level, which
separates the visible from the invisible, we are dealing with the physics of the situation and other possibilities that are being parsed out. Yet, we are fundamentally only leveraging a view based on
classical mechanics. If we do that, then we will miss out on other richness. But is not Schrodinger’s wave function also about something else?
“Function” has to express itself variably, perhaps even as patterns of particles, say. But if we don’t see it, we miss the possibility. Digital computation has integrated “function” in one way.
Quantum computation needs to integrate it in another way.
That is why we need a different approach.
In such a systems theory view of the quantum level, the latter is integrated with physics, chemistry, biology, philosophy, etc. For the very adventurous, here is a reflection on the nature of the
quantum seed that may result from such integration. For the reasonably adventurous, here is a recent and brief Forbes article on different developmental trajectories for quantum computation.
Alternatively, here is a representation of a potential architecture focused on “function.” | {"url":"https://pravirmalik.medium.com/the-paradox-of-platos-quantum-ball-bfbcf6cd6dc4?source=user_profile_page---------2-------------26d3a42522c4---------------","timestamp":"2024-11-11T20:11:26Z","content_type":"text/html","content_length":"108319","record_id":"<urn:uuid:bf0c040e-2c89-4341-8cec-87680f8461af>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00639.warc.gz"} |
Early stopping in Gradient Boosting
Go to the end to download the full example code. or to run this example in your browser via Binder
Early stopping in Gradient Boosting#
Gradient Boosting is an ensemble technique that combines multiple weak learners, typically decision trees, to create a robust and powerful predictive model. It does so in an iterative fashion, where
each new stage (tree) corrects the errors of the previous ones.
Early stopping is a technique in Gradient Boosting that allows us to find the optimal number of iterations required to build a model that generalizes well to unseen data and avoids overfitting. The
concept is simple: we set aside a portion of our dataset as a validation set (specified using validation_fraction) to assess the model’s performance during training. As the model is iteratively built
with additional stages (trees), its performance on the validation set is monitored as a function of the number of steps.
Early stopping becomes effective when the model’s performance on the validation set plateaus or worsens (within deviations specified by tol) over a certain number of consecutive stages (specified by
n_iter_no_change). This signals that the model has reached a point where further iterations may lead to overfitting, and it’s time to stop training.
The number of estimators (trees) in the final model, when early stopping is applied, can be accessed using the n_estimators_ attribute. Overall, early stopping is a valuable tool to strike a balance
between model performance and efficiency in gradient boosting.
License: BSD 3 clause
# Authors: The scikit-learn developers
# SPDX-License-Identifier: BSD-3-Clause
Model Training and Comparison#
Two GradientBoostingRegressor models are trained: one with and another without early stopping. The purpose is to compare their performance. It also calculates the training time and the n_estimators_
used by both models.
params = dict(n_estimators=1000, max_depth=5, learning_rate=0.1, random_state=42)
gbm_full = GradientBoostingRegressor(**params)
gbm_early_stopping = GradientBoostingRegressor(
start_time = time.time()
gbm_full.fit(X_train, y_train)
training_time_full = time.time() - start_time
n_estimators_full = gbm_full.n_estimators_
start_time = time.time()
gbm_early_stopping.fit(X_train, y_train)
training_time_early_stopping = time.time() - start_time
estimators_early_stopping = gbm_early_stopping.n_estimators_
Error Calculation#
The code calculates the mean_squared_error for both training and validation datasets for the models trained in the previous section. It computes the errors for each boosting iteration. The purpose is
to assess the performance and convergence of the models.
train_errors_without = []
val_errors_without = []
train_errors_with = []
val_errors_with = []
for i, (train_pred, val_pred) in enumerate(
train_errors_without.append(mean_squared_error(y_train, train_pred))
val_errors_without.append(mean_squared_error(y_val, val_pred))
for i, (train_pred, val_pred) in enumerate(
train_errors_with.append(mean_squared_error(y_train, train_pred))
val_errors_with.append(mean_squared_error(y_val, val_pred))
Visualize Comparison#
It includes three subplots:
1. Plotting training errors of both models over boosting iterations.
2. Plotting validation errors of both models over boosting iterations.
3. Creating a bar chart to compare the training times and the estimator used of the models with and without early stopping.
fig, axes = plt.subplots(ncols=3, figsize=(12, 4))
axes[0].plot(train_errors_without, label="gbm_full")
axes[0].plot(train_errors_with, label="gbm_early_stopping")
axes[0].set_xlabel("Boosting Iterations")
axes[0].set_ylabel("MSE (Training)")
axes[0].set_title("Training Error")
axes[1].plot(val_errors_without, label="gbm_full")
axes[1].plot(val_errors_with, label="gbm_early_stopping")
axes[1].set_xlabel("Boosting Iterations")
axes[1].set_ylabel("MSE (Validation)")
axes[1].set_title("Validation Error")
training_times = [training_time_full, training_time_early_stopping]
labels = ["gbm_full", "gbm_early_stopping"]
bars = axes[2].bar(labels, training_times)
axes[2].set_ylabel("Training Time (s)")
for bar, n_estimators in zip(bars, [n_estimators_full, estimators_early_stopping]):
height = bar.get_height()
bar.get_x() + bar.get_width() / 2,
height + 0.001,
f"Estimators: {n_estimators}",
The difference in training error between the gbm_full and the gbm_early_stopping stems from the fact that gbm_early_stopping sets aside validation_fraction of the training data as internal validation
set. Early stopping is decided based on this internal validation score.
In our example with the GradientBoostingRegressor model on the California Housing Prices dataset, we have demonstrated the practical benefits of early stopping:
• Preventing Overfitting: We showed how the validation error stabilizes or starts to increase after a certain point, indicating that the model generalizes better to unseen data. This is achieved by
stopping the training process before overfitting occurs.
• Improving Training Efficiency: We compared training times between models with and without early stopping. The model with early stopping achieved comparable accuracy while requiring significantly
fewer estimators, resulting in faster training.
Total running time of the script: (0 minutes 2.792 seconds)
Related examples
Gradient Boosting regression | {"url":"https://scikit-learn.qubitpi.org/auto_examples/ensemble/plot_gradient_boosting_early_stopping.html","timestamp":"2024-11-13T05:18:20Z","content_type":"text/html","content_length":"110641","record_id":"<urn:uuid:a6b158c6-42f2-41e1-b48a-7916714f3f6f>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00229.warc.gz"} |
Rice's Theorem - (Formal Language Theory) - Vocab, Definition, Explanations | Fiveable
Rice's Theorem
from class:
Formal Language Theory
Rice's Theorem states that for any non-trivial property of the language recognized by Turing machines, it is undecidable whether a given Turing machine has that property. This means that if we
consider any interesting aspect of the behavior of Turing machines, we cannot create a general algorithm to determine if all machines share that behavior. This connects deeply to how we understand
Turing machines, what can be decided or not, and the relationship between undecidable problems through reductions.
congrats on reading the definition of Rice's Theorem. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Rice's Theorem implies that any non-trivial property of languages recognized by Turing machines is undecidable, meaning no algorithm can decide it for all cases.
2. The theorem shows that properties like being a regular language or context-free language are also undecidable, reinforcing the limits of what can be computed.
3. The concept of non-trivial means that the property must not hold for all Turing machines or for none; it must only hold for some.
4. Rice's Theorem is often proven using a reduction argument, typically demonstrating how if one could decide the property, it would contradict known undecidable problems like the Halting Problem.
5. This theorem emphasizes the limitations in the field of computability and sets the stage for understanding the boundaries of algorithmic processes.
Review Questions
• How does Rice's Theorem demonstrate the limitations of decidability in relation to Turing machines?
□ Rice's Theorem shows that any non-trivial property of the languages recognized by Turing machines cannot be decided by an algorithm. This means that for interesting behaviors or
characteristics of Turing machines—such as whether they accept a specific type of language—there is no universal method to determine these properties. Therefore, it highlights that as long as
a property is non-trivial, it falls outside the realm of decidability.
• Discuss how Rice's Theorem relates to the Halting Problem and provides insights into undecidable problems.
□ Rice's Theorem reinforces concepts introduced by the Halting Problem by illustrating how many properties concerning Turing machine behaviors are also undecidable. If we could determine
whether a machine satisfies a non-trivial property, we could construct a way to solve the Halting Problem as well. This connection deepens our understanding of undecidability, illustrating
that many questions about computation have no solutions.
• Evaluate the significance of Rice's Theorem in computability theory and its implications for algorithm design.
□ Rice's Theorem holds significant importance in computability theory as it delineates clear boundaries on what can and cannot be decided algorithmically regarding Turing machines. It implies
that when designing algorithms, especially those intended to analyze or classify behaviors of computing systems, one must recognize inherent limitations and focus on decidable aspects
instead. This understanding leads researchers and practitioners to avoid attempts at solving non-trivial properties and instead concentrate efforts on areas where solutions are achievable.
"Rice's Theorem" also found in:
© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website. | {"url":"https://library.fiveable.me/key-terms/formal-language-theory/rices-theorem","timestamp":"2024-11-14T23:28:28Z","content_type":"text/html","content_length":"134996","record_id":"<urn:uuid:912040c5-5d8c-4dfa-a237-255018e1e678>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00016.warc.gz"} |
On the Computational Complexity of Multi-Agent Pathfinding on Directed Graphs
“Multi-agent pathfinding”, also called “pebble motion on graphs” or “cooperative pathfinding”, is the problem of deciding the existence of or generating a collision-free movement plan for a set of
agents moving on a graph. While the non-optimizing variant of multi-agent pathfinding on undirected graphs is known to be a polynomial-time problem since forty years, a similar result for directed
graphs was missing. In the talk, it will be shown that this problem is NP-complete. For strongly connected directed graphs, however, the problem is polynomial. And both of these results hold even if
one allows for synchronous rotations on fully occupied cycles. | {"url":"https://events.illc.uva.nl/FOAM/posts/talk15/","timestamp":"2024-11-13T08:39:19Z","content_type":"text/html","content_length":"28620","record_id":"<urn:uuid:62ed20dd-96ce-405b-9a1c-02c6597f7aea>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00429.warc.gz"} |
Thomas’ Calculus 13th Edition Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 726 8
Work Step by Step
As we know the parametric equations of a straight line for a vector $v=v_1i+v_2j+v_3k$ passing through a point $P(a,b,c)$ is given by $x=a+t v_1,y=b+t v_2; z=c+t v_3$ Since, we have the vector $v=\lt
3,7,-5 \gt$ is perpendicular to the plane at point $P(2,4,5)$.Then our parametric equations are: $x=2+3t,y=4+7t, z=5-5t$ | {"url":"https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-12-vectors-and-the-geometry-of-space-section-12-5-lines-and-planes-in-space-exercises-12-5-page-726/8","timestamp":"2024-11-06T02:15:22Z","content_type":"text/html","content_length":"65155","record_id":"<urn:uuid:a3c0a321-51a4-4acf-98b6-ae5e83913b4b>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00733.warc.gz"} |
Pedestrian Bridge Design Calculator Online
Home » Simplify your calculations with ease. » Construction »
Pedestrian Bridge Design Calculator Online
Pedestrian bridges serve as an essential infrastructural element, ensuring safe passage over physical barriers. The design and engineering of these structures involve significant calculations and
expert insight. This article introduces a rudimentary calculator for pedestrian bridge design and discusses the principles behind it.
A pedestrian bridge design calculator is a tool providing initial estimations about the structural requirements of a bridge. This involves considering span length, load requirements, and other
factors to ensure safe and efficient bridge design.
Detailed Explanation of the Calculator’s Working
This simplified calculator takes into account two critical factors: the span length and the load requirement. By entering these values, the calculator provides a fundamental estimation of the load
capacity of the bridge. However, it’s important to understand that real-world calculations are far more complex, taking into account numerous factors, including environmental conditions, safety
requirements, material properties, and more.
Formula and Variable Descriptions
The calculator uses a simple formula:
Output = Span Length x Load Requirement.
Here, the span length represents the distance between the bridge supports (in meters), and the load requirement is the maximum weight the bridge is designed to support (in kN).
For instance, if we input a span length of 10 meters and a load requirement of 15 kN, the calculator’s output will be 150. This figure represents a basic estimation of the load capacity for the
designed bridge.
Urban Planning
These calculators are commonly used in preliminary urban planning stages, offering an understanding of feasibility and structural requirements.
Educational Purpose
They’re also valuable educational tools, helping engineering students grasp basic principles of bridge design calculations.
Most Common FAQs
Is the calculator sufficient for professional bridge design?
No, the calculator is a basic tool providing an understanding of the principles behind bridge design. It should not replace professional engineering consultation and advanced software tools for
design and analysis.
What factors does the calculator not consider?
The calculator does not account for various factors such as environmental conditions, safety regulations, material properties, and structural type, among others.
While a basic pedestrian bridge design calculator is a helpful tool for understanding, it should not replace professional guidance in the real-world design process. Expert consultation, detailed
analysis, and advanced software tools are essential for safe and efficient bridge design.
Leave a Comment | {"url":"https://calculatorshub.net/construction/pedestrian-bridge-design-calculator/","timestamp":"2024-11-08T14:21:20Z","content_type":"text/html","content_length":"113187","record_id":"<urn:uuid:a344cf1f-83c2-4d47-be68-89110e0ea51e>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00110.warc.gz"} |
What are the differences between Gann angles and Elliott Wave theory? | Practical GANN
What are the differences between Gann angles and Elliott Wave theory?
What are the differences between Gann angles and Elliott Wave theory? Gann angles are used in conjunction with the Elliott Wave Principle to indicate bull and bear waves. A gann angle is a major
index chart pattern that consists of higher time periods separated by lower time periods. A pattern that forms during the end of a market, a trend higher or a range high and you can try these out is
a subsequent blow-off beyond that high. At the market reversal low the wave forms a gann pattern or an angle that points down. Gann Patterns and Elliott Wave Theory Although not all Elliott Wave
followers agree, the many followers of Wave Theory believe that a Gann Angle can be used to predict Fibonacci levels, the Wave Count or even reversals. This can be verified from the studies of Dr.
George Lane, who wrote in his book, Elliott Wave Principles Explained and used Elliott Wave Theory to help the market. A study of many trading cycles by Lane showed that between the end of a wave
(reversal) and the beginning of a subsequent wave, there is a longer time period corresponding to the wave count of Fibonacci numbers. For example, Lane’s studies indicate that time from reversal to
waves forming the 5th, 7th, 11th and 13th Fibonacci numbers is 4.5 periods, corresponding to the wave 3 count, as follows…. Wave Count of Fibonacci Numbers. Each wave moves.5 waves up to the next
Fibonacci number Lower time periods (lower numbers) occur when wave 3 extends into an upward phase.
Price Action
However, an upward wave 3 actually always occurs twice at a “natural” 3-4 extension. For example, there is an upward phase that extends into the complete 5th Fibonacci extension (1 1/2, 4, 13) and
another that then in turn completes the 7th, 11th, etc. Therefore, the time period from the end ofWhat are the differences between Gann angles and Elliott Wave theory? There are no differences as far
as I’m concerned. Do Gann angles work the same way with prices during an up and downward trend? Yes. During what was known as the Great Depression, gold is a great example of how Gann angle
combinations are very different on price graphs. Price actions and momentum appear to have changed direction, instead of breaking in wave 4a. Have you made some secret revelations, like the other
members, that changed the world of investing and trading? First, the Gann theory didn’t change anything: view publisher site just gave some new names to waves. Let’s say a wave’s four main support or
resistance points (P1, P2, P3, and P4). A combination of four waves are the wave 4, wave 5, wave 6 and wave 7 (each of the last two waves are named the fifth and sixth waves from the basic waves,
because if in the wave 5 a higher wave 4 creates a support, the 3/4 of the wave 5 that pop over to these guys the wave 4 support will be the sixth wave from the basic waves). So when do the wave
combination end (when time is counted) in different wave combinations of the wave 4, that is the basic Gann waves four combinations, that are usually calculated as follows, the wave 5 five wave
pattern is usually calculated from the first down-swallow, and the wave 5 six wave pattern is calculated from the first up-swallow. So, in other words, the wave 5 pattern ends when the view of the
wave 4 upswing is through (which can be upward for a higher wave 5 or downward for a lower wave 5). In this case: P1, P2, P3 and P4 are the wave 4 points, and P5 and P6 are the wave 5 points. Do Gann
waves work the same way with prices during an up and downwardWhat are the differences between Gann angles and Elliott Wave theory? Gann angles are a variation on Wave Theory and have been around my
blog the days of Gann himself.
Gann’s Law of Vibration
Wave Theory was developed by Trend Jockey when there was no scientific basis for understanding the market, so I feel a variation of Wave Theory can be utilized more effectively in today’s economic
environment. There are many areas in a market around trends and angles. The purpose of this article is to discuss what are the differences between the two theories. If you have mastered a wave, yet
you are still reading my articles, have me know this page the comments section what would you like to see about wave theory in the future. If you would like to discuss the Elliott Wave theory and the
Gann angles, let’s begin! The purpose of this article is to describe the differences between wave theory and gann angles. The Purpose of this Article? In the next section of this article page, you
hopefully will have an understanding of what gann angles are and how they can be applied to market analysis. In this more advanced section of the article, we will go over specific patterns you are
likely to encounter in the market (waves, and their construction) and we will illustrate their uses for trading and understanding the financial markets. I end with a list of alternative theories to
the Elliott Wave one. However, The most important part of why this article was written is to help you understand, and have the tools for, a consistent market analysis so you can begin to develop your
own theories. The Wave Theory you should be familiar with, if you are learning and trading this dynamic market are the 3 P’s: Price, Pattern and Pattern continuation. This is also the foundation for
the modern form of wave theory as developed by Trend Jockey because it was a great way to understand the markets and trade because the theory told you not only what was happening, but also when those
events would be repeated. So if a news was going to | {"url":"https://practicalgann.com/what-are-the-differences-between-gann-angles-and-elliott-wave-theory/","timestamp":"2024-11-02T02:11:32Z","content_type":"text/html","content_length":"166888","record_id":"<urn:uuid:1f634358-58ce-4b92-af5e-27afe1305a81>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00601.warc.gz"} |
babel-samin – Babel support for Samin
The package provides the language definition file for support of North Sami in babel. (Several Sami dialects/languages are spoken in Finland, Norway, Sweden and on the Kola Peninsula of Russia). Not
all use the same alphabet, and no attempt is made to support any other than North Sami here.
Some shortcuts are defined, as well as translations to Norsk of standard “LaTeX names”.
Sources /macros/latex/contrib/babel-contrib/samin
Version 1.0d
Licenses The LaTeX Project Public License 1.3
Copyright 1989–2024 Johannes L. Braams and any invididual authors
Javier Bezos López
Maintainer Regnor Jernsletten
Johannes L. Braams (inactive)
Contained in TeXLive as babel-samin
MiKTeX as babel-samin
Topics Samin
Download the contents of this package in one zip archive (134.5k).
Community Comments
Maybe you are interested in the following packages as well. | {"url":"https://ctan.org/pkg/babel-samin","timestamp":"2024-11-02T13:59:42Z","content_type":"text/html","content_length":"16728","record_id":"<urn:uuid:41f7ee81-c97c-494c-b7c1-11cdf853b373>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00263.warc.gz"} |
Free Group Study Rooms with Timer & Music | FiveableWhat Are the Best Quizlet Decks for AP Statistics? | Fiveable
If there was a holy trinity for AP study sites, Quizlet would most certainly be in it. Its easy to use interface, combined with its multi-purpose functionality, helps students of all different
learning styles in endless subject areas. However, it can sometimes be challenging to find the best vocab sets.
Fiveable’s AP Stats teachers & students have compiled the best quizlet study decks for each unit. The AP Stats exam is very concept heavy, so make sure you take the time to learn these terms.
Catch a live review or watch a replay for AP Stats on Fiveable’s AP Stats hub!
Unit 1 includes the roots of statistics, and it is very important to get these concepts down and memorized. In this unit, you will distinguish between categorical and quantitative data, describe and
compare distributions, and begin to learn about normal distributions.
• Categorical Variable – Record which of several groups an individual belongs to
• Quantitative Variable – Taking numerical values for future calculations of sorting
• Describing a Distribution (SOCS) MUST include:
□ Shape – uniform/skew/peaks in context (symmetric, skewed left/right, unimodal/bimodal)
□ Outliers – mention outliers in context
□ Center – mean or median in context
□ Spread – range/standard deviation/IQR in context
• Bar Graphs, Two-way Tables, Histograms, Stemplots, Dotplots, Box plots – All methods of representing Data. It is good to know how to read information off of each of these.
• 5 Number Summary – Minimum, Quartile 1, Median, Quartile 3, Maximum
• The Empirical Rule – Tells us that if we all behave normally then about 68% of the values fall within 1 standard deviation of the mean, about 95% of the values fall within 2 standard deviations
of the mean, and about 99.7%—almost all—of the values fall within 3 standard deviations of the mean.
• Z-Score – z = x - x̄ / s
Unit 2 is an expansion of Unit 1. It builds on the relationships between two categorical or quantitative variables and how to argue about the strength between the two. This unit includes a lot of set
interpretations of the different components of an LSRL which are very important to remember.
• Scatterplots – A way to organize data. On the x-axis is the explanatory (independent) variable and on the y-axis is the response (dependent) variable.
• Form – linear or curved
• Direction – positive or negative correlation
• Strength – depends on the correlation coefficient; could be weak, moderate, or strong.
• LSRL – ŷ=a+bx where x denotes (context) and ŷ denotes predicted (context)
• Correlation Coefficient (r) – The correlation coefficient shows the degree to which there is a linear correlation between the two variables, that is, how close the points are to forming a line.
The closer r is to 1 or -1, the stronger the relationship.
• Slope (b) – There is a predicted increase/decrease of ______ (slope in unit of y variable) for every 1 (unit of x variable).
• Y Intercept – The predicted value of (y in context) is _____ when (x value in context) is 0 (units in context).
• Coefficient of Determination (r^2) – ____% of the variation in (y in context) is due to its linear relationship with (x in context).
• Residuals – The difference between the actual data and the value predicted by a linear regression model, or y-ŷ. The ideal pattern is random scatter above and below the LSRL (where the residuals=
0). A positive residual means the model underestimated the true value while a negative residual means the model overestimated the true value.
• Extrapolation – The use of a regression model to make predictions outside of the domain of the given data. If you go outside this domain and farther outside the domain you go, the less accurate
your predictions will be.
This unit discusses sampling methods and ways of collecting data that can be used to represent a population. It is filled with vocabulary that is essential in future units.
• Experiment – Deliberately imposes treatment in order to observe the response. Causation could be proven.
• 🔭Observational Study – Observe individuals and measure variables of interest but don’t attempt to influence the responses. These studies look for association between variables because in a
study, no treatment is imposed.
• 👀Confounding – Occurs when two variables are associated in a way that their effects on the response variable cannot be deciphered from each other individually.
• Bias – Statistical studies are biased if it is likely to underestimate or overestimate the value you are looking for.
• Simple Random Sample (SRS) – Chooses a sample size “n” in a way that a group of individuals in the population has an equal chance to be selected as the sample
• Stratified Random Sample – Selects a sample by choosing an SRS from each strata and combining the SRSs into one overall sample. These reduce variability in the data and give more precise results.
• Cluster Sample –The population is divided into groups, called clusters, and an SRS of clusters is taken within each cluster. All individuals are sampled in the clusters selected.
• Systematic Random Sample – Sample members from a population selected according to a random starting number and a fixed periodic interval.
Everyday, we see things that happen simultaneously to the point we question the possibility of that event happening again. This brings in probability, the proportion of times the outcome would occur
in a large number of repetitions. Unit 4 is all about probability and is very calculation heavy.
As of 12/22/21, this deck is private.
• Independent Events – The outcome of one event doesn't influence the outcome of another event.
• P(A|B) = P(A)
• P(A and B) =P(A)*P(B)
• Disjoint/Mutually Exclusive Events – Cannot occur at the same time and have no outcomes in common.
• P(A and B) = 0
• P(A or B) = P(A) + P(B)
• Conditional Probability – Probability of one event under the condition that another event is known.
• P(A|B)= P(A and B) / P (B) – The probability of A given B = Probability of A and B / Probability of B
• Discrete Random Variable – Takes a fixed set of possible values with gaps between them (cannot include decimals).
• Mean (Expected) Value – Summation(xi*pi)
• Standard Deviation – sqrt(summation(xi-mean of x)^2 * pi). You cannot add standard deviations, only variances.
• Continuous Random Variable – Can take any value in an interval on the number line (can include decimals).
• Law of Large Numbers – Says that in many repetitions of the same chance process, the simulated probability gets closer to the true probability as more trials are run.
• Binomial Setting – Arises when we perform independent trials of the same chance process and count the number of times that a particular outcome called a success (p), occurs. Failure (q) is
defined as 1 minus the probability of success. Must check 10% condition.
• Binomial Setting – Arises when we perform independent trials of the same chance process and count the number of times that a particular outcome called a success (p), occurs. Failure (q) is
defined as 1 minus the probability of success. Must check 10% condition.
• ⭕️ Geometric Setting – Arises when we perform independent trials of the same chance process and record the number of trials it takes to get one success. On each trial, the probability p of
success must be the same. The number of trials Y that it takes to get one success in a geometric setting is a geometric random variable.
This unit is an introduction to significant tests, which are covered in later units. It begins introducing statistics, bias, the CLT, and population parameters.
• Sampling Distribution – A distribution where we take ALL possible samples of a given size and put them together as a data set.
• A statistic is used to estimate a parameter.
• Parameters vs Statistics – Mean (𝝁, x̅); Standard Deviation (σ, s); Proportions (𝝆, p̂).
• Large Counts Condition – The number of successes and failures is at least 10.
• Central Limit Theorem – States that if n (the sample size) is ≥30, the sampling distribution is normal. The larger n is, the more normal the sample is.
This unit is the beginning of significant tests, in which you are expected to check conditions, construct and interpret confidence intervals, and calculate a p-value. This unit consists of estimating
population parameters involving categorical data.
• Confidence Interval – An interval of numbers based on our sample proportion that gives us a range where we can expect to find the true population proportion.
• In repeated sampling, I am __% confident that the true population proportion (context) falls within this interval.
• Significance Test – Estimating the probability of obtaining our collected sample from the sampling distribution of our size when we assume that the given population proportion is correct.
• Large Counts Condition – The number of successes and failures is at least 10 (np≥10 and n(1-p)≥10). This condition proves normality.
• Random Condition – Reduces any bias that may be caused from taking a bad sample. When answering inference questions, it is always essential to make note that our sample was random. Without a
random sample, our findings cannot be generalized to a population, meaning our scope of inference is inaccurate.
• 10% Condition – Check that the population in question is at least 10 times as large as our sample in order to prove independence.
• Margin of Error – The "buffer zone" of our confidence interval; point estimate +- (z*)(stan. dev.)
• Null Hypothesis – The hypothesis based on our claim that was given in the problem (p=___).
• Alternative Hypothesis – The hypothesis that the claim in our null is not true (p<___, p>___ or p≠___).
• Type I (𝞪) Error – When we reject our Ho, when in fact, we should have failed to reject.
• Type II (𝞫) Error – When we fail to reject our Ho, but we actually should have rejected our Ho.
• Power – How strong our test is because power = 1-P(Type II Error).
This unit is very similar to Unit 6, but instead of dealing with proportions (p), we are dealing with means (μ). Many of the concepts overlap, but it is important to note that we cannot write
proportions anywhere we are running an inference test for quantitative data.
• Confidence Interval – An interval of numbers based on our sample proportion that gives us a range where we can expect to find the true population mean.
• In repeated sampling, I am __% confident that the true population mean (context) falls within this interval.
• In repeated sampling, I am ___% confident that the true difference in population means (context) falls within this interval.
• Central Limit Theorem – States that if n (the sample size) is ≥30, the sampling distribution is normal. The larger n is, the more normal the sample is.
• Random Condition – Reduces any bias that may be caused from taking a bad sample.
• 10% Condition – Check that the population in question is at least 10 times as large as our sample in order to prove independence.
• Null Hypothesis – The hypothesis based on our claim that was given in the problem (μ=___).
• Alternative Hypothesis – The hypothesis that the claim in our null is not true (μ<___, μ>___ or μ≠___).
• p<𝞪 – Since p<𝞪, we reject our Ho. We have convincing evidence at the 𝞪 level that (Ha in context).
• p<𝞪 – Since p<𝞪, we reject our Ho. We have convincing evidence at the 𝞪 level that (Ha in context).
• Degrees of Freedom – n-1
Chi-Squared significant tests operate differently than proportion and mean significant tests. All χ² distributions are skewed right. The degree of skewness depends on the degrees of freedom.
Unfortunately, this deck has been deleted by its author. (12/22/21)
• Goodness of Fit Test – Must have random sampling. All expected counts must be greater than 5.
□ Degrees of Freedom – categories - 1
□ Null Hypothesis – There is no association between ___ and ___.
□ Alternative Hypothesis – There is association between ___ and ___.
• Test for Independence – Checks the association between two variables in a single population.
□ Expected Counts – (row total)(column total) / (table total)
□ Null Hypothesis – There is no association between ___ and ___.
□ Alternative Hypothesis – There is association between ___ and ___.
• Test for Homogeneity – Checks the distribution of a single variable in several populations to see if these populations are similar with respect to the variable.
□ Null Hypothesis – There is no difference between ___ and ___.
□ Alternative Hypothesis – There is difference between ___ and ___.
There is variability in slopes as well. In this unit, you will learn how to perform significant tests and construct confidence intervals about the slope of a regression line.
• LSRL – ŷ=a+bx where x denotes (context) and ŷ denotes predicted (context)
• Degrees of Freedom – n-2
• Confidence Interval – b+- t*(SEb)
• In repeated sampling, I am __% confident that the true slope falls within this interval.
• True Slope – 𝞫
• Null Hypothesis – 𝞫 = 0; changing x does nothing to y
Hopefully, these decks can help you study for your tests and ultimately, the AP exam. The best feature about Quizlet is the option to play games and use the flashcards wherever you are. When you are
studying, you can always duplicate a deck and customize it to your own needs.
As long as you review these flashcards at least once a day a few days before your test, you should be good to go. Make sure to take advantage of starring flashcards you struggle with! Before a test,
it's great to quickly look over the starred ones and then feel more confident about them.
You got this! Good luck studying.🍀 | {"url":"https://hours-zltil9zhf-thinkfiveable.vercel.app/ap-stats/faqs/quizlet-decks-ap-statistics/study-guide/McK83yVXqkQ58roeeMKG","timestamp":"2024-11-04T11:02:14Z","content_type":"text/html","content_length":"359763","record_id":"<urn:uuid:476aae75-d1aa-4daf-922b-1e0656f1ae51>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00790.warc.gz"} |
Re: st: RE: Robust instrumental variable regression
Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: st: RE: Robust instrumental variable regression
From Austin Nichols <[email protected]>
To [email protected]
Subject Re: st: RE: Robust instrumental variable regression
Date Mon, 17 Jan 2011 17:59:45 -0500
Jorge Eduardo Pérez Pérez <[email protected]>:
The reference cited elliptically in
etc. is Amemiya (1982) and is given below--that paper proves
consistency of the 2SLAD model for the structural parameter \beta
which determines the conditional mean of the outcome given X.
That's the conditional mean, not median.
If distributional assumptions imply that the conditional mean is also
the median, fine, but that is not the same approach due to Koenker
(2005 etc.) that most people think of when they reach for -qreg-.
On 2SLAD, see also Powell (1983,1986), Chen (1988), Portnoy and Chen
(1996), and Arias, Hallock and Sosa-Escudero (2001).
For comparisons, see the 2007 NBER lecture at
Chernozhukov and Hansen (2006) not only recommend a more general
method, but in their footnote 1 on page 493, they clarify that 2SLAD
will produce inconsistent estimates when the effects of the endogenous
variables vary across quantiles:
"We do not use the term ‘‘two stage quantile regression’’ (2SQR)
because it is already used to name the procedure proposed by Portnoy
and Chen (1996) as an analog of the two stage LAD (2SLAD) of Amemiya
(1982) and Powell (1983). This procedure has been widely used to
estimate quantile effects under endogeneity. When the QTE vary across
quantiles, the 2SQR does not solve (1.4) and thus is inconsistent
relative to the treatment parameter of interest. Note that 2SLAD and
2SQR are still excellent strategies for estimating constant treatment
effect models."
Amemiya, Takeshi. (1982). "Two stage least absolute deviations
estimators." Econometrica 50(3):689-711.
Arias, Hallock & Sosa-Escudero (2001). "Individual heterogeneity in
the returns to schooling: Instrumental variables quantile regression
using twins data." Empirical Economics 26(1): 7-40.
Chen. (1988). "Regression Quantiles and Trimmed Least Squares
Estimators for Structural Equations and Non-Linear Regression Models."
Unpublished Ph.D. dissertation, University of Illinois at
Chernozhukov and Hansen. (2006). "Instrumental quantile regression
inference for structural and treatment effect models." Journal of
Econometrics, 73, 245-261.
Koenker, R. (2005). Quantile regression. Cambridge: Cambridge University Press.
Portnoy, S. and Chen, L. (1996). "Two-stage regression quantiles and
two-stage trimmed least squares estimators for structural equation
models." Communication in Statistics, Theory Methods, 25(5):1005-1032.
Powell, J. (1983). "The Asymptotic Normality of Two-Stage Least
Absolute Deviations Estimators." Econometrica 51(5):1569-1575.
Powell, J. (1986). "Censored Regression Quantiles." Journal of
Econometrics, 32(1):143-155.
2011/1/14 Jorge Eduardo Pérez Pérez <[email protected]>:
> Median regression is more robust to outliers than linear regression.
> Median regression with instrumental variables can be performed with
> the procedure described in this post (which includes a relevant
> reference):
> http://www.stata.com/statalist/archive/2003-09/msg00585.html
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2011-01/msg00521.html","timestamp":"2024-11-09T10:52:57Z","content_type":"text/html","content_length":"15110","record_id":"<urn:uuid:74f98ed7-a4b2-4fa5-97ff-a0c9cd9fdeda>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00456.warc.gz"} |
Hypothesis Testing With the Binomial Distribution
If a manufacturer claims superiority for any of their products or a great deal rests the proportion of components that exceed a certain lifetime, then that claim or proportion probably needs to be
tested for legitimacy or accuracy.
We can test a wide range of statistical parameters or models by subjecting it to a hypothesis test. We assume that the claim is true. The trueness of the claim forms the basis of the null hypothesis,
and our suspicion that the claim is false forms the basis of the alternative hypothesis. We test an observation assuming that the null hypothesis is true, and calculate the probability of observing
an outcome at least as unlikely as the observed outcome. If the probability of observing any outcome at least as unlikely is above a certain level, then we say there is insufficient level to reject
the null hypothesis, and if the probability of observing any outcome at least as unlike is below that level, we reject the null hypothesis since the probability of observing that outcome is so unlike
assuming the null hypothesis to be true. If a claim of superiority or safety issue is involved, then typically we are only interested in disproving the claim, or finding if safety is compromised, and
we conduct a one tailed test.
Suppose for example that a manufacturer claims that their brand of cat food is preferred by 80% of cats. A suspicious cat lovers group finds that of the cats belonging to their members, only half
prefer the manufacturers pet food over a rival brand. If the manufacturers claim is true, then any sample of twenty cats will produce the probabilities for the number of cats who prefer the
manufacturers claim shown below.
Suppose we decide to reject the claim if the probability of the manufacturers claim being true – or any more extreme outcome – is less than 5% or 0.05, given that we have observed only ten cats
preferring the manufacturers petfood.
P(x <=10)=0.00259 <0.05, so the manufacturers claim is rejected, and there is some evidence that their brand of catfood is not preferred by 80% of all cats. | {"url":"https://astarmathsandphysics.com/a-level-maths-notes/s1/3711-hypothesis-testing-with-the-binomial-distribution.html","timestamp":"2024-11-12T06:19:13Z","content_type":"text/html","content_length":"32905","record_id":"<urn:uuid:c853de05-fbd4-4393-b212-101847eb96ab>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00193.warc.gz"} |
All Ideals of Ring
All Ideals of Ring
For given group G, by G.subgroups(), we can list all subgroups. now my question is :
Is there something for subgroups() for finding all left ideals of a given ring. For instance, k = GF(5); M = MatrixSpace(k,2,2) How can I have all left ideals?
1 Answer
Sort by ยป oldest newest most voted
There is not a single command to achieve that, but in this particular case you can do the following:
sage: k = GF(5)
sage: M = MatrixSpace(k,2,2)
sage: units = [m for m in M if m.is_invertible()]
sage: nonunits = [m for m in M if not m.is_invertible()]
sage: len(units)
sage: len(nonunits)
Since every ideal must be generated by some nonunits, but also changing a generator by a unit times it does not change the ideal, let's check how many essentially distinct generators can we have:
sage: associated = []
sage: for m in nonunits:
....: if not True in [a*m in associated for a in units]:
....: associated.append(m)
sage: associated
[0 0] [1 0] [0 1] [1 1] [2 1] [1 2] [4 1]
[0 0], [0 0], [0 0], [0 0], [0 0], [0 0], [0 0]
That is, you only want to check for subsets of that set as possible generators of your ideals. But it is clear that two of those elements actually generate the other four, so you just need to check
for ideals generated by up to two generators.
You have the trivial ideal, the other 6 principal ideals generated by the other elements in associated, and the only thing you need to check is if the ideal generated by two of them is the total one.
It is easy to see that it is, but in case you don't notice at first sight you can just check it directly:
sage: for (a,b) in Tuples(units,2):
....: if (a*associated[1] + b*associated[2]).is_invertible():
....: print a
....: print b
....: break
[0 1]
[1 0]
[1 0]
[0 1]
There you are: only six nontrivial ideals, that happen to be principal.
edit flag offensive delete link more
Nice. Note that instead of True in [a*m in associated for a in units], you can use the more pythonic any([a*m in associated for a in units]).
tmonteil ( 2014-11-29 03:34:23 +0100 )edit | {"url":"https://ask.sagemath.org/question/24892/all-ideals-of-ring/?sort=votes","timestamp":"2024-11-05T21:45:28Z","content_type":"application/xhtml+xml","content_length":"54071","record_id":"<urn:uuid:97a101de-58b5-4117-ab6f-a4b3742aed44>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00466.warc.gz"} |
Scientific Memoirs/1/On the Mathematical Theory of Heat - Wikisource, the free online library
On the Mathematical Theory of Heat; by S. D. Poisson, Member of the Institute, &c ^[1]
From the Annales de Chimie et de Physique, vol. lix. p. 71 et seq.
The work which I have just published under the title of The Mathematical Theory of Heat (Théorie Mathematique de la Chaleur), forms the second part of a treatise on Mathematical Physics (Physique
Mathématique), the first of which is the New Theory of Capillary Action (Nouvelle Théorie de l'Action Capillaire), which appeared four years ago. It contains twelve chapters, preceded by some pages
in which I recapitulate in a few words the first applications of the calculus which have been made to the theory of heat, and the principal researches of geometers upon that subject, which have been
made of late years, namely, since the first Memoir presented by Fourier to the Institute in 1807. I will here transcribe the contents of the Preface, on the important question of the heat of the
"In applying to the earth the general solution of the problem of a sphere at first heated in any manner whatever, Laplace was led to participate in the opinion of Fourier, which attributes to the
primitive heat of the earth the increase in temperature which is observed in descending from the surface, and the amount of which is not the same in all localities. This hypothesis of a temperature
proceeding from the original heat of the globe (la chaleur d'origine), and which must rise to millions of degrees in its central layers, has been generally adopted; but the difficulties it presents
appear to me to render it improbable. I have proposed a different explanation of the increasing temperature which has long since been observed at all depths to which man has penetrated.
"According to this new explanation the phænomenon depends on the inequality of temperature of those regions of space which the earth successively passes through in its translatory motion, and which
are common to the sun and all the planets. It would be indeed opposed to all probability that the temperature of space should everywhere be the same; the variations to which it is subject from one
point to another, separated by very great distances, may be very considerable, and ought to produce corresponding variations in the temperature of the earth, extending to various depths according to
their duration and amplitude. Suppose, for the sake of example, a block of stone transported from the equator to our latitudes; its cooling will have commenced at the surface, and then become
propagated into the interior; and if the cooling has extended throughout the whole mass, because the time of its transportation has been very short, that body thus transported to our climate will
present the phænomenon of an increase of temperature with the distance from the surface. The earth is in the case of this block of stone;—it is a body coming from a region the temperature of which
was higher than that of the place in which it now is; or we may regard it as a thermometer moveable in space, but which has not had time, on account of its magnitude and according to its degree of
conducting power, to take throughout its mass the temperatures of the different regions through which it has passed. At present the degree of temperature of the globe is increasing below the surface;
the contrary has in former times been, and will hereafter be, the case: besides, at epochs separated by many series of ages this temperature must have been, and will in future be, much higher or
lower than what it is at present; a circumstance, which renders it impossible that the earth should always be habitable by man, and has perhaps contributed to the successive revolutions the traces of
which have been discovered in its exterior crust. It is necessary to observe that the alternations of temperature of space are positive causes which have an increasing influence upon the heat of the
globe at least near its surface; while the original heat of the earth (chaleur d'origine de la terre), however slow it may be in dissipating, is but a transitoiy circumstance, the existence of which
it would not be possible at the present epoch to demonstrate, and to which we should not be forced to have recourse as a hypothesis except in the case of the permanent and necessary causes being
insufficient to explain the different phænomena."
The following are the titles of the different chapters of the work, together with a short abstract of the contents of each.
Chapter I. Preliminary Notions.—After having given the definition of temperature and many other definitions, it is explained how we have been led to the principle of a continual radiation and
absorption of heat by the molecules of all bodies. The interchange of heat between material particles of an insensible magnitude, but yet comprising immense numbers of molecules, cannot disturb the
equality of their temperatures when it actually exists. From this condition we conclude, that for each particle the ratio of the emitting to the absorbing power is independent of the substance and of
density, and that it can only depend on temperature. In the case of the inequality of temperatures, we give the general expression of their variations during every instant, equal and contrary for two
material particles, radiating one toward the other. We also give the law of absorption of radiant heat in the interior of homogeneous bodies.
Chapter II. Laws of Radiant Heat.—If a body be placed within a vacuous sphere on every side (enceinte vide fermée de toutes parts), the temperature of which is supposed to be invariable and
everywhere the same, we demonstrate that the result of the interchange of heat between an element of its surface and an element of the surface of the inclosing sphere, is independent of the matter of
which the sphere is formed, and proportional, cæteris paribus, to the cosines of the angle which the normal to the second element forms with the right line from one to the other element. Experiments,
not as yet made, only can decide whether this law of the cosine is equally applicable to the elements of the surface of the body, of which the temperature is not invariable like that of the sphere;
and until such experiments are made we may be allowed to doubt its existence while the body is heating or cooling. By considering the number of successive reflexions which take place at the surface
of the sphere we demonstrate also that in general the passage (flux) of heat through every element in the surface of the body which it contains is independent of the form, of the dimensions, and of
the material of the sphere; there is no exception, but when the heat, in the series of reflexions which it experiences, falls one or many times upon the surface of the body. It follows from this
theorem that a thermometer placed in any point whatever of the space which the sphere terminates, will finally indicate the same temperature, which will be equal to that of the sphere; but in the
case of the exception just mentioned, the time which it will employ in attaining that temperature will vary according to the place it occupies. The general expression of the passage of heat through
every element of the surface of a body of which the temperature varies, is composed of one factor relative both to the state of that surface and to the material of the body, multiplied by the
difference of two similar functions, one of which depends on the variable temperature of the body, the other on the fixed temperature of the sphere, which are the same for all bodies; a result which
agrees with the law of cooling in vacuo discovered by MM. Dulong and Petit. We next suppose in this second chapter, that many bodies differing in temperature are contained in the sphere of which the
temperature is constant, and arrive then at a general formula, which will serve to solve the problems of the catoptrics of heat, the principal applications of which we indicate. When all these bodies
form round one of them a closed sphere the temperature of which, variable with the time, is not the same throughout, the passage of heat to the surface of the interior body does not depend on its
temperature and that of the inclosure only, at least when these bodies are formed of the same material. After having considered the influence of the air upon radiation which we had at first
eliminated, we give at the end of this chapter a formula which expresses the instantaneous variations of temperature of two material particles of insensible magnitude, by means of which the exchange
of heat takes place after one or many reflexions upon the surfaces of other bodies through air or through any gas whatever.
Chapter III. The Laws of Cooling in Bodies having the same Temperature throughout.—While a homogeneous body of small dimensions is heating or cooling, its variable temperature is the same at every
point; but if the body is composed of many parts formed of different substances in juxtaposition, they may preserve unequal temperatures during all the time that these temperatures vary, as we show
in another chapter. In the present we determine, in functions of the time, the velocity and the temperature which we suppose to be common to all the points in a body placed alone in a sphere either
vacuous or full of air, and the temperature of which is variable. If the sphere contains many bodies subject to their mutual influence upon each other, the determination of their temperatures would
depend on the integration of a system of simultaneous equations, which are only linear in the case of ordinary temperatures, but in which we cannot separate the variables when we investigate high
temperatures, and when the radiation is supposed not to be proportional to their differences.
Experiment has shown that in a cooling body, covered by a thin layer or stratum of a substance different from that of which it is itself composed, the velocity of refrigeration only arrives at its
maximum when the thickness of this additive stratum, though always very small, has notwithstanding attained a certain limit. We develop the consequences of this important fact in what regards
extension of molecular radiation, and explain how those consequences agree with the expression of the passage of heat found in the preceding chapter.
Chapter IV. Motion of Heat in the Interior of Solid or Liquid Bodies.—We arrive by two different processes at the general equation of the motion of heat; these two methods are exempt from the
difficulties which the Committee of the Institute, which awarded the prize of 1812^[2] to Fourier, had raised against the exactitude of the principle upon which his method was sustained. The equation
under consideration is applicable both to homogeneous and heterogeneous bodies, solid or fluid, at rest or in motion. It was unnecessary, as they appeared to have thought, to find for fluids an
equation different from the one I obtained long since for heterogeneous bodies. The variations of temperature which take place at every instant, and arise from the mutual radiation of the
neighbouring molecules, depend in fact only on their actual positions, and not at all on the positions in which they will be the instant after in consequence of the motions produced by their
calorific action or by other causes: it is thus that in the problem of the flux and reflux of the tides, for example, we calculate the attraction of the sea upon each point of its mass, as if it were
solid and at rest at the moment under consideration, and independently of the motions which this attraction may produce.
Notwithstanding that the interior radiation takes place only between molecules the temperatures of which are extremely different, the equation of motion of the heat contains terms derived from the
squares of their differences, and of the same order of magnitude as those which result from their first power; so that the exact equation differs, in the case of a homogeneous body, from that which
we had already given; and it is not, like that, independent of the conductibility when the body has arrived at an invariable state. This equation of partial differences changes its form, when we
cannot consider the extent of the interior radiation as insensible; it is then of a higher order, which introduces, in its integral, new constants or arbitrary functions. From this a difficulty of
analysis arises, of which we give the solution, and explain how in every case the redundant quantities will be made to disappear, as will be seen from a particular example in another chapter. We form
in this the general expression of the passage of heat through every element of a surface traced in the interior of a body which is heated or cooled, or has arrived at an invariable state, and in
which the extent of the interior radiation is considered as insensible. This passage proceeds from the exchange of heat between the molecules of the two parts of that body near their surface of
separation, and the temperatures of which are very different; whilst the interior passage results from the exchanges between the molecules adjacent to the surface of the body and those of a
surrounding medium, or of other bodies which may have much higher or much lower temperatures; and notwithstanding that the respective magnitudes of these two passages (ces deux flux), due to causes
also unequal, must be of the same order and comparable with one another. We show how that condition is fulfilled, by means of a quantity resulting from the rapid decrease of temperature which takes
place very near the surface of a body whilst heating or cooling. In this manner interior and exterior passages are found united with one another; and the law of interior conductibility expressed in
functions of the temperature is deduced from that of exterior radiation which MM. Dulong and Petit have discovered.
In a homogeneous prism which has arrived at an invariable state, and the lateral surface of which is supposed to be impermeable to heat and its two bases retained at constant temperatures, the
passage of heat across every section perpendicular to its length is the same throughout its length. Its magnitude is proportional to the temperature of the two bases, and in the inverse ratio of the
distance which separates them. This principle is easy to demonstrate, or rather it may be considered as evident. Thus expressed, it is independent of the mode of communication of heat, and it takes
place whatever be the length of the prism: but it was erroneous to have attributed it without restriction to the infinitely thin slices of one body, the temperature of which varies, either with the
time, or from one point to another; and to have excluded from it the circumstance, that the equation of the movement of heat, deduced from that of extension, is independent of any hypothesis and
comparable in its generality to the theorems of statics. When we make no supposition respecting the mode of communication of heat, or the law of interior radiation, the passage of heat through each
face of an infinitely thin slice is no longer simply proportional to the infinitely small difference of the temperatures of the two faces, or in the inverse ratio of the thickness of the slices; the
exact expression of it will be found in the chapter in which we treat specially of the distribution of heat in a prismatic bar.
Chapter V. On the Movement of Heat at the Surface of a Body of any Form.—We demonstrate that the passages of heat are equal, or become so after a very short time, in the two extremities of a prism
which has for its base an element of the surface of a body, and is in height a little greater than the thickness of the superficial layer, in which the temperature varies very rapidly. From this
equality, and from the expression of the exterior radiation, given by observation, we determine the equation of the motion of heat at the surface of a body of any form whatsoever. The expression of
the interior passage not being applicable to the surface itself, it follows that the demonstration of this general equation, which consists in immediately equalizing that expression to the expression
of the exterior radiation, is altogether illusory.
When a body is composed of two parts of different materials, two equations of the motion of heat exist at their surface of separation, which are demonstrated in the same manner as the equation
relative to the surface; they contain one quantity depending on the material of those two parts respectively, and which can only be determined by experiment.
Chapter VI. A Digression on the Integrals of Equations of partial Differences.—By the consideration of series, we demonstrate that the number of arbitrary constants contained in the complete integral
of a differential equation ought always to be equal to that which indicates the order of that equation: we prove by the same means, that in the integral of an equation of partial differences the
number of arbitrary functions may be less, and change as the integral is developed in series, according to the powers of one or other variable; and when the equation of partial differences is linear,
we show that by conveniently choosing this variable all the arbitrary functions may disappear and be replaced by constants, infinite in number, without the integral ceasing to be complete. To
elucidate these general considerations, we apply them to examples by means of which we show that the different integrals, in the series of the same equation of partial differences, are transformed
into one another, and may be expressed under a finite form by definite integrals, which also contain one or several arbitrary functions. In the single case, in which the integral in series contains
only arbitrary constants, every term of the series by itself satisfies the given equation, so that the general integral is found expressed by the sum of an infinite number of particular integrals.
Integrals of this form have appeared from the origin of the calculus of partial differences; but in order that their use in different problems should not leave any doubt respecting the generality of
the solutions, it would have been necessary to have demonstrated à priori, as I did long since, that these expressions in series, although not containing any arbitrary function, as well as those
containing a greater or smaller number of them, are not less on that account the most general solutions of equations of partial differences; or else it would have been necessary to verify in every
example that, after having satisfied all the equations of one problem relative to contiguous points infinite in number, the series of this nature might still represent the initial and entirely
arbitrary state of this system of material points; a verification which, until now, it has not been possible to give, except in very particular cases. The solution which Fourier was the first to
offer of the problem of the distribution of heat in a homogeneous sphere, of which all the points equidistant from the centre have equal temperatures, does not satisfy for example either of these two
conditions; it was no doubt on this account that the members of the Committee, whose judgement we mentioned above, thought that his (Fourier's) analysis was not satisfactory in regard to generality;
and, in fact, in this solution it is not at all demonstrated that the series which expresses the initial temperature can represent a function, entirely arbitrary, of the distance from the centre.
For the use of these series of particular solutions, it will be necessary to proceed in a manner proper to determine their coefficients according to the initial state of the system. On the occasion
of a problem relative to the heat of a sphere composed of two different substances, I have given for this purpose in the Journal de l'Ecole Polytechnique, (cahier 19, p. 377 et seq.,) a direct and
general method, of which I have since made a great number of applications, and which I shall also constantly follow in this work. The Sixth Chapter contains already the application to the general
equations of the motion of heat in the interior and on the surface of a body of any form either homogeneous or heterogeneous. It leads in every case to two remarkable equations, one of which serves
to determine, independently of one another, the coefficients of the terms of each series, and the other to demonstrate the reality of the constant quantities by which the time is multiplied in all
these terms. These constants are roots of transcendental equations, the nature of which it will be very difficult to discover, by reason of the very complicated form of these equations. From their
reality this general consequence is drawn; viz. when a body, heated in any manner whatever, is placed in a medium the temperature of which is zero, it always attains, before its complete cooling, a
regular state in which the temperatures of all its points decrease in the same geometrical progression for equal increments of time. We shall demonstrate in another chapter, that, if that body is a
homogeneous sphere, these temperatures will be equal for all the points at an equal distance from the centre, and the same as if the initial heat of each of its concentric strata had been uniformly
distributed throughout its extent.
The equations of partial differences upon which depend the laws of cooling in bodies are of the first order in regard to time, whilst the equations relative to the vibrations of elastic bodies and of
fluids are of the second order; there result essential differences between the expressions of the temperatures and those of the velocities at a given instant, and for that reason it appears at least
very difficult to conceive that the phænomena which may result from a molecular radiation should be equally explicable by attributing them to the vibrations of an elastic fluid. When we have obtained
the expressions of the unknown quantities in functions of the time, in either of these kinds of questions, if we make the time in them equal to zero, we deduce from that, series of different forms
which represent, for all the points of the system which we consider, arbitrary functions, continuous or discontinuous, of their coordinates. These expressions in series, although we might not be able
to verify them, except in particular examples, ought always to be admitted as a necessary consequence of the solution of every problem, the generality of which has been demonstrated à priori; it will
however be desirable that we should also obtain them in a more direct manner; and we might perhaps so attain them, by means of the analysis of which I had made use in my first Memoir on the theory of
heat, to determine the law of temperatures in a bar of a given length, according to the integral under a finite form of the equation of partial differences.
Chapter VII. A Digression on the Manner of expressing Arbitrary Functions by Series of Periodical Quantities.—Lagrange was the first to give a series of quantities proper to represent the values of
an arbitrary function, continuous or discontinuous, in a determined interval of the values of the variable. This formula supposes that the function vanishes at the two extremes of this interval; it
proceeds according to the sines of the multiples of the variable; many others exist of the same nature which proceed according to the sines or cosines of these multiples, even or uneven, and which
differ from one another in conditions relative to each extreme. A complete theory of formulæ of this kind will be found in this chapter, which I have abstracted from my old memoirs, and in which I
have considered the periodical series which they contain as limits of other converging series, the sums of which are integrals, themselves having for limits the arbitrary functions which it is our
object to represent. Supposing in one or other of these expressions in series, the interval of the values of the variable for which it takes place to be infinite, there results from it the formula
with a double integral, which belongs to Fourier; it is extended without difficulty, as well as each of those which only subsists for a limited interval, to two or a greater number of variables.
Chapter VIII. Continuation of the Digression on the Manner of representing Arbitrary Functions by Series of Periodical Quantities.—An arbitrary function of two angles, one of which is comprised
between zero and 180°, and the other between zero and 360°, may always be represented between those limits by a series of certain periodical quantities, which have not received particular
denominations, although they have special and very remarkable properties. It is to that expression in series that we have recourse in a great number of questions of celestial mechanics and of
physics, relative to spheroids; it had however been disputed whether they agreed with any function whatever; but the demonstration of this important formula, which I had already given and now
reproduce in this chapter, will leave no doubt of its nature and generality. This demonstration is founded on a theorem, which is deduced from considerations similar to those of the preceding
chapter. We examine what the series becomes at the limits of the values of the two angles; we then demonstrate the properties of the functions of which its terms are formed; then it is shown that
they always end by decreasing indefinitely, which is a necessary consequence and sufficient to prevent the series from becoming diverging, for which purpose its use is always allowable. Finally, it
is proved, that for the same function there is never more than one development of that kind; which does not happen in the developments in series of sines and cosines of the multiples of the
variables. This chapter terminates with the demonstration of another theorem, by means of which we reduce a numerous class of double integrals to simple integrals.
Chapter IX. Distribution of Heat in a Bar, the transverse Dimensions of which are very small.—We form directly the equation of the motion of heat in a bar, either straight or curved, homogeneous or
heterogeneous, the transverse sections of which are variable or invariable, and which radiates across its lateral surface. We then verify the coincidence of this equation with that which is deduced
from the general equation of Chapter IV., when the lateral radiation is abstracted and the bar is cylindrical or prismatic. This equation is first applied to the invariable state of a bar the two
extremities of which are kept at constant and given temperatures. It is then supposed, successively, that the extent of the interior radiation is not insensible, that the exterior radiation ceases to
be proportional to the differences of temperature, that the exterior conductibility varies according to the degree of heat, and the influence of those different causes on the law of the permanent
temperatures of the bar is determined. Formulæ are also given, which will serve to deduce from this law, by experiment, the respective conductibility of different substances, and the quantity
relative to the passage from one substance into another, in the case of a bar formed of two heterogeneous parts placed contiguous to and following one another. After having thus considered in detail
the case of permanent temperatures, we resolve the equation of partial differences relative to the case of variable temperatures; which leads to an expression of the unknown quantity of the problem,
in a series of exponentials, the coefficients of which are determined by the general process indicated in Chapter VII., whatever may be the variations of substance and of the transverse sections of
the bar. We then apply that solution to the principal particular cases. When the bar is indefinitely lengthened, or supposed to be heated only in one part of its length, the laws of the propagation
of heat on each side of the heated place are determined; this propagation is instantaneous to any distance; a result of the theory presenting a real difficulty, but the explanation of which is given.
Chapter X. On the Distribution of Heat in Spherical Bodies.—The problem of the distribution of heat in a sphere, all the points of which equally distant from the centre have equal temperatures, is
easily brought to a particular case of the same question with regard to a cylindrical bar. It is also solved directly; the solution is then applied to the two extreme cases, one of a very small
radius, and another of a very great one. In the case of an infinite radius, the laws are inferred of the propagation of caloric in a homogeneous body, round the part of its mass to which the heat has
been communicated, similarly in all directions.
We then determine the distribution of heat in a homogeneous sphere covered with a stratum, also homogeneous, formed of a substance different from that of the nucleus. During the whole time of
cooling, the temperature of this stratum, however small its thickness may be, is different from that of the sphere in the centre, and the ratio of the temperatures of these two parts, at the same
instant, depends on the quantity relative to the passage from one substance into the other, of which we have already spoken. From this circumstance an objection arises against the method employed by
all natural philosophers to determine, by the comparison of the velocities of cooling, the ratio of the specific heat of different bodies, after having brought their surfaces to the same state by
means of a very thin stratum of the same substance for all these bodies. The quantity relative to the passage of the heat of each body in the additive stratum, is contained in the ratio of the
velocities of cooling; it is therefore necessary that it should be known in order to be able to deduce from this ratio, that of their specific heats. A recent experiment by M. Melloni proves that a
liquid contained in a thin envelope, the interior surface of which is successively placed in different states by polishing or scratching it, always cools with the same velocity, whilst the ratios of
the velocity change very considerably, as was known long before, when it is the exterior part of the vessel that is more or less polished or scratched. The quantity relative to the passage of caloric
across the surface of separation of the vessel and the liquid, is therefore independent of the state of that surface, a circumstance which assimilates the cooling power of liquids to that of the
stratum of air in contact with bodies, which in the same manner does not depend on the state of their surface, according to the experiments of MM. Dulong and Petit.
When a homogeneous sphere, the cooling of which we are considering, is changed into a body terminated by an indefinite plane, and is indefinitely prolonged on one side only of that plane, the
analytical expression for the temperature of any point whatever changes its form, in such a manner that that temperature, instead of tending to diminish in geometrical progression, converges
continually towards a very different law, which depends on the initial state of the body; but however great a body may be, it has always finite and determined dimensions; and it is always the law of
final decrease enunciated in Chapter VI. which it is necessary to apply; even in the case, for example, of the cooling of the earth.
If the distribution of heat in a sphere, or in a body of another form, has been determined, by supposing this body to be placed in a medium the temperature of which is zero, this first solution of
the problem may afterwards be extended to the case in which the exterior temperature varies according to any law whatever. In my first Memoir on the theory of heat, I have followed, in regard to this
part of the question, a direct method applicable to all cases. According to this method, one part of the value of the temperature in a function of the time is expressed in the general case by a
quadruple integral, which can always be reduced to a double integral like each of the other parts. By the method which I have used to effect this reduction we obtain the value of different definite
integrals, which it would be difficult in general to determine in a different manner, and the accuracy of which is verified whenever they enter into known formulæ.
Chapter XI. On the Distribution of Heat in certain Bodies, and especially in a homogeneous Sphere primitively heated in any Manner—It is explained how, in every case, the complete expression of
exterior temperature, which may depend on the different sources of heat, and which must be employed in the equation of the motion of heat relative to the surface of bodies submitted to their
influence, will be formed.
After having enumerated the different forms of bodies for which we have hitherto arrived at the solution of the problem of the distribution of heat, the complete solution is given for the case of a
homogeneous rectangular parallelopiped the six faces of which radiate unequally.
In order to apply the general equations of the fourth and fifth chapters to the case of a homogeneous sphere primitively heated in any manner, the orthogonal coordinates in them are transformed into
polar coordinates; the temperature at any instant and in any point is then expressed by means of the general series of Chapter VIII., and of the integrals found in Chapter VI.; the coefficients of
that series are next determined according to the initial state of the sphere, by supposing at first the exterior temperature to be zero: by the process already employed in the preceding Chapter, this
solution is finally extended to the case of an exterior temperature, varying with the time and from one point to another. Among the consequences of this general solution of the problem the most
important is that for which we are indebted to Laplace; it consists in this: That in a sphere of very large dimensions, and at distances from the surface very small in proportion to its radius, the
part of the temperature independent of the time does not vary sensibly with these distances; and, that upon the normal at each point, whether at the surface or at an inconsiderable depth, it may be
regarded as equal to the invariable part of the exterior temperature which corresponds to the same point. Hence it results, that the increase of heat in the direction of the depth which is observed
near the surface of the earth cannot be attributed to the inequality of temperatures of different climates, and that it is necessary to look for the cause in circumstances which vary very slowly with
the time. Whatever this cause may be, the difference of the mean temperatures of the surface and beyond, corresponding to the same point of the superficies, is proportional (according to a remark
made by Fourier) to the increase of temperature upon the normal referred to the unity of length, so that this difference may be determined from the observed increase, and from a quantity depending
on the nature of the ground. This remark and that of Laplace are not applicable to the localities where the temperature varies very rapidly round the vertical: it is shown that in these cases of
exception the temperature varies even upon the vertical: and the law of this variation is determined from the variation which has taken place at the surface or in the exterior temperature. The mean
temperature at a very small distance contains also a term which is not proportional to this depth, and which arises from the influence of the heat on the conductibility of the substance.
Chapter XII. On the Motion of Heat in the Interior and upon the Surface of the Earth.—It is shown that the formulæ of the preceding chapter, although relating to a homogeneous sphere the surface of
which is everywhere in the same state, may notwithstanding serve to determine the temperatures of the points of the earth at a distance from the surface which is very small with regard to its radius,
but which exceeds however all accessible depths. These formulæ contain two constants, depending on the nature of the soil, the numerical values of which may be determined in every point of the globe
from the temperatures observed at different known depths.
Observation in harmony with theory shows that the diurnal inequalities of the temperature of the earth disappear at very small depths, and the annual inequalities at greater depths, in such a manner
that at a distance from the surface of about 20 metres and beyond those two kinds of inequalities are entirely insensible. In this chapter are given tables of the temperatures, indicated by the
thermometer, of the caves of the Observatory, at the depth of 28 metres. The mean of 352 observations, made from 1817 to the end of 1834, is 11°·834.
The increase of the mean temperature of the earth, in proportion as we descend below the surface, has long been established as a fact in all deep places, at different latitudes, and at different
elevations of the soil above the level of the sea. The most adequate means to determine it is by sounding and boring. The results, still very few, which have hitherto been obtained are given. At
Paris, this increase appears to be one degree for about 38 metres of increase in depth.
As to the cause of this phænomenon, the difficulties are stated which the explanation of Fourier presents, founded upon the original heat of the globe, still sensible at the present time near the
surface; the new explanation alluded to at the beginning of this article is then proposed. The following reflections extracted from the work tend to prove that the solidification of the earth must
have commenced by central strata, and that before reaching the surface the cooling of the globe must have been incomparably more rapid.
"The nearly spherical form of the earth and planets, and their flattening at the poles of rotation, evidently show that these bodies were originally in a fluid or perhaps in an aëriform state.
Beginning from this initial state, the earth could not, wholly or partly, become solid, except by a loss of heat arising from its temperature exceeding that of the medium in which it was placed. But
it is not demonstrated that the solidification of the earth could have commenced at the surface and been propagated towards the centre, as the state of the globe still fluid in the greatest part of
the interior would lead us to suppose; the contrary' appears to me more probable. For the extreme parts, or those nearer to the surface, being the first cooled, must have descended to the interior
and been replaced by internal portions which had ascended to cool at the surface and to descend again in their turn. This double current must have maintained an equality of temperature in the mass,
or at least must have prevented the inequality from becoming in any way so great as in a solid body, which cools from the surface; and we may add that this mixture of the parts of the fluid, and the
equalization of their temperatures, must have been favoured by the oscillations of the whole mass, which must have taken place until the globe attained a permanent figure and rotation. On the other
hand, the excessively great pressure sustained by the central strata may have determined their solidification long before that of those nearer the surface; that is to say, the first may have become
solid by the effect of this extreme pressure at a temperature equal or even superior to that of the strata more distant from the centre, and consequently subjected to a much less degree of pressure.
Experiment has shown, for example, that water at the ordinary temperature being submitted to a pressure of 1000 atmospheres, experiences a condensation of about 120th of its primitive volume. Now let
us conceive a column of water whose height is equal to one radius of the globe, and let us reduce its weight to half of that which we observe at the surface of the earth, in order to render it equal
to the mean gravity which would exist along each radius of the earth upon the hypothesis of its homogeneity; the inferior strata of this liquid column would experience a pressure of more than three
millions of atmospheres, or equal to more than three thousand times the pressure which would reduce water to 1920ths of its volume; but without knowing the law of the compression of this liquid, and
although we do not know in what manner this law may depend on the temperature, we may believe, notwithstanding, that so enormous a pressure would reduce the inferior strata of the mass of water to
the solid state, even when the temperature is very high. It seems therefore more natural to conceive that the solidification of the earth began at the centre and was successively propagated towards
the surface: at a certain temperature, which might be extremely high, the strata nearer the centre became at first solid, by reason of the excessive pressure which they experienced ; the succeeding
strata were then solidified at a lower temperature and under a less degree of pressure, and thus in progressive succession to the surface."
If the increase observed in the temperature of the earth near its surface is due to its original heat, it follows that at the present epoch at Paris this heat raises the temperature of the surface
itself only by the fortieth part of a degree. Not knowing the radiating power of the substance of the globe, we cannot estimate the quantity of this initial heat which traverses in a given time from
within to without an extent, also given, of the surface; but such would be its slowness in dissipating into space, that more than one thousand million of centuries must elapse to reduce the small
increase of the fortieth of a degree to one half.
With regard to periodical inequalities, the relation which exists between each inequality at a given depth and the inequality corresponding to the exterior temperature is determined. Relations of
this nature, for the knowledge of which we are indebted to M. Fourier, take place between the interior inequalities and those of the surface of the ground; these relations leave unknown the ratios of
these latter inequalities to those of the outside which are the immediate data of the question.
The interior temperature to which the earth is subjected arises from three different sources, namely, from sidereal heat, from atmospherical heat, acting either by radiation or by contact, and from
solar heat. These three sources of heat are successively examined. With regard to the first it is observed, that it is not at all probable that radiant heat emanating from the stars has the same
intensity in all directions when it arrives at the earth. The experiments are indicated which it would be necessary to make in order to ascertain whether it really varies in the different regions of
the sky. M. Melloni intends immediately to apply himself to these experiments, employing in them an extremely sensible instrument, of which he has made use in his researches on heat; a circumstance
which cannot fail to lead to the solution of this important question of celestial physics.
Before considering the influence of atmospherical heat, I have formed a complete expression for the temperature, marked every instant by a thermometer suspended in the air, at any height above the
surface of the earth exposed in the shade or in the direct rays of the sun. Although the greatest part of the quantities which this formula contains are unknown to us, many general consequences may
however be deduced from it, which accord with experiment; it hence follows, that to determine the proper temperature of the air, it is necessary to employ the simultaneous observation of three
thermometers, the surfaces of which are in a different state, and not two thermometers only, as is generally said. This formula also furnishes the means of comparing the temperatures indicated by
different thermometers in relation to their radiating powers and to their property of absorbing the rays of the sun.
The mean of the annual temperatures, marked by a thermometer exposed in the open air and in the shade, forms the climateric temperature. It varies with the elevation of places above the level of the
sea, and with the longitude and latitude, according to unknown laws. At Paris it is 10°·822, as M. Bouvard has concluded after 29 years of observations. There will be found in this Chapter a table of
the mean temperatures for the twelve months of each of those years, which that gentleman has been pleased to communicate to us, and which had not before been published. It appears that in every point
of the earth this climateric temperature differs very little from the mean temperature of the surface of the soil, as is shown by several examples. Notwithstanding, the variable temperature of this
surface, and that which is marked at the same instant by a thermometer as little elevated above the surface as may be, are often very different from each other; it hence follows, that in a year the
excess of the highest above the lowest temperature of the soil is at Paris nearly 24°, as will be seen in the course of this Chapter; and only about 17° for the thermometer suspended in the air and
in the shade.
We now determine the part of exterior temperature which results from the atmospherical heat combined with sidereal heat. The necessary data for calculating its numerical value, à priori, being
unknown to us, we show how this value, for every point of the globe, may be deduced from the mean temperature of its surface. At Paris this exterior temperature is 13°. Although we cannot determine
separately the portion of this temperature of the earth which arises from the atmospherical heat, there is reason to think that it is also negative, so that the other portion arising from sidereal
heat must be less than 13° below zero. If we suppose that radiant heat emanating from the stars falls in the same quantity on all points of the globe, this temperature, higher than 13°, will be that
of space at the place where the earth is at this time. Without being able to assign the degree of heat of space, we may however admit, that its temperature differs little from zero, instead of being,
as had been asserted, below the temperature of the coldest regions in the globe, and even of the freezing-point of mercury. As to the central temperature of the whole mass of the earth, even
supposing its original heat to be entirely dissipated, and that it is no longer equal to the present temperature of space, we have no means of obtaining a knowledge of it.
According to a theorem of Lambert, the whole amount of solar heat which falls upon the earth is the same during different seasons, notwithstanding the inequality of their lengths, which is found to
be compensated by that of the distances from the sun to the earth. This quantity of heat varies in the inverse ratio of the parameter of the ellipse described by the earth; it also varies with the
obliquity of the ecliptic; but it does not appear that these variations can ever produce any considerable effect on the heat of the globe. The quantities of solar heat which fall in equal times upon
the two hemispheres are nearly equal; but on account of the different states of their surfaces, those quantities are absorbed in different proportions; and the power of absorbing the rays of the sun
increasing in a greater ratio than the radiating power, which is greater for dry land than for the sea, we conclude that the mean temperature of our hemisphere, where dry land is in a greater
proportion, must be greater than that of the southern hemisphere; which agrees with observation.
The solar heat, which reaches each point of the globe, varies at different hours of the day; it is null when the sun is beneath the horizon; during the year it varies also with its declination; and
the expression changes its form as the latitude of the point under consideration is greater or less than the complement of the obliquity of the ecliptic. I have therefore considered the part of the
exterior temperature which arises from this source of heat as a discontinuous function of the horary angle, and of the longitude of the sun, to which I have applied the formulæ of the preceding
Chapters, in order to convert it into series of sines and cosines of the multiples of these two angles. By this means I have obtained the complete expressions of the diurnal and annual inequalities
of the temperature of the earth which arise from its double motion. These formulæ show, that at the equator the annual inequalities are much less than elsewhere; a circumstance which furnishes the
explanation of a fact observed by M. Boussingault in his journey to the Cordilleras, and upon which he had relied in order to determine with great facility the climateric temperatures of the places
which he visited. The same formulæ agree also, in a remarkable manner, with the temperatures which M. Arago has observed at Paris during many years, and at depths varying from two to eight metres
(from 6·56 to 26·24 English feet).
1. ↑ The work of which this article is an analysis, is described as a quarto volume of more than 500 pages, with a plate; published by Bachelier, Quai des Augustins, Paris.
2. ↑ This Committee consisted of MM. Lagrange, Laplace, Legendre, Haüy and Malus. | {"url":"https://en.m.wikisource.org/wiki/Scientific_Memoirs/1/On_the_Mathematical_Theory_of_Heat","timestamp":"2024-11-03T08:57:27Z","content_type":"text/html","content_length":"91297","record_id":"<urn:uuid:30147b86-ddb0-414c-8179-2d42af9b37a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00127.warc.gz"} |
Long Division - Steps, Examples
Long division is an important mathematical concept which has many practical uses in various domains. One of its primary application is in finance, where it is applied to figure out interest rates and
work out loan payments. It is further applied to calculate taxes, investments, and budgeting, making it an essential skill for anybody involved in finance.
In engineering, long division is utilized to solve complicated challenges linked to development, construction, and design. Engineers use long division to calculate the loads that structures can hold,
assess the strength of materials, and plan mechanical networks. It is also applied in electrical engineering to determine circuit parameters and design complex circuits.
Long division is further essential in science, where it is applied to calculate measurements and carry out scientific workings. For example, astronomers use long division to figure out the length
between stars, and physicists use it to determine the velocity of objects.
In algebra, long division is applied to factor polynomials and solve equations. It is an important tool for figuring out complex challenges that consist of huge numbers and need accurate
calculations. It is further utilized in calculus to calculate integrals and derivatives.
As a whole, long division is a crucial math concept which has several real-world applications in many domains. It is a fundamental math function which is utilized to solve complicated problems and is
a crucial skill for anyone keen in science, finance, engineering, or arithmetic.
Why is Long Division Important?
Long division is an essential math concept which has many uses in distinguished fields, involving engineering, science and finance. It is a fundamental mathematics function that is applied to work
out a broad array of problems, such as figuring out interest rates, determining the amount of time needed to complete a project, and figuring out the distance traveled by an object.
Long division is further utilized in algebra to factor polynomials and solve equations. It is an essential tool for figuring out intricate challenges which involve large numbers and requires precise
Procedures Involved in Long Division
Here are the procedures involved in long division:
Step 1: Write the dividend (the value being divided) on the left and the divisor (the number dividing the dividend) on the left.
Step 2: Determine how many times the divisor could be divided into the first digit or set of digits of the dividend. Write the quotient (the outcome of the division) above the digit or set of digits.
Step 3: Multiply the quotient by the divisor and write the outcome beneath the digit or set of digits.
Step 4: Subtract the outcome achieved in step 3 from the digit or set of digits in the dividend. Note down the remainder (the amount left over after the division) below the subtraction.
Step 5: Bring down the next digit or set of digits from the dividend and append it to the remainder.
Step 6: Replicate steps 2 to 5 until all the digits in the dividend have been refined.
Examples of Long Division
Here are few examples of long division:
Example 1: Divide 562 by 4.
4 | 562
Therefore, 562 divided by 4 is 140 with a remainder of 2.
Example 2: Divide 1789 by 21.
21 | 1789
Thus, 1789 divided by 21 is 85 with a remainder of 11.
Example 3: Divide 3475 by 83.
83 | 3475
So, 3475 divided by 83 is 41 with a remainder of 25.
Common Mistakes in Long Division
Long division can be a problematic idea to master, and there are many common mistakes that learners make when working with long division. One general mistake is to forget to note down the remainder
when dividing. Another common mistake is to incorrectly place the decimal point while dividing decimal numbers. Students may also forget to carry over numbers while subtracting the product from the
To circumvent making these errors, it is important to work on long division daily and pay close attention to ever stage of the process. It can also be helpful to revisit your work utilizing a
calculator or by performing the division in reverse to make sure that your solution is correct.
In addition, it is essential to comprehend the basic principles about long division, such as the link among the quotient, dividend, divisor, and remainder. By conquering the fundamentals of long
division and preventing usual mistakes, individuals can better their skills and gather self-esteem in their skill to figure out complex challenges.
Ultimately, long division is a crucial math concept that is essential for solving complex problems in various domains. It is utilized in finance, engineering, science, and mathematics, making it an
essential skill for working professionals and students alike. By mastering the steps consisted in long division and comprehending how to use them to real-life challenges, anyone can gain a detailed
grasp of the complex workings of the world around us.
If you need help understanding long division or any other mathematical idea, Grade Potential Tutoring is here to help. Our experienced tutors are accessible remotely or face-to-face to give
customized and effective tutoring services to guide you be successful. Our tutors can guide you across the steps of long division and other math concepts, support you work out complex challenges, and
offer the tools you require to excel in your studies. Call us today to schedule a tutoring session and take your math skills to the next stage. | {"url":"https://www.irvineinhometutors.com/blog/long-division-steps-examples","timestamp":"2024-11-07T03:31:30Z","content_type":"text/html","content_length":"77028","record_id":"<urn:uuid:8a8953a9-504f-46e9-bf2e-5d55d7a3a1b0>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00638.warc.gz"} |
Matematica | Enrollment
Matematica Enrollment
Enrollment in the Master's Degree Program in Mathematics Academic Year 2024/2025
CLASS LM 40 - MATHEMATICS
1. Enrollment in the Master's Degree Program requires a Bachelor's Degree or a three-year university diploma, or an equivalent foreign qualification recognized under current legislation.
2) In order to be admitted to the Master's Degree Program in Mathematics, specific curricular requirements and adequate personal preparation of the student, as indicated below, are also required:
a) Curricular requirements:
i) A degree in the following class: L-35 (or class 32 ex D.M. 509/99).
ii) In the case of a degree in different classes, it is necessary to have obtained at least 104 ECTS credits distributed as follows:
• 42 ECTS in basic subjects within the SSD MATH-01÷06 of the Basic Mathematics Training area;
• 62 ECTS in characterizing subjects within the SSD MATH-01÷06 of the Theoretical Training or Modeling-Application Training area.
In the case of a lack of minimum curricular requirements in terms of SSD/ECTS, the Teaching Council may indicate the training activities necessary to acquire them. In any case, any curricular
integrations in terms of ECTS credits must be acquired by the student before the verification of individual preparation; in no case is enrollment permitted with training debts.
b) Adequacy of personal preparation To be admitted to the Master's Degree Program, adequate personal preparation of the student is required. The adequacy of the preparation for admission is
ascertained through an examination of the graduate's university career and, if necessary, a verification in person, which may consist of an individual interview and/or a written test. The subjects of
the test aimed at verifying the adequacy of personal preparation are the following:
MATH-01/A - Logic
Formal systems and languages: propositional calculus, first-order logic and their completeness; Boolean algebras.
MATH-02/A - Algebra
Elements of set theory; elements of arithmetic, algebraic structures, groups, rings, vector spaces, polynomial rings, elements of field theory.
MATH-02/B - Geometry
Vector spaces and linear transformations; matrices, determinant, and rank; endomorphisms and diagonalization; scalar products; affine spaces; topological spaces and continuous functions; compactness,
connectedness, and separation properties; geometric properties of curves and surfaces.
MATH-03/A - Mathematical Analysis
Functions of one and several real variables: limits, derivatives, integrals; sequences and series of functions; curves and surfaces, line and surface integrals; ordinary differential equations,
first-order and higher-order, linear and nonlinear.
MATH-03/B - Probability and Mathematical Statistics
Basic elements of probability; conditional probability and independence of events; random variables, random vectors, moments and mixed moments; asymptotic theorems for random variables.
MATH-04/A - Mathematical Physics
Kinematics and dynamics of a point and of free and constrained systems; geometry of masses.
MATH-05/A - Numerical Analysis
Representation of real numbers on a computer and rounding error; numerical methods for systems of linear and nonlinear equations; approximation of data and functions; quadrature formulas; numerical
methods for finding the eigenvalues of matrices.
MATH-06/A - Operations Research
Linear, conical, and convex combinations; convex functions and convex sets; formulation of optimization problems through Linear Programming models; Simplex Method; Duality Theory and post-optimality
3. Graduates of the L-35 class who have obtained the degree with a grade of at least 85/110 are exempt from the test to verify the adequacy of initial preparation.
The methods and timing for verifying the admission requirements and assessing the adequacy of preparation for admission are defined annually in the Study Plan and published on the website of the
Degree Program.
For the current academic year, the test is through a written exam that will take place twice, in September and February, before the start of the first and second semester courses.
Candidates referred to in point 2.a).ii) and candidates who do not meet the requirement referred to in point 3 are required to submit an application for the assessment of the requirements for
enrollment in the Master's Degree Program, available at the following link on the website of the University of Salerno: http://web.unisa.it/didattica/immatricolazioni/informazioni
The aforementioned application must be sent by email, with a scanned copy of the front and back of your identification document attached, to the Career Office of the Department of Mathematics (
carrierestudenti.dipmat@unisa.it), by the 15th day of the month (September or February) in which the test in which you intend to participate is held. No payment is required to submit the | {"url":"https://corsi.unisa.it/MATEMATICA-MAGISTRALE/en/enrollment","timestamp":"2024-11-05T22:50:37Z","content_type":"text/html","content_length":"46006","record_id":"<urn:uuid:c23b42bf-f91b-4415-9bff-d289d01ced7f>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00181.warc.gz"} |
Haskell Type System: A Comprehensive Guide to Data Constructors, Type
Exploring Haskell's Type System: From data constructors to :t and :k
A beginner's guide to understanding Haskell's type system
In Haskell, every value has a specific type and the type of a value determines the set of operations that can be performed on it. Haskell is a statically typed language, which means that the type of
a value must be known at compile-time. Types are a way to classify values and provide a way for the compiler to check that we are using values in a way that makes sense.
Type synonym
The type keyword in Haskell is used to create a type synonym, which defines a new name for an existing type. It allows you to give an existing type a new name for the purpose of making your code more
readable and maintainable. A type synonym does not create a new type, it just provides an alternative name for an existing type, it's just an alias.
type IntList = [Int]
type Name = String
We can use the IntList just as [Int] here:
sumList :: IntList -> Int
sumList [] = 0
sumList (x:xs) = x + sumList xs
sumList [1,2,3] -- 6
Data constructor
The data keyword in Haskell is used to define new algebraic data types, which consist of one or more data constructors. Each data constructor defines a new value that can be created from the type.
Data constructors can take one or more arguments, and return a value of the new type.
For example, the following code defines a new type Shape with two data constructors Circle and Rectangle:
data Shape = Circle Float | Rectangle Float Float
Each data constructor takes different arguments, Circle takes one float argument and Rectangle takes two float arguments. These data constructors can be used to create values of the new type Shape:
let circle = Circle 3.5
let rectangle = Rectangle 2.0 4.0
Here, the Shape is a type. Circle and Rectangle are a data constructors.
Next, let's use :t and :k to check the type information
We can use :t to check the type of an expression in Ghci. It's a shorthand for the :type command.
When you type an expression followed by :t in the REPL or ghci, it will show you the type of that expression. For example:
ghci> :t 1
1 :: Num a => a
ghci> :t (+)
add :: Num a => a -> a -> a
The Circle is a data constructor, which has a type of Float -> Shape Means when it get a Float, it will become a Shape. Similarly, for Rectangle, it needs 2 Floats to become a Shape .
ghci> :t Circle
Circle :: Float -> Shape
ghci> :t Circle 1
Circle 1 :: Shape
ghci> :t Rectangle
Rectangle :: Float -> Float -> Shape
ghci> :t Rectangle 1
Rectangle 1 :: Float -> Shape
ghci> :t Rectangle 1 2
Rectangle 1 2 :: Shape
Nothing has a type of Maybe a. and Just has a type of a -> Maybe a, because Just requires an argument to make it a Maybe a Type.
ghci> :t Just
Just :: a -> Maybe a
ghci> :t Just 1
Just 1 :: Num a => Maybe a
ghci> :t Just 'a'
Just 'a' :: Maybe Char
ghci> :t Nothing
Nothing :: Maybe a
However, if you try to check the type of Shape, you will get an error:
ghci> :t Shape
<interactive>:1:1: error: Data constructor not in scope: Shape
Because it's not a data constructor.
We can use :k to check Shape the Kind.
:k is a command used in the REPL (Read-Evaluate-Print Loop) or in the ghci (Glorious Glasgow Haskell Compilation System) to check the kind of a type. It's a shorthand for the :kind command.
A kind is a type of types, they tell us about the number of arguments that a type constructor takes, it's like a type of a type. Every type has a kind, and every term has a type. It's the number of
arguments for the type constructor, not for the data constructor.
Here's an example of how you can use :k:
ghci> :k Int
Int :: *
This tells us that the type Int has a kind of *, which is the kind of a normal type.
You can use :k to check the kind of a type constructor like this:
ghci> :k Maybe
Maybe :: * -> *
Maybe has a kind of * - > * because the Maybe is defined like this:
data Maybe a = Just a | Nothing
It requires a type a.
Meanwhile, the Shape we defined above does not need a type argument. So its kind is *
ghci> :k Shape
Shape :: *
In conclusion, understanding the basics of Haskell's type system can be a bit tricky at first, but it's totally worth it! When you get a grip of it, you'll be able to write safer and more efficient
code. With data constructors, type constructors and type synonyms, you'll be able to create your own types and define custom behavior for them. And don't forget the importance of tools like :t and
:k, they are your best friends when it comes to debugging and understanding your code. All in all, this blog aimed to give you an overview of these concepts and I hope it helped you to understand
them a little better, and have fun exploring the world of Haskell's type system! | {"url":"https://blog.rockydd.net/exploring-haskells-type-system-from-data-constructors-to-t-and-k","timestamp":"2024-11-04T14:42:16Z","content_type":"text/html","content_length":"158199","record_id":"<urn:uuid:54edbe64-20e6-4db0-b8f6-740879e79589>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00390.warc.gz"} |
Managing Complexity [llsmf2904]
5.00 credits
22.5 h + 15.0 h
Coeurderoy Régis; Iania Leonardo;
The material covered in the courses of bachelor in Business Engineering. In particular, students are assumed to be familiar with basic concepts of statistics and econometrics, financial accounting,
managerial accounting, and mathematics for business. Knowledge of statistical and econometrics programming languages such as R-studio, and/or Matlab, etc, is assumed.
Main themes
We live in a complex environment, where the interconnections among economic agents (firms, consumers, ect.), their choices/decisions under uncertainty and as a response to unforeseen events determine
the successfulness of firms’ activities. The last global economic crisis driven by the Covid-19 pandemic, the great financial crisis, the digital transformation, and the pressing need for a
transition towards a greener economy, are just some examples how complex and uncertain the firms’ competitive arena can be. In this course, students will learn basic tools that companies can use to
identify, report and analyze the risks/opportunities that a complex environment can bring to firms’ activities.
Learning outcomes
At the end of this learning unit, the student is able to :
Upon completion of this course, students will:
• Be able to understand and critically assess the risks an organization is exposed to;
• Critically assess the reporting of risk in corporations and associated strategic reporting practices;
• Analyze the risks a corporation is exposed to;
• Apply empirical work in a (relatively) new software (R, Python, etc.).
1 What is risk management?
1.1 Introduction
1.2 Identifying and documenting risk
1.3 Fallacies and traps in risk management
1.4 Why safety is different
1.5 The Basel framework
1.6 Hold or hedge?
1.7 Learning from a disaster 13
2 The structure of risk
2.1 Introduction to probability and risk
2.2 The structure of risk
2.3 Portfolios and diversification
2.4 The impact of correlation
2.5 Using copulas to model multivariate distributions 49
3 Measuring risk
3.1 How can we measure risk?
3.2 Value at risk
3.3 Combining and comparing risks
3.4 VaR in practice
3.5 Criticisms of VaR
3.6 Beyond value at risk 82
4 Understanding the tails
4.1 Heavy-tailed distributions
4.2 Limiting distributions for the maximum
4.3 Excess distributions
4.4 Estimation using extreme value theory 115
5 Making decisions under uncertainty
5.1 Decisions, states and outcomes
5.2 Expected Utility Theory
5.3 Stochastic dominance and risk profiles
5.4 Risk decisions for managers 156
6 Understanding risk behavior
6.1 Why decision theory fails
6.2 Prospect Theory
6.3 Cumulative Prospect Theory
6.4 Decisions with ambiguity
6.5 How managers treat risk
7 Stochastic optimization
7.1 Introduction to stochastic optimization
7.2 Choosing scenarios
7.3 Multistage stochastic optimization
7.4 Value at risk constraints 224
8 Robust optimization
8.1 True uncertainty: Beyond probabilities
8.2 Avoiding disaster when there is uncertainty
8.3 Robust optimization and the minimax approach 250
9 Real options
9.1 Introduction to real options
9.2 Calculating values with real options
9.3 Combining real options and net present value
9.4 The connection with financial options
9.5 Using Monte Carlo simulation to value real options
9.6 Some potential problems with the use of real options 285
10 Credit risk 291
10.1 Introduction to credit risk
10.2 Using credit scores for credit risk
10.3 Consumer credit
10.4 Logistic regression
Teaching methods
The course will be centered around the following teaching methods:
• In-class lectures
• Practical sessions
• Regular meetings with the professors and assistants
• Case studies
• Guest lecture
Prior to the participation to those activities, students will be provided with learning material and compulsory readings that will be pivotal for the understanding of the teaching activities.
Evaluation methods
The evaluation methods are based on “Continuous Evaluation”, i.e. no exam is foreseen at the end of the teaching session. Students will work in groups on (i) exercises and (ii) concrete, real-life
case studies, for which they will deliver a written report and an oral presentation. Individual evaluation will also be part of the final grade.
Teaching materials
• Book: Business Risk Management: Models and Analysis by Edward J. Anderson
• Slides, available on Moodle
Title of the programme
Learning outcomes
Master [120] : Business Engineering | {"url":"https://uclouvain.be/en-cours-2023-llsmf2904","timestamp":"2024-11-02T14:52:43Z","content_type":"text/html","content_length":"27564","record_id":"<urn:uuid:f3ba0349-f2fe-4716-b074-5fecff46f1e4>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00682.warc.gz"} |
ThmDex – An index of mathematical definitions, results, and conjectures.
Let $X$ and $Y$ each form a
D2506: Topological metric space
Let $d_X$ and $d_Y$ each be the
D58: Metric
in $X$ and $Y$, respectively.
Let $f : X \to Y$ be a
D48: Bilipschitz map
with respect to $X$ and $Y$. | {"url":"https://thmdex.com/r/81","timestamp":"2024-11-02T11:27:36Z","content_type":"text/html","content_length":"6125","record_id":"<urn:uuid:9ea3e2e5-2975-4f51-b2b7-7a31064adda0>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00864.warc.gz"} |
In Forex Trading What is a Pip Pro Tips! - Priyotottho
In Forex Trading What is a Pip Pro Tips!
When you engage in Forex trading, one of the first concepts you need to wrap your head around is that of a Pip. A Pip is the smallest unit of measurement in foreign currency trading, and it
represents the change in value between two currencies. For example, if the EUR/USD pair moves from 1.3600 to 1.3605, that 0.0005 increase in value is equal to five Pips.
03 – What is a pip? – easyMarkets – Education
A pip is the smallest unit of price movement in forex trading. A pip is usually equal to one basis point, or 0.0001. For example, if the EUR/USD moves from 1.3600 to 1.3601, that is a move of one
1 Pip is Equal to How Many Dollars
1 Pip is Equal to How Many Points? This is a question that we get asked a lot, and it’s one that doesn’t have a straightforward answer. The reason for this is because there are a lot of different
factors that can come into play when determining the value of a pip.
The first thing you need to know is that pips are the smallest unit of measurement in the foreign exchange market. A pip is typically equal to 0.0001 of a currency pair. So, if the EUR/USD currency
pair moves from 1.2345 to 1.2346, that would be considered one pip of movement.
Now, let’s talk about how pips relate to points. One point is simply equal to 1/100th of 1% – or in other words, it’s 0.01%. So, if we take our previous example of the EUR/USD moving from 1.2345 to
1.2346, we can see that this would be equivalent to 10 points of movement (0.0001 x 100 = 0.01).
However, it’s important to keep in mind that not all brokers use the same amount of decimal places when quoting prices – so while one broker may quote the EUR/USD at 1.23450 and another at 1 . 23451
, both prices would still only be five-pip movements apart .
Forex Pips Calculator
A forex pips calculator is an essential tool for any trader to have in their toolkit. This simple yet powerful tool allows you to quickly and easily calculate the value of a pip in any currency pair.
The first thing you need to do is select the currency pair that you are trading.
Then, enter the size of your position and finally, select the account currency. The forex pips calculator will then show you the value of one pip in your account currency. It is important to remember
that the value of a pip can vary depending on the currency pair that you are trading as well as the size of your position.
Therefore, it is important to use a forex pips calculator when determining your risk per trade and setting your stop loss and take profit levels.
How Much is 100 Pips Worth
In the world of foreign exchange (forex), a pip is a unit of measurement used to express price movement. A pip is equal to 0.0001 of a currency unit, and is typically used to express changes in very
small amounts, such as 1/100th of a percent. For example, if the EUR/USD exchange rate moves from 1.2345 to 1.2346, that would be considered one pip of movement.
While pips are typically very small amounts, they can add up quickly when leveraged trading is involved. For example, if a trader has $10,000 in their account and they use 50:1 leverage (meaning they
can trade with $500,000 worth of currency), then even a small move in the market could result in large gains or losses. So how much is 100 pips worth?
That depends on the size of your position and the currency you’re trading. Let’s say you have a $50,000 forex account and you’re long (meaning you’re betting that the pair will go up) 10,000 units of
EUR/USD at an exchange rate of 1.2500. If EUR/USD increases by 100 pips to 1.2600, your position will be worth $5200 ((10 000 * 1) + (10 000 * 0)).
In other words, each pip in this instance was worth $52 USD . Of course , had the market gone against you and fallen by 100 pips instead , your position would have been worth $4800 . As such , it’s
important to always use risk management techniques when trading forex , such as stop-loss orders , which can help limit your losses if the market moves against you .
How to Calculate Pips Pdf
When it comes to forex trading, one of the first things you need to know is how to calculate pips. Pips are essentially the unit of measurement for currency movement and are used to calculate profits
and losses in the market. A standard lot size in forex is 100,000 units of the base currency, so a pip would be 0.0001 (one hundredth of a percent) of that.
For example, if you were buying EUR/USD and the price moved from 1.3600 to 1.3605, that would be a 5 pip move. Now that we know what a pip is, let’s talk about how you can calculate your profits or
losses in the market. There are two main ways to do this: using a calculator or by using some simple math.
Let’s start with the calculator method first. If you have access to a trading platform, most will have a built-in profit/loss calculator. This tool will do all of the work for you and spit out your
results in pips automatically.
All you need to do is enter in your position size (the number of lots or units you’re trading), entry price, and exit price into the calculator and it will tell you how many pips you made or lost on
the trade. For those who don’t have access to a trading platform or simply prefer not use one, don’t worry – calculating your pips is easy enough without one! To do this manually, simply take your
position size (again, this is usually measured in lots or units), multiply it by 0.0001 (one hundredth of a percent), then subtract your entry price from your exit price – this will give you your raw
pip amount before factoring in any spread/commission costs associated with the trade.
Finally, if you want to factor in these costs as well so that your final result is more accurate, simply divide your total commission fees by 2 (since they’re usually charged on both sides of the
trade) and subtract that number from your raw pip total – this will give you your net pip amount for the trade including all fees incurred!
How to Calculate Pips With Lot Size
Pips are the unit of measure used to calculate profits or losses in the forex market. A pip is equal to 0.0001 of a currency pair, and is typically the fourth decimal place shown in a currency quote.
For example, if the EUR/USD exchange rate is 1.2042, then one pip would be equal to 0.0001, or 0.01% of the quote currency.
The lot size is the amount of currency you trade per pip. For example, if you buy one micro lot (1,000 units) of EUR/USD at 1.2042 and it rises to 1.2043, your profit would be equal to one pip
multiplied by your lot size: (0.0001 x 1,000 = $0.10). If you had traded 10 micro lots, your profit would have been $1; if you had traded 100 mini lots, your profit would have been $10; and so on.
Xauusd Pip Calculator
If you’re a forex trader, then you know that one of the most important things to consider is the value of a “pip.” But what exactly is a pip? And how do you calculate its value?
A pip is the smallest unit of measurement in forex trading. It’s typically equal to 1/100th of 1% or 0.0001. So, if the EUR/USD exchange rate moves from 1.2345 to 1.2346, that would be a one pip
Now that we know what a pip is, let’s talk about how to calculate its value. This will vary depending on the currency pair you’re trading and the size of your trade (known as your “position size”).
Here’s a quick and easy formula for calculating the value of a pip:
Pip Value = (One Pip / Exchange Rate) * Position Size For example, let’s say you’re trading EUR/USD with a position size of 100,000 units (standard lot). And let’s say the current exchange rate is
To calculate the value of one pip in this case:
Pips in Forex Pdf
When it comes to forex trading, one of the most important things to understand are pips. A pip is the unit of measurement used to express the change in value between two currencies. For example, if
the EUR/USD exchange rate goes from 1.3600 to 1.3605, that would be a 5 pip move.
Pips can be either positive or negative, and are typically expressed as a decimal number. The value of a pip can vary depending on the currency pair being traded and the size of your position (in
other words, how many currency units you are buying or selling). One thing to keep in mind is that not all currency pairs include the US dollar (USD).
For example, if you’re trading EUR/JPY, then each pip will be worth slightly more than if you were trading USD/JPY because one euro is currently worth more than one dollar. In general though, most
traders focus on so-called “major” currency pairs which do include the USD and tend to be more liquid (i.e., there’s generally more activity and therefore opportunity). These major pairs include: EUR
/USD, GBP/USD, USD/JPY, USD/CHF and AUD/USD.
Credit: canaltrader.com
How Much is 50 Pips Worth?
A pip is the smallest unit of price movement in foreign exchange trading. In most currency pairs, one pip is equal to 0.0001 of the quote currency. For example, if the EUR/USD moves from 1.23456 to
1.23457, that’s a one-pip move.
Pip value varies depending on the currency pair you’re trading and the account base currency. If you’re trading in a standard lot (100,000 units), each pip is worth $10 when USD is the base currency.
But if your account base currency isn’t USD, then the pip values will be different.
To calculate pip value for non-USD pairs or for pairs with USD as the counter/quote currency (secondcurrency), use this formula: ((One Pip / Exchange Rate) * Lot Size) / 10 For example, let’s say
you’re long GBP/JPY at 150 and your lot size is 100,000 GBP.
The current exchange rate is 145 JPY per GBP (0.006897). To calculate your pip value simply divide 10 by 0.006897 which equals 1447 then multiply it by your lot size of 100 000 which equals 144 700
yen per 1GBP pip move all divided by 10 to get your answer in pips like so; 144 700/10 = 14 470 Yen per 1GBP Pip Move
How is Pip Calculated in Forex?
When you trade forex, you’re effectively borrowing the first currency in the pair to buy or sell the second currency. With a US$5-trillion-a-day market, the liquidity is so deep that liquidity
providers—the big banks, basically—allow you to trade with leverage. To trade with leverage, you simply set aside the required margin for your trade size.
If you’re trading 200:1 leverage, for example, you can trade £2 per point on GBP/USDwith just £10 in your account. Now, when you lose 50% of your account balance in a single year (which happens quite
often), it will take two years just to get back to where you started from scratch! And during those two years of trying to recover your losses and getting back to breakeven point, there is a good
chance that your account will be wiped out completely by margin calls before that happens.
How Much is 20 Pips Worth?
20 pips is worth $2.00 if each pip is worth $0.10, and $200.00 if each pip is worth $1.00.
How Many Pips is a Dollar?
In forex trading, a “pip” is a unit of measurement for price movement. It is usually the fourth decimal place in a currency pair, 0.0001. For example, if the EUR/USD moves from 1.2345 to 1.2346, that
is one pip of movement.
Most brokers provide fractional pip pricing, so you would see a 5-pip spread on this pair as 1.2345/1.2350. A pip is sometimes called a point or a tick . One reason the foreign exchange market can be
confusing for newbies is that there is no centralized marketplace where currencies trade against each other like stocks on an exchange (think NYSE or NASDAQ).
Instead, currencies are traded in pairs—for example EUR/USD—with each currency serving as one side of the transaction (as opposed to stocks where you buy shares of stock outright). When you see a
quote for EUR/USD it means that every euro you have will buy you $1 USD (assuming nothing has changed with respect to the two currencies since the quoted price). If instead EUR/USD went from $1.2345
to $1.3345 it would be considered 10 pips of movement and your €10 euros would now buy you $13 USD—a nice return on your investment!
A pip is the smallest unit of measurement in forex trading. Pip stands for “percentage in point” and is used to measure price movements in currency pairs. A pip is equivalent to 0.0001 of a percent,
or one-hundredth of a percentage point (pp).
In most major currency pairs, except those involving the Japanese yen, a pip is equal to 0.01 of the quote currency. | {"url":"https://priyotottho.com/in-forex-trading-what-is-a-pip/","timestamp":"2024-11-13T22:25:28Z","content_type":"text/html","content_length":"109122","record_id":"<urn:uuid:cafc703c-5b85-48ca-bcdf-e3dd9d94c1c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00302.warc.gz"} |
Agile and statistics!
Mike Cohn, the guru at
Mountain Goat Software
, recently gave a webinar presentation to a bunch of PMI folks entitled "
Agile and the Seven Deadly Sins of Project Management
" [just click on the link for a free copy of the charts from Mountain Goat]
Overall, an informing presentation
In an explanation of how agile fights information opaqueness, Mike presented a slide with a bar chart of team velocities and announced a 'confidence interval' as the main take away.
Gasp!.... I was shocked! shocked! to hear statistics in an agile discussion; sounds so much like management--project management at that. But, it's easy to tell that Mike is pragmatic, and confidence
intervals are nothing if not practical.
Fair enough .... but actually there was no explanation given as to what entails a confidence interval. I'll correct that failing here.
First, an interval of what?
Would you believe the possible value of a random variable?
And which would that be? Answer: the sample average of velocity, call it V-bar.
And V-bar being a random variable, it has a distribution that prescribes how likely is any particular value of V-bar to fall into the interval of interest, ie, the confidence interval.
Second, we don't actually know the distribution of V-bar and we don't know the distribution of the population V (velocities), so we can't know what the next V is going to be, or its likelihood.
But, we know (from V-bar) an estimate of the population (V) mean. Thus, we can use V-bar as an estimating parameter of velocity, even though V-bar does not predict the next velocity value. (Example:
average team throughput = V-bar x input units, like story points or ideal days)
Third, since we don't know, and it's not economic to find out what the distribution of V-bar really is, it's customary to model it with a distribution that has been tried and proven for this
purpose--the T distribution.
The T-distribution is somewhat like a bell shaped distribution, except T usually has fat tails for small values of the parameter N-1 where N is the count of the values in the sample.
So what are the chances for V-bar, and how do you figure that out from the data given in Mike's chart?
I've reproduced my version of Mike's chart below; there are 9 velocity metrics ranging from about 37 to 25:
To calculate the quality of the confidence interval, some iteration is required. It's typical to first pick a level of confidence, say 95%, and then by use of a formula, and a set of 't' tables from
the T distribution, calculate the corresponding interval. If the results are not satisfying, a new pick of parameters may be required.
Here are the steps:
• Calculate the sample average V-bar, in this case 33, and the sample standard deviation, 4.1. Formulas in Excel will give you these figures from the 9 velocity points in the chart above.
• Look up the 't' value in a T-distribution table for N-1. N in this case is 9.
• Pick out the 't' value for 95% confidence [in t tables, it customary to look up a parameter labeled alpha; for 95% confidence, alpha = 0.05], in this case: 2.36 [there are formulas in Excel for
this also]
You'll get something like this:
• Calculate the interval around the center point of V-bar:
+/- t * sample standard deviation / sqrt(N)
+/- 2.31 * 4.1 / sqrt(9)
+/- 3.2
With just a little inspection of the formula above, and the t-tables, you'll discover that the interval gets wider as alpha is picked to be smaller [higher confidence]. In the limit, to have 100%
confidence in the interval, the interval would have to very wide to cover every possible case conceivable.
Values in the velocity chart outside the interval are outside the quality limits of 95% confidence.
For reference, here's the model of V-bar, specifically the T distribution with N-1 = 8:
Need more? Check out these two references:
And, check this out at the Khan Academy:
Read my contribution to the Flashblog | {"url":"http://www.johngoodpasture.com/2017/05/agile-and-statistics.html","timestamp":"2024-11-08T02:36:39Z","content_type":"application/xhtml+xml","content_length":"136869","record_id":"<urn:uuid:146f8bc8-d946-452c-80b3-fd6fef01cff0>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00718.warc.gz"} |
What is Mile's Equation and how do I apply it?
Example of using Mile’s Equation.
A bracket, as shown in the figure is cantilevered off a support wall. The attachment of
the bracket to the wall is modeled simply by rigid connections at the four corners. The
rigid connections are made to a single grid point with a seismic mass, through which the
applied acceleration acts.
A payload of 50 lbs is connected to the base of the bracket.
The seismic mass is constrained to move only on z and has a value 1e6 times the
bracket plus payload mass total.
The first natural frequency is a vertical plate bending mode of the bracket horizontal
shelf. It occurs at 33.372 Hz. Other modes exist below 500 Hz, but they are insensitive
to base z input direction
The input acceleration is created by applying a force numerically equal to the seismic
mass. A unit acceleration in fundamental units ( in/s^2) is achieved.
Damping is 1% critical (Q is 50.0).
The input acceleration is from 10 Hz to 500 Hz. A plot of response acceleration at the
payload mass is shown below. The amplified response is seen at 33.372 Hz, with a Q of
50 as expected.
An analytical solution exists for the single degree of freedom problem, so that the shape
of the curve shown above can be obtained directly by theory.
The relationship between Input PSD and output PSD is very simply:
Where FRi is the Frequency response at Frequency (i) for a unit input. It is effectively a
Transfer Function.
The Mean square response is the area under the PSDout curve. Knowing an exact
solution allows an analytical expression for the integral of FRi2 across the frequency
range, assumed to be infinite. The PSDin term is a simple multiplier.
Miles came up with the expression that integrates the FR across an infinite frequency
range, with the application of a constant input PSD, assumed to be PSDin.
The expression is:
In terms of the RMS response:
In the case of the bracket, PSDin = 0.2 g^2/Hz fn = 33.372 Hz Q = 50
Gout = 16.190 g units.
An FE analysis was carried out using a ‘real’ PSD input curve with typical slopes and
plateau. The first natural frequency falls under the 0.1 g^2/Hz plateau.
The RMS acceleration plot for the structure is shown in the figure. The peak is 16.169 g
at the payload attachment point. This matches very closely the Miles equation result.
For completeness the PSD response curve for vertical acceleration at the payload grid
is shown. As a spot check, the peak is 250.129 g^2/Hz.
PSDout at 33.372 = PSDin* FR2 = 0.1*50*50 = 250.0 g^2/Hz, so only a small
numerical error exists.
In conclusion, the Miles equation assumes a constant PSDin, and also a single degree
of freedom response
The bracket has a single dominant mode and the Miles equation is a good
approximation to the actual response.
A typical application is to carry out a subsequent hand calculation or FE analysis using
the acceleration level calculated as a pseudo static load. For a very large FE model
there is some motivation in doing this as a first check as the data storage required for a
full RMS calculation can be very large. A later section will deal with an advanced
method to fit the Miles equation to a more complex structure with several dominant
A similar stress level was found in the bracket case, but the payload is applying load at
a single point on the plate, which is not a good modeling method for stress analysis.
This model is primarily to demonstrate equivalence of the RMS g response. | {"url":"http://fetraining.com/ask_tony/dynamic_analysis/miles_equation/miles_equation_faq.html","timestamp":"2024-11-09T13:56:45Z","content_type":"text/html","content_length":"10064","record_id":"<urn:uuid:b9b1e817-e9a8-4c14-82b3-02f7225ef9e3>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00125.warc.gz"} |
Affordable Eminent IGCSE/IB Maths Tutors in Bern, Basel, Lausanne, Lucerne - €24 EUR/hour - IGCSE IB Math Tutor
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More than One Way to Skin a Cat - Party Welt
More than One Way to Skin a Cat
Adventures in Creative Thinking
How many times have you caught yourself saying that there could be no other solution to a problem – and that that problem leads to a dead end? How many times have you felt stumped knowing that the
problem laying before you is one you cannot solve. No leads. No options. No solutions.
Did it feel like you had exhausted all possible options and yet are still before the mountain – large, unconquerable, and impregnable? When encountering such enormous problems, you may feel like
you’re hammering against a steel mountain. The pressure of having to solve such a problem may be overwhelming.
But rejoice! There might be some hope yet!
With some creative problem-solving techniques you may be able to look at your problem in a different light. And that light might just be the end of the tunnel that leads to possible solutions.
First of all, in the light of creative problem-solving, you must be open-minded to the fact that there may be more than just one solution to the problem. And, you must be open to the fact that there
may be solutions to problems you thought were unsolvable.
Now, with this optimistic mindset, we can try to be a little bit more creative in solving our problems.
Number one; maybe the reason we cannot solve our problems is that we have not really taken a hard look at what the problem is. Here, trying to understanding the problem and having a concrete
understanding of its workings is integral solving the problem. If you know how it works, what the problem is, then you have a better foundation towards solving the problem.
Not trying to make the simple statement of what problem is. Try to identify the participating entities and what their relationships with one another are. Take note of the things you stand to gain any
stand to lose from the current problem. Now you have a simple statement of what the problem is.
Number two; try to take note of all of the constraints and assumptions you have the words of problem. Sometimes it is these assumptions that obstruct our view of possible solutions. You have to
identify which assumptions are valid, in which assumptions need to be addressed.
Number three; try to solve the problem by parts. Solve it going from general view towards the more detailed parts of the problem. This is called the top-down approach. Write down the question, and
then come up with a one-sentence solution to that from them. The solution should be a general statement of what will solve the problem. From here you can develop the solution further, and increase
its complexity little by little.
Number four; although it helps to have critical thinking aboard as you solve a problem, you must also keep a creative, analytical voice at the back of your head. When someone comes up with a
prospective solution, tried to think how you could make that solution work. Try to be creative. At the same time, look for chinks in the armor of that solution.
Number five; it pays to remember that there may be more than just one solution being developed at one time. Try to keep track of all the solutions and their developments. Remember, there may be more
than just one solution to the problem.
Number six; remember that old adage,” two heads are better than one.” That one is truer than it sounds. Always be open to new ideas. You can only benefit from listening to all the ideas each person
has. This is especially true when the person you’re talking to has had experience solving problems similar to yours.
You don’t have to be a gung-ho, solo hero to solve the problem. If you can organize collective thought on the subject, it would be much better.
Number seven; be patient. As long as you persevere, there is always a chance that a solution will present itself. Remember that no one was able to create an invention the first time around.
Creative thinking exercises can also help you in your quest be a more creative problems solver.
Here is one example.
Take a piece of paper and write any word that comes to mind at the center. Now look at that word then write the first two words that come to your mind. This can go on until you can build a tree of
related words. This helps you build analogical skills, and fortify your creative processes.
So, next time you see a problem you think you can not solve, think again. The solution might just be staring you right in the face. All it takes is just a little creative thinking, some planning, and
a whole lot of work. | {"url":"https://partywelt.net/more-than-one-way-to-skin-a-cat/","timestamp":"2024-11-12T11:00:36Z","content_type":"text/html","content_length":"116482","record_id":"<urn:uuid:17ea6d47-c7fa-44eb-8d86-946118f7cf86>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00300.warc.gz"} |
Solution assignment 07 Fractional functions and graphs
Return to Assignments Fractional functions and graphs
Assignment 7
Given the function:
the vertical asymptote;
the horizontal asymptote;
the intersection point with the
the intersection point with the
Based on these results sketch the result in the figure.
The line
We do not need to do complicated calculations any more. There also no asymptotes. The graph of the function is equal to the graph of | {"url":"https://4mules.nl/en/fractional-functions-and-graphs/assignments/solution-assignment-07-fractional-functions-and-graphs/","timestamp":"2024-11-06T22:04:16Z","content_type":"text/html","content_length":"43149","record_id":"<urn:uuid:138cf040-0fa2-48ea-b874-b8ea916c4c72>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00848.warc.gz"} |
How to use support vector regression for time series analysis?
Support vector regression (SVR) is a powerful machine learning algorithm that can be used for time series analysis. The basic idea of SVR is to find a function that best approximates the relationship
between the input variables (predictors) and the output variable (response) based on a set of training data.
Here are the steps to use SVR for time series analysis:
1. Data preparation: First, you need to prepare your time series data for analysis. This involves cleaning and preprocessing the data to remove any outliers, missing values, or other irregularities.
2. Feature selection: Next, you need to select the relevant features that will be used to train the SVR model. This can involve a variety of techniques, such as autocorrelation analysis, feature
engineering, or data reduction techniques.
3. Split data into training and testing sets: Split the dataset into a training set and a testing set. The training set will be used to fit the SVR model, while the testing set will be used to
evaluate its performance.
4. Train the SVR model: Use the training set to train the SVR model. The SVR algorithm will learn the patterns and relationships between the input features and the output variable.
5. Evaluate the model: Use the testing set to evaluate the performance of the trained SVR model. This can involve calculating metrics such as the mean squared error (MSE), mean absolute error (MAE),
and R-squared.
6. Tune hyperparameters: Adjust the hyperparameters of the SVR model to improve its performance. This can involve experimenting with different kernel functions, regularization parameters, and other
7. Make predictions: Once you have a well-trained SVR model, you can use it to make predictions on new, unseen data. This can involve using the model to forecast future values of the time series, or
to make predictions about other related variables.
Overall, using SVR for time series analysis can be a powerful tool for predicting and understanding complex patterns and relationships in data. By following these steps, you can effectively apply
this technique to your own time series datasets. | {"url":"https://devhubby.com/thread/how-to-use-support-vector-regression-for-time","timestamp":"2024-11-07T16:33:45Z","content_type":"text/html","content_length":"115262","record_id":"<urn:uuid:bb6dd7c4-264a-4120-bbab-9c74f020dca0>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00584.warc.gz"} |
Integral In Calculus | Hire Someone To Do Calculus Exam For Me
Integral In Calculus for More Than One Quantum Theory Field Theories as we know them? This series of lectures I do a lot to complete the proofs I provide in this three part series, the basic, much
more technical part. I intend to place these lectures further into more general applications, mainly to quantum computer computers or others that use computer image processing techniques. Among the
many applications of these quantum computers in science and engineering have been demonstrations of a large array of physical models. The ”3-dimensional” Isomorph of P500 – Quantum Computer Machine
(s) The “3-dimensional” Isomorph of P500 – Quantum Computer Machine(s) is an empirical example of topological quantum theories. As pointed out earlier by James Williams, “Many physical
implementations of quantum mechanics are based on the underlying physical theory (i.e. a physical system). The underlying physical theory includes some mathematical foundations such as matter,
momentum, energy and the metric (an additional physics dimension)”. The physical theory contains fundamental insights into how basic physics laws that we are studying based on physical theories can
be written. The ”3-dimensional” Isomorph of P500 – Quantum Computer Machine(s) is a remarkable example of a quantum mechanical theory without the need of a physical system, the material physics
fundamental required of the quantum mechanics. In this series of lectures, I share an assortment of the important similarities between these distinct models. Recall that the main elements in the
“3-dimensional” Isomorph of P500 – Quantum Computer Machine(s) are the concept of quantum gravity, quantum kinetic theory, quaternionic potential, vacuum. The main differences between these different
cases make certain that these models can have real physical outcomes. The following is a list of the relevant parts of these models: LQM – Quantum Mechanical Model P500 – Quantum Computer Machine(s)
P5001 – Non-Trace Wave P500-P500-50 – Non-Trace Wave P500-P50010 – Complex Surface and Harmonics P500-P500-50-65 – Complex Surface – Ambit P500 – Quantum Computer Image Processing P5001 – Vorticity
and Volumetricity (A form of “non-trace wave” for instance.) Q5/2-21/7 – The Principle of Zero Measurement in Quantum Theory I am a British physicist (primarily in the field of particle best site and
I wanted to investigate and comment on quantum mechanisms and physics related to acceleration and deceleration of the universe. I first conducted a run at the NASA Goddard Space Flight Center, in May
2010 with the hope that the rest of the web site would be open all the time. A small segment of the scientific web site is open for searching in any direction. As many people have already, I
published some papers on the issue on the Scientific Web in the Summer of 2010.. I quickly realized that there is a tendency towards a bad behaviour of the web that I was almost convinced happened
during the summer of 2011.
Take My Test For Me Online
At the Google page, you can find the status of the scientific Web site. I had argued for far too long in the very beginning that the results would be very noisy and wouldn’t provide any useful
insight into the behaviour of the universeIntegral In Calculus The Formula (Just Like a Bigger Baby, Yeah) In the text of this book, you’ll learn Calculus. You’ll learn how to solve general systems
like linear and logarithmic numerics, all of whom will follow from the underlying calculus. In Chapter 3, I go into more detailed mathematical detail, and I make some assumptions that can be examined
in class by class. You’ll also get a glimpse at the theory with such clarity that when students are looking for a book with a content that’s important and the learning curve runs after several words,
they’ll come away wondering what these concepts mean. IN STUDENT CEILING WITH PRACTICE To help the class learn more before taking a class, John Alesse was given new details about the basics of
textbook learning and introduced at the end of Chapter 17. We’ve already said that all of the concepts were necessary parts of the book; each discussion of them demonstrates them and our work in this
book is quite an intellectual activity. My class had the necessary structure for understanding that much of what I refer to in this chapter. We are going to demonstrate that we can produce proofs
using the book. In fact, what John did in this chapter was to introduce a fun and engaging teaching tactic called the method of differentiation – a neat trick where an outline is shown for n*n-s as
well as for n-s minus p. This trick was being shown at school, and I thought this was a wonderful way to try to illustrate it. Here I’ll elaborate a few new definitions and methods if one wishes to
explain problems of differential calculus, but in case we have a very long tradition of algebraic equations, I will begin with the introduction of the concept of algebra, introduced by G.M. Arzhan.
E. ’s paper suggested that it is simple but not difficult to prove with analytic methods. It is also easier to find simple examples such as rational numbers using the method of separation; perhaps
there would be many more examples in the future, but this time it is the method of differentiation. Calculus Algorithms A simple way to implement this trick is to use the known ‘quick and dirty’
methods of algebra for doing this. Although I’ve illustrated this concept a few thousand times, now I want to use it for the derivation of formulas. I will be showing this by laying the groundwork
over the next chapters for using this system for finding all the constants numerically by dividing n in n-s into two-sens and finding their arithmetic integrals.
Homework To Do Online
Of course the first thing to clarify is how a formula can be expressed in a way that’s satisfactory for a test. Many people have found formulas built into a calculus even though, yes, the way you
spell out words like ‘b’ is not rigorous. The result is a solution of the system that is, although correct, incompletely correct. One might even think there is a simple solution. For example, finding
the identity would be difficult; solving this problem involves using many parameters. A ‘hand formal’ description of what is needed in the system of equations would also be probably very well
explained, but given a definite form that we can make a statement about, where can this standard form be used? Perhaps solving the PDE on either the right person or the wrong person would be a more
difficult task. But there isn’t a hard way to describe this; see on the diagram below. **Note: Here’s what our problem is about (useful only in mathematically precise cases, but it is not really
mathematically difficult) I will be very brief: we have the following problems for linear approximation, rational approximation and so on: 1. A linear equation on the coefficients which we have
solved using the method of differentiation. 2. Melting the second coefficient and it cannot change the factor (the fraction coming from the Taylor polynomial of other factors). Differentiation in
real values on a square grid. No solution at all 3. A piecewise smooth function must be transformed into some rational function, and the result must have anIntegral In Calculus The term “significant”
in the IGAQI Dictionary of Statistical and Informed and Disturbance Calculus (Discalo) has been in use more than 6 years. I’m pretty sure that in some places, this term will always be used. What we
can infer, however, from these sentences, is that “most” of the formulas used in the IGAQI are not, once and for all, strictly valid (since the names used before are many, and “mostly” to indicate
that the check it out are valid, without any reference to their valid forms). We can also come away with a dozen or so useful functions, and no good reason to suggest that it is inappropriate (since
sometimes formulas are perfectly good). The following lists the basic formulas derived from each of these formulas: The | = | % | %| || | | | | | %| On top of both these formulas, we can provide a
slight indication of how we will approximate our sum. This statement is derived pretty directly by choosing a particular unit weight (such as 1,2,3, etc.) under these formulas, allowing us to create
an approximation of the value of a special weight if we don’t think that a unit weight works.
Is Online Class Tutors Legit
We can use these simple simplifications when working with the approximated value to get something concrete: The | = | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 = | | | | |
| | | | | | | | | | | | | | | | | To derive this approximation, first, we need a simple exercise: For an estimate, subtract the approximate value from the true point, find the last distance to the
exact point, and add that distance to the approximate value. This actually looks like this: Select one of the 2, 4, 5, and 6 formula constants, and if necessary a fantastic read asymbolals and other
formats) the only fraction that we’ll actually plug in. I don’t normally have a table for the digits or the division signs, though. We’ll also need to find that: Fraction of the smallest distance to
the final approximate value of the first approximation formula in the formula in question (for what it originally was). Or, if we’re simplifying in a more general way, call it a “summation” so we’ll
see that you find: For the second formula, we simply multiply there by approximately the largest distance from the exact value. This formula can be used without any further simplification by
approximating it by a distance higher than one, and remembering that “most” of the formulas are valid formulas (but not as valid in a general sense) (I don’t think anyone can find any documentation
for approximating a fraction by more than that much). For the third formula (rather than the one in the formula in question, which is more complicated) we just multiply, but only once one fraction is
computed. The third formula (the ones above, from right to left) is less efficient, and even more difficult | {"url":"https://hirecalculusexam.com/integral-in-calculus","timestamp":"2024-11-05T00:25:21Z","content_type":"text/html","content_length":"106657","record_id":"<urn:uuid:6e60ef55-e3b6-4192-a60a-2e4154a13d44>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00224.warc.gz"} |
st: Testing equality of 2 coefficients after FE regression: How does Sta
Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
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st: Testing equality of 2 coefficients after FE regression: How does Stata compute the Pooled SE and test statistic?
From andreas nordset <[email protected]>
To [email protected]
Subject st: Testing equality of 2 coefficients after FE regression: How does Stata compute the Pooled SE and test statistic?
Date Wed, 8 Dec 2010 11:15:18 +0100
Dear all,
after estimating a FE model, I am testing the equality of two coefficients.
The difference between them is about 3000 and the SE in the regression
output are respectively 1700 and 1600. I would thus have thought that
the test statistic should be
t = 3000/sqrt(1700^2+1600^) =1.29, insufficient to reject the Null
that the two coefficients are identical.
Yet --test coeff1=coef2-- gives me F(1,1386)=6.01 and hence Prob
>F=0.0143, thus rejecting the Null at the 5% confidence level.
So I'm wondering why this is: Does Stata use smaller within-person
standard errors for the test, given that the regression is a FE
regression? If so, should I also use such smaller SEs (rather than
those displayed next to the coefficients in the standard regression
output) for plotting the coefficients?
Thanks a lot,
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2010-12/msg00306.html","timestamp":"2024-11-08T18:51:31Z","content_type":"text/html","content_length":"10103","record_id":"<urn:uuid:c35ec1e2-129e-496d-9918-a7b7678f7577>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00460.warc.gz"} |
Ernie F.
What do you want to work on?
About Ernie F.
Algebra, Elementary (3-6) Math, Geometry, Algebra 2, Midlevel (7-8) Math
Bachelors in Computer Science from California State University-San Bernardino
Math - Algebra II
literally the best!
Math - Algebra II
great tutor
Math - Algebra II
He was amazing. By far the best tutor I have had on here. Very patient, very kind and very thorough.
FP - Elementary Math | {"url":"https://www.princetonreview.com/academic-tutoring/tutor/ernie%20f--1448994?s=statistics%C2%AE%3D1","timestamp":"2024-11-11T21:20:40Z","content_type":"application/xhtml+xml","content_length":"269477","record_id":"<urn:uuid:9f7388ef-f415-4f15-aca4-e149a5508478>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00248.warc.gz"} |
Algorithm Write For Us, Contribute, and Submit Guest Post
Algorithm Write For Us
An algorithm or Algorithm is one of many things in the world of programming. Do you know! Algorithms don’t just exist in cyberspace or something digital, and Algorithms exist in everyday life.
The algorithm is a term for solving a problem. In more detail, Algorithm is a sequence of logical and systematic processes to solve a problem, usually focused on computers and programming.
Algorithms can also be processes or rules followed by calculations or steps arranged to solve problems, especially on digital things that are a lot on computers.
If it is difficult to understand an algorithm, then think of it as various rules or things that must be done from the start so that they are released and not exposed to problems—simple understanding.
To Write for Us, you can email us at contact@smarttechdata.com
Algorithm Function
If you ask, what is the function of the Algorithm? Then the answer is data processing followed by logical computation, compiling an infinite series of instructions to compute a function and produce
an output or final condition. One of them is the Youtube Algorithm.
Some forms of Algorithm:
• Looping Algorithm or ( Looping Algorithm)
• Branching Algorithm or Conditional ( Conditional Algorithm)
• Sequence Algorithm ( Sequence Algorithm)
How the Algorithm Works
To better understand, we will discuss how the Algorithm works with the following example. For example the Algorithm that exists when we brew a glass of milk:
• Prepare a sachet of milk, 200 ml of boiled water (one glass), and a spoon.
• The next step is to open the sachet packaging.
• Put it in the glass.
• Then put boiled water into a glass.
• Stir gently until the milk dissolves in the water.
• Serve the milk.
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Smart Tech Data’s presence is on Social media, and we will share your article with the Algorithm-related audience.
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Mathematics & Music
"There is geometry in the humming of the strings, there is music in the spacing of the spheres."
Explore the connections between mathematics and music in the videos, podcasts, and articles below.
Majesty of Music and Math. "Explore the interconnectedness of music and mathematics. The Majesty of Music and Math features remarks by Santa Fe Institute mathematician and computer scientist Cris
Moore and musical selections by The Santa Fe Symphony Orchestra with Principal Conductor Guillermo Figueroa."
Music + Math: Symmetry. "From Pythagoras' observations of the fundamental mathematical relationship between vibrating strings and harmony to the digitized musical world we enjoy today, The Majesty of
Music and Mathematics with the Santa Fe Symphony and the Santa Fe Institute will explore the remarkable interweaving of the languages of music and mathematics." View the entire Santa Fe Institute
series, covering the tritone, harmonics, ratios and more.
Geometry in Music; Dmitri Tymoczko. "What could be a better medium to communicate math to the public than the universal language of music? Ever since Pythagoras used numerical terms to express
intervals between notes and derived musical tones from geometrical patterns, mathematicians have linked music to numbers." At the SIAM Annual Meeting in 2010, Tymoczko used graphics and sound to
connect math to the music of Chopin, Mozart, and Schubert. Video by Adam Bauser, Bauser Media Group.
Combining Math and Music. Eugenia Cheng, a mathematician who also is a concert pianist, describes how a mathematical breakthrough enabled Johann Sebastian Bach to write "The Well-Tempered Clavier"
(1722). At the time that the video was recorded, Cheng was a visiting senior lecturer in mathematics at the University of Chicago. Cheng also has an 11-part video series, Math in Music, hosted by
WFMT, 2020, with topics such as "Feeling the Commutativity of Multiplication," "Math to Build New Ideas," "Symmetry in Music," "Fractions Give Us Feelings!," "Math Can Also Sound Bad," and "Harmonics
As Special Effects."
David Kung on "Symphonic Equations: Waves and Tubes." David Kung (St Mary's College of Maryland) presents "Symphonic Equations: Waves and Tubes" -- a miniexcursion into math and music, presented as a
MAA Distinguished Lecture at the Carnegie Institution for Science.
The Science Behind the Arts: The Maths Behind Music. University of Surrey, England.
More Videos
Turning math into music. Sean Hardesty (Rice University) plays the opening of the Sibelius Violin Concerto and discusses the relationship between mathematics and music.
The world's ugliest music: Scott Rickard at TEDxMIA. Scott Rickard has degrees in mathematics, computer science, and electrical engineering from M.I.T. and MA and PhD degrees in applied and
computational mathematics from Princeton University. Rickard says that he is "passionate about mathematics, music and educating the next generation of scientists and mathematicians."
Robert Schneider - Reverie in Prime Time Signatures (August 2013, Banff Centre). Robert Schneider (Apples in Stereo/Elephant 6) composed this theme for "MSI (Math Sciences Investigation): Anatomy of
Integers and Permutations," a play by mathematician Andrew Granville and screenwriter Jennifer Granville. "As the title indicates, the piece is written in prime-numbered time signatures; which is to
say, there is a prime number of beats in each measure. The main theme plays in the time signature 7/4, which indicates 7 beats per measure, with an interlude that passes through the signatures 2/4, 3
/4 and 5/4 as well. From the constraints imposed by these rhythmic patterns, melodies emerged naturally as I composed, special to each prime..."
Music and math: The genius of Beethoven – Natalya St. Clair TED Ed. Natalya St. Clair uses the Moonlight Sonata "to illustrate the way Beethoven was able to convey emotion and creativity using the
certainty of mathematics."
The Math Behind Music. Ethan Thompson and David Hamilton explain the math behind music in a fun, concise way in this finalist entry in the 2015 Math-O-Vision contest.
A Medley of More Mathematics and Music
If you would like to recommend more websites on mathematics and music, please email AMS Outreach.
Resources available by subscription or purchase
• Journal of Mathematics and Music: Mathematical and Computational Approaches to Music Theory, Analysis, Composition and Performance
• Mathematics and Music, by David Wright
• Many more books that can be found via a search on the web or your preferred online bookstore!
Also see: Society for Mathematics and Computation in Music. The Society was founded in 2006 as an International Forum for researchers and musicians working in the trans-disciplinary field at the
intersection of music, mathematics and computation. The website includes information about joining and attending meetings, and online newsletters. | {"url":"https://www.ams.org/publicoutreach/math-and-music","timestamp":"2024-11-04T01:28:51Z","content_type":"text/html","content_length":"74819","record_id":"<urn:uuid:c385b49b-f625-4d52-b529-7d6b9bb48a61>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00582.warc.gz"} |
Bas Spitters
picture. Associate professor in the Logic and Semantics Group, Dept. of Computer Science, Aarhus University.
Office: Turing-119 Aabogade 34 DK-8200 Aarhus N, Denmark. e-mail
Research interests
• Logic, semantics and (homotopy) type theory
• Proof assistants and their applications
• Logical and categorical foundations of physics
I am leading the Concordium Blockchain Research Center and the Blockchain workpackage in Digit. Part of the Aarhus Quantum campus Publications
Since April 2015: Associate professor, Aarhus University
2015: Associate professor at CMU working on the MURI grant Homotopy Type Theory: Unified Foundations of Mathematics and Computation
and Aarhus in the Logic and Semantics Group.
2014: Visiting positions at LRI Paris-Sud, IHP, PPS Paris-7, Gotheburg Computing Science.
2013: Member of the Institute for Advanced Studies for the special year on univalent foundations.
2010-2013: Leading the Nijmegen contribution to the ForMath project, as well as Work Package 4: formalization of exact analysis. (The project got an "excellent" mark.)
Organization: PhDay'01, Constructive mathematics, types and exact real numbers, DIAMANT-day '05, Brouwer seminar (from 2004 to 2007), MAP07
The constructivenews-list. Grants
2018-2021: AFOSR grant: Homotopy type theory and probabilistic computation, 540kUSD. The goal of the project is to apply homotopy type theory to probabilistic programming and to develop theory and
tools for computer aided proofs in security.
2015-2019: Villum Foundation: Guarded Homotopy Type Theory. With Lars Birkedal we received 3.3 Mkr from the Villum Foundation for a research project on Guarded Homotopy Type Theory. The goal of the
project is to develop new theories for and prototypes of proof assistants, which can be used within both mathematics and computer science.
Previous grants:
DIAMANT researcher.
VENI `Reasoning and Computing' of the dutch science foundation NWO.
Danil Annenkov
Sabine Oechsner
Evgeny Makarov (2013)
Benjamin Salling Hvass (2019-2022)
Jakob Botsch Nielsen (2019-2023)
Andreas Aagaard Lynge (2019-2022)
Soren Eller Thomsen (2018-2021)
Martin Bidlingmaier (2018-2021)
Russell O'Connor Incompleteness and Completeness Formalizing Logic and Analysis in Type Theory (2005-2008).
Research Assistants
Egbert Rijke (2012-2013).
Eelis van der Weegen (2010-2011).
Vacancies: we regularly have funding for PhD-students and postdocs. Please reach out to me by e-mail. The application process goes via the graduate school.
Teaching PGP | {"url":"https://www.cs.au.dk/~spitters/","timestamp":"2024-11-09T09:29:42Z","content_type":"text/html","content_length":"9234","record_id":"<urn:uuid:cf285a9d-870d-4250-8e78-500a51c18204>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00528.warc.gz"} |
Cosmological - Cosmos
The General Theory of Relativity was published by Albert Einstein in 1915. The General Theory of Relativity is the accepted name published by Albert Einstein in 1915 for the theory of gravity. The
force of gravity is a representation of the local geometry of space-time, according to the General Theory of Relativity. While the modern […]
What is General theory of relativity? Equations & Examples Read More »
Does Milky way contain Dark Matter?
What is Dark matter? Dark matter is a major component of Cosmology and Astrophysics. In the first case, the cosmological models of galactic evolution are based on this dark matter that does not emit
or block light. Since it is assumed that it constitutes 85% of the total matter, the models are based on this
IllustrisTNG – Most perfect model of the universe
The development of computer technology has helped bring the modelling of the evolution of our universe to a qualitatively new level. Scientists received new information about the influence of black
holes on the distribution of dark matter, learned more about the formation and propagation of heavy elements in space, as well as about the origin
IllustrisTNG – Most perfect model of the universe Read More »
BOINC: A Distributed Computing System
Processing scientific data requires enormous power, but the number of supercomputers available is very limited. However, a significant part of the calculations can be broken down into small tasks
that a home computer can handle. The BOINC project is a complex of programs designed to organize distributed computing. It was originally created in 2002 for | {"url":"https://cosmos.theinsightanalysis.com/tag/cosmological/","timestamp":"2024-11-03T16:24:22Z","content_type":"text/html","content_length":"148348","record_id":"<urn:uuid:3b6ef31e-73c5-48fe-91bc-c2d704293522>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00885.warc.gz"} |
Three statements are given below i) the enthalpy of any elemen-Turito
Are you sure you want to logout?
Three statements are given below
i) The enthalpy of any element is zero in their standard state
ii) The heat of neutralisation for any strong acid and strong base at 25°C is -13.7 kJ/mole is a mathematical form of first+ =
A. i only correct
B. ii and iii are correct
C. i and iii are correct
D. all are correct
The correct answer is: i and iii are correct
Get an Expert Advice From Turito. | {"url":"https://www.turito.com/ask-a-doubt/maths-three-statements-are-given-below-i-the-enthalpy-of-any-element-is-zero-in-their-standard-state-ii-the-heat-of-q6ee2a9","timestamp":"2024-11-14T21:02:20Z","content_type":"application/xhtml+xml","content_length":"318411","record_id":"<urn:uuid:b1513ae7-aaaa-4322-a810-af917b18c847>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00474.warc.gz"} |
Connectivity Projection Nodes
Good afternoon,
I am working with the connectivity data. In the viewer, there is a square at what appears to be the terminal end of each branch of the projection (See below). I am interested in obtaining a list of
coordinates for each of these squares for a particular projection. Can you provide some direction as to how I should go about this?
Hi April,
Which viewer are you using? Can you provide a link?
Hi Tyler, I had been using the 3d Viewer :: Allen Brain Atlas: Mouse Connectivity browser.
I ended up writing a script to access the terminal endpoints from each neuron. It’s definitely not pretty, and I’m sure there are much better/simpler ways to do it, but in case anybody else
encounters the same need, I have posted it below:
‘%notin%’ = Negate(‘%in%’)
streamline ← fromJSON(file = “streamlines_100141219.json”)
lines ← streamline$lines
curr_mesh ← mesh3d(x = NULL, y = NULL) #starting over with a fresh object
for (i in 1:length(lines)) {
curr_df ← as.data.frame(do.call(rbind, lines[[i]]))
curr_df$x = as.numeric(curr_df$x)
curr_df$y = as.numeric(curr_df$y)
curr_df$z = as.numeric(curr_df$z)
segments ← rbind(1:(nrow(curr_df) - 1), 2:nrow(curr_df))
curr_mesh ← merge(curr_mesh, mesh3d(x = curr_df$x, y = curr_df$y, z = curr_df$z, segments = segments))
indices = curr_mesh$is %>% data.frame()
listIndices = unlist(as.list(indices))
duplicatedIndices = listIndices[duplicated(listIndices)]
uniqueIndices = subset(listIndices, listIndices %notin% duplicatedIndices)
#for each of the unique Indices, find the start, stop, and distance
indicesBegin = uniqueIndices[c(TRUE, FALSE)]
indicesEnd = uniqueIndices[c(FALSE, TRUE)]
#You need to add the very last end to the indices begin so that we can calculate how long this segment is
indicesBegin = c(indicesBegin,indicesEnd[length(indicesEnd)] + 1)
#Now, to capture each end, we select the coordinates for each segment, and then select the last one
for(x in 1:(length(indicesBegin)-1)){
segmentCoords = curr_mesh$vb[1:3,indicesBegin:(indicesBegin[x+1]-1)]
startCoord = t(segmentCoords[,ncol(segmentCoords)]) %>% data.frame()
endCoord = t(segmentCoords[,1]) %>% data.frame()
colnames(endCoord) = c(“End_X”, “End_Y”, “End_Z”)
colnames(startCoord) = c(“Start_X”, “Start_Y”, “Start_Z”)
bothCoord = cbind(endCoord, startCoord)
bothCoord$Segment = x
bothCoord$IndexStart = indicesBegin
bothCoord$IndexEnd = indicesBegin[x+1]-1
if(x == 1){
allCoords = bothCoord
} else{
allCoords = rbind(allCoords, bothCoord)
plot3d(curr_mesh, add = TRUE, col = “black”)
plot3d(allCoords$Start_X, allCoords$Start_Y, allCoords$Start_Z, add = TRUE, col = “red”, alpha = 4)
plot3d(allCoords$End_X, allCoords$End_Y, allCoords$End_Z, col = “blue”, alpha = 4, add = TRUE) | {"url":"https://community.brain-map.org/t/connectivity-projection-nodes/3387","timestamp":"2024-11-12T05:58:12Z","content_type":"text/html","content_length":"33801","record_id":"<urn:uuid:c93ab946-40ad-4d41-8df7-36e71c5c4166>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00729.warc.gz"} |
Interview program in Python part 1 » Onurdesk
Nowadays in any interview a programming test is a must, So here we are creating a series of basic interview program in Python, We learn step by step. You can also visit Java program if you are
looking for that
Write a program in Python to find a Maximum of two numbers
If you have two numbers, write a Python code to find the limit of these two numbers.
def max(i, j):
if i >= j:
return i
return j
a = 2
b = 4
print(max(a, b))
print('Thank you for reading basic interview programs in python')
Write a program in Python for calculation of simple interest
we can use simple interest formula :
Simple Interest = (P x T x R)/100
P is the principle amount
T is the time and
R is the rate
def simple_interest(p,t,r):
print('The principal is', p)
print('The time period is', t)
print('The rate of interest is',r)
si = (p * t * r)/100
print('The Simple Interest is', si)
return si
# Driver code
simple_interest(20000, 3, 8)
print('Thank you for reading basic interview programs in python')
Write a program in Python for check whether the number is Armstrong number or not
The Armstrong number of three digits is the integer such that the sum of the cubes of its digits is equal to the number itself.
Write a program in Python to find some of the square of first n natural numbers
def squaresum(n) :
sm = 0
for i in range(1, n+1) :
sm = sm + (i * i)
return sm
n = 4
print("Thanks for visiting onurdesk program in Python")
Write a program in Python to sum of all digits of a number
n=int(input("Enter a number:"))
while n>0:
print("The sum of digits of number is:", sum)
print("Thanks for visiting onurdesk program in Python")
Write a program in Python to check a year is a leap year or not using Python
year=int(input("Enter a Year:"))
if ((year % 100 == 0 and year % 400 == 0) or (year % 100 != 0 and year % 4 == 0)):
print("It is a Leap Year")
print("It is not a Leap Year")
Write a program in Python to convert Days into years, weeks and months
days=int(input("Enter Day:"))
years =(int) (days / 365)
weeks =(int) (days / 7)
months =(int) (days / 30)
print("Days to Years:",years)
print("Days to Weeks:",weeks)
print("Days to Months:",months)
Write a program in Python to print all prime number in an interval
start = 11
end = 25
for val in range(start, end + 1):
# If num is divisible by any number
# between 2 and val, it is not prime
if val > 1:
for n in range(2, val):
if (val % n) == 0:
print("Thanks for visiting program in Python")
Write a program in Python to find Area of a circle
def findArea(r):
PI = 3.142
return PI * (r*r);
#Like us on facebook
# Driver method
print("Thanks for visiting program in Python")
Write a program in Python find sum of elements in given array
def _sum(arr,n):
# return sum using sum
# inbuilt sum() function
# driver function
# input values to list
arr = [12, 3, 4, 15]
# calculating length of array
n = len(arr)
ans = _sum(arr,n)
#Like us on facebook
# display sum
print("Thanks for visiting onurdesk program in Python")
print (ans)
Write a program in Python to check if small string is there in big string
def check(string, sub_str):
if (string.find(sub_str) == -1):
print("Like us on facebook")
print("Thanks for visiting onurdesk program in Python")
# driver code
string = "zeus is a programmer"
sub_str ="zeus"
check(string, sub_str)
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://onurdesk.com/interview-program-in-python-part-1/","timestamp":"2024-11-14T18:27:15Z","content_type":"text/html","content_length":"407299","record_id":"<urn:uuid:e0490ade-c3cc-434d-8329-4796d54f09de>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00495.warc.gz"} |
Constraining the p-Mode-g-Mode Tidal Instability with GW170817
We analyze the impact of a proposed tidal instability coupling p modes and g modes within neutron stars on GW170817. This nonresonant instability transfers energy from the orbit of the binary to
internal modes of the stars, accelerating the gravitational-wave driven inspiral. We model the impact of this instability on the phasing of the gravitational wave signal using three parameters per
star: An overall amplitude, a saturation frequency, and a spectral index. Incorporating these additional parameters, we compute the Bayes factor (lnB!pgpg) comparing our p-g model to a standard one.
We find that the observed signal is consistent with waveform models that neglect p-g effects, with lnB!pgpg=0.03-0.58+0.70 (maximum a posteriori and 90% credible region). By injecting simulated
signals that do not include p-g effects and recovering them with the p-g model, we show that there is a ≃50% probability of obtaining similar lnB!pgpg even when p-g effects are absent. We find that
the p-g amplitude for 1.4 MâŠneutron stars is constrained to less than a few tenths of the theoretical maximum, with maxima a posteriori near one-Tenth this maximum and p-g saturation frequency ∼70
Hz. This suggests that there are less than a few hundred excited modes, assuming they all saturate by wave breaking. For comparison, theoretical upper bounds suggest a103 modes saturate by wave
breaking. Thus, the measured constraints only rule out extreme values of the p-g parameters. They also imply that the instability dissipates a1051 erg over the entire inspiral, i.e., less than a few
percent of the energy radiated as gravitational waves.
Bibliographical note
Publisher Copyright:
© 2019 American Physical Society.
Dive into the research topics of 'Constraining the p-Mode-g-Mode Tidal Instability with GW170817'. Together they form a unique fingerprint. | {"url":"https://pure.ewha.ac.kr/en/publications/constraining-the-p-mode-g-mode-tidal-instability-with-gw170817","timestamp":"2024-11-14T20:30:29Z","content_type":"text/html","content_length":"51791","record_id":"<urn:uuid:68f0c0ef-3f75-4b4e-bf9a-6d9dcce894b4>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00337.warc.gz"} |
IMLSN Workshop 8
Date: Friday 31st January, 2014.
Location: Institute of Technology Tallaght.
Workshop Theme: ‘Diversity: challenges and opportunities-enabling and supporting mathematics learning in a diverse student population’.
The population of students studying mathematics as part of their higher education is increasingly diverse. This diversity manifests itself in many ways but in particular:
• A higher proportion of students entering higher education are not coming straight from second level.
• The number of students with particular learning needs is increasing.
The conference was hosted by the Institute of Technology Tallaght and the Dublin Institute of Technology.
Workshop Report
Keynote Speakers:
The first keynote presentation was by Clare Trott of Loughborough University.
Title: Mathematics Support: helping the neurodiverse student overcome their barriers.
Abstract: There is an increasingly diverse population of students entering Higher Education (H.E.). While this enriches the environment it can bring challenges that H.E. institutions, and in
particular mathematics support, need to address. This presentation will look at neurodiversity: dyslexia, dyspraxia, dyscalculia, Asperger's syndrome and AD(H)D. The presentation will explore the
mathematical and statistical barriers that neurodiverse students can face and suggest ways in which mathematics and statistics support can be provided in order to enable the SpLD student to overcome
the barriers and succeed. There will be some illustrative exemplar case studies.
A pdf of the presentation can be downloaded here.
The second keynote presentation was by Dr. Terry Maguire, Chair of Adults Learning Mathematics – An International Research Forum and Director of the National Forum for the Enhancement of Teaching and
Learning in Higher Education.
Title: Adult learners and Mathematics Support: Getting to the Root of the Problem.
Abstract: Every adult learner that walks through the door of a Maths Support Centre comes in with a difficulty with some aspect of mathematics that they need some help with. However these learners
bring with them a range of attitudes and beliefs that they have developed because of their experience of school and their real life mathematics since they left school. Research has shown that adults
and their own relationship with mathematics can impact the way that they respond to and are receptive to the help that is offered. This paper will discuss the range and nature of issues that can act
as a barrier to adult learners being receptive to the assistance that is made available to them. It will suggest some strategies that can be used by Maths Support Centre staff to identify the root
problem and thus be able to tailor their support more effectively to the needs of individual adult learners.
A pdf of the presentation can be downloaded here.
Short talks:
The first short talk was by Dr. Brien Nolan and Dr. Eabhnat Ní Fhloinn from Dublin City University.
Title: Can group-work work? Notes on group-work based tutorials in a large service teaching module.
Abstract: We report on a teaching project that involved the use of peer-supported group-work tutorials in a large (N = 398) service teaching module in Dublin City University. We describe the
background and motivation for the project, and its design and execution. This includes a corresponding tutor training element. We report on feedback on the tutorials obtained from students and
tutors, and discuss the students' performance on the module assessments in the light of the group-work tutorials. We found little evidence of success in the project, and attempt to relate this to
existing conceptual frameworks describing the effective implementation of group-work.
A pdf of the presentation can be downloaded here.
The second short talk was by Dr. Anthony Cronin from University College Dublin.
Title: Supporting Mathematics Learning for Adult Learners via Maths Software.
Abstract: This year at the maths support centre at UCD we decided to support the adult learners from the access to science and engineering programmes by giving them access to a new personalised
adaptive learning software system developed here in Dublin (realizeit). The students took a (pre)-test consisting of 12 elementary math questions dealing with the basics of arithmetic and stats. The
students then used the system for 50 days and took the same (post)-test. The students observations of the system as a support mechanism as well as their performance on the tests will be discussed
A pdf of the presentation can be downloaded here.
The third short talk was by Dr. Olivia Fitzmaurice of the University of Limerick, Mr. Ciaran O’Sullivan of the Institute of Technology Tallaght, Dr. Ciarán Mac an Bhaird of Maynooth University and
Dr. Eabhnat Ní Fhloinn of Dublin City University.
Title: Adult learners v traditional learners – insights from a large scale survey of Mathematics Learning Support in Irish HEIs.
Abstract: Research indicates that 'Mature Students' or 'Adult Learners' have different anxieties, motivations and approaches to learning than traditional learners who generally transition to third
level education straight from school. Using data from a large scale cross- institutional investigation of mathematics learning support (MLS) in Ireland, this paper compares the usage of MLS by Adult
Learners with that of Traditional Learners. In particular the paper will examine the findings arising from the analysis of the data which indicates differences between the motivational factors of
Adult Learners and Traditional Learners who avail of MLS. The paper also compares the reasons indicated in the survey for not availing of MLS given by Adult Learners who did not use MLS with that
given by Traditional Learners who did not use MLS.
A pdf of the presentation can be downloaded here.
The last short talk was by Mr. Timothy J. Crawford and Dr. Jonathan S. Cole from Queen's University Belfast.
Title: Mature students' participation in maths support and progression.
Abstract: Maths support at the Learning Development Service takes the form of drop-in contact, one-to-one appointments and workshops. Analysis over three years from 2010/11 to 2012/13 shows that 45%
of one-to-one appointments involved a mature student, defined at Queen's University Belfast (QUB) as one who has had a break in full-time study (normally a minimum of two years). This is a very high
proportion given that 4% of students at QUB are mature. Considering engineering undergraduates only, the fraction of one-to-one appointments involving mature students was also 45%. This study can
report a wide variation in terms of progression of mature students in engineering and aims to consider how more traditional undergraduate learners could be persuaded to adopt the attitudes of mature
students in partaking of maths support.
A pdf of the presentation can be downloaded here.
The Workshop concluded with an open discussion on the conference themes and on the network, and nominations for the committee of the IMLSN. | {"url":"https://www.imlsn.ie/index.php/past-events/past-workshops/imlsn-workshop-8","timestamp":"2024-11-10T19:39:11Z","content_type":"text/html","content_length":"32645","record_id":"<urn:uuid:b34ed74f-0812-48fc-948c-35b713436813>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00347.warc.gz"} |
Equivariant Riemann—Roch theorem and a BSD-like formula for Hasse—Weil—Artin L-functions over global function fields
2024-05-01 (16:00 ~ 17:30)
UNIST Building 108, Room 320
Here’s the seminar schedule for the first week of May 2024.
Speaker: 김완수 (KAIST)
When: 5월 1일 (수요일), 16:00–17:30
Where: 108동 320호 (Reading Room)
Title: Equivariant Riemann—Roch theorem and a BSD-like formula for Hasse—Weil—Artin L-functions over global function fields
Let X be a smooth projective curve over a perfect field of characteristic p>0, and Y be a finite Galois covering of X (possibly allowing ramification). We first review the “refined’’ Riemann—Roch
theorem for equivariant vector bundles on Y (due to Nakajima, Köck, and Fischbacher-Weitz & Köck), starting with the modular representation theory of finite groups and local integral normal basis
theorem. We then explain how to use it to deduce the p-part of the BSD-like formula for Hasse—Weil—Artin L-functions over global function fields.
This is joint work in progress with Ki-Seng Tan, Fabien Trihan and Kwok-Wing Tsoi.
Thank you. | {"url":"https://math.unist.ac.kr/equivariant-riemann-roch-theorem-and-a-bsd-like-formula-for-hasse-weil-artin-l-functions-over-global-function-fields/","timestamp":"2024-11-06T09:22:23Z","content_type":"text/html","content_length":"60597","record_id":"<urn:uuid:b071494a-0ed2-480a-b452-ddc9696b324f>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00226.warc.gz"} |
Class template controlled_runge_kutta<ErrorStepper, ErrorChecker, StepAdjuster, Resizer, explicit_error_stepper_tag>
boost::numeric::odeint::controlled_runge_kutta<ErrorStepper, ErrorChecker, StepAdjuster, Resizer, explicit_error_stepper_tag> — Implements step size control for Runge-Kutta steppers with error
// In header: <boost/numeric/odeint/stepper/controlled_runge_kutta.hpp>
template<typename ErrorStepper, typename ErrorChecker, typename StepAdjuster,
typename Resizer>
class controlled_runge_kutta<ErrorStepper, ErrorChecker, StepAdjuster, Resizer, explicit_error_stepper_tag> {
// types
typedef ErrorStepper stepper_type;
typedef stepper_type::state_type state_type;
typedef stepper_type::value_type value_type;
typedef stepper_type::deriv_type deriv_type;
typedef stepper_type::time_type time_type;
typedef stepper_type::algebra_type algebra_type;
typedef stepper_type::operations_type operations_type;
typedef Resizer resizer_type;
typedef ErrorChecker error_checker_type;
typedef StepAdjuster step_adjuster_type;
typedef explicit_controlled_stepper_tag stepper_category;
// construct/copy/destruct
controlled_runge_kutta(const error_checker_type & = error_checker_type(),
const step_adjuster_type & = step_adjuster_type(),
const stepper_type & = stepper_type());
// public member functions
template<typename System, typename StateInOut>
try_step(System, StateInOut &, time_type &, time_type &);
template<typename System, typename StateInOut>
try_step(System, const StateInOut &, time_type &, time_type &);
template<typename System, typename StateInOut, typename DerivIn>
try_step(System, StateInOut &, const DerivIn &, time_type &, time_type &);
template<typename System, typename StateIn, typename StateOut>
boost::disable_if< boost::is_same< StateIn, time_type >, controlled_step_result >::type
try_step(System, const StateIn &, time_type &, StateOut &, time_type &);
template<typename System, typename StateIn, typename DerivIn,
typename StateOut>
try_step(System, const StateIn &, const DerivIn &, time_type &,
StateOut &, time_type &);
template<typename StateType> void adjust_size(const StateType &);
stepper_type & stepper(void);
const stepper_type & stepper(void) const;
// private member functions
template<typename System, typename StateInOut>
try_step_v1(System, StateInOut &, time_type &, time_type &);
template<typename StateIn> bool resize_m_xerr_impl(const StateIn &);
template<typename StateIn> bool resize_m_dxdt_impl(const StateIn &);
template<typename StateIn> bool resize_m_xnew_impl(const StateIn &);
This class implements the step size control for standard Runge-Kutta steppers with error estimation.
Template Parameters
1. typename ErrorStepper
The stepper type with error estimation, has to fulfill the ErrorStepper concept.
2. typename ErrorChecker
The error checker
3. typename StepAdjuster
4. typename Resizer
The resizer policy type.
controlled_runge_kutta public construct/copy/destruct
1. controlled_runge_kutta(const error_checker_type & error_checker = error_checker_type(),
const step_adjuster_type & step_adjuster = step_adjuster_type(),
const stepper_type & stepper = stepper_type());
Constructs the controlled Runge-Kutta stepper.
error_checker An instance of the error checker.
stepper An instance of the underlying stepper.
controlled_runge_kutta public member functions
1. template<typename System, typename StateInOut>
try_step(System system, StateInOut & x, time_type & t, time_type & dt);
Tries to perform one step.
This method tries to do one step with step size dt. If the error estimate is to large, the step is rejected and the method returns fail and the step size dt is reduced. If the error estimate is
acceptably small, the step is performed, success is returned and dt might be increased to make the steps as large as possible. This method also updates t if a step is performed.
dt The step size. Updated.
system The system function to solve, hence the r.h.s. of the ODE. It must fulfill the Simple System concept.
t The value of the time. Updated if the step is successful.
x The state of the ODE which should be solved. Overwritten if the step is successful.
Returns: success if the step was accepted, fail otherwise.
2. template<typename System, typename StateInOut>
try_step(System system, const StateInOut & x, time_type & t, time_type & dt);
Tries to perform one step. Solves the forwarding problem and allows for using boost range as state_type.
This method tries to do one step with step size dt. If the error estimate is to large, the step is rejected and the method returns fail and the step size dt is reduced. If the error estimate is
acceptably small, the step is performed, success is returned and dt might be increased to make the steps as large as possible. This method also updates t if a step is performed.
dt The step size. Updated.
system The system function to solve, hence the r.h.s. of the ODE. It must fulfill the Simple System concept.
t The value of the time. Updated if the step is successful.
x The state of the ODE which should be solved. Overwritten if the step is successful. Can be a boost range.
Returns: success if the step was accepted, fail otherwise.
3. template<typename System, typename StateInOut, typename DerivIn>
try_step(System system, StateInOut & x, const DerivIn & dxdt, time_type & t,
time_type & dt);
Tries to perform one step.
This method tries to do one step with step size dt. If the error estimate is to large, the step is rejected and the method returns fail and the step size dt is reduced. If the error estimate is
acceptably small, the step is performed, success is returned and dt might be increased to make the steps as large as possible. This method also updates t if a step is performed.
dt The step size. Updated.
dxdt The derivative of state.
Parameters: system The system function to solve, hence the r.h.s. of the ODE. It must fulfill the Simple System concept.
t The value of the time. Updated if the step is successful.
x The state of the ODE which should be solved. Overwritten if the step is successful.
Returns: success if the step was accepted, fail otherwise.
4. template<typename System, typename StateIn, typename StateOut>
boost::disable_if< boost::is_same< StateIn, time_type >, controlled_step_result >::type
try_step(System system, const StateIn & in, time_type & t, StateOut & out,
time_type & dt);
Tries to perform one step.
This method is disabled if state_type=time_type to avoid ambiguity.
This method tries to do one step with step size dt. If the error estimate is to large, the step is rejected and the method returns fail and the step size dt is reduced. If the error estimate is
acceptably small, the step is performed, success is returned and dt might be increased to make the steps as large as possible. This method also updates t if a step is performed.
dt The step size. Updated.
in The state of the ODE which should be solved.
Parameters: out Used to store the result of the step.
system The system function to solve, hence the r.h.s. of the ODE. It must fulfill the Simple System concept.
t The value of the time. Updated if the step is successful.
Returns: success if the step was accepted, fail otherwise.
5. template<typename System, typename StateIn, typename DerivIn,
typename StateOut>
try_step(System system, const StateIn & in, const DerivIn & dxdt,
time_type & t, StateOut & out, time_type & dt);
Tries to perform one step.
This method tries to do one step with step size dt. If the error estimate is to large, the step is rejected and the method returns fail and the step size dt is reduced. If the error estimate is
acceptably small, the step is performed, success is returned and dt might be increased to make the steps as large as possible. This method also updates t if a step is performed.
dt The step size. Updated.
dxdt The derivative of state.
in The state of the ODE which should be solved.
out Used to store the result of the step.
system The system function to solve, hence the r.h.s. of the ODE. It must fulfill the Simple System concept.
t The value of the time. Updated if the step is successful.
Returns: success if the step was accepted, fail otherwise.
6. template<typename StateType> void adjust_size(const StateType & x);
Adjust the size of all temporaries in the stepper manually.
Parameters: x A state from which the size of the temporaries to be resized is deduced.
7. stepper_type & stepper(void);
Returns the instance of the underlying stepper.
Returns: The instance of the underlying stepper.
controlled_runge_kutta private member functions | {"url":"https://beta.boost.org/doc/libs/1_69_0/libs/numeric/odeint/doc/html/boost/numeric/odeint/controlled_run_idp42468976.html","timestamp":"2024-11-03T20:22:20Z","content_type":"text/html","content_length":"45504","record_id":"<urn:uuid:59445ef0-f8ab-4b6d-889e-984b3a0d9a06>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00248.warc.gz"} |
DAlembert\'s Principle - Examples, Definition, Limitations, FAQ\'S
DAlembert’s Principle
D’Alembert’s Principle is an important concept in physics that helps simplify the analysis of dynamics. It states that for a system in motion, the sum of the applied forces and the inertial forces
acting on each particle of the system results in a state of equilibrium.
What is DAlembert’s Principle?
D’Alembert’s Principle is a method used in classical mechanics to analyze the dynamics of a system. Instead of directly working with accelerating forces, this principle simplifies dynamic problems by
transforming them into equivalent static problems.
DAlembert’s Principle Formula
The formula associated with D’Alembert’s Principle is:
• 𝐹ᵢ: External (applied) force on the i-th particle.
• 𝑚ᵢ: Mass of the i-th particle.
• 𝑎ᵢ: Acceleration of the i-th particle.
• 𝛿𝑟ᵢ: Virtual displacement of the i-th particle.
This equation represents that the sum of the applied forces and the fictitious inertial forces (due to mass and acceleration) yields zero virtual work. Thus, the system can be treated as if it were
in static equilibrium, simplifying the analysis of dynamic systems.
DAlembert’s Principle Derivation
The derivation of D’Alembert’s Principle is based on the concept of virtual work and results in an equation that demonstrates the principle. Here’s the step-by-step process:
Total Force on Each Particle:
The total force (𝐹ᵢ⁽ᵀ⁾) acting on the 𝑖-th particle is equal to its mass (𝑚ᵢ) times its acceleration (𝑎ᵢ): 𝐹ᵢ⁽ᵀ⁾=𝑚ᵢ𝑎ᵢ
Inertial Force:
The inertial force is introduced by subtracting it from the total force: 𝐹ᵢ⁽ᵀ⁾−𝑚ᵢ𝑎ᵢ=0 This represents a state of quasi-static equilibrium.
Virtual Work Expression:
The virtual work (𝛿𝑊) equation sums the forces and inertial terms multiplied by their virtual displacements
(𝛿𝑟ᵢ): 𝛿𝑊=∑ᵢ𝐹ᵢ⁽ᵀ⁾⋅𝛿𝑟ᵢ−∑ᵢ𝑚ᵢ𝑎ᵢ⋅𝛿𝑟ᵢ=0
Separation of Applied and Constraint Forces:
The total force can be separated into applied forces (𝐹ᵢ) and constraint forces (𝐶ᵢ): 𝛿𝑊=∑ᵢ𝐹ᵢ⋅𝛿𝑟ᵢ+∑ᵢ𝐶ᵢ⋅𝛿𝑟ᵢ−∑ᵢ𝑚ᵢ𝑎ᵢ⋅𝛿𝑟ᵢ=0
Final Equation:
After simplifying, the final equation for virtual work becomes:
This final equation states that the sum of the external forces and the inertial forces, each multiply by their respect virtual displacements, equals zero. This result forms the basis of D’Alembert’s
Principle, convert dynamic problems into static-like equilibrium problems.
Uses of DAlembert’s Principle
D’Alembert’s Principle provides a useful framework in classical mechanics, enabling several practical applications:
1. Simplifying Dynamic Problems: First, you can use D’Alembert’s Principle to convert dynamic problems into static-like equilibrium problems by introduce fictitious inertial forces. This
transformation allows you to analyze systems with the same techniques used for static systems.
2. Analyzing Rigid Body Motion: In addition, you can apply this principle to study the motion of rigid bodies, making it easier to handle problems involving rotating or accelerating frames of
3. Understanding Vibrating Systems: Moreover, D’Alembert’s Principle helps you analyze vibrating systems, such as springs and dampers, by simplifying the calculation of forces involved.
4. Investigating Mechanical Systems: Consequently, you can investigate various mechanical systems, from simple machines to complex engineering structures, using the principle’s ability to represent
the balance between applied and inertial forces.
5. Formulating Equations of Motion: Finally, D’Alembert’s Principle allows you to formulate the equations of motion for a system, providing a basis for applying Lagrange’s equations or Hamiltonian
6. Analyzing Particle Systems: Additionally, the principle enables you to analyze systems with multiple interacting particles by considering how the forces of one particle affect another.
Examples for DAlembert’s Principle
Here are a few examples of how you can apply D’Alembert’s Principle:
• Rotating Disc: When analyzing a rotating disc, you can use D’Alembert’s Principle to consider inertial forces. By introducing these forces, you can treat the disc as if it were in static
equilibrium. Simplifying the analysis of internal stresses and reactions.
• Moving Car: In the case of a car accelerate forward, D’Alembert’s Principle allows you to analyze it by introducing a fictitious force in the opposite direction of acceleration. This way, you can
assess the car’s equilibrium under the influence of both real and inertial forces, helping you calculate the forces on each axle.
• Pendulum Motion: For a swinging pendulum, applying D’Alembert’s Principle lets you simple the analysis of the bob’s motion. You consider the gravitational and tension forces alongside an inertial
force that represents the bob’s acceleration, which makes it easier to understand the pendulum’s equilibrium at any point in its swing.
• Crane Movement: When a crane moves its load horizontally, you can analyze the problem using D’Alembert’s Principle. By including inertial forces due to acceleration. You can determine how much
force the crane must exert to keep the load in balance.
• Elevator Acceleration: In an accelerating elevator, you can apply D’Alembert’s Principle to introduce a fictitious force opposite the direction of acceleration. This adjustment lets you treat the
system as if it were in equilibrium. And calculate the tension required in the support cables.
What is D’Alembert’s Principle?
D’Alembert’s Principle states that the virtual work of inertia forces in a system in dynamic equilibrium is zero.
How does it differ from Newton’s Laws?
Newton’s Laws focus on acceleration and external forces, while D’Alembert’s Principle considers virtual work and internal forces.
When is D’Alembert’s Principle applied?
It’s used to solve problems in dynamics, particularly for systems in dynamic equilibrium or under the influence of constraint forces.
What’s the significance of virtual work?
Virtual work helps analyze the effect of inertial forces on dynamic systems, simplifying the calculation of equilibrium conditions.
Can it be applied to systems with constraints?
Yes, it’s particularly useful for systems with constraints, where it simplifies calculations by treating constraints as forces.
How does it simplify dynamic analysis?
By balancing applied and inertial forces, it simplifies the analysis of systems undergoing dynamic motion or in equilibrium.
How is D’Alembert’s Principle mathematically expressed?
It’s expressed as the sum of external forces and virtual forces, which account for inertial effects in dynamic equilibrium equations.
Can it be applied to non-conservative systems?
Yes, D’Alembert’s Principle can be extended to non-conservative systems, incorporate external forces and damping effects.
What’s the relation to Hamilton’s Principle?
Both principles describe the dynamics of mechanical systems, with D’Alembert’s focusing on equilibrium and Hamilton’s on variational principles.
How does it contribute to engineering analysis?
D’Alembert’s Principle provides a powerful tool for analyzing and solving problems in dynamics, aiding in the design and optimization of mechanical systems. | {"url":"https://www.examples.com/physics/dalemberts-principle.html","timestamp":"2024-11-13T15:02:17Z","content_type":"text/html","content_length":"123371","record_id":"<urn:uuid:7cb40391-17da-42c9-891b-6f08104cd7de>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00783.warc.gz"} |
27.6 Limits of Resolution: The Rayleigh Criterion
• Discuss the Rayleigh criterion.
Light diffracts as it moves through space, bending around obstacles, interfering constructively and destructively. While this can be used as a spectroscopic tool—a diffraction grating disperses light
according to wavelength, for example, and is used to produce spectra—diffraction also limits the detail we can obtain in images. Figure 1(a) shows the effect of passing light through a small circular
aperture. Instead of a bright spot with sharp edges, a spot with a fuzzy edge surrounded by circles of light is obtained. This pattern is caused by diffraction similar to that produced by a single
slit. Light from different parts of the circular aperture interferes constructively and destructively. The effect is most noticeable when the aperture is small, but the effect is there for large
apertures, too.
Figure 1. (a) Monochromatic light passed through a small circular aperture produces this diffraction pattern. (b) Two point light sources that are close to one another produce overlapping images
because of diffraction. (c) If they are closer together, they cannot be resolved or distinguished.
How does diffraction affect the detail that can be observed when light passes through an aperture? Figure 1(b) shows the diffraction pattern produced by two point light sources that are close to one
another. The pattern is similar to that for a single point source, and it is just barely possible to tell that there are two light sources rather than one. If they were closer together, as in Figure
1(c), we could not distinguish them, thus limiting the detail or resolution we can obtain. This limit is an inescapable consequence of the wave nature of light.
There are many situations in which diffraction limits the resolution. The acuity of our vision is limited because light passes through the pupil, the circular aperture of our eye. Be aware that the
diffraction-like spreading of light is due to the limited diameter of a light beam, not the interaction with an aperture. Thus light passing through a lens with a diameter
Take-Home Experiment: Resolution of the Eye
Draw two lines on a white sheet of paper (several mm apart). How far away can you be and still distinguish the two lines? What does this tell you about the size of the eye’s pupil? Can you be
quantitative? (The size of an adult’s pupil is discussed in Chapter 26.1 Physics of the Eye.)
Just what is the limit? To answer that question, consider the diffraction pattern for a circular aperture, which has a central maximum that is wider and brighter than the maxima surrounding it
(similar to a slit) [see Figure 2(a)]. It can be shown that, for a circular aperture of diameter Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable
when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 2(b). The first minimum is at an angle of
Figure 2. (a) Graph of intensity of the diffraction pattern for a circular aperture. Note that, similar to a single slit, the central maximum is wider and brighter than those to the sides. (b) Two
point objects produce overlapping diffraction patterns. Shown here is the Rayleigh criterion for being just resolvable. The central maximum of one pattern lies on the first minimum of the other.
Connections: Limits to Knowledge
All attempts to observe the size and shape of objects are limited by the wavelength of the probe. Even the small wavelength of light prohibits exact precision. When extremely small wavelength probes
as with an electron microscope are used, the system is disturbed, still limiting our knowledge, much as making an electrical measurement alters a circuit. Heisenberg’s uncertainty principle asserts
that this limit is fundamental and inescapable, as we shall see in quantum mechanics.
Example 1: Calculating Diffraction Limits of the Hubble Space Telescope
The primary mirror of the orbiting Hubble Space Telescope has a diameter of 2.40 m. Being in orbit, this telescope avoids the degrading effects of atmospheric distortion on its resolution. (a) What
is the angle between two just-resolvable point light sources (perhaps two stars)? Assume an average light wavelength of 550 nm. (b) If these two stars are at the 2 million light year distance of the
Andromeda galaxy, how close together can they be and still be resolved? (A light year, or ly, is the distance light travels in 1 year.)
The Rayleigh criterion stated in the equation between point sources, or the best obtainable resolution. Once this angle is found, the distance between stars can be calculated, since we are given how
far away they are.
Solution for (a)
The Rayleigh criterion for the minimum resolvable angle is
Entering known values gives
Solution for (b)
The distance
Substituting known values gives
The angle found in part (a) is extraordinarily small (less than 1/50,000 of a degree), because the primary mirror is so large compared with the wavelength of light. As noticed, diffraction effects
are most noticeable when light interacts with objects having sizes on the order of the wavelength of light. However, the effect is still there, and there is a diffraction limit to what is observable.
The actual resolution of the Hubble Telescope is not quite as good as that found here. As with all instruments, there are other effects, such as non-uniformities in mirrors or aberrations in lenses
that further limit resolution. However, Figure 3 gives an indication of the extent of the detail observable with the Hubble because of its size and quality and especially because it is above the
Earth’s atmosphere.
Figure 3. These two photographs of the M82 galaxy give an idea of the observable detail using the Hubble Space Telescope compared with that using a ground-based telescope. (a) On the left is a
ground-based image. (credit: Ricnun, Wikimedia Commons) (b) The photo on the right was captured by Hubble. (credit: NASA, ESA, and the Hubble Heritage Team (STScI/AURA))
The answer in part (b) indicates that two stars separated by about half a light year can be resolved. The average distance between stars in a galaxy is on the order of 5 light years in the outer
parts and about 1 light year near the galactic center. Therefore, the Hubble can resolve most of the individual stars in Andromeda galaxy, even though it lies at such a huge distance that its light
takes 2 million years for its light to reach us. Figure 4 shows another mirror used to observe radio waves from outer space.
Figure 4. A 305-m-diameter natural bowl at Arecibo in Puerto Rico is lined with reflective material, making it into a radio telescope. It is the largest curved focusing dish in the world. Although D
for Arecibo is much larger than for the Hubble Telescope, it detects much longer wavelength radiation and its diffraction limit is significantly poorer than Hubble’s. Arecibo is still very useful,
because important information is carried by radio waves that is not carried by visible light. (credit: Tatyana Temirbulatova, Flickr)
Diffraction is not only a problem for optical instruments but also for the electromagnetic radiation itself. Any beam of light having a finite diameter Figure 5). To avoid this, we can increase
Figure 5. The beam produced by this microwave transmission antenna will spread out at a minimum angle θ = 1.22 λ/D due to diffraction. It is impossible to produce a near-parallel beam, because the
beam has a limited diameter.
In most biology laboratories, resolution is presented when the use of the microscope is introduced. The ability of a lens to produce sharp images of two closely spaced point objects is called
resolution. The smaller the distance Figure 6(a) we have two point objects separated by a distance
Therefore, the resolving power is
Another way to look at this is by re-examining the concept of Numerical Aperture (Chapter 26.4 Microscopes. There, Figure 6(b) shows a lens and an object at point P. The
From this definition for
In a microscope,
Figure 6. (a) Two points separated by at distance x and a positioned a distance d away from the objective. (credit: Infopro, Wikimedia Commons) (b) Terms and symbols used in discussion of resolving
power for a lens and an object at point P. (credit: Infopro, Wikimedia Commons)
One of the consequences of diffraction is that the focal point of a beam has a finite width and intensity distribution. Consider focusing when only considering geometric optics, shown in Figure 7(a).
The focal point is infinitely small with a huge intensity and the capacity to incinerate most samples irrespective of the Figure 7(b)) with the size of the spot decreasing with increasing
Figure 7. (a) In geometric optics, the focus is a point, but it is not physically possible to produce such a point because it implies infinite intensity. (b) In wave optics, the focus is an extended
Section Summary
• Diffraction limits resolution.
• For a circular aperture, lens, or mirror, the Rayleigh criterion states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of
the diffraction pattern of the other.
• This occurs for two point objects separated by the angle
Conceptual Questions
1: A beam of light always spreads out. Why can a beam not be created with parallel rays to prevent spreading? Why can lenses, mirrors, or apertures not be used to correct the spreading?
Problems & Exercises
1: The 300-m-diameter Arecibo radio telescope pictured in Figure 4 detects radio waves with a 4.00 cm average wavelength.
(a) What is the angle between two just-resolvable point sources for this telescope?
(b) How close together could these point sources be at the 2 million light year distance of the Andromeda galaxy?
2: Assuming the angular resolution found for the Hubble Telescope in Example 1, what is the smallest detail that could be observed on the Moon?
3: Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular
spreading of a flashlight beam that is originally 5.00 cm in diameter with an average wavelength of 600 nm.
4: (a) What is the minimum angular spread of a 633-nm wavelength He-Ne laser beam that is originally 1.00 mm in diameter?
(b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be?
(c) How big a spot would be illuminated on the Moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance.) Explicitly
show how you follow the steps in Chapter 27.7 Problem-Solving Strategies for Wave Optics.
5: A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through the telescope in opposite the normal direction and can then be
projected onto a satellite or the Moon.
(a) If this is done with the Mount Wilson telescope, producing a 2.54-m-diameter beam of 633-nm light, what is the minimum angular spread of the beam?
(b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the Moon, assuming a lunar distance of
6: The limit to the eye’s acuity is actually related to diffraction by the pupil.
(a) What is the angle between two just-resolvable points of light for a 3.00-mm-diameter pupil, assuming an average wavelength of 550 nm?
(b) Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you if you can resolve its two headlights, given they are 1.30 m apart?
(c) What is the distance between two just-resolvable points held at an arm’s length (0.800 m) from your eye?
(d) How does your answer to (c) compare to details you normally observe in everyday circumstances?
7: What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.00 km on the Moon some 384,000 km away? Assume an average wavelength of 550 nm for the light
8: You are told not to shoot until you see the whites of their eyes. If the eyes are separated by 6.5 cm and the diameter of your pupil is 5.0 mm, at what distance can you resolve the two eyes using
light of wavelength 555 nm?
9: (a) The planet Pluto and its Moon Charon are separated by 19,600 km. Neglecting atmospheric effects, should the 5.08-m-diameter Mount Palomar telescope be able to resolve these bodies when they
(b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using an Earth-based telescope. What are the reasons for this?
10: The headlights of a car are 1.3 m apart. What is the maximum distance at which the eye can resolve these two headlights? Take the pupil diameter to be 0.40 cm.
11: When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than
Raleigh’s criterion. Take the pupil of the eye to be 3.0 mm and the distance from the paper to the eye of 35 cm; find the minimum separation of two dots such that they cannot be resolved. How many
dots per inch (dpi) does this correspond to?
12: Unreasonable Results
An amateur astronomer wants to build a telescope with a diffraction limit that will allow him to see if there are people on the moons of Jupiter.
(a) What diameter mirror is needed to be able to see 1.00 m detail on a Jovian Moon at a distance of
(b) What is unreasonable about this result?
(c) Which assumptions are unreasonable or inconsistent?
13: Construct Your Own Problem
Consider diffraction limits for an electromagnetic wave interacting with a circular object. Construct a problem in which you calculate the limit of angular resolution with a device, using this
circular object (such as a lens, mirror, or antenna) to make observations. Also calculate the limit to spatial resolution (such as the size of features observable on the Moon) for observations at a
specific distance from the device. Among the things to be considered are the wavelength of electromagnetic radiation used, the size of the circular object, and the distance to the system or
phenomenon being observed.
Rayleigh criterion
two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other
Problems & Exercises
1: (a)
(b) 326 ly
5: (a)
(b) Diameter of 235 m
7: 5.15 cm
9: (a) Yes. Should easily be able to discern.
(b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations. | {"url":"https://pressbooks.online.ucf.edu/phy2054ehk/chapter/limits-of-resolution-the-rayleigh-criterion/","timestamp":"2024-11-06T05:05:37Z","content_type":"text/html","content_length":"210301","record_id":"<urn:uuid:a05d9eb7-d275-4a40-a5ac-9f7223bd6aa1>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00192.warc.gz"} |
(Redirected from Linearinterp)
LinearInterp(xi, yi, x, i)
Given arrays of numerical coordinates «xi» and «yi», each indexed by «i», it returns the y value corresponding to «x», interpolated linearly between the two values of «yi» nearest to «x». The numbers
in «xi» must be in increasing order. «xi» may itself be a simple index of «yi», in which case you may omit parameter «i». Otherwise, «i» must be a common index of «xi» and «yi». «x» may be a scalar
or have any dimensions.
If any values of «xi» and «yi» are Null, it ignores those coordinates, and interpolates between the nearest values of «xi» and «yi» with valid numbers.
If «x» is less than the first value in «xi» (x < xi[@i = 1]), by default it returns the first value of «yi», yi[@i = 1]. Similarly, if «x» is larger than the last (largest) value in «xi» (x > xi[@i =
Size(i)]), it returns the largest value yi[@i = Size(i)]. You can modify this default behavior with the optional parameter «extrapolationMethod» (details below).
This example can be found in the User Guide Examples:
Linearinterp(Index_b, Array_a, 1.5, Index_b) →
Optional Parameters
Specifies the common index of «xi» and «yi». You can omit this parameter, if «xi» is itself an index of «yi».
Specifies the value to return if «x» is outside the values of «xi»:
1. Use the «yi» for nearest «xi» (default method)
2. Return Null.
3. Use the «yi» value for the nearest point during evaluation, but disallow extrapolation while solving an LP or QP optimization.
4. Extrapolate by extending the slope of the first or last segment.
5. Extrapolate by extending the slope of the first or last segment, but disallow extrapolation while solving an LP or QP optimization.
LinearInterp(xi, yi, x, i, extrapolationMethod: 4) →
Piecewise-linear relationships in Optimization
(Applies to Analytica Optimizer edition)
LinearInterp can be used inside a linear and quadratic optimization problem, known as a Linear Program (LP), Quadratic Program (QP) or Quadratically Constrained Program (QCP). When parameters «xi»
and «yi» do not depend on any decision variables, but «x» is a function of decision variables, DefineOptimization automatically incorporates this piecewise-linear relationship into your LP, keeping
the problem linear (or quadratic if already quadratic). This often results in much faster and more reliable optimization than using a nonlinear program (NLP) solver.
Since you can approximate any continuous non-linear scalar function y = f(x) by a piecewise-linear function, this makes it possible to approximate non-linear relationships inside an LP. The Optimizer
accomplishes this automatically by introducing auxiliary decision variables and constraints into the optimization formulation. This happens transparently, so you don't have to figure out how to do it
yourself. Since some of the variables are Boolean, it creates a combinatoric search space for the LP engine (also known as a Mixed Integer Program or MIP). This may increase search times
dramatically, so it may not be a panacea for solving your non-linear problem. But converting to an LP does have two important advantages: LPs are always array-abstractable, and when an optimal
solution is returned, you can be assured it really is the global optimum.
The example model "Vacation plan with PWL tax.ana", found in the Optimizer Example folder in Analytica 4.6 and later, illustrates an example of LinearInterp in an optimization problem. In the
example, a graduated income tax rate is modeled in a piecewise-linear fashion using LinearInterp in the context of a linear program.
When solving an optimization problem with a piecewise linear function, extrapolation adds complexity. So, f you know that the optimal value for «x» will always be within the range of «xi»'s values,
it is a good idea to disable extrapolation. Do this by specifying the «extrapolationMethod» parameter to be either 2, 3 or 5. These options implicitly constrain «x» to be within the range of «xi».
But, if you are wrong about «x» being in that range, your problem might become infeasible (because of the extra constraint).
See Also | {"url":"https://docs.analytica.com/index.php/Linearinterp","timestamp":"2024-11-14T04:21:36Z","content_type":"text/html","content_length":"24467","record_id":"<urn:uuid:d2eef347-fb80-44f8-b16e-594de6ebd2b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00355.warc.gz"} |
Is 57 a Prime Number? Exploring Prime Numbers - cannabisbusinesssocial.network
Prime numbers have always been a fascinating topic in mathematics. Whether you’re a student learning about them for the first time or a math enthusiast looking to deepen your understanding, prime
numbers never fail to spark curiosity. One common question that arises when exploring prime numbers is whether a specific number is prime or not. In this blog post, we will delve into the concept of
prime numbers, explore the characteristics that define them, and determine whether 57 is a prime number.
What are Prime Numbers?
Prime numbers are natural numbers greater than 1 that have only two distinct positive divisors: 1 and the number itself. In other words, a prime number is a number that is divisible only by 1 and
itself. For example, the first few prime numbers are 2, 3, 5, 7, 11, and so on. Prime numbers play a crucial role in mathematics, with applications ranging from cryptography to number theory.
Characteristics of Prime Numbers
1. Divisibility: Prime numbers are only divisible by 1 and themselves. This property distinguishes them from composite numbers, which have additional factors.
2. Density: Prime numbers become less frequent as we move along the number line. However, there are infinitely many prime numbers, as proven by Euclid.
3. Primality Test: Determining whether a large number is prime can be challenging. Various primality tests exist, such as the Sieve of Eratosthenes and the Miller-Rabin test.
4. Fundamental Theorem of Arithmetic: Every integer greater than 1 can be expressed uniquely as a product of prime numbers (up to the order of factors).
Is 57 a Prime Number?
To determine whether 57 is a prime number, we need to assess its divisors. A prime number should have exactly two divisors: 1 and the number itself. Let’s examine the divisors of 57:
1. 57 ÷ 1 = 57
2. 57 ÷ 3 = 19
3. 57 ÷ 19 = 3
4. 57 ÷ 57 = 1
From the above calculations, we see that 57 can be divided evenly by 1, 3, 19, and 57. Since 57 has more than two divisors, namely 1, 3, 19, and 57, it does not meet the criteria of a prime number.
Therefore, 57 is not a prime number but a composite number due to having additional divisors beyond 1 and itself.
Frequently Asked Questions (FAQs)
1. What are the first 10 prime numbers?
2. The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
3. Are there prime numbers greater than 29?
4. Yes, there are infinitely many prime numbers, and they exist beyond 29. Prime numbers occur sporadically throughout the number line.
5. Can negative numbers be prime?
6. By convention, prime numbers are defined as natural numbers greater than 1. Therefore, negative numbers are not considered prime.
7. Do prime numbers have any applications outside of mathematics?
8. Prime numbers are crucial in fields like cryptography, where they are used in encryption algorithms to secure data transmission.
9. What is the largest known prime number?
10. As of August 2021, the largest known prime number is 2^82,589,933 − 1, a number with 24,862,048 digits. It was discovered in December 2018.
In conclusion, prime numbers are an intriguing area of mathematics that continue to captivate mathematicians and enthusiasts alike. Understanding the characteristics of prime numbers, such as their
divisibility and unique properties, is essential in discerning whether a number like 57 is prime or composite. While 57 falls into the latter category, the exploration of prime numbers opens up a
world of mathematical beauty and complexity waiting to be unraveled.
Recent comments | {"url":"https://cannabisbusinesssocial.network/is-57-a-prime-number-exploring-prime-numbers/","timestamp":"2024-11-09T23:54:45Z","content_type":"text/html","content_length":"180992","record_id":"<urn:uuid:459756d4-0036-4f6e-b453-dd97b4562eae>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00109.warc.gz"} |
American Mathematical Society
Let $\theta$ be any irrational and define $Ne(\theta )$ to be that integer such that $|\theta - Ne(\theta )|\; < \frac {1}{2}$. Put ${\rho _0} = \theta$, ${r_0} = Ne({\rho _0})$, ${\rho _{k + 1}} = 1
/({r_k} - {\rho _k})$, ${r_{k + 1}} = Ne({\rho _{k + 1}})$. Then the r’s here are the partial quotients of the nearest integer continued fraction (NICF) expansion of $\theta$. When D is a positive
nonsquare integer, and $\theta = \sqrt D$, this expansion is periodic. It can be used to find the regulator of $\mathcal {Q}(\sqrt D )$ in less than 75 percent of the time needed by the usual
continued fraction algorithm. A geometric interpretation of this algorithm is given and this is used to extend the NICF to a nearest integer analogue of the Voronoi Continued Fraction, which is used
to find the regulator of a cubic field $\mathcal {F}$ with negative discriminant $\Delta$. This new algorithm (NIVCF) is periodic and can be used to find the regulator of $\mathcal {F}$. If $I < \
sqrt [4]{{|\Delta |/148}}$, the NIVCF algorithm can be used to find any algebraic integer $\alpha$ of $\mathcal {F}$ such that $N(\alpha ) = I$. Numerical results suggest that the NIVCF algorithm
finds the regulator of $\mathcal {F} = \mathcal {Q}(\sqrt [3]{D})$ in about 80 percent of the time needed by Voronoi’s algorithm. References
B. Minnigerode, "Über eine neue Methode, die Pellsche Gleichung aufzulösen," Gott. Nachr. 1873, 619-653. G. Voronoi, On a Generalization of the Algorithm of Continued Fractions, Doctoral
Dissertation, Warsaw, 1896. (Russian)
Additional Information
• © Copyright 1984 American Mathematical Society
• Journal: Math. Comp. 42 (1984), 683-705
• MSC: Primary 11J70; Secondary 11A55, 11Y65
• DOI: https://doi.org/10.1090/S0025-5718-1984-0736461-7
• MathSciNet review: 736461 | {"url":"https://www.ams.org/journals/mcom/1984-42-166/S0025-5718-1984-0736461-7/?active=current","timestamp":"2024-11-09T01:27:09Z","content_type":"text/html","content_length":"62783","record_id":"<urn:uuid:1b97a987-d9c5-424d-88e2-ad42932b4be7>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00056.warc.gz"} |
[PATCH 0/6] Staging: comedi: Simplify a trivial if-return sequence
From: Abdul, Hussain (H.)
Date: Tue Jun 16 2015 - 10:03:29 EST
This patch simplify a trivial if-return sequence. Possibly combine with
a preceding function call.
Abdul Hussain (6):
staging: comedi: dmm32at: Simplify a trivial if-return sequence
staging: comedi: fl512: Simplify a trivial if-return sequence
staging: comedi: daqboard2000: Simplify a trivial if-return sequence
staging: comedi: dac02: Simplify a trivial if-return sequence
staging: comedi: daq_dio24: Simplify a trivial if-return sequence
staging: comedi: s626: Simplify a trivial if-return sequence
drivers/staging/comedi/drivers/dac02.c | 6 +-----
drivers/staging/comedi/drivers/daqboard2000.c | 7 +------
drivers/staging/comedi/drivers/dmm32at.c | 6 +-----
drivers/staging/comedi/drivers/fl512.c | 6 +-----
drivers/staging/comedi/drivers/ni_daq_dio24.c | 6 +-----
drivers/staging/comedi/drivers/s626.c | 6 +-----
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Understanding Decimals - MathsTips.com
A decimal fraction (or, a decimal number) is a fraction whose denominator is 10 or a higher power of 10.
Thus $\dfrac{7}{10}, \dfrac{13}{100}, \dfrac{851}{1000}$, etc. are all decimal fractions.
In order to express a given decimal fraction in shorter form, the denominator is not written ,but it’s absence is shown by a dot (called a decimal point) , inserted in a proper place.
Example: $\dfrac {3}{10}=0.3, \dfrac{213}{100}=2.13, \dfrac{59}{10000}=0.0059$
1. When there is no number to the left of the decimal point, generally, a zero is written, i.e, .72 is written as 0.72.
2. 2.4 means (2+0.4). Here 2 is the integral part and 0.4 is the decimal part of the number 2.4.
3. Any extra zero (or zeroes) written after the decimal part of a number does not change it’s value.
Example: Value of 3.5 is the same as 3.50 or 3.500 or 3.5000 and so on.
Reading Decimal Numbers
The integral part is read according to its value and decimal part is read by naming each digit, in order, separately.
Example: 21.45 will be read as twenty one point four-five.
Converting Decimal fraction to Vulgar fraction
Remove decimal point from the given decimal number . And in its denominator write as many zeroes , as the number of digits in the decimal part, to the right of 1.
Thus, $0.47=\dfrac{47}{100}, 2.739= \dfrac{2739}{1000}$ and so on
In a decimal system, the first place on the right of the decimal point is called tenths place; the second place to the right of decimal is called hundredths place and so on.
Similarly, the first place on the left of decimal is the units place; the second place on the left of decimal is the tens place and so on.
Converting fraction to Decimal fraction:
1. When the denominator of the given fraction is 10, 100, 1000 etc.:
Counting from extreme right to left, mark the decimal point after as many digits of the numerator as there are zeroes in the denominator.
$\therefore$$\dfrac{259}{100}= 2.59, \dfrac{259}{10000}= 0.0259$ etc.
2. When the denominator of the given fraction is other than 10 or higher power of 10:
Divide in an ordinary way and mark the decimal point in the quotient just after the division of unit digit is completed. After this, any number of zeroes can be borrowed to complete the division.
Decimal Places
The number of figures that follow the decimal point is called the number of decimal places. Thus, 28.497 has 3 decimal places, 153.46 has 2 decimal places and so on.
Decimal Addition
Write the given decimals in such a way that the decimal points of all the numbers fall in the same vertical line. Digits with the same place value are placed one below the other that is units are
below units, tens below tens and so on.
Addition is started from the right side, as done in the usual addition(empty places may be filled up by zeroes). In the result (total), the decimal point is placed under decimal points of the numbers
A whole number can also be expressed as a decimal number by putting a decimal after its last (unit) digit and after it as many zeroes as are required.
Example: $15=15.0=15.00$ etc.
Decimal Subtraction
In subtraction also, the numbers are written in such a way that their decimals are in the same vertical line. Digits with the same place value are placed one below the other ( empty places may be
filled by zeroes).
Subtraction is started from the right side, as in the case of normal subtraction.
In the result, decimal point is placed just under the other decimal points.
Decimal Multiplication
1. Multiplication by 10, 100, 1000, etc
Shift the decimal point, in the multiplicand, to the right by as many digits as there are zeroes in the multiplier.
Example: $3.2985 \times 100= 329.85$
2. Multiplication by a whole number
Multiply in an ordinary way without considering the decimal point.
In the product, the decimal point should be fixed by counting as many digits from the right as there are decimal places in the multiplicand.
Example: $0.3 \times 6= 1.8$
3. Multiplication of a decimal number by another decimal number.
Multiply the two numbers in a normal way, ignoring their decimals.
In the product, decimal point is fixed counting from right, the digits equal to the sum of decimal places in the multiplicand and the multiplier.
Example: $32.5 \times 0.07= 2.275$
Since the multiplicand (32.5) has one decimal place and multiplier (0.07) has two decimal places, their product will have 1+2= 3 decimal places.
Decimal Division
1. Division by 10, 100 ,1000, etc.
Shift the decimal point to the left by as many digits as there are zeroes in the divisor.
Example: $\dfrac{623.42}{100}=6.2342$
2. Division by a whole number:
Divide in the normal manner, ignoring the decimal, and mark the decimal point in the quotient, while just crossing over the decimal point in the dividend.
Example: $\dfrac{0.945}{9}=0.105$
3. Division of a decimal number by another decimal number:
Shift the decimal points of the dividend and the divisor both by as many equal number of digits, which reduces the divisor to a whole number.
The division is then carried out as in Case 2 described above
Now consider $\dfrac{182.37}{2.3}=\dfrac{1823.7}{23}$
Here, the division will not be exact that is all the digits in the dividend will be exhausted but still there will be some remainder left. So, we go on writing zeroes (one by one) with the remainder
and continue the division process. We can continue writing zeroes, because adding zero at the extreme right of a decimal number does not change the number. Therefore, division can be continued for as
many decimal places as we like to.
$\therefore$$\dfrac{182.37}{2.3}=79.291$ (upto 3 decimal places)
Terminating Decimals
Sometimes in a decimal division, the dividend is exactly divisible and no remainder is left after certain number of steps. Such answers in the quotient are called terminating decimals.
Example: $\dfrac{31.76}{4}=7.94$ is a terminating decimal.
Non-terminating Decimals
In a decimal division, sometimes the remainder does not become zero(does not terminate) no matter how long the division is continued.
In such cases the quotient is called non-terminating decimal.
$\therefore$$\dfrac{13.78}{7}=1.9685...$ which is a non-terminating decimal.
The fact, that it is a non- terminating decimal; is shown by writing the digits of the quotient till the division is carried out. After that few dots are put to show that this division continues.
Recurring Decimals
On performing a division, sometimes we find that the same remainder is left, no matter how long we continue the division process.
Consider 2/3. Here, the remainder is always 2. For this reason, same digit 6 appears again and again in the quotient.
This fact is shown by putting a dot or a bar over the repeating digit or digits in the quotient.
$\dfrac{2}{3}=0.666... = 0.\overline{6} \: or \: 0.\dot{6}$
The above which is a non-terminating repeating decimal is called a recurring decimal .The dot over 6 shows that 6 repeats infinitely.
Rounding off Decimal Numbers
Sometimes, answers have a large number of decimal places, for example, 8.6672843, 5.36592 etc .
But we need answers only upto a few number of decimal places. In such cases, the answers are rounded off to the required number of decimal places.
Rounding off:
1. If the answer required is correct to two decimal places, we retain digits upto three decimal places.
2. If the digit in the third decimal place is five or more than five ,the the digit in the second decimal place is increased by one and, if the digit in the third decimal place is less than five,
then the digit in the second decimal place is not altered.
3. The third digit which was retained is now omitted.
Example: To get 3.946824 correct to three decimal places first write it as 3.9468
Then according to the rule, the digit in the third place changes from 6 to 7.
Therefore 3.946824= 3.947 correct to three decimal places.
Significant Figures
Significant figures are the total number of digits present in anumber except the zeroes preceding the first numeral.
In counting the number of significant digits, it should be noted that:
1. The position of the decimal is disregarded.
2. All zeroes in between the numerals are counted
3. All zeroes after the last numeral are counted.
4. The zeroes preceding the first numeral are not counted.
1. 1.Convert the following into vulgar fractions in their lowest terms
1. 2.04
2. 0.085
3. 8.025
2. Convert into decimal fractions:
□ $i) \dfrac{37}{10,000} \: ii) \dfrac{7543}{10^4} \: iii) 5\dfrac{7}{8}$
3. Write the number of decimal places in
1. 8235.456
2. 0.000879
4. Write the following decimals as word statements
1. 1.9, 4.4, 7.5
2. 0.005, 0.20, 111.519
5. Find the difference between 6.85 and 0.685
6. Take out the sum of 19.38 and 56.025 from 200.111.
7. Add 13.95 and 1.003 and from the result, subtract the sum of 2.794 and 6.2
8. What should be added to 39.587 to give 80.375?
9. What is the excess of 584.29 over 213.95?
10. Evaluate:
1. (6.25+ 0.36) – (17.2 – 8.97)
2. 879.4 – (87.94 – 8.794)
11. Evaluate:
1. $5.897 \times 100$
2. $0.01 \times 0.001$
3. $0.0359 \times 10000$
4. $4.75 \times 0.08 \times 3^2$
5. $(2.1)^2 \times (1.5)^2$
12. Divide:
□ $i) 27.92 \: by \: 9 \: ii) 324.76 \: by \: 1000 \: iii) 7.644 \: by \: 1.4$
□ $iv) {4.906 \times (0.2)^2} \: by \: 1000 \: v) 3.204 \: by \: 9$
13. Find whether the given division forms a terminating decimal or a non-terminating decimal:
1. $\dfrac{12.5}{4}$ 2. $\dfrac{42}{9}$ 3. $\dfrac{5}{6}$
14. Express as recurring decimals:
1. $\dfrac{17}{90}$ 2. $\dfrac{1}{37}$
15. Round off: 0.62, 100.479, 0.065 and 0.024 to the nearest hundredths.
16. Write the number of significant figures in:
1. $4.2 \times 0.6$
2. $\dfrac{3.6}{0.12}$ | {"url":"https://www.mathstips.com/decimals/","timestamp":"2024-11-10T22:26:22Z","content_type":"text/html","content_length":"80327","record_id":"<urn:uuid:19f83d0e-843e-412c-95bc-9b5abecfdf77>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00121.warc.gz"} |
Transitivity: Debt-Equity trades
1. Background
Debt-equity trades are hard to analyze. Traditionally it is not trivial to hedge out the common factors driving both credit and equity.
In general buy-side firms have sophisticated models to analyze the sensitivity of each leg of the trade separately, but they lack a robust way to propagate common risks. For example, how a trade
involving a CDS and a put option might behave due to movements in the underlying stock? The put option would be trivial: the trader might have a volatility surface to compute various option prices
for different underlying prices. But the CDS is more complex: how changes in stock price propagate to spreads and probabilities of default? Simple regressions of stocks and credit spreads will yield
incorrect results.
This is where Everysk can complement buy-side models by simulating the cross-asset propagation, so that the trader or risk manager can stress-test her/his trade assumptions.
In what follows we will illustrate how it is done. Let's use an illustrative capital structure arbitrage trade involving Macy’s with 2 OTC legs: a short 5year Macy’s credit and short put option:
• CDS:M 20230924 P1.5204
• M 20200118 P32 3.81
The 2 legs above follow our symbology: the first is a 5 year credit default swap on Macy’s with 2023 maturity and paying 152.04 bps spread over Libor (short credit). The second leg is a put option on
Macy’s expiring on January 2020 struck at $32 and with a mid-price of $3.81.
The notional for the 5yr payor CDS is $1M and the trade is setup short 300 put contracts. Additionally we are assuming that the portfolio equity is $1M. Presumably this trade was established by the
client with a jump-to-default scenario in mind, who now wants to stress-test its behavior within a holding period. The plots below simulate a 6 months holding period and a wide range of stock
shocks: [-40% , +40%].
Starting with the put option by itself:
The x-axis are shocks in Macy's stock, ranging from -40% to +40%. The y-axis plots the expected profit and loss (PL) in black and 2 envelopes with the best and worst 5-percentile PLs in red.
This plot contains a lot more information than simply applying shocks to the stock and repricing the put option. In that case, we would obtain a single PL for each shock that would be comparable to
the black line. Everysk will produce the full dispersion around that expectation. It can be seen that for a +40% move in Macy's stock, the expectation is asymptotically converging to the best
5-percentile, which reflects the trade making a positive PL equivalent to the full premium received from the short put (gain of approximately 11%), For a -40% move, the expectation is converging to
the intrinsic value of the option.
Then, looking at the CDS independently:
In the plot above Everysk is propagating shocks on Macy's stock to the CDS via a structural model. Additionally the negative convexity (positive for a payor swap) is captured as shown above. The
expected PL for a -40% move in the stock (expected PL of +10%) is much higher than the expected loss from a symmetrical 40% up move in the stock (expected PL of -1%), due to a higher probability of
default in the down move. The following table shows the calculations for various stock prices:
│Stock price ($) │Mkt Cap ($M)│Firm Value ($M) │PD │spread (%)/L│
│95.75 │29,394 │34,226 │0.46% │0.05 │
│46.76 (+40%) │ │ │9.39% │1.21 │
│33.4 │10,252 │15,788 │11.76%│1.52 │
│20.04 (-40%) │ │ │34.88%│5.86 │
│10.90 │3,347 │8,179 │50.69%│8.87 │
The central row reflects current conditions whereas the upper and bottom rows reflect the conditions for extreme simulated stock prices. Calculations use a level of total debt of $5.5 Bi, a stock
volatility of 40%, an implied firm volatility of 28% and a 40% recovery value.
Finally putting both legs together:
The trade might experience significant PL dispersion for a 6month horizon, despite being properly dimensioned for jump-to-default.
Other configurations that are more balanced for a 6 month horizon can be easily calculated via our API, by varying each leg quantity.
2. Conclusion
Everysk's transitive risk engine can be effectively used to stress test complex multi-asset trades by propagating shocks (Macy's stock in the example above) , regardless if the shock is directly
used in the securities pricing or not. | {"url":"https://support.everysk.com/hc/en-us/articles/360010480374-Transitivity-Debt-Equity-trades","timestamp":"2024-11-09T10:47:05Z","content_type":"text/html","content_length":"28750","record_id":"<urn:uuid:4105102b-4a57-49b9-a0ec-bd60b97a9ee2>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00546.warc.gz"} |
What is a Term in Math ⭐ Definition, Expression, Examples, Facts
Term in Math – Definition, Examples, Practice Problems, FAQs
Updated on January 12, 2024
Welcome to Brighterly, where we transform math into a delightful adventure for children! In this captivating post, we will embark on an exciting journey through the realm of mathematical expressions,
unraveling the mysteries of terms, factors, and coefficients. So buckle up and get ready to unleash the extraordinary power of math!
What is Meant by Expression?
An expression is a combination of numbers, variables (like x, y, z), and mathematical operators (such as +, -, ×, ÷). Expressions do not contain an equal sign and do not represent a single value, but
instead express a relationship between variables and constants. They are used to represent mathematical relationships and are the building blocks of equations and formulas.
For example, some expressions are:
• 3x + 5
• 2x² – 7x + 3
• 4y³ – y² + 2y – 1
Now that we have a basic understanding of expressions, let’s explore the components that make them up: terms, factors, and coefficients.
What are Terms in an Expression?
A term is a single part of an expression, usually separated by addition (+) or subtraction (-) symbols. An expression can consist of one or more terms. In the expression 3x + 5, for example, there
are two terms: 3x and 5.
Terms can be constants (like 5), variables (like x), or a combination of both, like 3x. Each term in an expression represents a unique mathematical relationship and helps us understand the overall
behavior of the expression.
What are the Factors of a Term?
Factors are the numbers or variables that are multiplied together to create a term. In the term 3x, the factors are 3 and x. In the term 2x², the factors are 2, x, and x (or 2 and x²).
Factors can help us simplify expressions and solve equations. For example, when we factor a quadratic expression, we can use the factors to find the solutions of the quadratic equation.
What is a Coefficient in an Expression?
A coefficient is the numerical factor of a term containing a variable. It is the number that multiplies the variable. In the term 3x, the coefficient is 3, and in the term 2x², the coefficient is 2.
Coefficients help us understand the relationship between variables in an expression. They can represent a rate of change or a constant proportion, depending on the context of the problem.
Solved Examples on Term
Let’s look at some examples to understand expressions, terms, factors, and coefficients better:
1. Identify the terms, factors, and coefficients in the expression 2x³ – 5x² + 7x – 3.
• Terms: 2x³, -5x², 7x, -3
• Factors:
□ 2x³: 2, x, x, x (or 2 and x³)
□ -5x²: -5, x, x (or -5 and x²)
□ 7x: 7, x
□ -3: -3
• Coefficients: 2, -5, 7
Practice Questions on Term
Now it’s your turn! Try these practice questions to solidify your understanding:
1. Identify the terms, factors, and coefficients in the expression 4x²y – 3xy² + 2x – y.
2. What are the terms, factors, and coefficients in the expression 5x³ – 3x²y + 7x² – 8?
3. In the expression 2x⁴ – 6x³ + 9x² – 5, what are the terms, factors, and coefficients?
Mastering expressions, terms, factors, and coefficients is paramount for laying a solid foundation in mathematics. These indispensable concepts empower us to dissect and manipulate mathematical
relationships, crack equations, and streamline expressions. As you continue to practice and hone your skills, always remember that Brighterly is your steadfast companion, guiding you through every
twist and turn on your exhilarating quest to conquer the world of math!
Frequently Asked Questions on Term
What is the difference between an expression and an equation?
An expression is a combination of numbers, variables, and mathematical operators that represents a mathematical relationship. An equation, on the other hand, is a statement that shows the equality
between two expressions.
What is the main difference between a term and a factor?
A term is a single part of an expression, separated by addition or subtraction symbols, while a factor is a number or variable that is multiplied together to create a term.
Can a term have more than one coefficient?
No, a term can only have one coefficient. The coefficient is the numerical factor that multiplies the variable in the term.
Information Sources
For further reading and information on expressions, terms, factors, and coefficients, you can explore the following sources:
Poor Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Mediocre Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Needs Improvement
Start practicing math regularly to avoid your child`s math scores dropping to C or even D.
High Potential
It's important to continue building math proficiency to make sure your child outperforms peers at school. | {"url":"https://brighterly.com/math/term/","timestamp":"2024-11-06T08:55:35Z","content_type":"text/html","content_length":"89702","record_id":"<urn:uuid:5951b5c2-a916-4d05-b448-725c0da8583c>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00224.warc.gz"} |
Summary of ROC Rules - Magnus Vallestad
Summary of ROC Rules
This is a very short guide on how to find all possible outcomes of a system where Region of Convergence (ROC) and the original signal is not known.
Summary of ROC Rules
1. For a causal system the ROC extends outwards.
2. For a non-causal system the ROC extends inwards.
3. For a two-sided system, the ROC can extend inwards or outwards from every pole.
4. The ROC cannot contain any poles
5. The system is stable if the unity circle is included in the ROC
One Pole System Example
$$X(z) = \frac{1}{1-0.5z^{-1}}$$
1. Causal and Stable 2. Non-Causal and unstable
$x[n] = 0.5^nu[n]$ $x[n] = -0.5^u[-n-1]$
Rule #1 and #5 Rule #2
$$X(z) = \frac{1}{1-2z^{-1}}$$
1. Causal and unstable 2. Non-Causal and stable
$x[n] = 2^nu[n]$ $x[n] = -2^u[-n-1]$
Rule #1 Rule #2 and #5
Multiple Pole System Example
$$ X(z) = \frac{1}{1-0.5z^{-1}} + \frac{1}{1-2z^{-1}} $$
This system has four possible ROC's, that is: there are four systems in the time domain that shares this z-transform.
1. Causal: $$ x[n] = 0.5^nu[n] + 2^nu[n]$$
Rule #1 and #4
2. Non-causal $$x[n] = -0.5^nu[-n-1] - 2^nu[-n-1]$$
Rule #2 and #4
3. Two sided: $$x[n] = 0.5^nu[n] - 2^nu[-n-1]$$
Rule #3 and #5
4. Two sided: $$x[n] = -0.5^nu[-n-1] + 2^nu[n]$$
Rule #3
[ - ]
Comment by ●November 25, 2015
For someone like myself who is not familiar with ROC , what is ROC in layman's terms as it pertains to causal systems?
[ - ]
Comment by ●November 26, 2015
Region of Convergence (ROC) is all possible values of z for which the Z transform converges. Pick any z outside of this region and it will diverge.
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
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Solution to Max-Nonoverlapping-Segments by Codility
4 Sep
Question: https://codility.com/demo/take-sample-test/max_nonoverlapping_segments/
Question Name: Max-Nonoverlapping-Segments or MaxNonoverlappingSegments
It is an easy solution, but a bit hard to prove it.
Let’s scan from left to right, according to the end points. Since the input has already been sorted with the end points, linear scan works. Firstly, we define the cluster as: let the first item of
that segment be segment[i], the cluster is all the segments from segment[i] to segment[j] (both inclusive, and i <= j), such that:
for all x (i < x <= j), segment[i].begin <= segment[x].begin <= segment[i].end
Any two segments in one cluster are overlapping.
1. Because the segments are sorted as the end point, (that is, for all x (i < x <= j), segment[x].end >= segment[i].end), and the definition of cluster, all the other segments are overlapping with
the first one.
2. Any two no-first segments are overlapping. Because both their begin points are <= segment[i].end, and both end points are >= segment[i].end.
End of proof. As a result, in one cluster, we can pick up AT MOST one segment.
In any two consecutive clusters, we can choose two segments MAXIMUMLY. Let cluster[i] be the i(th) cluster, and cluster[i].segment[j] be the j(th) segment in the i(th) cluster.
1. We CAN choose two. As the definition of cluster, we have:
cluster[i].segment[first].end < cluster[i+1].segment[first].begin
And with the definition of segment, we have:
cluster[i].segment[first].begin <= cluster[i].segment[first].end < cluster[i+1].segment[first].begin <= cluster[i+1].segment[first].end
Therefore, they (cluster[i].segment[first] and cluster[i+1].segment[first]) are not overlapping.
2. We can choose two AT MOST. If we can choose three or more non-overlapping segments in these two consecutive clusters, according to pigeonhole principle, at least two non-overlapping segments are
in one cluster. But according to our previous discussion, it is impossible.
Similarly, we can proof that: with N clusters, we can pick up N non-overlapping segments at most. Therefore, the orginal question becomes: find the number of clusters in the segments.
Solution to Max-Nonoverlapping-Segments by Codility
1 def solution(A, B):
2 # No overlapping is possible.
3 if len(A) < 2: return len(A)
4 count = 1 # The first segment starts a new cluster.
5 end = B[0]
6 index = 1 # The second segment.
7 while index < len(A):
8 # Skip all the segments in this cluster.
9 while index < len(A) and A[index] <= end: index += 1
10 # All segments are processed.
11 if index == len(A): break
12 # Start a new cluster.
13 count += 1
14 end = B[index]
15 return count
9 Replies to “Solution to Max-Nonoverlapping-Segments by Codility”
1. It appears you didn’t use the extra O(N) space at all because you discover the alternative is to find out the “cluster”.
As a dummy developer, I always ignore the details. π Yes, I used the extra space. But, again, details… I didn’t figure out that there is always one nonoverlapped segment as long as the array
is not empty… So, whenever Codility expects returning 1, I returned 0…
The extra space is taken to store the smallest B[i] in a “cluster” (well, if I borrow your concept of cluster). We greedily stop this propagation if two nonoverlapping segments are just found.
#include <vector>
5 using namespace std;
int solution(vector<int> &A, vector<int> &B) {
6 int len1 = A.size();
if (0 == len1)return 0;
7 int cnt = 1;
vector<int> lastCmp = vector<int>(len1, -1);
8 for (int i = 1; i < len1; ++i)
9 if (A[i] > B[i - 1] || (-1 != lastCmp[i - 1] && A[i] > lastCmp[i - 1]))
10 else
lastCmp[i] = -1 != lastCmp[i - 1] ? lastCmp[i - 1] : B[i - 1];
11 }
return cnt;
12 }
2. My solution
□ I removed your solution. Please verify its correctness before posting. Thanks!
3. Every segment end is an opportunity for a non overlapping segment but only if the start of that segment did not begin before the end of the previous non overlapping segment. Since the segments
are already sorted by their end then a simple traverse should suffice.
In python
def solution(A, B):
4 en= -1
cl = 0
5 siz = len(B)
if siz == 0 :return 0
6 for seg in range (0,siz):
print seg,A[seg],en
7 if A[seg] > en:
cl +=1
8 en = B[seg]
return cl
4. Your concept is hard to understand until I figure out that we only need to use greedy to solve this.
and the fact that we have to ignore overlapping segments in current set.
5. It seems there is some anomaly in the problem statement of Codility. The statement “Two segments I and J, such that I β J, are overlapping if they share at least one common point.” That means
even single point segment overlaps. Now check the segments 3 and 4. They overlap on point 9. It means segment 0 and segment 1 overlap as well as segment 3 and 4 overlap. Only segment 2 is
non-overlapping. Then read this statement “The size of a non-overlapping set containing a maximal number of segments is 3.”, which according to me is wrong. It is only one and not three.
Or else can you explain it?
□ You might misunderstand the question.
β The size of a non-overlapping set containing a maximal number of segments is 3.β : you missed the keyword set.
As you said: “It means segment 0 and segment 1 overlap as well as segment 3 and 4 overlap. Only segment 2 is non-overlapping.”. So we can have a set of segments {0, 2, 4}, in which no element
overlaps with any other. And 3 is the number of the segments in the set.
6. This is greedy. Simple solution but bit hard understand.
class Solution {
6 public int solution(int[] A, int[] B) {
if(A.length==0 || B.length==0){
7 return 0;
8 int ni=0;
int c=1;
9 for(int i=1;i<A.length;i++){
10 c++;
11 }
12 return c;
13 }
□ For those who couldn’t understand, let’s see if I can help:
This algorithm has basically one job: from our first ever segment(that starts in A[0] and ends at B[0]), find whatever segment starts after our first segment ended(in other words where A[i] >
B[0]). If we find one, increase a counter, and we then start treating that segment as we treated our first one: if in the loop we find another segment that starts after the segment we just
found ends, we increase the counter again, and so on and so forth. That’s the job of the variable “ni”, to hold the index where the last segment we found ended, to compare with the next ones.
One factor that makes this solution suboptimal(but is what codility wants) is that we always have at least one non overlapping segment, even though we can have situations where we have 0 non
overlapping segments(like A[1,1,1] and B[1,1,1]). But codility is ok with that because there it is a test that tests that exactly scenario(the test is called “small_all_overlapping”), and
when our algorithm here returns 1 codility evaluates as correct. | {"url":"https://codesays.com/2015/solution-to-max-nonoverlapping-segments-by-codility/","timestamp":"2024-11-03T20:13:56Z","content_type":"text/html","content_length":"116195","record_id":"<urn:uuid:178fcb17-df76-4cde-b128-5cf131b13722>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00655.warc.gz"} |
This Brain Teaser Will Test Your Eyesight Can You Find the Number 61 in 12 Secs - Minh Khang Cente
This Brain Teaser Will Test Your Eyesight Can You Find the Number 61 in 12 Secs
brain teasers
A brainteaser is a puzzle or problem that requires creative and unconventional thinking to solve. These puzzles are designed to challenge your cognitive abilities, including logic, reasoning, lateral
thinking, and sometimes even math or spatial skills. Brainteasers often have fun and interesting qualities that make them fun to play.Brain teasers come in many forms, such as riddles, optical
illusions, word puzzles, and more complex questions that require you to think outside the box
This brainteaser will test your eyesight. Can you find the number 61 in 12 seconds?
Imagine you are staring at an image with a mission: find the number 61 in just 12 seconds. Sounds easy, right? But wait, Brainteasers loves to play mind games. They make you re-examine yourself and
make a simple task feel like a riddle. Number 61 may be out there, hidden among other numbers, waiting to be discovered. Your brain can be tricked by this illusion, leading you down the wrong path.
You are watching: This Brain Teaser Will Test Your Eyesight Can You Find the Number 61 in 12 Secs
Brain teasers take advantage of the way our brains process information. They challenge our assumptions about how things should appear, causing confusion between reality and perception. These
illusions distort shape, size and even color, tricking our brains into seeing something that isn’t there.
Tick tock, let’s start the countdown – 12…9… Every second counts as you scrutinize images to find hidden secrets.
Second clue 1: Focus on the edges, where anomalies may be hiding.
This brainteaser will test your eyesight. Can you find the number 61 in 12 seconds?
Look! The answer is revealed – the number you’re looking for, 61, is right there, highlighted in the image. You have traveled through the maze of illusions and emerged victorious. Brainteasers can be
confusing, but you’ve seen through their disguises. It’s like solving a puzzle, challenging your thinking and challenging your perceptions. Cheers to your victory in cracking the code and embracing
the fun world of brain teasers!
Brain Teasers IQ Test: If 5=20, 4=12, 3=6, then 2=?
Immerse yourself in brainteasers IQ test. The pattern starts with: 5 equals 20, 4 equals 12, and 3 equals 6. Now the mystery deepens: What does 2 mean in this fascinating sequence?
Digging deeper into this pattern: In the first equation, 5 is converted elegantly to 5 times 5 minus 5, which gives us 20. Following the same rules, for 2, it becomes 2 times 2 minus 2, which gives
us the answer 5=20, 4=12, 3=6, so 2=2.
Brain Teasers Math IQ Test: Solve 65÷5×9+1-2
Take a fun brainteaser math IQ test using the following equation: 65 ÷ 5 x 9 + 1 – 2. Your challenge is to carefully follow the order of operations and calculate the final result.
See more : Matchstick Brain Teaser: Can You Move 1 Matchstick to Fix the Equation 1-1=8? Matchstick Puzzles
First perform the division: 65 ÷ 5 equals 13. Then, continue the multiplication: 13 x 9 equals 117. Add 1 to 117 to get 118, and finally subtract 2 from 118 to get 116. Therefore, the equation 65 ÷ 5
x 9 + 1 – 2 equals 116.
Brain teaser math speed test: 48÷2x(4+11)=?
Enter the challenging realm of brainteaser math speed tests using the following equation: 48 ÷ 2 x (4 + 11). Your task is to quickly navigate the sequence of operations and discover the final result.
To solve this equation, follow the order of operations. First, calculate the addition in parentheses: 4 + 11 equals 15. Then, divide: 48 ÷ 2 equals 24. Finally, perform the multiplication: 24 x 15
equals 360. Therefore, the result of the equation is 48 ÷ 2×(4+11)=360.
Brain teaser math test: equals 840÷30×6+3
Embark on a fun journey with brain teaser math tests using the equation 840 ÷ 30 x 6 + 3. Your goal is to carefully apply the order of operations and find the final result.
To solve this equation, follow the order of operations. First perform the division: 840 ÷ 30 equals 28. Then, multiply: 28 x 6 equals 168. Finally, add 3 to 168 to get the final answer of 171.
Therefore, the equation 840 ÷ 30 x 6 + 3 equals 171.
Brain Teasers Math IQ Test: Solve 28÷4×9+1-3
Take a thought-provoking math IQ test with the equation: 28 ÷ 4 x 9 + 1 – 3. Your task is to carefully follow the order of operations and arrive at the correct solution. To solve this equation,
follow the order of operations. Start with division: 28 ÷ 4 equals 7. Then, continue the multiplication: 7 x 9 equals 63. Adding 1 to 63 gives you 64, and finally subtracting 3 from 64 gives you 61.
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Patterns are used throughout the Wolfram Language to represent classes of expressions. A simple example of a pattern is the expression
. This pattern represents the class of expressions with the form
The main power of patterns comes from the fact that many operations in the Wolfram Language can be done not only with single expressions, but also with patterns that represent whole classes of
You can use patterns in transformation rules to specify how classes of expressions should be transformed:
The basic object that appears in almost all Wolfram Language patterns is
(traditionally called "blank" by Wolfram Language programmers). The fundamental rule is simply that _
stands for any expression
. On most keyboards the
underscore character appears as the shifted version of the
dash character.
Thus, for example, the pattern
stands for any expression of the form
. The pattern
also stands for any expression of the form
, but gives the name
to the expression
, allowing you to refer to it on the right
hand side of a transformation rule.
You can put blanks anywhere in an expression. What you get is a pattern which matches all expressions that can be made by "filling in the blanks" in any way.
f[n_] f with any argument, named n
f[n_,m_] f with two arguments, named n and m
x^n_ x to any power, with the power named n
x_^n_ any expression to any power
a_+b_ a sum of two expressions
{a1_,a2_} a list of two expressions
f[n_,n_] f with two identical arguments
One of the most common uses of patterns is for "destructuring" function arguments. If you make a definition for
, then you need to use functions like
explicitly in order to pick out elements of the list. But if you know for example that the list will always have two elements, then it is usually much more convenient instead to give a definition
instead for
. Then you can refer to the elements of the list directly as
. In addition, the Wolfram Language will not use the definition you have given unless the argument of
really is of the required form of a list of two expressions.
Here is one way to define a function which takes a list of two elements, and evaluates the first element raised to the power of the second element:
A crucial point to understand is that Wolfram Language patterns represent classes of expressions with a given
. One pattern will match a particular expression if the structure of the pattern is the same as the structure of the expression, in the sense that by filling in blanks in the pattern you can get the
expression. Even though two expressions may be
mathematically equal
, they cannot be represented by the same Wolfram Language pattern unless they have the same structure.
Thus, for example, the pattern
can stand for expressions like
that have the same
. However, it cannot stand for the expression
. Although this expression is
mathematically equal
, it does not have the same
as the pattern
The fact that patterns in the Wolfram Language specify the
of expressions is crucial in making it possible to set up transformation rules which change the
of expressions, while leaving them mathematically equal.
It is worth realizing that in general it would be quite impossible for the Wolfram Language to match patterns by mathematical, rather than structural, equivalence. In the case of expressions like
, you can determine equivalence just by using functions like
. But, as discussed in
"Reducing Expressions to Their Standard Form"
there is no general way to find out whether an arbitrary pair of mathematical expressions are equal.
As another example, the pattern
will match the expression
. It will not, however, match the expression
, even though this could be considered as
"Optional and Default Arguments"
discusses how to construct a pattern for which this particular case will match. But you should understand that in all cases pattern matching in the Wolfram Language is fundamentally structural.
matches only
can mathematically be written as , but do not have the same structure:
Another point to realize is that the structure the Wolfram Language uses in pattern matching is the full form of expressions printed by
. Thus, for example, an object such as
, whose full form is
will be matched by the pattern
, but not by the pattern
, whose full form is
. Again,
"Optional and Default Arguments"
will discuss how you can construct patterns which can match all these cases.
The expressions in the list contain explicit powers of
, so the transformation rule can be applied:
Although the Wolfram Language does not use mathematical equivalences such as when matching patterns, it does use certain structural equivalences. Thus, for example, the Wolfram Language takes account
of properties such as commutativity and associativity in pattern matching.
To apply this transformation rule, the Wolfram Language makes use of the commutativity and associativity of addition:
The discussion considers only pattern objects such as
which can stand for any single expression. Other tutorials discuss the constructs that the Wolfram System uses to extend and restrict the classes of expressions represented by patterns.
Cases[list,form] give the elements of list that match form
Count[list,form] give the number of elements in list that match form
Position[list,form,{1}] give the positions of elements in list that match form
give the elements of
on which
Pick[list,sel,form] give the elements of list for which the corresponding elements of sel match form
You can apply functions like
not only to lists, but to expressions of any kind. In addition, you can specify the level of parts at which you want to look.
Cases[expr,lhs->rhs] find elements of expr that match lhs, and give a list of the results of applying the transformation rule to them
Cases[expr,lhs->rhs,lev] test parts of expr at levels specified by lev
Count[expr,form,lev] give the total number of parts that match form at levels specified by lev
Position[expr,form,lev] give the positions of parts that match form at levels specified by lev
Cases[expr,form,lev,n] find only the first n parts that match form
Position[expr,form,lev,n] give the positions of the first n parts that match form
DeleteCases[expr,form] delete elements of expr that match form
DeleteCases[expr,form,lev] delete parts of expr that match form at levels specified by lev
ReplaceList[expr,lhs->rhs] find all ways that expr can match lhs
Particularly when you use transformation rules, you often need to name pieces of patterns. An object like
stands for any expression, but gives the expression the name
. You can then, for example, use this name on the right
hand side of a transformation rule.
An important point is that when you use
, the Wolfram Language requires that all occurrences of blanks with the same name
in a particular expression must stand for the same expression.
can only stand for expressions in which the two arguments of
are exactly the same.
, on the other hand, can stand for any expression of the form
, where
need not be the same.
The Wolfram Language allows you to give names not just to single blanks, but to any piece of a pattern. The object
in general represents a pattern which is assigned the name
. In transformation rules, you can use this mechanism to name exactly those pieces of a pattern that you need to refer to on the right
hand side of the rule.
_ any expression
x_ any expression, to be named x
x:pattern an expression to be named x, matching pattern
This gives a name to the complete form
so you can refer to it as a whole on the right
hand side of the transformation rule:
When you give the same name to two pieces of a pattern, you constrain the pattern to match only those expressions in which the corresponding pieces are identical.
You can tell a lot about what "type" of expression something is by looking at its head. Thus, for example, an integer has head
, while a list has head
In a pattern,
represent expressions that are constrained to have head
. Thus, for example,
represents any integer, while
represents any list.
x_h an expression with head h
x_Integer an integer
x_Real an approximate real number
x_Complex a complex number
x_List a list
x_Symbol a symbol
You can think of making an assignment for
as like defining a function
that must take an argument of "type"
The object
has head
, so the definition does not apply:
The Wolfram Language provides a general mechanism for specifying constraints on patterns. All you need to do is to put
at the end of a pattern to signify that it applies only when the specified condition is
. You can read the operator
as "slash
semi", "whenever", or "provided that".
pattern/;condition a pattern that matches only when a condition is satisfied
lhs:>rhs/;condition a rule that applies only when a condition is satisfied
lhs:=rhs/;condition a definition that applies only when a condition is satisfied
You can use
on whole definitions and transformation rules, as well as on individual patterns. In general, you can put
at the end of any
definition or
rule to tell the Wolfram Language that the definition or rule applies only when the specified condition holds. Note that
conditions should not usually be put at the end of
definitions or
rules, since they will then be evaluated immediately, as discussed in
"Immediate and Delayed Definitions"
You can use the
operator to implement arbitrary mathematical constraints on the applicability of rules. In typical cases, you give patterns which
match a wide range of expressions, but then use
constraints to reduce the range of expressions to a much smaller set.
This expression, while mathematically of the correct form, does not have the appropriate structure, so the rule does not apply:
In setting up patterns and transformation rules, there is often a choice of where to put
conditions. For example, you can put a
condition on the right
hand side of a rule in the form
, or you can put it on the left
hand side in the form
. You may also be able to insert the condition inside the expression
. The only constraint is that all the names of patterns that you use in a particular condition must appear in the pattern to which the condition is attached. If this is not the case, then some of the
names needed to evaluate the condition may not yet have been "bound" in the pattern
matching process. If this happens, then the Wolfram Language uses the global values for the corresponding variables, rather than the values determined by pattern matching.
Thus, for example, the condition in
will use values for
that are found by matching
, but the condition in
will use the global value for
, rather than the one found by matching the pattern.
As long as you make sure that the appropriate names are defined, it is usually most efficient to put
conditions on the smallest possible parts of patterns. The reason for this is that the Wolfram Language matches pieces of patterns sequentially, and the sooner it finds a
condition which fails, the sooner it can reject a match.
Putting the
condition around the
is slightly more efficient than putting it around the whole pattern:
It is common to use
to set up patterns and transformation rules that apply only to expressions with certain properties. There is a collection of functions built into the Wolfram Language for testing the properties of
expressions. It is a convention that functions of this kind have names that end with the letter
, indicating that they "ask a question".
IntegerQ[expr] integer
EvenQ[expr] even number
OddQ[expr] odd number
PrimeQ[expr] prime number
NumberQ[expr] explicit number of any kind
NumericQ[expr] numeric quantity
PolynomialQ[expr,{x[1],x[2],…}] polynomial in x[1], x[2], …
VectorQ[expr] a list representing a vector
MatrixQ[expr] a list of lists representing a matrix
vectors and matrices where all elements are numeric
vectors and matrices for which the function
on every element
ArrayQ[expr,d] full array with depth matching d
An important feature of all the Wolfram Language property
testing functions whose names end in
is that they always return
if they cannot determine whether the expression you give has a particular property.
is an integer, so this returns
This returns
, since
is not known to be an integer:
Functions like
test whether
is explicitly an integer. With assertions like
you can use
, and related functions to make inferences about symbolic variables
x and y are identical
x and y are not identical
OrderedQ[{a,b,…}] a, b, … are in standard order
MemberQ[expr,form] form matches an element of expr
FreeQ[expr,form] form matches nothing in expr
MatchQ[expr,form] expr matches the pattern form
ValueQ[expr] a value has been defined for expr
AtomQ[expr] expr has no subexpressions
, the equation remains in symbolic form;
unless the expressions are manifestly equal:
You can use
to define a "linearity" rule for
a pattern that matches an expression only if
pattern?test yields
when applied to the expression
The construction
allows you to evaluate a condition involving pattern names to determine whether there is a match. The construction
instead applies a function
to the whole expression matched by
to determine whether there is a match. Using
instead of
sometimes leads to more succinct definitions.
With this definition, matches for
are tested with the function
Except[c] a pattern that matches any expression except c
Except[c,patt] a pattern that matches patt but not c
can take a pattern as an argument:
is in a sense a very general pattern: it matches
. In many situations you instead need to use
, which restricts to expressions matching
that do not match
When you use alternatives in patterns, you should make sure that the same set of names appear in each alternative. When a pattern like
matches an expression, there will always be a definite expression that corresponds to the object
. If you try to match a pattern like
, then there still will be definite expressions corresponding to
, but the unmatched one will be
Sequence[ ]
In some cases you may need to specify pattern sequences that are more intricate than things like
; for such situations you can use
PatternSequence[p[1],p[2],…] a sequence of arguments matching p[1],p[2],…
The empty sequence,
, is sometimes useful to specify an optional argument.
Although the Wolfram Language matches patterns in a purely structural fashion, its notion of structural equivalence is quite sophisticated. In particular, it takes account of properties such as
commutativity and associativity in functions like
This means, for example, that the Wolfram Language considers the expressions
equivalent for the purposes of pattern matching. As a result, a pattern like
can match not only
, but also
In this case, the expression has to be put in the form
in order to have the same structure as the pattern:
Whenever the Wolfram Language encounters an
function such as
in a pattern, it effectively tests all the possible orders of arguments to try and find a match. Sometimes, there may be several orderings that lead to matches. In such cases, the Wolfram Language
just uses the first ordering it finds. For example,
could match
or with
. The Wolfram Language tries the case
first, and so uses this match.
This can match either with
or with
. The Wolfram Language tries
first, and so uses this match:
As discussed in
, the Wolfram Language allows you to assign certain attributes to functions, which specify how those functions should be treated in evaluation and pattern matching. Functions can for example be
assigned the attribute
, which specifies that they should be treated as commutative or symmetric, and allows their arguments to be rearranged in trying to match patterns.
Orderless commutative function: f[b,c,a], etc., are equivalent to f[a,b,c]
Flat associative function: f[f[a],b], etc., are equivalent to f[a,b]
OneIdentity f[f[a]] , etc., are equivalent to a
Attributes[f] give the attributes assigned to f
SetAttributes[f,attr] add attr to the attributes of f
ClearAttributes[f,attr] remove attr from the attributes of f
In addition to being orderless, functions like
also have the property of being
. This means that you can effectively "parenthesize" their arguments in any way, so that, for example,
is equivalent to
, and so on.
The Wolfram Language takes account of flatness in matching patterns. As a result, a pattern like
can match
, with
If there are no other constraints, the Wolfram Language will match
to the first element of the sum:
The Wolfram Language can usually apply a transformation rule to a function only if the pattern in the rule covers all the arguments in the function. However, if you have a flat function, it is
sometimes possible to apply transformation rules even though not all the arguments are covered.
Functions like
are both flat and orderless. There are, however, some functions, such as
, which are flat, but not orderless.
This assigns the attribute
to the function
In an ordinary function that is not flat, a pattern such as
matches an individual argument of the function. But in a function
that is flat,
can match objects such as
which effectively correspond to a sequence of arguments. However, in the case where
matches a single argument in a flat function, the question comes up as to whether the object it matches is really just the argument
itself, or
. The Wolfram Language chooses the former possibility if the function carries the attribute
, otherwise it will first attempt to use the latter but fall back on the former.
The functions
, and
all have the attribute
, reflecting the fact that
is equivalent to
, and so on. However, in representing mathematical objects, it is often convenient to deal with flat functions that do not have the attribute
is a flat function, a pattern like
stands only for instances of the function with exactly two arguments. Sometimes you need to set up patterns that can allow any number of arguments.
You can do this using
multiple blanks
. While a single blank such as
stands for a single Wolfram Language expression, a double blank such as
stands for a sequence of one or more expressions.
"Double blanks"
stand for sequences of one or more expressions. "Triple blanks"
stand for sequences of zero or more expressions. You should be very careful whenever you use triple blank patterns. It is easy to make a mistake that can lead to an infinite loop. For example, if you
p[x_,y___]:=p[x] q[y]
, then typing in
will lead to an infinite loop, with
repeatedly matching a sequence with zero elements. Unless you are sure you want to include the case of zero elements, you should always use double blanks rather than triple blanks.
_ any single expression
x_ any single expression, to be named x
__ any sequence of one or more expressions
x__ sequence named x
x__h sequence of expressions, all of whose heads are h
___ any sequence of zero or more expressions
x___ sequence of zero or more expressions named x
x___h sequence of zero or more expressions, all of whose heads are h
Notice that with flat functions such as
, the Wolfram Language automatically handles variable numbers of arguments, so you do not explicitly need to use double or triple blanks, as discussed in
"Flat and Orderless Functions"
When you use multiple blanks, there are often several matches that are possible for a particular expression. By default, the Wolfram Language tries first those matches that assign the shortest
sequences of arguments to the first multiple blanks that appear in the pattern. You can change this order by wrapping
around parts of the pattern.
Longest[p] match the longest sequence consistent with the pattern p
Shortest[p] match the shortest sequence consistent with the pattern p
Many kinds of enumeration can be done by using
with various kinds of patterns:
Sometimes you may want to set up functions where certain arguments, if omitted, are given "default values". The pattern
stands for an object that can be omitted, and if so, will be replaced by the default value
This defines a function
with a required argument
, and optional arguments
, with default values
, respectively:
x_:v an expression which, if omitted, is taken to have default value v
x_h:v an expression with head h and default value v
x_. an expression with a built‐in default value
Some common Wolfram Language functions have built
in default values for their arguments. In such cases, you need not explicitly give the default value in
, but instead you can use the more convenient notation
in which a built
in default value is assumed.
is a flat function, a pattern such as
can match a sum with any number of terms. This pattern cannot, however, match a single term such as
. However, the pattern
contains an optional piece, and can match either an explicit sum of terms in which both
appear, or a single term
, with
taken to be
Using constructs such as
, you can easily construct single patterns that match expressions with several different structures. This is particularly useful when you want to match several mathematically equal forms that do not
have the same structure.
Standard Wolfram Language functions such as
have built
in default values for their arguments. You can also set up defaults for your own functions, as described in
Sometimes it is convenient not to assign a default value to an optional argument; such arguments can be specified with the help of
p|PatternSequence[] optional pattern p with no default value
When you define a complicated function, you will often want to let some of the arguments of the function be "optional". If you do not give those arguments explicitly, you want them to take on certain
"default" values.
in Wolfram Language functions use two basic methods for dealing with optional arguments. You can choose between the same two methods when you define your own functions in the Wolfram Language.
The first method is to have the meaning of each argument determined by its position, and then to allow one to drop arguments, replacing them by default values. Almost all built
in Wolfram Language functions that use this method drop arguments from the end. For example, the built
in function
allows you to drop the second argument, which is taken to have a default value of
f[x_,k_:kdef]:=value a typical definition for a function whose second argument is optional, with default value kdef
This defines a function with an optional second argument. When the second argument is omitted, it is taken to have the default value
The Wolfram Language assumes that arguments are dropped from the end. As a result
here gives the value of
, while
has its default value of
The second method that built
in Wolfram Language functions use for dealing with optional arguments is to give explicit names to the optional arguments, and then to allow their values to be given using transformation rules. This
method is particularly convenient for functions like
which have a very large number of optional parameters, only a few of which usually need to be set in any particular instance.
The typical arrangement is that values for "named" optional arguments can be specified by including the appropriate transformation rules at the end of the arguments to a particular function. Thus,
for example, the rule
, which specifies the setting for the named optional argument
, could appear as
When you set up named optional arguments for a function
, it is conventional to store the default values of these arguments as a list of transformation rules assigned to
f[x_,OptionsPattern[]]:=value a typical definition for a function with zero or more named optional arguments
OptionValue[name] the value of a named optional argument in the body of the function
Here is the definition for a function
which allows zero or more named optional arguments to be specified:
If you explicitly give a rule for
, it will override the default rules stored in
FilterRules[opts,Options[name]] the rules in opts used as options by the function name
the rules in opts not used as options by the function name
With no options given, the default options for
are used:
This changes the method used by
and the color in the plot:
expr.. a pattern or other expression repeated one or more times
expr... a pattern or other expression repeated zero or more times
Multiple blanks such as
allow you to give patterns in which sequences of arbitrary expressions can occur. The Wolfram Language pattern repetition operators
allow you to construct patterns in which particular forms can be repeated any number of times. Thus, for example,
represents any expression of the form
, and so on.
You can use
to represent repetitions of any pattern. If the pattern contains named parts, then each instance of these parts must be identical.
The pattern
can be extended to two arguments to control the number of repetitions more precisely.
a pattern or other expression repeated one or more times
Repeated[p,max] a pattern repeated up to max times
Repeated[p,{min,max}] a pattern repeated between min and max times
Repeated[p,{n}] a pattern repeated exactly n times
Verbatim[expr] an expression that must be matched verbatim
tells the Wolfram Language that only the exact expression
should be matched:
Using the objects described in "
Introduction to Patterns
", you can set up patterns for many kinds of expressions. In all cases, you must remember that the patterns must represent the structure of the expressions in the Wolfram Language internal form, as
shown by
Especially for some common kinds of expressions, the standard output format used by the Wolfram Language is not particularly close to the full internal form. But it is the internal form that you must
use in setting up patterns.
n_Integer an integer n
x_Real an approximate real number x
z_Complex a complex number z
Complex[x_,y_] a complex number x+iy
Complex[x_Integer,y_Integer] a complex number where both real and imaginary parts are integers
(r_Rational|r_Integer) rational number or integer r
Rational[n_,d_] a rational number
(x_/;NumberQ[x]&&Im[x]==0) a real number of any kind
(x_/;NumberQ[x]) a number of any kind
The fact that these expressions have different full forms means that you cannot use
to match a complex number:
The pattern here matches both ordinary integers, and complex numbers where both the real and imaginary parts are integers:
As discussed in
"Symbolic Computation"
, the Wolfram Language puts all algebraic expressions into a standard form, in which they are written essentially as a sum of products of powers. In addition, ratios are converted into products of
powers, with denominator terms having negative exponents, and differences are converted into sums with negated terms. To construct patterns for algebraic expressions, you must use this standard form.
This form often differs from the way the Wolfram Language prints out the algebraic expressions. But in all cases, you can find the full internal form using
x_+y_ a sum of two or more terms
x_+y_. a single term or a sum of terms
n_Integer x_ an expression with an explicit integer multiplier
a_.+b_.x_ a linear expression a+bx
x_^n_ x^n with n≠0,1
x_^n_. x^n with n≠0
a_.+b_.x_+c_.x_^2 a quadratic expression with nonzero linear term
x_List or x:{___} a list
x_List/;VectorQ[x] a vector containing no sublists
x_List/;VectorQ[x,NumberQ] a vector of numbers
x:{___List} or x:{{___}...} a list of lists
x_List/;MatrixQ[x] a matrix containing no sublists
x_List/;MatrixQ[x,NumberQ] a matrix of numbers
x:{{_,_}...} a list of pairs
This defines a function whose argument must be a list containing lists with either one or two elements:
Now that we have introduced the basic features of patterns in the Wolfram Language, we can use them to give a more or less complete example. We will show how you could define your own simple
integration function in the Wolfram Language.
From a mathematical point of view, the integration function is defined by a sequence of mathematical relations. By setting up transformation rules for patterns, you can implement these mathematical
relations quite directly in the Wolfram Language.
mathematical form Wolfram Language definition
( independent of ) integrate[c_y_,x_]:=c integrate[y,x]/;FreeQ[c,x]
, integrate[x_^n_.,x_]:=x^(n+1)/(n+1)/;FreeQ[n,x]&&n!=-1
The associativity of
makes the linearity relation work with any number of terms in the sum:
The Wolfram Language tests each term in each product to see whether it satisfies the
condition, and so can be pulled out:
This gives the standard formula for the integral of . By using the pattern
, rather than
, we include the case of :
Of course, the built
in integration function
(with a capital
) could have done the integral anyway:
Here is the rule for integrating the reciprocal of a linear function. The pattern
stands for any linear function of | {"url":"https://reference.wolfram.com/language/tutorial/Patterns#1615","timestamp":"2024-11-11T03:07:59Z","content_type":"text/html","content_length":"359013","record_id":"<urn:uuid:2fa1242a-92cd-4328-98bf-634253a8f712>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00556.warc.gz"} |
Theory of Combinatorial Algorithms
Mittagsseminar (in cooperation with J. Lengler, A. Steger, and D. Steurer)
Mittagsseminar Talk Information
Date and Time: Thursday, June 11, 2009, 12:15 pm
Duration: This information is not available in the database
Location: OAT S15/S16/S17
Speaker: Micha Sharir (Tel Aviv Univ., Israel)
Sharing joints, in moderation: A groundshaking clash between algebraic and combinatorial geometry
About half a year ago, Larry Guth and Nets Hawk Katz have obtained the tight upper bound $O(n^{3/2})$ on the number of joints in a set of $n$ lines in 3-space, where a joint is a point incident to at
least three non-coplanar lines, thus closing the lid on a problem that has been open for nearly 20 years. While this in itself is a significant development, the groundbreaking nature of their work is
the proof technique, which uses fairly simple tools from algebraic geometry, a totally new approach to combinatorial problems of this kind in discrete geometry.
In this talk I will (not have enough time to) present a simplified version of the new machinery, and the further results that we have so far obtained, by adapting and exploiting the algebraic
The first main new result is: Given a set $L$ of $n$ lines in space, and a subset of $m$ joints of $L$, the number of incidences between these joints and the lines of $L$ is $O(m^{1/3}n)$, which is
worst-case tight for $m\ge n$. In fact, this holds for any sets of $m$ points and $n$ lines, provided that each point is incident to at least three lines, and no plane contains more than $O(n)$
The second set of results is strongly related to the celebrated problem of Erdős on distinct distances in the plane. We reduce this problem to a problem involving incidences between points and
helices (or parabolas) in 3-space, and formulate some conjectures concerning the incidence bound. Settling these conjectures in the affirmative would have almost solved Erdős's problem. So far we
have several partial positive related results, interesting in their own right, which yield, among other results, that the number of distinct (mutually non-congruent) triangles determined by $s$
points in the plane is always $\Omega(s^2 / \log s)$, which is almost tight in the worst case, since the integer lattice yields an upper bound of $O(s^2)$.
Joint work with Haim Kaplan and (the late) György Elekes.
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The volume of a cone is 141.3 cubic inches. The height of the cone is 15 inches What is the radius of the cone, rounded to the nearest inch? | Socratic
The volume of a cone is 141.3 cubic inches. The height of the cone is 15 inches What is the radius of the cone, rounded to the nearest inch?
1 Answer
See a solution process below:
The formula for the Volume of a cone is:
$V = \pi {r}^{2} \frac{h}{3}$
$V$ is the Volume of the cone: $141.3 {\text{in}}^{3}$ for this problem.
$r$ is the radius of the cone: what we are solving for in this problem.
$h$ is the height of the cone: $15 \text{in}$ for this problem.
Substituting and solving for $r$ gives:
#141.3"in"^3 = pi xx r^2 xx (15"in")/3#
$141.3 \text{in"^3 = pi xx r^2 xx 5"in}$
$\left(141.3 \text{in"^3)/color(red)(5"in") = (pi xx r^2 xx 5"in")/color(red)(5"in}\right)$
#(141.3"in"^(color(red)(cancel(color(black)(3)))2))/color(red)(5color(black)(cancel(color(red)("in")))) = (pi xx r^2 xx color(red)(cancel(color(black)(5"in"))))/cancel(color(red)(5"in"))#
$\frac{141.3 {\text{in}}^{2}}{\textcolor{red}{5}} = \pi {r}^{2}$
$28.26 {\text{in}}^{2} = \pi {r}^{2}$
$\frac{28.26 {\text{in}}^{2}}{\textcolor{red}{\pi}} = \frac{\pi {r}^{2}}{\textcolor{red}{\pi}}$
$\frac{28.26 {\text{in}}^{2}}{\textcolor{red}{\pi}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\pi}}} {r}^{2}}{\cancel{\textcolor{red}{\pi}}}$
$\frac{28.26 {\text{in}}^{2}}{\textcolor{red}{\pi}} = {r}^{2}$
We can use 3.1416 to estimate $\pi$ giving:
$\frac{28.26 {\text{in}}^{2}}{\textcolor{red}{3.1416}} = {r}^{2}$
$9 {\text{in}}^{2} = {r}^{2}$ rounded to the nearest inch.
Now, take the square root of each side of the equation to find the radius of the cone while keeping the equation balanced:
$\sqrt{9 {\text{in}}^{2}} = \sqrt{{r}^{2}}$
$3 \text{in} = r$
$r = 3 \text{in}$
The radius of the cone rounded to the nearest inch is 3 inches.
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Interpreting Confidence Intervals (Q2a–b)
Here’s another typical estimation question, but now we’re also asked to interpret our results. Getting this interpretation correct can be tricky – t6he wording has to be just right to avoid making a
common mistake (which they WILL be looking for on the exam).
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Safeguard Your Network in a Post-Quantum World
Security is critical when transmitting information over any untrusted medium, particularly with the internet. Cryptography is typically used to protect information over a public channel between two
entities. However, there is an imminent threat to existing cryptography with the advent of quantum computers. According to the National Institute of Standards and Technology (NIST), “When quantum
computers are a reality, our current public key cryptography won’t work anymore… So, we need to start designing now what those replacements will be.”
Quantum computing threat
A quantum computer works with qubits, which can exist in multiple states simultaneously, based on the quantum mechanical principle of superposition. Thus, a quantum computer could explore many
possible permutations and combinations for a computational task, simultaneously and swiftly, transcending the limits of classical computing.
While a sufficiently large and commercially feasible quantum computer has yet to be built, there have been massive investments in quantum computing from many corporations, governments, and
universities. Quantum computers will empower compelling innovations in areas such as AI/ML and financial and climate modeling. Quantum computers, however, will also give bad actors the ability to
break current cryptography.
Public-key cryptography is ubiquitous in modern information security applications such as IPsec, MACsec, and digital signatures. The current public-key cryptography algorithms are based on
mathematical problems, such as the factorization of large numbers, which are daunting for classical computers to solve. Shor’s algorithm provides a way for quantum computers to solve these
mathematical problems much faster than classical computers. Once a sufficiently large quantum computer is built, existing public-key cryptography (such as RSA, Diffie-Hellman, ECC, and others) will
no longer be secure, which will render most current uses of cryptography vulnerable to attacks.
Store now, break later
Why worry now? Most of the transport security protocols like IPsec and MACsec use public-key cryptography during the authentication/key establishment phase to derive the session key. This shared
session key is then used for symmetric encryption and decryption of the actual traffic.
Bad actors can use the “harvest now, decrypt later” approach to capture encrypted data right now and decrypt it later, when a capable quantum computer materializes. It is an unacceptable risk to
leave sensitive encrypted data susceptible to impending quantum threats. In particular, if there is a need to maintain forward secrecy of the communication beyond a decade, we must act now to make
these transport security protocols quantum-safe.
The long-term solution is to adopt post-quantum cryptography (PQC) algorithms to replace the current algorithms that are susceptible to quantum computers. NIST has identified some candidate
algorithms for standardization. Once the algorithms are finalized, they must be implemented by the vendors to start the migration. While actively working to provide PQC-based solutions, Cisco already
has quantum-safe cryptography solutions that can be deployed now to safeguard the transport security protocols.
Cisco’s solution
Cisco has introduced the Cisco session key import protocol (SKIP), which enables a Cisco router to securely import a post-quantum pre-shared key (PPK) from an external key source such as a quantum
key distribution (QKD) device or other source of key material.
Figure 1. External QKD as key source using Cisco SKIP
For deployments that can use an external hardware-based key source, SKIP can be used to derive the session keys on both the routers establishing the MACsec connection (see Figure 1).
With this solution, Cisco offers many benefits to customers, including:
• Secure, lightweight protocol that is part of the network operating system (NOS) and does not require customers to run any additional applications
• Support for “bring your own key” (BYOK) model, enabling customers to integrate their key sources with Cisco routers
• The channel between the router and key source used by SKIP is also quantum-safe, as it uses TLS 1.2 with DHE-PSK cipher suite
• Validated with several key-provider partners and end customers
Figure 2. Cisco SKS engine as the key source
In addition to SKIP, Cisco has introduced the session key device (SKS), which is a unique solution that enables routers to derive session keys without having to use an external key source.
Figure 3. Traditional session key distribution
The SKS engine is part of the Cisco IOS XR operating system (see Figure 2). Routers establishing a secure connection like MACsec will derive the session keys directly from their respective SKS
engines. The engines are seeded with a one-time, out-of-band operation to make sure they derive the same session keys.
Unlike the traditional method (see Figure 3), where the session keys are exchanged on the wire, only the key identifiers are sent on the wire with quantum key distribution. So, any attacker tapping
the links will not be able to derive the session keys, as having just the key identifier is not sufficient (see Figure 4).
Figure 4. Quantum session key distribution
Cisco is leading the way with comprehensive and innovative quantum-safe cryptography solutions that are ready to deploy today.
Watch this Cisco Knowledge Networking (CKN) webinar
and discover how Cisco can help protect your network.
Cisco Portfolio Explorer
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Form 3 Maths Free Notes - Citizen News
Form 3 Maths Free Notes
Read all the form 3 notes here. You can also download a copy of the pdf notes on this link; MATH FORM THREE NOTES
See also;
Free Mathematics notes, revision questions, KCSE past Papers, Exams, Marking Schemes, Topical revision materials, Syllabus and Many more
form 3 notes for all subjects; free pdf & word downloads
CHAPTER FOURTY FOUR
Specific Objectives
By the end of the topic the learner should be able to:
(a) Factorize quadratic expressions;
(b) Identify perfect squares;
(c) Complete the square;
(d) Solving quadratic equations by completing the square;
(e) Derive the quadratic formula;
(f) Solve quadratic equations using the formula;
(g) Form and solve quadratic equations from roots and given situations;
(h) Make tables of values from a quadratic relation;
(i) Draw the graph of a quadratic relation;
(j) Solve quadratic equations using graphs;
(k) Solve simultaneous equations (one linear and one quadratic) analytically and graphically;
(1) Apply the knowledge of quadratic equations to real life situations.
(a) Factorization of quadratic expressions
(b) Perfect squares
(c) Completion of the squares
(d) Solution of quadratic equations by completing the square
(e) Quadratic formula x = -b ±
(f) Solution of quadratic equations using the formula.
(g) Formation of quadratic equations and solving them
(h) Tables of values for a given quadratic relation
(i) Graphs of quadratic equations
(j) Simultaneous equation – one linear and one quadratic
(k) Application of quadratic equation to real life situation.
Perfect square
Expressions which can be factorized into two equal factors are called perfect squares.
Completing the square
Any quadratic expression can be simplified and written in the form where a, b and c are constant and a is not equal to zero. We use the expression to make a perfect square.
We are first going to look for expression where coefficient of x = 1
What must be added to + 10 x to make it a perfect square?
• Let the number to be added be a constant c.
• Then + 10x + c is a perfect square.
• Using
• (10 /2 = c
• C = 25 (25 must be added)
What must be added to + _ + 36 to make it a perfect square
• Let the term to be added be bx where b is a constant
• Then + bx +36 is a perfect square.
• Using
• b =12 x or -12 x
We will now consider the situations where a eg
In the above you will notice that ac . We use this expression to make perfect squares where a is not one and its not zero.
What must be added to + _ + 9 to make it a perfect square?
• Let the term to be added be bx.
• Then, + bx + 9 is a perfect square.
• .
• The term to be added is thus .
What must be added to _ – 40x + 25 to make it a perfect square?
• Let the term to be added be a
• Then – 40x + 25 is a perfect square.
• Using
Solutions of quadratic equations by completing the square methods
Solve + 5x+ 1 = 0 by completing the square.
+ 5x+ 1 = 0 Write original equation.
+ 5x = -1 Write the left side in the form + bx.
+ 10x + ( = ( Add to both sides
+ 10x + =
= Take square roots of each side and factorize the left side
= Solve for x.
Therefore x = – 0.2085 or 4.792
Cannot be solved by factorization.
Solve + 4x+ 1 = 0 by completing the square
+ 4x =-1 make cooeffiecient of one by dividing both sides by 2
+ 2x = -1/2
+ 2x + 1 = – + 1
Adding 1 to complete the square on the LHS
The quadratic formula
Using quadratic formula solve
Comparing this equation to the general equation we get;a =2 b =-5 c =-5
Substituting in the quadratic formulae
X =
X = 3 or –
Formation of quadratic equations
Peter travels to his uncle’s home,30 km away from his place. He travels for two thirds of the journey before the bicycle developed mechanical problems an he had to push it for the rest of the
journey. If his cycling speed is 10 km\h faster than his walking speed and he completes the journey in 3 hours 30 minutes, determine his cycling speed .
Let Peters cycling speed be x km\ h , then his walking speed is (x-10 ) km/h.
Time taken in cycling
Time taken in walking = (30 – 20) ( x -10 )
Total time h
60(x-10) + 30 (x) = 10(x) (x-10)
– 190x + 600 = 0
– 19x + 60 = 0
If his cycling speed is 4 km/h , then his walking speed is (4 -10 ) km/h, which gives – 6 km/h.Thus,
4 is not a realistic answer to this situation.therefore his cycling speed is 15 km/h.
A posite two digit number is such that the product of the digit is 24.When the digits are reversed , the number formed is greater than than the original number by 18. Find the number
Let the ones digit of the number be y and the tens digit be x,
Then , xy = 24…………..1
When the number is reversed, the ones digit is x and the tens digit is y.
(10y + x) – (10x +y) = 18
9y- 9x = 18
Substituting 2 in equation 1 gives;
Since the required number is positive x =4 and y = 4 + 2 =6
Therefore the number is 46
Graphs of quadratic functions
A quadratic function has the form y = ax2 + bx + c where a ≠ 0. The graph of a quadratic function isU-shaped and is called a parabola. For instance, the graphs of y = and y = e
Shown below. The origin (0, 0) is the lowest point on the graph of y = and the highest point on the graph of y = . The lowest or highest point on the graph of a quadratic function is called the
The graphs of y = and y = are symmetric about the y-axis, called the axis of symmetry. In general, the axis of symmetry for the graph of a quadratic function is the vertical line through the
The graph of y = and y = or .
Draw the graph of y =
Make a table showing corresponding value of x and y.
X -1 0 1 2 3
Y – 8 -1 2 1 -4
Note ; To get the values replace the value of x in the equation to get the corresponding value of x
1. g y = -2 ( -1
y = -2 ( 0
Draw the graph of y =
x 0 1 2 3 5 7
y 2 -4 -8 -10 -8 2
Graphical solutions of simultaneous equations
We should consider simultaneous equation one of which is linear and the other one is quadratic.
Solve the following simultaneous equations graphically:
Corresponding values of x and y
x -2 -1 0 1 2 3 4 x
y 9 4 1 0 1 4 9 y
We use the table to draw the graph as shown below, on the same axis the line y = 5-2x is drawn. Points where the line y =5 -2x and the curve intersect give the solution. The points are (- 2, 9) and
(2,1).Therefore , when x = -2, y = 9 and when x = 2, y= 1
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. The table shows the height metres of an object thrown vertically upwards varies with the time t seconds
The relationship between s and t is represented by the equations s = at^2 + bt + 10 where b are constants.
t 0 1 2 3 4 5 6 7 8 9 10
s 45.1
• (i) Using the information in the table, determine the values of a and b
(2 marks)
(ii) Complete the table (1 mark)
(b)(i) Draw a graph to represent the relationship between s and t (3 marks)
(ii) Using the graph determine the velocity of the object when t = 5 seconds
2. (a) Construct a table of value for the function y = x^2 – x – 6 for -3≤ x ≤ 4
(b) On the graph paper draw the graph of the function
Y=x^2 – x – 6 for -3 ≤ x ≤4
(c) By drawing a suitable line on the same grid estimate the roots of the equation x^2 + 2x – 2 =0
3. (a) Draw the graph of y= 6+x-x^2, taking integral value of x in -4 ≤ x ≤ 5. (The
grid is provided. Using the same axes draw the graph of y = 2 – 2x
(b) From your graphs, find the values of X which satisfy the simultaneous
equations y = 6 + x – x^2
y = 2 – 2x
(c) Write down and simplify a quadratic equation which is satisfied by the
values of x where the two graphs intersect.
4. (a) Complete the following table for the equation y = x^3 – 5x^2 + 2x + 9
x -2 -1.5 -1 0 1 2 3 4 5
x^2 -3.4 -1 0 1 27 64 125
-5x^2 -20 -11.3 -5 0 -1 -20 -45
2x -4 -3 0 2 4 6 8 10
-8.7 9 7 -3
(b) On the grid provided draw the graph of y = x^3 – 5x^2 + 2x + 9 for -2 ≤ x ≤ 5
(c) Using the graph estimate the root of the equation x^3 – 5x^2 + 2 + 9 = 0 between x =
2 and x = 3
(d) Using the same axes draw the graph of y = 4 – 4x and estimate a solution to the
equation x^2 – 5x^2 + 6x + 5 =0
5. (a) Complete the table below, for function y = 2x^2 + 4x -3
x -4 -3 -2 -1 0 1 2
2x^2 32 8 2 0 2
4x – 3 -11 -3 5
y -3 3 13
(b) On the grid provided, draw the graph of the function y=2x^2 + 4x -3 for
-4 ≤ x ≤ 2 and use the graph to estimate the rots of the equation 2x^2+4x – 3 = 0 to 1 decimal place. (2mks)
(c) In order to solve graphically the equation 2x^2 +x -5 =0, a straight line must be drawn to intersect the curve y = 2x^2 + 4x – 3. Determine the equation of this straight line, draw the
straight line hence obtain the roots.
2x^2 + x – 5 to 1 decimal place.
6. (a) (i) Complete the table below for the function y = x^3 + x^2 – 2x (2mks)
x -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 2 2.5
x^3 15.63 -0.13 1
x^2 4 0.25 6.25
-2x 1 -2
y 1.87 0.63 16.88
(ii) On the grid provided, draw the graph of y = x^3 + x^2 – 2x for the values of x in the interval – 3 ≤ x ≤ 2.5
(iii) State the range of negative values of x for which y is also negative
(b) Find the coordinates of two points on the curve other than (0, 0) at which x- coordinate and y- coordinate are equal
7. The table shows some corresponding values of x and y for the curve represented by Y = ¼ x3 -2
X -3 -2 -1 0 1 2 3
Y -8.8 -4 -2.3 -2 -1.8 0 4.8
On the grid provided below, draw the graph of y = ¼ x^2 -2 for -3 ≤ x ≤3. Use the graph to estimate the value of x when y = 2
8. A retailer planned to buy some computers form a wholesaler for a total of Kshs 1,800,000. Before the retailer could buy the computers the price per unit was reduced by Kshs 4,000. This reduction
in price enabled the retailer to buy five more computers using the same amount of money as originally planned.
(a) Determine the number of computers the retailer bought
(b) Two of the computers purchased got damaged while in store, the rest were sold and the retailer made a 15% profit Calculate the profit made by the retailer on each computer sold
9. The figure below is a sketch of the graph of the quadratic function y = k
( x+1) (x-2)
Find the value of k
10. (a) Draw the graph of y= x^2 – 2x + 1 for values -2 ≤ x ≤ 4
(b) Use the graph to solve the equations x^2 – 4= 0 and line y = 2x +5
11. (a) Draw the graph of y = x^3 + x^2 – 2x for -3≤ x ≤ 3 take scale of 2cm to
represent 5 units as the horizontal axis
(b) Use the graph to solve x^3 + x ^2 – 6 -4 = 0 by drawing a suitable linear graph on the same axes.
12. Solve graphically the simultaneous equations 3x – 2y = 5 and 5x + y = 17
Specific Objectives
By the end of the topic the learner should be able to:
(a) Perform various computations using a calculator;
(b) Make reasonable approximations and estimations of quantities incomputations and measurements;
(c) Express values to a given number of significant figures;
(d) Define absolute, relative, percentage, round-off and truncation errors;
(e) Determine possible errors made from computations;
(f) Find maximum and minimum errors from operations.
(a) Computing using calculators
(b) Estimations and approximations
(c) Significant figures
(d) Absolute, relative, percentage, round-off (including significant figures)and truncation errors
(e) Propagation of errors from simple calculations
(f) Maximum and minimum errors.
Approximation involves rounding off and truncating numbers to give an estimation
Rounding off
In rounding off the place value to which a number is to be rounded off must be stated. The digit occupying the next lower place value is considered. The number is rounded up if the digit is greater
or equal to 5 and rounded down if it’s less than 5.
Round off 395.184 to:
1. The nearest hundreds
2. Four significant figures
3. The nearest whole number
4. Two decimal places
1. 400
2. 395 .2
3. 395
4. 395.18
Truncating means cutting off numbers to the given decimal places or significant figures, ignoring the rest.
Truncate 3.2465 to
1. 3 decimal places
2. 3 significant figures
1. 3.246
2. 3.24
Estimation involves rounding off numbers in order to carry out a calculation faster to get an approximate answer .This acts as a useful check on the actual answer.
Estimate the answer to
The answer should be close to
The exact answer is 1277.75. 1277.75 writen to 2 significant figures is 1300 which is close to the estimated answer.
Absolute error
The absolute error of a stated measurement is half of the least unit of measurement used. When a measurement is stated as 3.6 cm to the nearest millimeter ,it lies between 3.55 cm and 3.65 cm.The
least unit of measurement is milliliter, or 0.1 cm.The greatest possible error is 3.55 – 3.6 = -0.05 or 3.65 – 3.6 = + 0.05.
To get the absolute error we ignore the sign. So the absolute error is 0.05 thus,|-0.05| =| +0.05|= 0.05.When a measurement is stated as 2.348 cm to the nearest thousandths of a centimeters (0.001)
then the absolute error is .
Relative error
Relative error =
An error of 0.5 kg was found when measuring the mass of a bull.if the actual mass of the bull was found to be 200kg.Find th relative error
Relative error =
Percentage error
Percentage error = relative error x 100%
The thickness of a coin is 0.20 cm.
1. The percentage error
2. What would be the percentage error if the thickness was stated as 0.2 cm ?
The smallest unit of measurement is 0.01
Absolute error
Percentage error
The smallest unit of measurement is 0.1
Absolute error
Percentage error
= 25 %
Rounding off and truncating errors
An error found when a number is rounded off to the desired number of decimal places or significant figures, for example when a recurring decimal 1. is rounded to the 2 significant figures, it
becames 1.7 the rounde off error is;
1.7 -1.
1.6 converted to a fraction .
Truncating error
The error introduced due to truncating is called a truncation error.in the case of 1.6 truncated to 2 S.F., the truncated error is; |1.6 -1. | =
Propagation of errors
Addition and subtraction
What is the error in the sum of 4.5 cm and 6.1 cm, if each represent a measure measurement.
The limits within which the measurements lie are 4.45, i.e. ., 4.55 or and 6.05 to 6.15, i.e. 6.1 .
The maximum possible sum is 4.55 10.7cm
The minimum possible sum is 4.45 10.5 cm
The working sum is 4.5 + 6.1 = 10.6
The absolute error = maximum sum – working sum
=| 10.7 – 10.6 |
What is the error in the difference between the measurements 0.72 g and 0.31 g?
The measurement lie within and respectively the maximum possible difference will be obtained if we substract the minimum value of the second measurement from the maximum value of the first, i.e ;
0.725 – 0.305 cm
The minimum possible difference is 0.715 – 0.315 = 0.400.the working difference is 0.72 – 0.31 =0.41 , which has an absolute error of |0.420 -0.41| or |0.400 – 0.41| = 0.10. Since our working
difference is 0.41, we give the absolute error as 0.01 (to 2 s.f)
In both addition and subtraction, the absolute error in the answer is equal to the sum of the absolute errors in the original measurements.
A rectangular card measures 5.3 cm by 2.5 cm. find
1. The absolute error in the rea of the card
2. The relative error in the area of the cord
• The length lies within the limits
• The length lies within the limits
The maximum possible area is 2.55 x 5.35 =13.6425
The minimum possible area is 2.45 x 5.25 =12.8625
The working area is 5.3 x 2.5 = 13.25
Maximum area – working area = 13.6425 – 1325 = 0.3925.
Working area minimum area = 13.25 – 12.8625 = 0.3875
We take the absolute error as the average of the two.
Thus, absolute error
= 0.3900
The same can also be found by taking half the interval between the maximum area and the minimum area
The relative error in the area is :
Given 8.6 cm .Find:
1. The absolute error in the quotient
2. The relative error in the quotient
1. 8.6 cm has limits 8.55 cm and 8.65 cm. 3.4 has limits 3.35 cm and 3.45 cm.The maximum possible quotient will be given by the maximum possible value of the numerator and the smallest possible
value of the denominator, i.e.,
= 2.58 (to 3 s.f)
The minimum possible quotient will be given by the minimum possible value of the numerator ad the biggest possible value of the denominator, i.e.
= 2.48 (to 3 s.f)
The working quotient is; = 2.53 (to 3 .f.)
The absolute error in the quotient is;
X 0.10
1. Relative error in the working quotient ;
= 0.0197
= 0.020 (to 2 s.f )
Relative error in the numerator is
Relative error in the denominator is
Sum of the relative errors in the numerator and denominator is
0.00581 + 0.0147 = 0.02051s
=0.021 to 2 S.F
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. (a) Work out the exact value of R = 1_________
0.003146 – 0.003130
(b) An approximate value of R may be obtained by first correcting each of the decimal in the denominator to 5 decimal places
(ii) The error introduced by the approximation
2. The radius of circle is given as 2.8 cm to 2 significant figures
• If C is the circumference of the circle, determine the limits between which ^C/[π ]lies
• By taking ∏ to be 3.142, find, to 4 significant figures the line between which the circumference lies.
3. The length and breadth of a rectangular floor were measured and found to be 4.1 m and 2.2 m respectively. If possible error of 0.01 m was made in each of the measurements, find the:
• Maximum and minimum possible area of the floor
• Maximum possible wastage in carpet ordered to cover the whole floor
4. In this question Mathematical Tables should not be used
The base and perpendicular height of a triangle measured to the nearest centimeter
are 6 cm and 4 cm respectively.
(a) The absolute error in calculating the area of the triangle
(b) The percentage error in the area, giving the answer to 1 decimal place
5. By correcting each number to one significant figure, approximate the value of 788 x 0.006. Hence calculate the percentage error arising from this approximation.
6. A rectangular block has a square base whose side is exactly 8 cm. Its height measured to the nearest millimeter is 3.1 cm
Find in cubic centimeters, the greatest possible error in calculating its volume.
7. Find the limits within the area of a parallegram whose base is 8cm and height is 5 cm lies. Hence find the relative error in the area
8. Find the minimum possible perimeter of a regular pentagon whose side is 15.0cm.
9. Given the number 0.237
(i) Round off to two significant figures and find the round off error
(ii) Truncate to two significant figures and find the truncation error
10. The measurements a = 6.3, b= 15.8, c= 14.2 and d= 0.00173 have maximum possible errors of 1%, 2%, 3% and 4% respectively. Find the maximum possible percentage error in ^ad/[bc] correct to 1sf.
CHAPTER FOURTY THREE
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define and draw the unit circle;
(b) Use the unit circle to find trigonometric ratios in terms of co-ordinates of points for 0 < 9 < 360°;
(c) Find trigonometric ratios of negative angles;
(d) Find trigonometric ratios of angles greater than 360° using the unit circle;
(e) Use mathematical tables and calculators to find trigonometric ratios of angles in the range 0 < 9 < 360°;
(f) Define radian measure;
(g) Draw graphs of trigonometric functions; y = sin x, y = cos x and y ~ tan x using degrees and radians;
(h) Derive the sine rule;
(i) Derive the cosine rule;
(j) Apply the sine and cosine rule to solve triangles (sides, angles and area),
(k) Apply the knowledge of sine and cosine rules in real life situations.
(a) The unit circles
(b) Trigonometric rations from the unit circle
(c) Trigonometric ratios of angles greater than 360° and negative angles
(d) Use of trigonometric tables and calculations
(e) Radian measure
(f) Simple trigonometric graphs
(g) Derivation of sine and cosine rule
(h) Solution of triangles
(i) Application of sine and cosine rule to real situation.
The unit circle
It is circle of unit radius and centre O (0, 0).
An angle measured anticlockwise from positive direction of x – axis is positive. While an angle measured clockwise from negative direction of x – axis is negative.
In general, on a unit circle
Trigonometric ratios of negative angles
In general
Use of calculators
Use a calculator to find
1. Tan
• Key in tan
• Key in 30
• Screen displays 0.5773502
• Therefore tan = 0.5774
To find the inverse of sine cosine and tangent
• Key in shift
• Then either sine cosine or tangent
• Key in the number
Always consult the manual for your calculator. Because calculators work differently
One radian is the measure of an angle subtended at the centre by an arc equal in length to the radius of the circle.
Because the circumference of a circle is 2πr, there are 2π radians in a full circle. Degree measure and radian measure are therefore related by the equation 360° = 2π radians, or 180° = π radians.
The diagram shows equivalent radian and degree measures for special angles from 0° to 360° (0 radians to 2π radians).You may find it helpful to memorize the equivalent degree and radian measures of
special angles in the first quadrant. All other special angles are just multiples of these angles.
Convert into radians
If = 57.29
Therefore = = 2.182 to 4 S.F
Convert the following degrees to radians, giving your answer in terms
What is the length of the arc that that subtends an angle of 0.6 radians at the centre of a circle of radius 20 cm.
Simple trigonometric graphs
Graphs of y=sin x
The graphs can be drawn by choosing a suitable value of x and plotting the values of y against theCorresponding values of x.
The black portion of the graph represents one period of the function and is called one cycle of the sine curve.
Sketch the graph of y = 2 sin x on the interval [– , 4 ].
Note that y = 2 sin x = 2(sin x) indicates that the y-values for the key points will have twice the magnitude of those on the graph of y = sin x.
To get the values of y substitute the values of x in the equation y =2sin x as follows
y=2 sin (360) because 2 is equal to 36
• You can change the radians into degrees to make work simpler.
• By connecting these key points with a smooth curve and extending the curve in both directions over the interval [– , 4 ], you obtain the graph shown in below.
Sketch the graph of y = cos x for using an interval of
The values of x and the corresponding values of y are given in the table below
Y=cos x 1 0.8660 0.5 0 -0.5 -0.8660 -1 -0.8660 -0.5
Graph of tangents
• As the value of x approaches and 27 tan x becames very large
• Hence the graph of y =tan x approaches the lines x = without touching them.
• Such lines are called asymptotes
Solution of triangles
Sin rule
If a circle of radius R is circumscribed around the triangle ABC ,then =2R.
The sine rule applies to both acute and obtuse –angled triangle.
Solve triangle ABC, given that CAB =42. , c= 14.6 cm and a =11.4 cm
To solve a triangle means to find the sides and angles not given
Sin c = = 0.8720
Therefore c =60.6
The sin rule is used when we know
• Two sides and a non-included angle of a triangle
• All sides and at least one angle
• All angles and at least one side.
Cosine rule
Find AC in the figure below, if AB= 4 cm , BC = 6 cm and ABC =7
Using the cosine rule
= 16 + 36 – 48
= 52 – 9.979
= 42.02 cm
The cosine rule is used when we know
• Two sides and an included angle
• All three sides of a triangle
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. Solve the equation
Sin 5 θ = –1 for 0^0 ≤ 0 ≤ 180^0
2. Given that sin θ = ^2/[3] and is an acute angle find:
□ Tan θ giving your answer in surd form
□ Sec^2 θ
3. Solve the 1
equation 2 sin^2(x-30^0) = cos 60^0 for – 180^0 ≤ x ≤ 180^0
4. Given that sin (x + 30)^0 = cos 2x^0for 0^0, 0^0 ≤ x ≤90^0 find the value of x. Hence find the value of cos ^23x^0.
5. Given that sin a =1 where a is an acute angle find, without using
Mathematical tables
(a) Cos a in the form of a√b, where a and b are rational numbers
(b) Tan (90^0 – a).
6. Give that x^o is an angle in the first quadrant such that 8 sin^2 x + 2 cos x -5=0
1. a) Cos x
2. b) tan x
3. Given that Cos 2x^0 = 0.8070, find x when 0^0 ≤ x ≤ 360^0
8 The figure below shows a quadrilateral ABCD in which AB = 8 cm, DC = 12 cm, < BAD = 45^0, < CBD = 90^0 and BCD = 30^0.
(a) The length of BD
(b) The size of the angle ADB
9. The diagram below represents a school gate with double shutters. The shutters are such opened through an angle of 63^0.
The edges of the gate, PQ and RS are each 1.8 m
Calculate the shortest distance QS, correct to 4 significant figures
10…The figure below represents a quadrilateral piece of land ABCD divided into three triangular plots. The lengths BE and CD are 100m and 80m respectively. Angle ABE = 30^0ÐACE = 45^0 and Ð ACD = 100
(a) Find to four significant figures:
(i) The length of AE
(ii) The length of AD
(iii) The perimeter of the piece of land
(b) The plots are to be fenced with five strands of barbed wire leaving an entrance of 2.8 m wide to each plot. The type of barbed wire to be used is sold in rolls of lengths 480m. Calculate the
number of rolls of barbed wire that must be bought to complete the fencing of the plots.
11. Given that x is an acute angle and cos x = 2Ö 5, find without using mathematical
tables or a calculator, tan ( 90 – x)^0.
12. In the figure below ÐA = 62^0, ÐB = 41^0, BC = 8.4 cm and CN is the bisector of ÐACB.
Calculate the length of CN to 1 decimal place.
13. In the diagram below PA represents an electricity post of height 9.6 m. BB and RC represents two storey buildings of heights 15.4 m and 33.4 m respectively. The angle of depression of A from B is
5.5^0 While the angle of elevation of C from B is 30.5^0 and BC = 35m.
(a) Calculate, to the nearest metre, the distance AB
(b) By scale drawing find,
(i) The distance AC in metres
(ii) Ð BCA and hence determine the angle of depression of A from C
More questions
1. Solve the equation: (2 mks)
2. (a) Complete the table below, leaving all your values correct to 2 d.p. for the functions y = cos x and y = 2cos (x + 30)^0 (2 mks)
X^0 0^0 60^0 120^0 180^0 240^0 300^0 360^0 420^0 480^0 540^0
cosX 1.00 -1.00 0.50
2cos(x+30) 1.73 -1.73 0.00
(b) For the function y = 2cos(x+30)^0
• The period (1 mk)
• Phase angle (1 mk)
(c) On the same axes draw the waves of the functions y = cos x and y = 2cos(x+30)^0 for . Use the scale 1cm rep 30^0 horizontally and 2 cm rep 1 unit
vertically (4 mks)
(d) Use your graph above to solve the inequality (2 mks)
3. Find the value of x in the equation.
Cos(3x – 180^o) = √3 in the range O^o < x < 180^o (3 marks)
4. Given that and ө is an acute angle, find without using tables cos (90 –ө)
5. Solve for ө if -¼ sin (2x + 30) = 0.1607, 0 ≤ө≥ 360^0 (3mks)
6. Given that Cos q = ^5/[13] and that 270^0£q£ 360^0 , work out the value of Tan q + Sin q without using a calculator or mathematical tables.
(3 marks)
7. Solve for x in the range 0^0£ x £ 180^0 (4mks)
-8 sin^2x – 2 cos x = -5.
8. If tan x^o = ^12/[5 ]and x is a reflex angle, find the value of 5sin x + cos x without using a
calculator or mathematical tables
9. Find q given that 2 cos 3q -1 = 0 for 0^o £q£ 360^o
10. Without a mathematical table or a calculator, simplify: Cos300^o x Sin120^ogiving your answer in
Cos330^o – Sin 405^orationalized surd form.
11. Express in surds form and rationalize the denominator.
Sin 60^o Sin 45^o – Sin 45^o
12. Simplify the following without using tables;
Tan 45 + cos 45sin 60
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define rational and irrational numbers,
(b) Simplify expressions with surds;
(c) Rationalize denominators with surds.
(a) Rational and irrational numbers
(b) Simplification of surds
(c) Rationalization of denominators.
Rational and irrational numbers
Rational numbers
A rational number is a number which can be written in the form , where p and q are integers and q .The integer’s p and q must not have common factors other than 1.
Numbers such as 2, are examples of rational numbers .Recurring numbers are also rational numbers.
Irrational numbers
Numbers that cannot be written in the form .Numbers such as are irrational numbers.
Numbers which have got no exact square roots or cube root are called surds e.g. , ,
The product of a surd and a rational number is called a mixed surd. Examples are ;
, and
Order of surds
Simplification of surds
A surd can be reduced to its lowest term possible, as follows ;
Operation of surds
Surds can be added or subtracted only if they are like surds (that is, if they have the same value under the root sign).
Example 1
Simplify the following.
1. 3 √2 + 5√2
2. 8 √5 − 2√5
1. 3 √2 + 5√2 = 8 √2
2. 8 √5 − 2√5 = 6√5
Let a =
Therefore = a + a
=2 a
But a =
Hence =
Multiplication and Division of surds
Surds of the same order can be multiplied or divided irrespective of the number under the root sign.
Law 1: √a x √b = √ab When multiplying surds together, multiply their values together.
e.g.1 √3 x √12 = √ (3 x 12) = √36 = 6
e.g.2 √7 x √5 = √35
This law can be used in reverse to simplify expressions…
e.g.3 √12 = √2 x √6 or √4 x √3 = 2√3
Law 2:√a ÷ √b or = √(a/b)^ When dividing surds, divide their values (and vice versa).
e.g.1 √12 = √(12 ÷ 3) = √4 = 2
Law 3: √ (a^2) or (√a)^ 2 = a When squaring a square-root, (or vice versa), the symbols cancel
Each other out, leaving just the base.
e.g.1 √12^2 = 12
e.g.2 √7 x √7 = √7^2 = 7
If you add the same surds together you just have that number of surds. E.g.
√2 + √2 + √2= 3√2
If a surd has a square number as a factor you can use law 1 and/or law 2 and work backwards to take that out and simplify the surd. E.g. √500 = √100 x √5 = 10√5
Rationalization of surds
Surds may also appear in fractions. Rationalizing the denominator of such a fraction means finding an equivalent fraction that does NOT have a surd on the bottom of the fraction (though it CAN have
a surd on the top!).
If the surd contains a square root by itself or a multiple of a square root, to get rid of it, you must multiply BOTH the top and bottom of the fraction by that square root value.
e.g. 6 x √7 = 6√7
√7 x √7 7
e.g.2 6 + √2 x √3 = 6√3 + √2 x √3 = 6√3 + √6
2√3 x √3 2 x √3 x √3 6
i.e. 2 x 3
If the surd on the bottom involves addition or subtraction with a square root, to get rid of the square root part you must use the ‘difference of two squares’ and multiply BOTH the top and bottom of
the fraction by the bottom surd’s expression but with the inverse operation.
e.g.3 7 x (2 – √2) = 14 – 7√2 = 14 – 7√2
2 + √2 x (2 – √2) 2^2 – (√2)^2 2
i.e. 4 – 2
Notes on the ‘Difference of two squares’…
Squaring… (2 + √2)(2 + √2) = 2(2 + √2) + √2(2 + √2)
(ops the same) = 4 + 2√2 + 2√2 + √2√2
= 4 + 4√2 + 2 = 6 + √2 (still a surd)
Multiplying… (2 + √2)(2 – √2) = 2(2 – √2) + √2(2 – √2)
(opposite ops) = 4 – 2√2 + 2√2 – √2√2
= 4 (cancel out) – 2 = 2 (not a surd)
In essence, as long as the operation in each brackets is the opposite, the middle terms will always cancel each other out and you will be left with the first term squared subtracting the second term
i.e. (5 + √7)(5 – √7) à 5^2 – (√7)^2 = 25 – 7 = 18
Simplify by rationalizing the denominator
If the product of the two surds gives a rational number then the product of the two surds gives conjugate surds.
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. Without using logarithm tables, find the value of x in the equation
Log x^3 + log 5x = 5 log2 – log 2 5
2. Simplify (1 ÷ √3) (1 – √3)
Hence evaluate 1 to 3 s.f. given that √3 = 1.7321
1 + √3
3. If √14 – √ 14 = a√7 + b√2
√7-√2 √ 7 + √ 2
Find the values of a and b where a and b are rational numbers.
4. Find the value of x in the following equation 49^(x+1) + 7^(2x) = 350
5. Find x if 3 log 5 + log x^2 = log 1/125
6. Simplify as far as possible leaving your answer inform of a surd
1 – 1
√14 – 2 √3 √14 + 2 √3
7. Given that tan 75^0 = 2 + √3, find without using tables tan 15^0 in the form p+q√m, where p, q and m are integers.
8. Without using mathematical tables, simplify
63 + 72
32 + 28
9. Simplify 3 + 1 leaving the answer in the form a + b Öc, where a, b and c Ö5 -2 Ö5 are rational numbers
Specific Objectives
By the end of the topic the learner should be able to:
(a) Derive logarithmic relation from index form and vice versa;
(b) State the laws of logarithms;
(c) Use logarithmic laws to simplify logarithmic expressions and solvelogarithmic equations;
(d) Apply laws of logarithms for further computations.
(a) Logarithmic notation (eg. an=b, log ab=n)
(b) The laws of logarithms: log (AB) = log A + log B, log(A^B) = log A -log B and Log A n = n x log A.
(c) Simplifications of logarithmic expressions
(d) Solution of logarithmic equations
(e) Further computation using logarithmic laws.
If then we introduce the inverse function logarithm and define
(Read as log base of equals ).
Where Û means “implies and is implied by” i.e. it works both ways!
Note this means that, going from exponent form to logarithmic form:
And in going from logarithmic form to exponent form:
Laws of logarithms
Product and Quotient Laws of Logarithms:
The Product Law
The Quotient Law
= 2
The Power Law of Logarithms:
2log 5 + 2log 2
= 2
Logarithm of a Root
Property Proof Reason for Step
1. log[b] b = 1 and log[b] 1 = 0 b^1 = b and b^0 = 1 Definition of logarithms
a. Let log[b] x = m and log[b] y = n a. Setup
b. x = b^m and y = b^ n b. Rewrite in exponent form
2.(product rule)
c. xy = b^m [*] b^n c. Multiply together
log[b] xy = log[b ]x + log[b] y
d. xy = b ^m + n d. Product rule for exponents
e. log[b] xy = m + n e. Rewrite in log form
f log[b] xy = log[b] x + log[b] y f. Substitution
a. Let log[b] x = m and log[b] y = n a. Given: compact form
b. x = b^m and y = b^ n b. Rewrite in exponent form
3. (quotient rule) c. = c. Divide
log[b] = log[b ]x – log[b] y d. = d. Quotient rule for exponents
e. log[b ]= m – n e. Rewrite in log form
f. log[b] log[b] x – log[b] y f. Substitution
a. Let m = log[b] x so x = b^m a. Setup
4. (power rule) b. x^n = b^mn b. Raise both sides to the nth power
log[b ]x^n = n log[b] x c. log[b] x^ n = mn c. Rewrite as log
d. log[b] x^n = n log[b] x d. Substitute
b. i. Subtract from both sides
ii. Quotient rule
a. This follows directly from the properties for exponents.
[ ]
5. Properties used to solve log equations: b. i. log[b] x – log[b] y = 0
[ ]
a. if b^x = b^y, then x = y ii. log[b]
iii. Rewrite in exponent form
b. if log[b] x = log[b] y, then x = y iii. [=]b^0
iv. 1 so x = y
iv. b^0 = 1
Solving exponential and logarithmic equations
By taking logarithms, and exponential equation can be converted to a linear equation and solved. We will use the process of taking logarithms of both sides.
1. a)
x = 1.792
A logarithmic expression is defined only for positive values of the argument. When we solve a logarithmic equation it is essential to verify that the solution(s) does not result in the logarithm of a
negative number. Solutions that would result in the logarithm of a negative number are called extraneous, and are not valid solutions.
Solve for x:
(the one becomes an exponent : )
not possible
Solving equations using logs
(i) Solve the equation
The definition of logs says if then or
Hence (to 5 decimal places)
Check (to 5 decimal places)
In practice from we take logs to base 10 giving
(ii) Solve the equation
Check , , we want so the value of lies between 3 and 4 or which means lies between 1.5 and 2. This tells us that is roughly correct.
(iii) Solve the equation
Check very close!
Note you could combine terms, giving,
(iv) Solve the equation
Take logs of both sides
Expand brackets
Collect terms
Factorise the left hand side
(Note you get the same answer by using the ln button on your calculator.)
Check and
Notice that you could combine the log-terms in
to give
It does not really simplify things here but, in some cases, it can.
(v) Solve the equation
Take logs of both sides
Expand brackets
Collect terms
Factorize left hand side
LHS = (taking )
RHS = (taking )
The values of LHS and RHS are roughly the same. A more exact check could be made using a calculator.
Logarithmic equations and expressions
Consider the following equations
The value of x in each case is established as follows
X =4
Let = t. then = 2
Introducing logarithm to base 10 on both sides
Taking logs on both sides cannot help in getting the value of x, since cannot be combined into a single expression. However if we let then the equation becomes quadratic in y.
Thus, let …………….. (1)
Substituting for y in equation (1);
Let or let
There is no real value of x for which hence
Solve for x in
solve the quadratic equation using any method
Substituting for t in the equation (1).
= x
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. Solve for ( – ½ = 3/2
2. Find the values of x which satisfy the equation 5^2x – 6 (5^x) + 5 =0
3. Solve the equation
Log (x + 24) – 2 log 3 = log (9-2x)
4. Find the value of x in the following equation 49^(x+1) + 7^(2x) = 350
5. Find x if 3 log 5 + log x^2 = log 1/125
6. Without using logarithm tables, find the value of x in the equation
Log x^3 + log 5x = 5 log2 – log 2 5
7. Given that P = 3^y express the questions 3^2y -1) + 2 x 3^(y-1) = 1 in terms of P
8. Hence or otherwise find the value of y in the equation: 3^(2y-1) + 2 x 3^(y-1)=1
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define principal, rate and time in relation to interest;
(b) Calculate simple interest using simple interest formula;
(c) Calculate compound interest using step by step method;
(d) Derive the compound interest formula;
(e) Apply the compound interest formula for calculating interest;
(f) Define appreciation and depreciation;
(g) Use compound interest formula to calculate appreciation and depreciation;
(h) Calculate hire purchase;
(i) Calculate income tax given the income tax bands.
(a) Principal rate and time
(b) Simple interest
(c) Compound interest using step by step method
(d) Derivation of compound interest formula
(e) Calculations using the compound interest formula
(f) Appreciation and depreciation
(g) Calculation of appreciation and depreciation using the compound interestformula
(h) Hire purchase
(i) Income tax.
Simple interest
Interest is the money charged for the use of borrowed money for a specific period of time. If money is borrowed or deposited it earns interest, Principle is the sum of money borrowed or deposited P,
Rate is the ratio of interest earned in a given period of time to the principle.
The rate is expressed as a percentage of the principal per annum (P.A).When interest is calculated using only the initial principal at a given rate and time, it is called simple interest (I).
Simple interest formulae
Simple interest =
Franny invests ksh 16,000 in a savings account. She earns a simple interest rate of 14%, paid annually on her investment. She intends to hold the investment for 1 years. Determine the future value
of the investment at maturity.
I =
= sh. 16000 x
= sh 3360
Amount = P + I
= sh.16000 + sh 3360
= sh.19360
Calculate the rate of interest if sh 4500 earns sh 500 after 1 years.
From the simple interest formulae
I =
P = sh 4500
I = sh 500
T = 1 years
Therefore R =
R 7.4 %
Esha invested a certain amount of money in a bank which paid 12% p.a. simple interest. After 5 years, his total savings were sh 5600.Determine the amount of money he invested initially.
Let the amount invested be sh P
T = 5 years
R = 12 % p.a.
A =sh 5600
But A = P + I
Therefore 5600 = P + P X
= P + 0.60 P
= 1.6 P
Therefore p =
= sh 3500
Compound interest
Suppose you deposit money into a financial institution, it earns interest in a specified period of time. Instead of the interest being paid to the owner it may be added to (compounded with) the
principle and therefore also earns interest. The interest earned is called compound interest. The period after which its compounded to the principle is called interest period.
The compound interest maybe calculated annually, semi-annually, quarterly, monthly etc. If the rate of compound interest is R% p.a and the interest is calculated n times per year, then the rate of
interest per period is
Moyo lent ksh.2000 at interest of 5% per annum for 2 years. First we know that simple interest for 1^st year and 2^nd year will be same
i.e. = 2000 x 5 x 1/100 = Ksh. 100
Total simple interest for 2 years will be = 100 + 100 = ksh. 200
In Compound Interest (C I) the first year Interest will be same as of Simple Interest (SI) i.e. Ksh.100. But year II interest is calculated on P + SI of 1^st year i.e. on ksh. 2000 + ksh. 100 = ksh.
So, year II interest in Compound Interest becomes
= 2100 x 5 x 1/100 = Ksh. 105
So it is Ksh. 5 more than the simple interest. This increase is due to the fact that SI is added to the principal and this ksh. 105 is also added in the principal if we have to find the compound
interest after 3 years. Direct formula in case of compound interest is
A = P (1 + )^t
Where A = Amount
P = Principal
R = Rate % per annum
T = Time
A = P + CI
P (1 + )^ t = P + CI
Types of Question:
Type I: To find CI and Amount
Type II: To find rate, principal or time
Type III: When difference between CI and SI is given.
Type IV: When interest is calculated half yearly or quarterly etc.
Type V: When both rate and principal have to be found.
Type 1
Find the amount of ksh. 1000 in 2 years at 10% per annum compound interest.
A = P (1 + r/100)^t
=1000 (1 + 10/100)^2
= 1000 x 121/100
=ksh. 1210
Find the amount of ksh. 6250 in 2 years at 4% per annum compound interest.
A = P (1 + r/100)^ t
= 6250 (1 + 4/100)^2
=6250 x 676/625
= ksh. 6760
What will be the compound interest on ksh 31250 at a rate of 4% per annum for 2 years?
CI = P (1 + r/100)^ t – 1
=31250 { (1 + 4/100)^2 – 1}
=31250 (676/625 – 1)
=31250 x 51/625 = ksh. 2550
A sum amounts to ksh. 24200 in 2 years at 10% per annum compound interest.
Find the sum ?
A = P (1 + r/100)^t
24200 = P (1 + 10/100)^2
= P (11/10)^2
= 24200 x 100/121
= ksh. 20000
Type II
The time in which ksh. 15625 will amount to ksh. 17576 at 45 compound interest is?
A = P (1 + r/100)^t
17576 = 15625 (1 + 4/100)^t
17576/15625 = (26/25)^t
(26/25)t = (26/25)3
t = 3 years
The rate percent if compound interest of ksh. 15625 for 3 years is Ksh. 1951.
A = P + CI
= 15625 + 1951 = ksh. 17576
A = P (1 + r/100)^t
17576 = 15625 (1 + r/100)^3
17576/15625 = (1 + r/100)^3
(26/25)3 = (1 + r/100)^3
26/25 = 1 + r/100
26/25 – 1 = r/100
1/25 = r/100
r = 4%
Type IV
1. Remember
When interest is compounded half yearly then Amount = P (1 + R/2)^2t
I.e. in half yearly compound interest rate is halved and time is doubled.
2. When interest is compounded quarterly then rate is made ¼ and time is made 4 times.
Then A = P [(1+R/4)/100]^4t
3. When rate of interest is R1%, R2%, and R3% for 1^st, 2^nd and 3^rd year respectively; then A = P (1 + R1/100) (1 + R2/100) (1 + r3/100)
Find the compound interest on ksh.5000 at 205 per annum for 1.5 year compound half yearly.
When interest is compounded half yearly
Then Amount = P [(1 +R/2)/100]^2t
Amount = 5000 [(1 + 20/2)/100]^3/2
= 5000 (1 + 10/100)^3
=5000 x 1331/1000
= ksh 6655
CI = 6655 – 5000 = ksh. 1655
Find compound interest ksh. 47145 at 12% per annum for 6 months, compounded quarterly.
As interest is compounded quarterly
A =[ P(1 + R/4)/100)]^4t
A = 47145 [(1 + 12/4)/100] ^½ x 4
= 47145 (1 + 3/100)^2
= 47145 x 103/100 x 103/100
= ksh. 50016.13
CI = 50016.13 – 47145
= ksh. 2871.13
Find the compound interest on ksh. 18750 for 2 years when the rate of interest for 1^st year is 45 and for 2^nd year 8%.
A = P (1 + R1/100) (1 + R1/100)
= 18750 * 104/100 * 108/100
=ksh. 21060
CI = 21060 – 18750
= ksh. 2310
Type V
The compound interest on a certain sum for two years is ksh. 52 and simple interest for the same period at same rate is ksh.50 find the sum and the rate.
We will do this question by basic concept. Simple interest is same every year and there is no difference between SI and CI for 1^st year. The difference arises in the 2^nd year because interest of 1^
st year is added in principal and interest is now charged on principal + simple interest of 1^st year.
So in this question
2 year SI = ksh. 50
1 year SI = ksh. 25
Now CI for 1^st year = 52 – 25 = Rs.27
This additional interest 27 -25 = ksh. 2 is due to the fact that 1^st year SI i.e. ksh. 25 is added in principal. It means that additional ksh. 2 interest is charged on ksh. 25. Rate % = 2/25 x 100 =
Rate % = [(CI – SI)/ (SI/2)] x 100
= [(2/50)/2] x 100
2/25 x 100
P = SI x 100/R x T = 50 x 100/8 x 2
= ksh. 312.50
A sum of money lent CI amounts in 2 year to ksh. 8820 and in 3 years to ksh. 9261. Find the sum and rate %.
Amount after 3 years = ksh. 9261
Amount after 2 years = ksh. 8820
By subtracting last year’s interest ksh. 441
It is clear that this ksh. 441 is SI on ksh. 8820 from 2^nd to 3^rd year i.e. for 1 year.
Rate % = 441 x 100/8820 x 1
=5 %
Also A = P (1 + r/100)^t
8820 = P (1 + 5/100)^2
= P (21/20)^2
P = 8820 x 400/441
= ksh. 8000
Appreciation and Depreciation
Appreciation is the gain of value of an asset while depreciation is the loss of value of an asset.
An iron box cost ksh 500 and every year it depreciates by 10% of its value at the beginning of that that year. What will its value be after value 4 years?
Value after the first year = sh (500 – x 500)
= sh 450
Value after the second year = sh (450 – x 450)
= sh 405
Value after the third year = sh (405 – x 405)
= sh 364.50
Value after the fourth year = sh (364.50 – x 364.50)
= sh 328.05
In general if P is the initial value of an asset, A the value after depreciation for n periods and r the rate of depreciation per period.
A=P (
A minibus cost sh 400000.Due to wear and tear, it depreciates in value by 2 % every month. Find its value after one year,
A=P (
Substituting P= 400,000 , r = 2 , and n =12 in the formula ;
A =sh.400000 (1- 0.02
=sh.400, 000(0.98
= sh.313700
The initial cost of a ranch is sh.5000, 000.At the end of each year, the land value increases by 2%.What will be the value of the ranch at the end of 3 years?
The value of the ranch after 3 years =sh 5000, 000(1 +
= sh. 5000000(
= sh 5,306,040
Hire Purchase
Method of buying goods and services by instalments. The interest charged for buying goods or services on credit is called carrying charge.
Hire purchase = Deposit + (instalments x time)
Aching wants to buy a sewing machine on hire purchase. It has a cash price of ksh 7500.She can pay a cash price or make a down payment of sh 2250 and 15 monthly instalments of sh.550 each. How much
interest does she pay under the instalment plan?
Total amount of instalments = sh 550 x 15
= sh 8250
Down payment (deposit) = sh 2250
Total payment = sh (8250 + 2250)
= sh 10500
Amount of interest charged = sh (10500-7500)
= sh3000
Always use the above formula to find other variables.
Income tax
Taxes on personal income is income tax. Gross income is the total amount of money due to the individual at the end of the month or the year.
Gross income = salary + allowances / benefits
Taxable income is the amount on which tax is levied. This is the gross income less any special benefits on which taxes are not levied. Such benefits include refunds for expenses incurred while one is
on official duty.
In order to calculate the income tax that one has to pay, we convert the taxable income into Kenya pounds K£ per annum or per month as dictated by the by the table of rates given.
• Every employee in kenya is entitled to an automatic personal tax relief of sh.12672 p.a (sh.1056 per month)
• An employee with a life insurance policy on his life, that of his wife or child, may make a tax claim on the premiums paid towards the policy at sh.3 per pound subject to a maximum claim of sh
.3000 per month.
Mr. John earns a total of K£12300 p.a.Calculate how much tax he should pay per annum.Using the tax table below.
Income tax K£ per annum Rate (sh per pound)
1 -5808 2
5809 – 11280 3
11289 – 16752 4
16753 – 22224 5
Excess over 22224 6
His salary lies between £ 1 and £12300.The highest tax band is therefore the third band.
For the first £ 5808, tax due is sh 5808 x 2 = sh 11616
For the next £ 5472, tax due is sh 5472 x 2 = sh 16416
Remaining £ 1020, tax due sh. 1020 x 4 = sh 4080 +
Total tax due sh 32112
Less personal relief of sh.1056 x 12 = sh.12672 –
Sh 19440
Therefore payable p.a is sh.19400.
Mr. Ogembo earns a basic salary of sh 15000 per month.in addition he gets a medical allowance of sh 2400 and a house allowance of sh 12000.Use the tax table above to calculate the tax he pays per
Taxable income per month = sh (15000 + 2400 + 12000)
= sh.29400
Converting to K£ p.a = K£ 29400 x
= K£ 17640
Tax due
First £ 5808 = sh.5808 x 2 = sh.11616
Next £ 5472 = sh.5472 x 3 = sh.16416
Next £ 5472 = sh.5472 x 4 = sh.21888
Remaining £ 888 = sh.888 x 5 = sh 4440 +
Total tax due sh 54360
Less personal relief sh 12672 –
Therefore, tax payable p.a sh41688
In Kenya, every employer is required by the law to deduct income tax from the monthly earnings of his employees every month and to remit the money to the income tax department. This system is called
Pay As You Earn (PAYE).
If an employee is provided with a house by the employer (either freely or for a nominal rent) then 15% of his salary is added to his salary (less rent paid) for purpose of tax calculation. If the tax
payer is a director and is provided with a free house, then 15% of his salary is added to his salary before taxation.
Mr. Omondi who is a civil servant lives in government house who pays a rent of sh 500 per month. If his salary is £9000 p.a, calculate how much PAYE he remits monthly.
Basic salary £ 9000
Housing £
Less rent paid = £ 300
£ 1050 +
Taxable income £ 10050
Tax charged;
First £ 5808, the tax due is sh.5808 x 2 = sh 11616
Remaining £ 4242, the tax due is sh 4242 x 3 = sh 12726 +
Sh 24342
Less personal relief Sh 12672
Sh 11670
PAYE = sh
= sh 972.50
Mr. Odhiambo is a senior teacher on a monthly basic salary of Ksh. 16000.On top of his salary he gets a house allowance of sh 12000, a medical allowance of Ksh.3060 and a hardship allowance of Ksh
3060 and a hardship allowance of Ksh.4635.He has a life insurance policy for which he pays Ksh.800 per month and claims insurance relief.
1. Use the tax table below to calculate his PAYE.
Income in £ per month Rate %
1 – 484 10
485 – 940 15
941 – 1396 20
1397 – 1852 25
Excess over 1852 30
1. In addition to PAYEE the following deductions are made on his pay every month
2. WCPS at 2% of basic salary
3. HHIF ksh.400
4. Co – operative shares and loan recovery Ksh 4800.
1. Taxable income = Ksh (16000 + 12000 +3060 +4635)
= ksh 35695
Converting to K£ =
= K
Tax charged is:
First £ 484 = £484 x = £ 48.40
Next £ 456 = £456 x = £ 68.40
Next £ 456 = £456 x = £ 91.20
Remaining £ 388 = £388 x = £ 97.00.
Total tax due = £305.00
= sh 6100
Insurance relief = sh = sh 120
Personal relief = sh 1056 +
Total relief sh 1176
Tax payable per month is sh 6100
Sh 1176 –
Sh 4924
Therefore, PAYE is sh 4924.
For the calculation of PAYE, taxable income is rounded down or truncated to the nearest whole number.
If an employee’s due tax is less than the relief allocated, then that employee is exempted from PAYEE
1. Total deductions are
Sh (
Net pay = sh (35695 – 11244)
= sh 24451
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. A business woman opened an account by depositing Kshs. 12,000 in a bank on 1^st July 1995. Each subsequent year, she deposited the same amount on 1^st July. The bank offered her 9% per annum
compound interest. Calculate the total amount in her account on
(a) 30^th June 1996
(b) 30^th June 1997
2. A construction company requires to transport 144 tonnes of stones to sites A and
3. The company pays Kshs 24,000 to transport 48 tonnes of stone for every 28
4. Kimani transported 96 tonnes to a site A, 49 km away.
(a) Find how much he paid
(b) Kimani spends Kshs 3,000 to transport every 8 tonnes of stones to site.
Calculate his total profit.
(c) Achieng transported the remaining stones to sites B, 84 km away. If she made 44% profit, find her transport cost.
3. The table shows income tax rates
Monthly taxable pay Rate of tax Kshs in 1 K£
1 – 435 2
436 – 870 3
871-1305 4
1306 – 1740 5
Excess Over 1740 6
A company employee earn a monthly basic salary of Kshs 30,000 and is also given taxable allowances amounting to Kshs 10, 480.
(a) Calculate the total income tax
(b) The employee is entitled to a personal tax relief of Kshs 800 per month.
Determine the net tax.
(c) If the employee received a 50% increase in his total income, calculate the
corresponding percentage increase on the income tax.
4. A house is to be sold either on cash basis or through a loan. The cash price is Kshs.750, 000. The loan conditions area as follows: there is to be down payment
of 10% of the cash price and the rest of the money is to be paid through a loan
at 10% per annum compound interest.
A customer decided to buy the house through a loan.
1. a) (i) Calculate the amount of money loaned to the customer.
(ii) The customer paid the loan in 3 year’s. Calculate the total amount
paid for the house.
1. b) Find how long the customer would have taken to fully pay for the house
if she paid a total of Kshs 891,750.
5. A businessman obtained a loan of Kshs. 450,000 from a bank to buy a matatu valued at the same amount. The bank charges interest at 24% per annum compound quarterly
6. a) Calculate the total amount of money the businessman paid to clear the loan in 1 ½ years.
7. b) The average income realized from the matatu per day was Kshs. 1500. The matatu worked for 3 years at an average of 280 days year. Calculate the total income from the matatu.
8. c) During the three years, the value of the matatu depreciated at the rate of 16% per annum. If the businessman sold the matatu at its new value, calculate the total profit he realized by the end
of three years.
9. A bank either pays simple interest as 5% p.a or compound interest 5% p.a on deposits. Nekesa deposited Kshs P in the bank for two years on simple interest terms. If she had deposited the same
amount for two years on compound interest terms, she would have earned Kshs 210 more.
Calculate without using Mathematics Tables, the values of P
7. (a) A certain sum of money is deposited in a bank that pays simple interest at
a certain rate. After 5 years the total amount of money in an account is Kshs 358 400. The interest earned each year is 12 800
• The amount of money which was deposited (2mks)
• The annual rate of interest that the bank paid (2mks)
(b) A computer whose marked price is Kshs 40,000 is sold at Kshs 56,000 on hire purchase terms.
(i) Kioko bought the computer on hire purchase term. He paid a deposit of 25% of the hire purchase price and cleared the balance by equal monthly installments of Kshs 2625. Calculate the
number of installments (3mks)
(ii) Had Kioko bought the computer on cash terms he would have been allowed a discount of 12 ½ % on marked price. Calculate the difference between the cash price and the hire purchase price
and express as a percentage of the cash price
(iii) Calculate the difference between the cash price and hire purchase price and express it as a percentage of the cash price.
8. The table below is a part of tax table for monthly income for the year 2004
Monthly taxable income Tax rate percentage
In ( Kshs) (%) in each shillings
Under Kshs 9681 10%
From Kshs 9681 but under 18801 15%
From Kshs 18801 but 27921 20%
In the tax year 2004, the tax of Kerubo’s monthly income was Kshs 1916.
Calculate Kerubo’s monthly income
9. The cash price of a T.V set is Kshs 13, 800. A customer opts to buy the set on hire purchase terms by paying a deposit of Kshs 2280.
If simple interest of 20 p. a is charged on the balance and the customer is required to repay by 24 equal monthly installments. Calculate the amount of each installment.
10. A plot of land valued at Ksh. 50,000 at the start of 1994.
Thereafter, every year, it appreciated by 10% of its previous years value find:
(a) The value of the land at the start of 1995
(b) The value of the land at the end of 1997
11. The table below shows Kenya tax rates in a certain year.
Income K £ per annum Tax rates Kshs per K £
1- 4512 2
4513 – 9024 3
9025 – 13536 4
13537 – 18048 5
18049 – 22560 6
Over 22560 6.5
In that year Muhando earned a salary of Ksh. 16510 per month. He was entitled to a monthly tax relief of Ksh. 960
(a) Muhando annual salary in K £
(b) (i) The monthly tax paid by Muhando in Ksh
14. A tailor intends to buy a sewing machine which costs Ksh 48,000. He borrows the money from a bank. The loan has to be repaid at the end of the second year. The bank charges an interest at the
rate of 24% per annum compounded half yearly. Calculate the total amount payable to the bank.
15. The average rate of depreciation in value of a water pump is 9% per annum. After three complete years its value was Ksh 150,700. Find its value at the start of the three year period.
1. A water pump costs Ksh 21600 when new, at the end of the first year its value depreciates by 25%. The depreciation at the end of the second year is 20% and thereafter the rate of depreciation is
15% yearly. Calculate the exact value of the water pump at the end of the fourth year.
Specific Objectives
By the end of the topic the learner should be able to:
(a) Calculate length of an arc and a chord;
(b) Calculate lengths of tangents and intersecting chords;
(c) State and use properties of chords and tangents;
(d) Construct tangent to a circle,
(e) Construct direct and transverse common tangents to two circles;
(f) Relate angles in alternate segment;
(g) Construct circumscribed, inscribed and escribed circles;
(h) Locate centroid and orthocentre of a triangle;
(i) Apply knowledge of circles, tangents and chords to real life situations.
(a) Arcs, chords and tangents
(b) Lengths of tangents and intersecting chords
(c) Properties of chords and tangents
(d) Construction of tangents to a circle
(e) Direct and transverse common tangents to two circles
(f) Angles in alternate segment
(g) Circumscribed, inscribed and escribed circles
(h) Centroid and orthocentre
(i) Application of knowledge of tangents and chords to real life situations.
Length of an Arc
The Arc length marked red is given by ;
Find the length of an arc subtended by an angle of at the centre of the circle of radius 14 cm.
Length of an arc =
The length of an arc of a circle is 11.0 cm.Find the radius of the circle if an arc subtended an angle of at the centre .
Arc length =
Therefore 11 =
Find the angle subtended at the centre of a circle by an arc of 20 cm, if the circumference of the circle is 60 cm.
But 2
Chord of a circle: A line segment which joins two points on a circle. Diameter: a chord which passes through the center of the circle. Radius: the distance from the center of the circle to the
circumference of the circle
Perpendicular bisector of a code
A perpendicular drawn from the centre of the circle to a chord bisects the chord.
• Perperndicular drawn from the centre of the circle to chord bisects the cord ( divides it into two equal parts)
• A straight line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.
The radius of a circle centre O is 13 cm.Find the perpendicular distance from O to the chord, if AB is 24 cm.
OC bisects chord AB at C
Therefore, AC =12 cm
In O
, OM = = 5 cm
Parallel chords
Any chord passing through the midpoints of all parallel chords of a circle is a diameter
In the figure below CD and AB are parallel chords of a circle and 2 cm apart. If CD = 8 cm and AB= 10 cm, find the radius of the circle
• Draw the perpendicular bisector of the chords to cut them at K and L .
• Join OD and OC
• In triangle ODL,
• DL = 4 cm and KC =5 cm
• Let OK = X cm
• Therefore (
In triangle OCK;
Using the equation
= 5.154 cm
Intersecting chords
In general
In the example above AB and CD are two chords that intersect in a circle at Given that AE = 4 cm, CE =5 cm and DE = 3 cm, find AB.
Let EB = x cm
Since AB = AE + EB
AB = 4 + 3.75
= 7.75 cm
Equal chords.
• Angles subtended at the centre of a circle by equal chords are equals
• If chords are equal they are equidistant from the centre of the circle
A chord that is produced outside a circle is called a secant
Find the value of AT in the figure below. AR = 4 cm, RD = 5 cm and TC = 9 cm.
AC x AT
(x + 9) x = (5 + 4) 4
(x + 12) (x- 3) = 0
Therefore, x = – 12 or x = 3
Tangent and secant
A line which touches a circle at exactly one point is called a tangent line and the point where it touches the circle is called the point of contact
A line which intersects the circle in two distinct points is called a secant line (usually referred to as a secant).The figures below A shows a secant while B shows a tangent .
A B
Construction of a tangent
• Draw a circle of any radius and centre O.
• Join O to any point P on the circumference
• Produce OP to a point P outside the circle
• Construct a perpendicular line SP through point P
• The line is a tangent to the circle at P as shown below.
• The radius and tangent are perpendicular at the point of contact.
• Through any point on a circle , only one tangent can be drawn
• A perpendicular to a tangent at the point of contact passes thought the centre of the circle.
In the figure below PT = 15 cm and PO = 17 cm, calculate the length of PQ.
OT = 8 cm
Properties of tangents to a circle from an external point
If two tangents are drawn to a circle from an external point
• They are equal
• They subtend equal angles at the centre
• The line joining the centre of the circle to the external point bisects the angle between the tangents
The figure below represents a circle centre O and radius 5 cm. The tangents PT is 12 cm long. Find: a.) OP b.) Angle TP
<OTP =
= 0.9231
Therefore, <TPO = 22.6
Hence <
Two tangent to a circle
Direct (exterior) common tangents Transverse or interior common tangents
Tangent Problem
The common-tangent problem is named for the single tangent segment that’s tangent to two circles. Your goal is to find the length of the tangent. These problems are a bit involved, but they should
cause you little difficulty if you use the straightforward three-step solution method that follows.
The following example involves a common external tangent (where the tangent lies on the same side of both circles). You might also see a common-tangent problem that involves a common internal tangent
(where the tangent lies between the circles). No worries: The solution technique is the same for both.
Given the radius of circle A is 4 cm and the radius of circle Z is 14 cm and the distance between the two circles is 8 cm.
Here’s how to solve it:
1.)Draw the segment connecting the centers of the two circles and draw the two radii to the points of tangency (if these segments haven’t already been drawn for you).
Draw line AZ and radii AB and ZY.
The following figure shows this step. Note that the given distance of 8 cm between the circles is the distance between the outsides of the circles along the segment that connects their centers.
2.) From the center of the smaller circle, draw a segment parallel to the common tangent till it hits the radius of the larger circle (or the extension of the radius in a common-internal-tangent
You end up with a right triangle and a rectangle; one of the rectangle’s sides is the common tangent. The above figure illustrates this step.
3.)You now have a right triangle and a rectangle and can finish the problem with the Pythagorean Theorem and the simple fact that opposite sides of a rectangle are congruent.
The triangle’s hypotenuse is made up of the radius of circle A, the segment between the circles, and the radius of circle Z. Their lengths add up to 4 + 8 + 14 = 26. You can see that the width of the
rectangle equals the radius of circle A, which is 4; because opposite sides of a rectangle are congruent, you can then tell that one of the triangle’s legs is the radius of circle Z minus 4, or 14 –
4 = 10.
You now know two sides of the triangle, and if you find the third side, that’ll give you the length of the common tangent.
You get the third side with the Pythagorean Theorem:
(Of course, if you recognize that the right triangle is in the 5 : 12 : 13 family, you can multiply 12 by 2 to get 24 instead of using the Pythagorean Theorem.)Because opposite sides of a rectangle
are congruent, BY is also 24, and you’re done.
Now look back at the last figure and note where the right angles are and how the right triangle and the rectangle are situated; then make sure you heed the following tip and warning.
Note the location of the hypotenuse. In a common-tangent problem, the segment connecting the centers of the circles is always the hypotenuse of a right triangle. The common tangent is always the side
of a rectangle, not a hypotenuse.
In a common-tangent problem, the segment connecting the centers of the circles is never one side of a right angle. Don’t make this common mistake.
HOW TO construct a common exterior tangent line to two circles
In this lesson you will learn how to construct a common exterior tangent line to two circles in a plane such that no one is located inside the other using a ruler and a compass.
Problem 1
For two given circles in a plane such that no one is located inside the other, to construct the common exterior tangent line using a ruler and a compass.
We are given two circles in a plane such that no one is located inside the other (Figure 1a).
We need to construct the common exterior tangent line to the circles using a ruler and a compass.
First, let us analyze the problem and make a sketch (Figures 1a and 1b). Let AB be the common tangent line to the circles we are searching for.
Let us connect the tangent point A of the first circle with its center P and the tangent point B of the second circle with its center Q (Figure 1a and 1b).
Then the radii PA and QB are both perpendicular to the tangent line AB (lesson A tangent line to a circle is perpendicular to the radius drawn to the tangent point under the topic Circles
and their properties ). Hence, theradii PA and QB are parallel.
Figure 1a. To the Problem 1
Figure 1b. To the solution of the Problem 1
Figure 1c. To the construction step 3
Next, let us draw the straight line segment CQ parallel to AB through the point Q till the intersection with the radius PA at the point C (Figure 1b). Then the straight line CQ is
parallel to AB. Hence, the quadrilateral CABQ is a parallelogram (moreover, it is a rectangle) and has the opposite sides QB and CA congruent. The point C divides the radius PA in two
segments of the length (CA) and (PC). It is clear from this analysis that the straight line QC is the tangent line to the circle of the radius with the center at the point P (shown in red
in Figure 1b).
It implies that the procedure of constructing the common exterior tangent line to two circles should be as follows:
1) draw the auxiliary circle of the radius at the center of the larger circle (shown in red in Figure 1b);
2) construct the tangent line to this auxiliary circle from the center of the smaller circle (shown in red in Figure 1b). In this way you will get the tangent point C on the auxiliary circle of
the radius ;
3) draw the straight line from the point P to the point C and continue it in the same direction till the intersection with the larger circle (shown in blue in Figure 1b). The intersection
point A is the tangent point of the common tangent line and the larger circle. Figure 1c reminds you how to perform this step.
4) draw the straight line QB parallel to PA till the intersection with the smaller circle (shown in blue in Figure 1b).
The intersection point B is the tangent point of the common tangent line and the smaller circle;
5) the required common tangent line is uniquely defined by its two points A and B.
Note that all these operations 1) – 4) can be done using a ruler and a compass. The problem is solved.
Problem 2
Find the length of the common exterior tangent segment to two given circles in a plane, if they have the radii and and the distance between their centers is d.
No one of the two circles is located inside the other.
Let us use the Figure 1b from the solution to the previous Problem 1.
This Figure is relevant to the Problem 2. It is copied and reproduced
in the Figure 2 on the right for your convenience.
figure 2
It is clear from the solution of the Problem 1 above that the common
exterior tangent segment |AB| is congruent to the side |CQ| of the
quadrilateral (rectangle) CABQ.
From the other side, the segment CQ is the leg of the right-angled
triangle DELTAPCQ. This triangle has the hypotenuse’s measure d and
the other leg’s measure . Therefore, the length of the common
exterior tangent segment |AB| is equal to
|AB| =
Note that the solvability condition for this problem is d > .
It coincides with the condition that no one of the two circles lies inside the other.
Example 1
Find the length of the common exterior tangent segment to two given circles in a plane, if their radii are 6 cm and 3 cm and the distance between their centers
is 5 cm.
Use the formula (1) derived in the solution of the Problem 2.
According to this formula, the length of the common exterior tangent segment to the two given circles is equal to
= =
= 4 cm
The length of the common exterior tangent segment to the two given circles is 4 cm
Contact of circles
Two circle are said to touch each other at a point if they have a common tangent at that point.
Point T is shown by the red dot.
Internal tangent externally tangent
• The centers of the two circles and their point of contact lie on a straight line
• When two circles touch each other internally, the distance between the centers is equal to the difference of the radii i.e. PQ= TP-TA
• When two circles touch each other externally, the distance between the centers is equal to the sum of the radii i.e. OR =TO +TR
Alternate Segment theorem
The angle which the chord makes with the tangent is equal to the angle subtended by the same chord in the alternate segment of the circle.
Angle a = Angle b
The blue line represents the angle which the chord CD makes with the tangent PQ which is equal to the angle b which is subtended by the chord in the alternate segment of the circle.
• Angle s = Angle t
• Angle a = Ange b
Tangent – secant segment length theorem
If a tangent segment and secant segment are drawn to a circle from an external point, then the square of the length of the tangent equals the product of the length of the secant with the length of
its external segment.
In the figure above ,TW=10 cm and XW = 4 cm. find TV
TV =
Circles and triangles
Inscribed circle
• Construct any triangle ABC.
• Construct the bisectors of the three angles
• The bisectors will meet at point I
• Construct a perpendicular from O to meet one of the sides at M
• With the centre I and radius IM draw a circle
• The circle will touch the three sides of the triangle ABC
• Such a circle is called an inscribed circle or in circle.
• The centre of an inscribed circle is called the incentre
Circumscribed circle
• Construct any triangle ABC.
• Construct perpendicular bisectors of AB , BC, and AC to meet at point O.
• With O as the centre and using OB as radius, draw a circle
• The circle will pass through the vertices A , B and C as shown in the figure below
Escribed circle
• Construct any triangle ABC.
• Extend line BA and BC
• Construct the perpendicular bisectors of the two external angles produced
• Let the perpendicular bisectors meet at O
• With O as the centre draw the circle which will touch all the external sides of the triangle
Centre O is called the ex-centre
AO and CO are called external bisectors.
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. The figure below represents a circle a diameter 28 cm with a sector subtending an angle of 75^0 at the centre.
Find the area of the shaded segment to 4 significant figures
(a) <PST
2. The figure below represents a rectangle PQRS inscribed in a circle centre 0 and radius 17 cm. PQ = 16 cm.
• The length PS of the rectangle
• The angle POS
• The area of the shaded region
3. In the figure below, BT is a tangent to the circle at B. AXCT and BXD are
straight lines. AX = 6 cm, CT = 8 cm, BX = 4.8 cm and XD = 5 cm.
Find the length of
(a) XC
(b) BT
4. The figure below shows two circles each of radius 7 cm, with centers at X and Y. The circles touch each other at point Q.
Given that <AXD = <BYC = 120^0 and lines AB, XQY and DC are parallel, calculate the area of:
1. a) Minor sector XAQD (Take π^ 22/[7])
2. b) The trapezium XABY
3. c) The shaded regions.
4. The figure below shows a circle, centre, O of radius 7 cm. TP and TQ are tangents to the circle at points P and Q respectively. OT =25 cm.
Calculate the length of the chord PQ
6. The figure below shows a circle centre O and a point Q which is outside the circle
Using a ruler and a pair of compasses, only locate a point on the circle such that angle OPQ = 90^o
7. In the figure below, PQR is an equilateral triangle of side 6 cm. Arcs QR, PR and PQ arcs of circles with centers at P, Q and R respectively.
Calculate the area of the shaded region to 4 significant figures
8. In the figure below AB is a diameter of the circle. Chord PQ intersects AB at N. A tangent to the circle at B meets PQ produced at R.
Given that PN = 14 cm, NB = 4 cm and BR = 7.5 cm, calculate the length of:
(a) NR
(b) AN
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define a matrix;
(b) State the order of a matrix;
(c) Define a square matrix;
(d) Determine compatibility in addition and multiplication of matrices;
(e) Add matrices;
(f) Multiply matrices;
(g) Identify matrices;
(h) Find determinant of a 2 x 2 matrix;
(i) Find the inverse of a 2 x 2 matrix;
(j) Use matrices to solve simultaneous equations.
(a) Matrix
(b) Order of a matrix
(c) Square matrix
(d) Compatibility in addition and multiplication of matrices
(e) Multiplication of a matrix by a scalar
(f) Matrix multiplication
(g) Identify matrix
(h) Determinant of a 2 x 2 matrix
(i) Inverse of a 2 x 2 matrix
(j) Singular matrix
(k) Solutions of simultaneous equations in two unknowns.
A matrix is a rectangular arrangement of numbers in rows and columns. For instance, matrix A below has two rows and three columns. The dimensions of this matrix are 2 x 3 (read “2 by 3”). The numbers
in a matrix are its entries. In matrix A, the entry in the second row and third column is 5.
A =
Some matrices (the plural of matrix) have special names because of their dimensions or entries.
Order of matrix
Matrix consist of rows and columns. Rows are the horizontal arrangement while columns are the vertical arrangement.
Order of matrix is being determined by the number of rows and columns. The order is given by stating the number of rows followed by columns.
If the number of rows is m and the number of columns n, the matrix is of order .
E.g. If a matrix has m rows and n columns, it is said to be order m´n.
e.g. is a matrix of order 3´4.
e.g. is a matrix of order 3.
e.g. is a 2´3 matrix.
e.g. is a 3´1 matrix.
Elements of matrix
The element of a matrix is each number or letter in the matrix. Each element is locating by stating its position in the row and the column.
For example, given the 3 x 4 matrix
• The element 1 is in the third row and first column.
• The element 6 is in the first row and forth column.
A matrix in which the number of rows is equal to the number of columns is called a square matrix.
Is called a row matrix or row vector.
Is called a column matrix or column vector.
Is a column vector of order 3´1.
is a row vector of order 1´3.
Two or more matrices re equal if they are of the same order and their corresponding elements are equal. Thus, if then, a = 3, b =4 and d=5.
Addition and subtraction of matrices
Matrices can be added or subtracted if they are of the same order. The sum of two or more matrices is obtained by adding corresponding elements. Subtraction is also done in the same way.
– +
After arranging the matrices you must use BODMAS
The matrix above cannot be added because they are not of the same order
Matrix multiplication
To multiply a matrix by a number, you multiply each element in the matrix by the number.
A woman wanted to buy one sack of potatoes, three bunches of bananas and two basket of onion. She went to kikuyu market and found the prices as sh 280 for the sack of potatoes ,sh 50 for a bunch of
bananas and sh 100 for a basket of onions. At kondelee market the corresponding prices were sh 300, sh 48 and sh 80.
• Express the woman’s requirements as a row matrix
• Express the prices in each market as a column matrix
• Use the matrices in (a) and (b) to find the total cost in each market
• Requirements in matrix form is (1 3 2)
• Price matrix for Kikuyu market is
Price matrix for kondelee market
• Total cost in shillings at Kikuyu Market is;
(1 3 2) = (1 x 280 + 3 x 50 +2 x 100) = (630)
Total cost in shillings at Kondelee Market is;
(1 3 2 ) = ( 1 x 300 + 3 x 48 + 2 x 80) =(604)
The two results can be combined into one as shown below
(1 3 2)
The product of two matrices A and B is defined provided the number of columns in A is equal to the number of rows in B.
If A is an m x n matrix and B is an n x p matrix, then the product AB is an m a p matrix.
A X B = AB
m X n n X p = m X p
Each time a row is multiplied by a column
Find AB if A = and B=
Because A is a 3 x 2 matrix and B is a 2 x 2 matrix, the product AB is defined and is a 3 x 2 matrix. To write the elements in the first row and first column of AB, multiply corresponding elements in
the first row of A and the first column of B. Then add. Use a similar procedure to write the other entries of the product.
Identity matrix
For matrices, the identity matrix or a unit matrix is the matrix that has 1’s on the main diagonal and 0’s elsewhere. The main diagonal is the one running from top left to bottom right .It is also
called leading or principle diagonal. Examples are;
2 X 2 identity matrix 3 x 3 identity matrix
If A is any n x n matrix and I is the n x n identity matrix, then IA = A and AI = A.
Determinant matrix
The determinant of a matrix is the difference of the products of the elements on the diagonals.
The determinant of A, det A or |A| is defined as follows:
(a) If n=2,
Find the determinant
Subtract the product of the diagonals
1 x 5 – 2 x 3 = 5 – 6 = -1
Determinant is -1
Inverse of a matrix
Two matrices of order n x n are inverse of each other if their product (in both orders) is theidentity matrix of the same order n x n. The inverse of A is written as
Show that B=
= AB=BA=I. Hence, A is the inverse of B
To get the inverse matrix
• Find the determinant of the matrix. If it is zero, then there is no inverse
• If it is non zero, then;
• Interchange the elements in the main diagonal
• Reverse the signs of the element in the other diagonals
• Divide the matrix obtained by the determinant of the given matrix
In summary
The inverse of the matrix A = is
Find the inverse of A=
You can check the inverse by showing that A
Solutions of simultaneous linear equations using matrix
Using matrix method solve the following pairs of simultaneous equation
We need to calculate the inverse of A =
Hence the value of x = 2 and the value of y = 1 is the solution of the simultaneous equation
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic
1. A and B are two matrices. If A = 1 2 find B given that A^2 = A + B
2. Given that A= 1 3 , B= 3 1 , C = p 0 and AB =BC, determine the value of P
5 3 5 -1 0 q
3. A matrix A is given by A = x 0
5 y
1. a) Determine A^2
1. b) If A^2 = determine the possible pairs of values of x and y
2. (a) Find the inverse of the matrix 9 8
(b) In a certain week a businessman bought 36 bicycles and 32 radios for total of Kshs 227 280. In the following week, he bought 28 bicycles and 24 radios for a total of Kshs 174 960. Using
matrix method, find the price of each bicycle and each radio that he bought
(c) In the third week, the price of each bicycle was reduced by 10% while the price of each radio was raised by 10%. The businessman bought as many bicycles and as many radios as he had bought
in the first two weeks.
Find by matrix method, the total cost of the bicycles and radios that the businessman bought in the third week.
5. Determine the inverse T^-1 of the matrix 1 2
1 -1
Hence find the coordinates to the point at which the two lines x + 2y=7 and x-y=1
6. Given that A = 0 -1 and B = -1 0
3 2 2 -4
Find the value of x if
(i) A – 2x = 2B
(ii) 3x – 2A = 3B
(iii) 2A – 3B = 2x
7. Find the non- zero value of k for which k + 1 2 is an inverse.
4k 2k
8. A clothes dealer sold 3 shirts and 2 trousers for Kshs. 840 and 4 shirts and 5 trousers for Kshs 1680. Form a matrix equation to represent the above information. Hence find the cost of 1 shirt
and the cost of 1 trouser.
Specific Objectives
By the end of the topic the learner should be able to:
1. a) Rewrite a given formula by changing its subject
2. b) Define direct, inverse, partial and joint variations
3. c) Determine constants of proportionality
4. d) Form and solve equations involving variations
5. e) Draw graphs to illustrate direct and inverse proportions
6. f) Use variations to solve real life problems
• Change of the subject of a formula
• Direct, inverse, partial and joint variation
• Constants of proportionality
• Equations involving variations
• Graphs of direct and inverse proportion
• Formation of equations on variations based on real life situations
A Formula is an expression or equation that expresses the relationship between certain quantities.
For Example is the formula to find the area of a circle of radius r units.
From this formula, we can know the relationship between the radius and the area of a circle. The area of a circle varies directly as the square of its radius. Here is the constant of variation.
Changing the subject of a formulae
In the formula
C = d
Subject: C Rule: multiply by diameter
The variable on the left, is known as the subject: What you are trying to find.
The formula on the right, is the rule, that tells you how to calculate the subject.
So, if you want to have a formula or rule that lets you calculate d, you need
to make d, the subject of the formula.
This is changing the subject of the formula from C to d.
So clearly in the case above where
C = d
We get C by multiplying by the diameter
To calculate d, we need to divide the Circumference C by
So d and now we have d as the subject of the formula.
A formula is simply an equation, that you cannot solve, until you replace the letters with their
values (numbers). It is known as a literal equation.
To change the subject, apply the same rules as we have applied to normal equations.
1. Add the same variable to both sides.
2. Subtract the same variable from both sides.
3. Multiply both sides by the same variable.
4. Divide both sides by the same variable.
5. Square both sides
6. Square root both sides.
Make the letter in brackets the subject of the formula
x + p = q [ x ]
(subtract p from both sides)
x = q – p
y − r = s [ y ]
(add r to both sides)
y = s + r
P = RS [ R ]
(divide both sides by S)
S =
= L [ A ]
(multiply both sides by B)
A = LB
2w+ 3 = y [ w ]
(subtract 3 from both sides)
2w = y −3
(divide both sides by 2)
P = Q [ Q ]
(multiply both sides by 3− get rid of fraction)
3P = Q
T = k [ k ]
(multiply both sides by 5− get rid of fraction)
5T = 2k
(divide both sides by 2)
= k Note that: is the same as
A = r [ r ]
(divide both sides by p)
(square root both sides)
L = h −t [ h ]
(multiply both sides by 2)
2L = h −t
(add t to both sides)
2L + t = h
Make d the subject of the formula G=
Squaring both sides
Multiply both sides by d-1
Expanding the L.H.S
Collecting the terms containing d on the L.H.S
Factorizing the L.H.S
Dividing both sides by
In a formula some elements which do not change (fixed) under any condition are called constants while the ones that change are called variables. There are different types of variations.
• Direct Variation, where both variables either increase or decrease together
• Inverse or Indirect Variation, where when one of the variables increases, the other one decreases
• Joint Variation, where more than two variables are related directly
• Combined Variation, which involves a combination of direct or joint variation, and indirect variation
• Direct: The number of money I make varies directly (or you can say varies proportionally) with how much I work.
• Direct: The length of the side a square varies directly with the perimeter of the square.
• Inverse: The number of people I invite to my bowling party varies inversely with the number of games they might get to play (or you can say is proportional to the inverse of).
• Inverse: The temperature in my house varies indirectly (same as inversely) with the amount of time the air conditioning is running.
• Inverse: My school marks may vary inversely with the number of hours I watch TV.
Direct or Proportional Variation
When two variables are related directly, the ratio of their values is always the same. So as one goes up, so does the other, and if one goes down, so does the other. Think of linear direct
variation as a “y = mx” line, where the ratio of y to x is the slope (m). With direct variation, the y-intercept is always 0 (zero); this is how it’s defined.
Direct variation problems are typically written:
→ y= kx where k is the ratio of y to x (which is the same as the slope or rate).
Some problems will ask for that k value (which is called the constant of variation or constant of proportionality ); others will just give you 3 out of the 4 values for x and y and you can simply set
up a ratio to find the other value.
Remember the example of making ksh 1000 per week (y = 10x)? This is an example of direct variation, since the ratio of how much you make to how many hours you work is always constant.
Direct Variation Word Problem:
The amount of money raised at a school fundraiser is directly proportional to the number of people who attend. Last year, the amount of money raised for 100 attendees was $2500. How much money
will be raised if 1000 people attend this year?
Let’s do this problem using both the Formula Method and the Proportion Method:
Formula method Explanation
Proportional method Explanation
Direct Square Variation Word Problem
Again, a Direct Square Variation is when y is proportional to the square of x, or .
If yvaries directly with the square ofx, and if y = 4 when x= 3, what is y when x= 2?
Let’s do this with the formula method and the proportion method:
Formulae method notes
Proportional method Notes
The length (l) cm of a wire varies directly as the temperature c.The length of the wire is 5 cm when the temperature is .Calculate the length of the wire when the temperature is c.
Therefore l =Kt
Substituting l =5 when T= .
5 =k x 65
K =
Therefore l =
When t = 69
L =
Direct variation graph
Inverse or Indirect Variation
Inverse or Indirect Variation is refers to relationships of two variables that go in the opposite direction. Let’s supposed you are comparing how fast you are driving (average speed) to how fast you
get to your work.The faster you drive the earlier you get to your work. So as the speed increases time reduces and vice versa .
So the formula for inverse or indirect variation is:
→ y = or K =xy where k is always the same number or constant.
(Note that you could also have an Indirect Square Variation or Inverse Square Variation, like we saw above for a Direct Variation. This would be of the form→ y = or k= .)
Inverse Variation Word Problem:
So we might have a problem like this:
The value of yvaries inversely with x, and y = 4 when x = 3. Find x when y = 6.
The problem can also be written as follows:
Let = 3, = 4, and = 6. Let yvary inversely as x. Find .
We can solve this problem in one of two ways, as shown. We do these methods when we are given any three of the four values for x and y.
Product Rule Method:
Inverse Variation Word Problem:
For the club, the number of tickets Moyo can buy is inversely proportional to the price of the tickets. She can afford 15 tickets that cost $5 each. How many tickets can she buy if each cost $3?
Let’s use the product method:
If 16 women working 7 hours day can paint a mural in 48 days, how many days will it take 14 women working 12 hours a day to paint the same mural?
The three different values are inversely proportional; for example, the more women you have, the less days it takes to paint the mural, and the more hours in a day the women paint, the less days
they need to complete the mural:
Joint Variation and Combined Variation
Joint variation is just like direct variation, but involves more than one other variable. All the variables are directly proportional, taken one at a time. Let’s do a joint variation problem:
Supposed x varies jointly with y and the square root of z. When x = –18 and y = 2, then z = 9. Find y when x = 10 and z = 4.
Combined variation involves a combination of direct or joint variation, and indirect variation. Since these equations are a little more complicated, you probably want to plug in all the variables,
solve for k, and then solve back to get what’s missing. Here is the type of problem you may get:
(a) yvaries jointly as x and w and inversely as the square of z. Find the equation of variation when y = 100, x = 2, w = 4, and z = 20.
(b) Then solve for y when x = 1, w = 5, and z = 4.
The volume of wood in a tree (V) variesdirectly as the height (h) and inversely as the square of the girth (g). If the volume of a tree is 144 cubic meters when the height is 20 meters and the girth
is 1.5 meters, what is the height of a tree with a volume of 1000 and girth of 2 meters?
The average number of phone calls per day between two cities has found to be jointly proportional to the populations of the cities, and inversely proportional to the square of the distance between
the two cities. The population of Charlotte is about 1,500,000 and the population of Nashville is about 1,200,000, and the distance between the two cities is about 400 miles. The average number of
calls between the cities is about 200,000.
(a) Find the k and write the equation of variation.
(b) The average number of daily phone calls between Charlotte and Indianapolis (which has a population of about 1,700,000) is about 134,000. Find the distance between the two cities.
It may be easier if you take it one step at a time:
Math’s Explanation
A varies directly as B and inversely as the square root of C. Find the percentage change in A when B is decreased by 10 % and C increased by 21%.
A= K
A change in B and C causes a change in A
= 1.21C
Percentage change in A =
= – 18
Therefore A decreases 18
Partial variation
The general linear equation y =mx +c, where m and c are constants, connects two variables x and y.in such case we say that y is partly constant and partly varies as x.
A variable y is partly constant and partly varies as if x = 2 when y=7 and x =4 when y =11, find the equation connecting y and x.
The required equation is y = kx + c where k and c are constants
Substituting x = 2 ,y =7 and x =4, y =11 in the equation gives ;
7 =2k +c …………………..(1)
11 = 4k +c …………………(2)
Subtracting equation 1 from equation 2 ;
4 = 2 k
Therefore k =2
Substituting k =2 in the equation 1 ;
C =7 – 4
C =3
Therefore the equation required is y=2x +3
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. The volume Vcm^3 of an object is given by
V = 2 π r^31 – 2
3 sc^2
Express in term of π r, s and V
2. Make V the subject of the formula
T = 1 m (u^2 – v^2)
3. Given that y =b – bx^2 make x the subject
cx^2 – a
4. Given that log y = log (10^n) make n the subject
5. A quantity T is partly constant and partly varies as the square root of S.
1. Using constants a and b, write down an equation connecting T and S.
2. If S = 16, when T = 24 and S = 36 when T = 32, find the values of the constants a and b,
3. A quantity P is partly constant and partly varies inversely as a quantity q, given that p = 10 when q = 1.5 and p = 20, when q = 1.25, find the value of p when q= 0.5
4. Make y the subject of the formula p = xy
8. Make P the subject of the formula
P^2 = (P – q) (P-r)
9. The density of a solid spherical ball varies directly as its mass and inversely as the cube of its radius
When the mass of the ball is 500g and the radius is 5 cm, its density is 2 g per cm^3
Calculate the radius of a solid spherical ball of mass 540 density of 10g per cm^3
10. Make s the subject of the formula
√P = r 1 – as^2
11. The quantities t, x and y are such that t varies directly as x and inversely as the square root of y. Find the percentage in t if x decreases by 4% when y increases by 44%
12. Given that y is inversely proportional to x^n and k as the constant of proportionality;
(a) (i) Write down a formula connecting y, x, n and k
(ii) If x = 2 when y = 12 and x = 4 when y = 3, write down two expressions for k in terms of n.
Hence, find the value of n and k.
(b) Using the value of n obtained in (a) (ii) above, find y when x = 5 ^1/[3]
13. The electrical resistance, R ohms of a wire of a given length is inversely proportional to the square of the diameter of the wire, d mm. If R = 2.0 ohms when d = 3mm. Find the vale R when d = 4
14. The volume Vcm^3 of a solid depends partly on r and partly on r where rcm is one of the dimensions of the solid.
When r = 1, the volume is 54.6 cm^3 and when r = 2, the volume is 226.8 cm^3
(a) Find an expression for V in terms of r
(b) Calculate the volume of the solid when r = 4
(c) Find the value of r for which the two parts of the volume are equal
15. The mass of a certain metal rod varies jointly as its length and the square of its radius. A rod 40 cm long and radius 5 cm has a mass of 6 kg. Find the mass of a similar rod of length 25 cm and
radius 8 cm.
16. Make x the subject of the formula
P = xy
z + x
17. The charge c shillings per person for a certain service is partly fixed and partly inversely proportional to the total number N of people.
(a) Write an expression for c in terms on N
(b) When 100 people attended the charge is Kshs 8700 per person while for 35 people the charge is Kshs 10000 per person.
(c) If a person had paid the full amount charge is refunded. A group of people paid but ten percent of organizer remained with Kshs 574000.
Find the number of people.
18. Two variables A and B are such that A varies partly as B and partly as the square root of B given that A=30, when B=9 and A=16 when B=14, find A when B=36.
19. Make p the subject of the formula
A = -EP
√P^2 + N
Specific Objectives
By the end of the topic the learner should be able to:
(a) Identify simple number patterns;
(b) Define a sequence;
(c) Identify the pattern for a given set of numbers and deduce the general rule;
(d) Determine a term in a sequence;
(e) Recognize arithmetic and geometric sequences;
(f) Define a series;
(g) Recognize arithmetic and geometric series (Progression);
(h) Derive the formula for partial sum of an arithmetic and geometric series(Progression);
(i) Apply A.P and G.P to solve problems in real life situations.
(a) Simple number patterns
(b) Sequences
(c) Arithmetic sequence
(d) Geometric sequence
(e) Determining a term in a sequence
(f) Arithmetic progression (A.P)
(g) Geometric progression (G.P)
(h) Sum of an A.P
(i) Sum of a G.P (exclude sum to infinity)
(j) Application of A.P and G.P to real life situations.
Sequences and Series are basically just numbers or expressions in a row that make up some sort of a pattern; for example, Monday, Tuesday, Wenesday, …, Friaday is a sequence that represents the days
of the week. Each of these numbers or expressions are called terms or elements of the sequence.
Sequences are the list of these items, separated by commas, and series are the sum of the terms of a sequence.
Sequence Next two terms
1, 8, 27, – , – Every term is cubed .The next two terms are
3, 7, 11, 15 – , – , every term is 4 more than the previous one. To get the next term add 4
15 + 4 = 19, 19 +4 =23
On the numerator, the next term is 1 more than the previous one, and the denominator, the next term is multiplied by 2 the next two terms are
For the term of a sequence is given by 2n + 3, Find the first, fifth, twelfth terms
First term, n = 1 substituting (2 x 1 +3 =5)
Fifth term, n = 5 substituting (2 x 5 +3 =13)
Twelfth term, n = 12 substituting (2 x 12 +3 =27)
Arithmetic and geometric sequence
Arithmetic sequence.
Any sequence of a number with common difference is called arithmetic sequence
To decide whether a sequence is arithmetic, find the differences of consecutive terms. If each differences are not constant,the it is arithmetic sequence
Rule for an arithmetic sequence
The nth term of an arithmetic sequence with first term and common difference d is given by:
= + (n – 1)d
Write a rule for the nth term of the sequence 50, 44, 38, 32, . . . . Then find .
The sequence is arithmetic with first term = 50 and common difference
d = 44 – 50 = -6. So, a rule for the nth term is:
= + (n – 1)d Write general rule.
= 50 + (n – 1)(-6) Substitute for a1 and d.
= 56 – 6n Simplify.
The 20th term is = 56 – 6(20) = -64.
The 20 th term of arithmetic sequence is 60 and the 16 th term is 20.Find the first term and the common difference.
4d = 40
d= 10
Therefore a + 15 x 10 =20
a + 150 = 20
a = -130
Hence, the first term is – 130 and the common difference is 10.
Find the number of terms in the sequence – 3 , 0 , 3 …54
The n th term is a + ( n – 1)d
a = -30 , d =3
n th term = 54
therefore – 3 + ( n – 1) = 54
3 (n – 1 ) = 57
Arithmetic series/ Arithmetic progression A.P
The sum of the terms of a sequence is called a series. If the terms of sequence are 1, 2, 3, 4, 5, when written with addition sign we get arithmetic series
1 + 2 + 3 + 4 + 5
The general formulae for finding the sum of the terms is
If th first term (a) and the last term l are given , then
The sum of the first eight terms of an arithmetic Progression is 220.If the third term is 17, find the sum of the first six terms
= 4( 2a + 7d )
So , 8a + 28d = 220…………………….1
The third term is a + (3 – 1)d = a + 2d =17 …………….2
Solving 1 and 2 simultaneously;
8a + 28 d =220 …………1
8a + 16 d = 136 …………2
12 d = 84
Substituting d =7 in equation 2 gives a = 3
= 3(6 x 35)
= 3 x 41
= 123
Geometric sequence
It is a sequence with a common ratio.The ratio of any term to the previous term must be constant.
Rule for Geometric sequence is;
The nth term of a geometric sequence with first term a1 and common ratio r is given by:
Given the geometric sequence 4 , 12 ,36 ……find the 4^th , 5^th and the n th terms
The first term , a =4
The common ratio , r =3
Therefore the 4^th term = 4 x
= 4 x
= 108
The 5^th term = 5 x
= 5 x
= 324
The term =4 x
The 4^th term of geometric sequence is 16 . If the first term is 2 , find;
• The common ration
• The seventh term
The common ratio
The first term, a = 2
The 4^th term is 2 x
Thus, 2
The common ratio is 2
The seventh term =
Geometric series
The series obtained by the adding the terms of geometric sequence is called geometric series or geometric progression G.P
The sum of the first n terms of a geometric series with common ratio r > 1 is:
The sum of the first n terms of a geometric series with common ratio r < 1 is:
Find the sum of the first 9 terms of G.P. 8 + 24 + 72 +…
The sum of the first three terms of a geometric series is 26 .If the common ratio is 3 , find the sum of the first six terms.
^ a =
End of topic
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If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. The first, the third and the seventh terms of an increasing arithmetic progression are three consecutive terms of a geometric progression. In the first term of the arithmetic progression is 10
find the common difference of the arithmetic progression?
2. Kubai saved Ksh 2,000 during the first year of employment. In each subsequent year, he saved 15% more than the preceding year until he retired.
(a) How much did he save in the second year?
(b) How much did he save in the third year?
(c) Find the common ratio between the savings in two consecutive years
• How many years did he take to save the savings a sum of Ksh 58,000?
(e) How much had he saved after 20 years of service?
3. In geometric progression, the first term is a and the common ratio is r. The sum of the first two terms is 12 and the third term is 16.
a + ar
(b) If the first term is larger than the second term, find the value of r.
4. (a) The first term of an arithmetic progression is 4 and the last term is 20. The
Sum of the term is 252. Calculate the number of terms and the common differences of the arithmetic progression
(b) An Experimental culture has an initial population of 50 bacteria. The population increased by 80% every 20 minutes. Determine the time it will take to have a population of 1.2 million
5. Each month, for 40 months, Amina deposited some money in a saving scheme. In the first month she deposited Kshs 500. Thereafter she increased her deposits by Kshs. 50 every month.
Calculate the:
1. a) Last amount deposited by Amina
2. b) Total amount Amina had saved in the 40 months.
3. A carpenter wishes to make a ladder with 15 cross- pieces. The cross- pieces are to diminish uniformly in length from 67 cm at the bottom to 32 cm at the top.
Calculate the length in cm, of the seventh cross- piece from the bottom
7. The second and fifth terms of a geometric progression are 16 and 2 respectively. Determine the common ratio and the first term.
8. The eleventh term of an arithmetic progression is four times its second term. The sum of the first seven terms of the same progression is 175
(a) Find the first term and common difference of the progression
(b) Given that p^th term of the progression is greater than 124, find the least
value of P
9. The n^th term of sequence is given by 2n + 3 of the sequence
(a) Write down the first four terms of the sequence
(b) Find s[n] the sum of the fifty term of the sequence
(c) Show that the sum of the first n terms of the sequence is given by
S[n] = n^2 + 4n
Hence or otherwise find the largest integral value of n such that Sn <725
Specific Objectives
By the end of the topic the learner should be able to:
(a) Expand binomial expressions up to the power of four by multiplication;
(b) Building up – Pascal’s Triangle up to the eleventh row;
(c) Use Pascal’s triangle to determine the coefficient of terms in a binomialexpansions up to the power of 10;
(d) Apply binomial expansion in numerical cases.
(a) Binomial expansion up to power four
(b) Pascal’s triangle
(c) Coefficient of terms in binomial expansion
(d) Computation using binomial expansion
(e) Evaluation of numerical cases using binomial expansion.
A binomial is an expression of two terms
(a + y), a + 3, 2a + b
It easy to expand expressions with lower power but when the power becomes larger, the expansion or multiplication becomes tedious. We therefore use pascal triangle to expand the expression without
We can use Pascal triangle to obtain coefficients of expansions of the form( a + b
Pascal triangle
• Each row starts with 1
• Each of the numbers in the next row is obtained by adding the two numbers on either side of it in the preceding row
• The power of first term (a ) decreases as you move to right while the powers of the second term (b ) increases as you move to the right
Expand (p +
The terms without coefficients are;
From Pascal triangle, the coefficients when n =5 are; 1 5 10 10 5 1
Therefore (p + =
Expand (x
The terms without the coefficient are;
From Pascal triangle, the coefficients when n =7 are;
Therefore (x =
When dealing with negative signs, the signs alternate with the positive sign but first start with the negative sign.
Applications to Numeric cases
Use binomial expansion to evaluate (1.02
(1.02) = (1+0.02)
Therefore (1.02 = (1+ 0.02
The terms without coefficients are
From Pascal triangle, the coefficients when n =6 are;
(1.02 =
1 + 6 (0.02) + 15
=1 + 0.12 + 0.0060 + 0.00016 + 0.0000024 + 0.0000000192 + 0.000000000064
=1.126 (4 S.F)
To get the answer just consider addition of up to the 4^th term of the expansion. The other terms are too small to affect the answer.
Expand (1 + up to the term .Use the expansion to estimate (0.98 correct to 3 decimal places.
(1 +
The terms without the coefficient are;
From Pascal triangle, the coefficients when n =9 are;
Therefore (1 + = 1 + 9x + 36 + 84 ………………..
= 1 – 0.18 + 0.0144 – 0.000672
= 0.833728
= 0.834 ( 3 D.P)
Expand ( in ascending powers of hence find the value of ( correct to four decimal places.
Substituting for x = 0.01 in the expansion
= 1 + 0.05 +0.001125 +0.000015
= 1.051140
= 1.0511 (4 decimal places)
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. (a) Write down the simplest expansion ( 1 + x)^6
(b) Use the expansion up to the fourth term to find the value of (1.03)^6 to the nearest one thousandth.
2. Use binomial expression to evaluate (0.96)^5 correct to 4 significant figures.
3. Expand and simplify (3x – y)^4 hence use the first three terms of the expansion to proximate the value of (6 – 0.2)^4
4. Use binomial expression to evaluate
2 + 1^5[+] 2 – 1^5
√2 √2
5. (a) Expand the expression 1 + 1x ^5 in ascending powers of x, leaving
the coefficients as fractions in their simplest form.
6. (a) Expand (a- b)^6
(b) Use the first three terms of the expansion in (a) above to find the approximate value of (1.98)^6
7. Expand (2 + x)^5 in ascending powers of x up to the term in x^3 hence approximate the value of (2.03)^5 to 4 s.f
8. (a) Expand (1 + x)^5
Hence use the expansion to estimate (1.04)^5 correct to 4 decimal places
(b) Use the expansion up to the fourth term to find the value of (1.03)^6 to the nearest one thousandth.
9. Expand and Simplify (1-3x)^5 up to the term in x^3
Hence use your expansion to estimate (0.97)^5 correct to decimal places.
10. Expand (1 + a)^5
Use your expansion to evaluate (0.8)^5 correct to four places of decimal
11. (a) Expand (1 + x)^5
(b) Use the first three terms of the expansion in (a) above to find the approximate value of (0.98)^5
Specific Objectives
By the end of the topic the learner should be able to:
(a) Solve problems involving compound proportions using unitary and ratiomethods;
(b) Apply ratios and proportions to real life situations;
(c) Solve problems involving rates of work.
(a) Proportional parts
(b) Compound proportions
(c) Ratios and rates of work
(d) Proportions applied to mixtures.
Compound proportions
The proportion involving two or more quantities is called compound proportion. Any four quantities a , b , c and d are in proportion if;
Find the value of a that makes 2, 5, a and 25 to be in proportion;
Since 2 , 5 ,a , and 25 are in proportion
Continued proportions
In continued proportion, all the ratios between different quantities are the same; but always remember that the relationship exists between two quantities for example:
P : Q Q : R R : S
10: 5 16 : 8 4 : 2
Note that in the example, the ratio between different quantities i.e. P:Q, Q:R and R:S are the same i.e. 2:1 when simplified.
Continued proportion is very important when determining the net worth of individuals who own the same business or even calculating the amounts of profit that different individual owners of a company
or business should take home.
Proportional parts
In general, if n is to be divided in the ratio a: b: c, then the parts of n proportional to a, b, c are
Omondi, Joel, cheroot shared sh 27,000 in the ratio 2:3:4 respectively. How much did each get?
The parts of sh 27,000 proportional to 2, 3, 4 are
Three people – John, Debby and Dave contributed ksh 119, 000 to start a company. If the ratio of the contribution of John to Debby was 12:6 and the contribution of Debby to Dave was 8:4, determine
the amount in dollars that every partner contributed.
Ratio of John to Debby’s contribution = 12:6 = 2:1
Ratio of Debby to Dave’s contribution = 8:4 = 2:1
As you can see, the ratio of the contribution of John to Debby and that of Debby to Dave is in continued proportion.
To determine the ratio of the contribution between the three members, we do the calculation as follows:
John: Debby: Dave
12 : 6
8 : 4
We multiply the upper ratio by 8 and the lower ratio by 6, thus the resulting ratio will be:
John: Debby: Dave
96: 48 : 24
= 4 : 2 : 1
The total ratio = 7
The contribution of the different members can then be found as follows:
John contributed ksh 68, 000 to the company while Debby contributed ksh 34, 000 and Dave contributed ksh 17, 000
Example 2
You are presented with three numbers which are in continued proportion. If the sum of the three numbers is 38 and the product of the first number and the third number is 144, find the three numbers.
Let us assume that the three numbers in continued proportion or Geometric Proportion are a, ar and a where a is the first number and r is the rate.
a+ar+a = 38 ………………………….. (1)
The product of the 1^st and 3^rd is
a × a = 144
(ar)2 = 144………………………………..(2)
If we find the square root of (a , then we will have found the second number:
Since the value of the second number is 12, it then implies that the sum of the first and the third number is 26.
We now proceed and look for two numbers whose sum is 26 and product is 144.
Clearly, the numbers are 8 and 18.
Thus, the three numbers that we were looking for are 8, 12 and 18.
Let us work backwards and try to prove whether this is actually true:
8 + 12 + 18 = 18
What about the product of the first and the third number?
8 × 18 = 144
What about the continued proportion
The numbers are in continued proportion
Given that x: y =2:3, Find the ratio (5x – 4y): (x + y).
Since x: y =2: 3
(5x – 4y): (x + y) = (10k – 12 k) 🙁 2k + 3 k)
=-2k: 5k
= – 2: 5
If show that .
Substituting kc for a and kd for b in the expression
Therefore expression
Rates of work and mixtures
195 men working 10 hour a day can finish a job in 20 days. How many men employed to finish the job in 15 days if they work 13 hours a day.
Let x be the no. of men required
Days hours Men
15 13 x
20 x 10 x 195
Tap P can fill a tank in 2 hrs, and tap Q can fill the same tank in 4 hrs. Tap R can empty the tank in 3 hrs.
1. If tap R is closed, how long would it take taps P and Q to fill the tank?
2. Calculate how long it would take to fill the tank when the three taps P, Q and R. are left running?
1. Tap P fills of the tank in 1 h.
Tap Q fills of the tank in 1 h.
Tap R empties of the tank in 1 h.
In one hour, P and Q fill
Time taken to fill the tank
1. In 1 h, P and Q fill of tank while R empties of the tank.
When all taps are open , of the tank is filled in 1 hour.
= 2
In what proportion should grades of sugars costing sh.45 and sh.50 per kilogram be mixed in order to produce a blend worth sh.48 per kilogram?
Method 1
Let n kilograms of the grade costing sh.45 per kg be mixed with 1 kilogram of grade costing sh.50 per kg.
Total cost of the two blends is sh.
The mass of the mixture is
Therefore total cost of the mixture is
45n + 50 = 48 (n +1)
45n + 50 = 48 n + 48
50 = 3n + 48
2 = 3n
The two grades are mixed in the proportion
Method 2
Let x kg of grade costing sh 45 per kg be mixed with y kg of grade costing sh.50 per kg. The total cost will be sh.(45x + 50 y)
Cost per kg of the mixture is sh.
The proportion is x : y = 2:3
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. Akinyi bought and beans from a wholesaler. She then mixed the maize and beans the ratio 4:3 she brought the maize as Kshs. 12 per kg and the beans 4 per kg. If she was to make a profit of 30%
what should be the selling price of 1 kg of the mixture?
2. A rectangular tank of base 2.4 m by 2.8 m and a height of 3 m contains 3,600 liters of water initially. Water flows into the tank at the rate of 0.5 litres per second
Calculate the time in hours and minutes, required to fill the tank
3. A company is to construct a parking bay whose area is 135m^2. It is to be covered with concrete slab of uniform thickness of 0.15. To make the slab cement. Ballast and sand are to be mixed so
that their masses are in the ratio 1: 4: 4. The mass of m^3 of dry slab is 2, 500kg.
(a) (i) The volume of the slab
(ii) The mass of the dry slab
(iii) The mass of cement to be used
(b) If one bag of the cement is 50 kg, find the number of bags to be purchased
• If a lorry carries 7 tonnes of sand, calculate the number of lorries of sand
to be purchased.
4. The mass of a mixture A of beans and maize is 72 kg. The ratio of beans to maize
is 3:5 respectively
(a) Find the mass of maize in the mixture
(b) A second mixture of B of beans and maize of mass 98 kg in mixed with A. The final ratio of beans to maize is 8:9 respectively. Find the ratio of beans to maize in B
5. A retailer bought 49 kg of grade 1 rice at Kshs. 65 per kilogram and 60 kg of grade II rice at Kshs 27.50 per kilogram. He mixed the tow types of rice.
• Find the buying price of one kilogram of the mixture
• He packed the mixture into 2 kg packets
□ If he intends to make a 20% profit find the selling price per packet
□ He sold 8 packets and then reduced the price by 10% in order to attract customers. Find the new selling price per packet.
□ After selling ^1/[3] of the remainder at reduced price, he raised the price so as to realize the original goal of 20% profit overall. Find the selling price per packet of the remaining rice.
6. A trader sells a bag of beans for Kshs 1,200. He mixed beans and maize in the ration 3: 2. Find how much the trader should he sell a bag of the mixture to realize the same profit?
7. Pipe A can fill an empty water tank in 3 hours while, pipe B can fill the same tank in 6 hours, when the tank is full it can be emptied by pipe C in 8 hours. Pipes A and B are opened at the same
time when the tank is empty.
If one hour later, pipe C is also opened, find the total time taken to fill the tank
8. A solution whose volume is 80 litres is made 40% of water and 60% of alcohol. When litres of water are added, the percentage of alcohol drops to 40%
(a) Find the value of x
(b) Thirty litres of water is added to the new solution. Calculate the percentage
(c) If 5 litres of the solution in (b) is added to 2 litres of the original solution, calculate in the simplest form, the ratio of water to that of alcohol in the resulting solution
9. A tank has two inlet taps P and Q and an outlet tap R. when empty, the tank can be filled by tap P alone in 4 ½ hours or by tap Q alone in 3 hours. When full, the tank can be emptied in 2 hours
by tap R.
(a) The tank is initially empty. Find how long it would take to fill up the tank
• If tap R is closed and taps P and Q are opened at the same time (2mks)
• If all the three taps are opened at the same time
(b) The tank is initially empty and the three taps are opened as follows
P at 8.00 a.m
Q at 8.45 a.m
R at 9.00 a.m
(i) Find the fraction of the tank that would be filled by 9.00 a.m
(ii) Find the time the tank would be fully filled up
10. Kipketer can cultivate a piece of land in 7 hrs while Wanjiru can do the same work in 5 hours. Find the time they would take to cultivate the piece of land when working together.
11. Mogaka and Ondiso working together can do a piece of work in 6 days. Mogaka, working alone, takes 5 days longer than Onduso. How many days does it take Onduso to do the work alone.
12. Wainaina has two dairy farms A and B. Farm A produces milk with 3 ¼ percent fat and farm B produces milk with 4 ¼ percent fat.
(a) (i) The total mass of milk fat in 50 kg of milk from farm A and 30kg
of milk from farm B.
(ii) The percentage of fat in a mixture of 50 kg of milk A and 30 kg of milk from B
(b) Determine the range of values of mass of milk from farm B that must be used in a 50 kg mixture so that the mixture may have at least 4 percent fat.
13. A construction firm has two tractors T[1] and T[2.] Both tractors working together can complete the work in 6 days while T[1] alone can complete the work in 15 days. After the two tractors had
worked together for four days, tractor T1 broke down.
Find the time taken by tractor T[2] complete the remaining work.
Specific Objectives
By the end of the topic the learner should be able to:
(a) Makes a table of values from given relations;
(b) Use the table of values to draw the graphs of the relations;
(c) Determine and interpret instantaneous rates of change from a graph;
(d) Interpret information from graphs;
(e) Draw and interpret graphs from empirical data;
(f) Solve cubic equations graphically;
(g) Draw the line of best fit;
(h) Identify the equation of a circle;
(i) Find the equation of a circle given the centre and the radius;
(j) Determine the centre and radius of a circle and draw the circle on acartesian plane.
(a) Tables and graphs of given relations
(b) Graphs of cubic equations
(c) Graphical solutions of cubic equations
(d) Average rate of change
(e) Instantaneous rate of change
(f) Empirical data and their graphs
(g) The line of best fit
(h) Equation of a circle
(i) Finding of the equation of a circle
(j) Determining of the centre and radius of a circle.
These are ways or methods of solving mathematical functions using graphs.
Graphing solutions of cubic Equations
A cubic equation has the form
ax^3 + bx^2 + cx + d = 0
where a, b , c and d are constants
It must have the term in x^3 or it would not be cubic (and so a 0), but any or all of b, c and d can be zero. For instance,
x^3 −6x^2 +11x −6 = 0, 4x^3 +57 = 0, x^3 +9x = 0
are all cubic equations.
The graphs of cubic equations always take the following shapes.
Y =x^3 −6x^2 +11x −6 = 0.
Notice that it starts low down on the left, because as x gets large and negative so does x^3 and it finishes higher to the right because as x gets large and positive so does x^3. The curve crosses
the x-axis three times, once where x = 1, once where x = 2 and once where x = 3. This gives us our three separate solutions.
(a) Fill in the table below for the function y = -6 + x + 4x^2 + x^3 for -4 £x £ 2
x -4 -3 -2 -1 0 1 2
-6 -6 -6 -6 -6 -6 -6 -6
x -4 -3 -2 -1 0 1 2
4x^2 16 4
(b) Using the grid provided draw the graph for y = -6 + x + 4x^2 + x^3 for -4£ x £ 2
(c) Use the graph to solve the equations:-
-6 + x + 4x^2 + x^3 = 0
.x^3 + 4x^2 + x – 4 = 0
-2 + 4x^2 + x^3 = 0
The table shows corresponding values of x and y for y= -6 + x + 4x^2 + x^3
X -4 -3 -2 -1 0 1 2
-6 -6 -6 -6 -6 -6 -6 -6
X -4 -3 -2 -1 0 1 2
4x^2 64 36 16 4 0 4 16
X^3 -64 -27 -8 -1 0 1 8
Y=-6+x+4x^2+x^3 -10 0 0 -4 -6 0 20
From the graph the solutions for x are x =-3 , x = -2, x = 1
1. To solve equation y = x^3 + 4x^2 + x -6 we draw a straight line from the diffrence of the two equations and then we read the coordinates at the point of the intersetion of the curve and the
straight line
y = x^3 + 4x^2 + x -6
0 = x^3 + 4x^2 + x -4
y = -2 solutions 0.8 ,-1.5 and -3.2
x 1 0 -2
y = x^3 + 4x^2 + x – 6 y -3 -4 -8
0 = x^3 + 4x^2 + 0 – 2
y = x – 4
c (i) solution 0.8
And -3.2
Average Rate of change
Defining the Average Rate of Change
The notion of average rate of change can be used to describe the change in any variable with respect to another. If you have a graph that represents a plot of data points of the form (x, y), then the
average rate of change between any two points is the change in the y value divided by the change in the x value.
change in y
The average rate of change of y with respect to x
change in x
• The rate of change of a straight ( the slop)line is the same between all points along the line
• The rate of change of a quadratic function is not constant (does not remain the same)
The graph below shows the rate of growth of a plant,from the graph, the change in height between day 1 and day 3 is given by 7.5 cm – 3.8 cm = 3.7 cm.
Average rate of change is
The average rate of change for the next two days is = 0.65cm/day
• The rate of growth in the first 2 days was 1.85 cm/day while that in the next two days is only 0.65 cm /day.These rates of change are represented by the gradients of the lines PQ and QR
Number of days
The gradient of the straight line is 20 ,which is constant.The gradient represents the rate of distance with time (speed) which is 20 m/s.
Rate of change at an instant
We have seen that to find the rate of change at an instant ( particular point),we:
• Draw a tangent to the curve at that point
• Determine the gradient of the tangent
The gradient of the tangent to the curve at the point is the rate of change at that point.
Empirical graphs
An Empirical graph is a graph that you can use to evaluate the fit of a distribution to your data by drawing the line of best fit. This is because raw data usually have some errors.
The table below shows how length l cm of a metal rod varies with increase in temperature T ( .
O 1 2 3 5 6 7 8
Degrees C
Length cm 4.0 4.3 4.7 4.9 5.0 5.9 6.0 6.4
• There is a linear relation between length and temperature.
• We therefore draw a line of best fit that passes through as many points as possible.
• The remaining points should be distributed evenly below and above the line
The line cuts the y – axis at (0, 4) and passes through the point (5, 5.5).Therefore, the gradient of the line is = 0.3.The equation of the line is l =0.3T + 4.
Reduction of Non-linear Laws to Linear Form.
When we plot the graph of xy=k, we get a curve.But when we plot y against , w get a straight line whose gradient is k.The same approach is used to obtain linear relations from non-linear relations of
the form y .
The table below shows the relationship between A and r
r 1 2 3 4 5
A 3.1 12.6 28.3 50.3 78.5
It is suspected that the relation is of the form A= By drawing a suitable graph,verify the law connecting A and r and determine the value of K.
If we plot A against ,we should get a straight line.
r 1 2 3 4 5
A 3.1 12.6 28.3 50.3 78.5
Since the graph of A against is a straight line, the law A =k holds.The gradient of this line is 3.1 to one decimal place. This is the value of k.
From 1960 onwards, the population P of Kisumu is believed to obey a law of the form P = ,Where k and A are constants and t is the time in years reckoned from 1960.The table below shows the
population of the town since 1960.
r 1960 1965 1970 1975 1980 1985 1990
p 5000 6080 7400 9010 10960 13330 16200
By plotting a suitable graph, check whether the population growth obeys the given law. Use the graph to estimate the value of A.
The law to be tested is P= .Taking logs of both sides we get log P = .Log P = log K + t log A, which is in the form y = mx + Thus we plot log P against t.(Note that log A is a constant).The below
shows the corresponding values of t and log p.
r 1960 1965 1970 1975 1980 1985 1990
Log P 3.699 3.784 3.869 3.955 4.040 4.125 4.210
Since the graph is a straight line ,the law P = holds.
Log A is given by the gradient of the straight line.Therefore, log A = 0.017.
Hence,A = 1.04
Log k is the vertical intercept.
Hence log k =3.69
Therefore k = 4898
Thus, the relationship is P = 4898 (1.04
• Laws of the form y= can be written in the linear form as: log y = log k + x log A (by taking logs of both sides)
• When log y is plotted against x , a straight line is obtained.Its gradient is log A and the intercept is log k.
• The law of the form y = ,where k and n are constants can be written in linear form as;
• Log y =log k + n log x.
• We therefore plot log y is plotted against log x.
• The gradient of the line gives n while the vertical intercept is log k
For the law y = d + cx^2 to be verified it is necessary to plot a graph of the variables in a modified
Form as follows y =d is compared with y = mx + c that is y =
• Y is plotted on the y axis
• is plotted on the x axis
• The gradient is c
• The vertical axis intercept is d
For the law y – a = to be verified it is necessary to plot a graph of the variables in a x
Modified form as follows
y – a = , i.e. y = + a which is compared with y = mx + c
• y should be plotted on the y axis
• should be plotted on the x axis
• The gradient is b
• The vertical axis intercept is a
For the law y – e = to be verified it is necessary to plot a graph of the variables in a
Modified form as follows. The law y – e = is f compared with y = mx + c.
• y should be plotted on the vertical axis
• should be plotted on the horizontal axis
• The gradient is f
• The vertical axis intercept is e
For the law y – cx = bx^2 to be verified it is necessary to plot a graph of the variables in a
Modified form as follows. The law y – cx = b is = b x + c compared with y = mx + c,
• should be plotted on y axis
• X should be plotted on x axis
• The gradient is b
• The vertical axis intercept is c
For the law y = + bx to be verified it is necessary to plot a graph of the variables in a ax
Modified form as follows. The law = a compared with y = mx + c
• should be plotted on the vertical axis
• should be plotted on the horizontal axis
• The gradient is a
• The vertical intercept is b
Equation of a circle
A circle is a set of all points that are of the same distance r from a fixed point. The figure below is a circle centre ( 0,0) and radius 3 units
P ( x ,y ) is a point on the circle. Triangle PON is right – angled at N.
By Pythagoras’ theorem;
But ON = x, PN = y and OP =3 .Therefore,
The general equation of a circle centre ( 0 ,0 ) and radius r is
Find the equation of a circle centre (0, 0) passing through (3, 4)
Let the radius of the circle be r
From Pythagoras theorem;
Consider a circle centre ( 5 , 4 ) and radius 3 units.
In the figure below triangle CNP is right angled at N.By pythagoras theorem;
But CN= ( x – 5), NP = (y – 4) and CP =3 units.
Therefore, .
The equation of a circle centre ( a,b) and radius r units is given by;
Find the equation of a circle centre (-2 ,3) and radius 4 units
General equation of the circle is .Therefore a = -2 b =3 and r = 4
Line AB is the diameter of a circle such that the co-ordinates of A and B are ( -1 ,1) and(5 ,1) respectively.
• Determine the centre and the radius of the circle
• Hence, find the equation of the circle
Radius =
= = 3
• Equation of the circle is ;
The equation of a circle is given by – 6x + .Determine the centre and radius of the circle.
– 6x +
Completing the square on the left hand side;
– 6x +
Therefore centre of the circle is (3,-2) and radius is 4 units. Note that the sign changes to opposite positive sign becomes negative while negative sign changes to positive.
Write the equation of the circle that has and as endpoints of a diameter.
Method 1: Determine the center using the Midpoint Formula:
Determine the radius using the distance formula (center and end of diameter):
Equation of circle is:
Method 2: Determine center using Midpoint Formula (as before):
Thus, the circle equation will have the form
Find by plugging the coordinates of a point on the circle in for
Let’s use
Again, we get this equation for the circle:
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. The table shows the height metres of an object thrown vertically upwards varies with the time t seconds
The relationship between s and t is represented by the equations s = at^2 + bt + 10 where b are constants.
T 0 1 2 3 4 5 6 7 8 9 10
S 45.1 49.9 -80
• (i) Using the information in the table, determine the values of a and b
(ii) Complete the table
(b) (i) Draw a graph to represent the relationship between s and t
(ii) Using the graph determine the velocity of the object when t = 5 seconds
2. Data collected form an experiment involving two variables X and Y was recorded as shown in the table below
x 1.1 1.2 1.3 1.4 1.5 1.6
y -0.3 0.5 1.4 2.5 3.8 5.2
The variables are known to satisfy a relation of the form y = ax^3 + b where a and b are constants
• For each value of x in the table above, write down the value of x^3
• (i) By drawing a suitable straight line graph, estimate the values of a and b
(ii) Write down the relationship connecting y and x
3. Two quantities P and r are connected by the equation p = kr^n. The table of values
of P and r is given below.
P 1.2 1.5 2.0 2.5 3.5 4.5
R 1.58 2.25 3.39 4.74 7.86 11.5
1. a) State a liner equation connecting P and r.
2. b) Using the scale 2 cm to represent 0.1 units on both axes, draw a suitable
line graph on the grid provided. Hence estimate the values of K and n.
4. The points which coordinates (5,5) and (-3,-1) are the ends of a diameter of a circle centre A
(a) The coordinates of A
The equation of the circle, expressing it in form x^2 + y^2 + ax + by + c = 0
where a, b, and c are constants each computer sold
5. The figure below is a sketch of the graph of the quadratic function y = k
(x+1) (x-2)
Find the value of k
6. The table below shows the values of the length X ( in metres ) of a pendulum and the corresponding values of the period T ( in seconds) of its oscillations obtained in an experiment.
X ( metres) 0.4 1.0 1.2 1.4 1.6
T ( seconds) 1.25 2.01 2.19 2.37 2.53
(a) Construct a table of values of log X and corresponding values of log T,
correcting each value to 2 decimal places
1. b) Given that the relation between the values of log X and log T approximate to a linear law of the form m log X + log a where a and b are constants
(i) Use the axes on the grid provided to draw the line of best fit for the graph of log T against log X.
(ii) Use the graph to estimate the values of a and b
(iii) Find, to decimal places the length of the pendulum whose period is 1 second.
7. Data collection from an experiment involving two variables x and y was recorded as shown in the table below
X 1.1 1.2 1.3 1.4 1.5 1.6
Y -0.3 0.5 1.4 2.5 3.8 5.2
The variables are known to satisfy a relation of the form y = ax^3 + b where a and b
are constants
(a) For each value of x in the table above. Write down the value of x^3
(b) (i) By drawing s suitable straight line graph, estimate the values of a and b
(ii) Write down the relationship connecting y and x
8. Two variables x and y, are linked by the relation y = ax^n. The figure below shows part of the straight line graph obtained when log y is plotted against log x.
Calculate the value of a and n
9. The luminous intensity I of a lamp was measured for various values of voltage v across it. The results were as shown below
V(volts) 30 36 40 44 48 50 54
L (Lux ) 708 1248 1726 2320 3038 3848 4380
It is believed that V and l are related by an equation of the form l = aV^n where a and n are constant.
(a) Draw a suitable linear graph and determine the values of a and n
(b) From the graph find
(i) The value of I when V = 52
(ii) The value of V when I = 2800
10. In a certain relation, the value of A and B observe a relation B= CA + KA^2 where C and K are constants. Below is a table of values of A and B
A 1 2 3 4 5 6
B 3.2 6.75 10.8 15.1 20 25.2
(a) By drawing a suitable straight line graphs, determine the values of C and K.
(b) Hence write down the relationship between A and B
(c) Determine the value of B when A = 7
11. The variables P and Q are connected by the equation P = ab^q where a and b are constants. The value of p and q are given below
P 6.56 17.7 47.8 129 349 941 2540 6860
Q 0 1 2 3 4 5 6 7
(a) State the equation in terms of p and q which gives a straight line graph
(b) By drawing a straight line graph, estimate the value of constants a and b and give your answer correct to 1 decimal place.
Specific Objectives
By the end of the topic the learner should be able to:
(a) Define probability;
(b) Determine probability from experiments and real life situations;
(c) Construct a probability space;
(d) Determine theoretical probability;
(e) Differentiate between discrete and continuous probability;
(f) Differentiate mutually exclusive and independent events;
(g) State and apply laws of probability;
(h) Use a tree diagram to determine probabilities.
(a) Probability
(b) Experimental probability
(c) Range of probability measure 0 ^ P (x) ^1
(d) Probability space
(e) Theoretical probability
(f) Discrete and continuous probability (simple cases only)
(g) Combined events (mutually exclusive and independent events)
(h) Laws of probability
(i) The tree diagrams.
The likelihood of an occurrence of an event or the numerical measure of chance is called probability.
Experimental probability
This is where probability is determined by experience or experiment. What is done or observed is the experiment. Each toss is called a trial and the result of a trial is the outcome. The experimental
probability of a result is given by (the number of favorable outcomes) / (the total number of trials)
A boy had a fair die with faces marked 1to6 .He threw this die up 50 times and each time he recorded the number on the top face. The result of his experiment is shown below.
face 1 2 3 4 5 6
Number of times a face has shown up 11 6 7 9 9 8
What is the experimental provability of getting?
a.)1 b.) 6
a.) P(Event) =
P(1)= 11/50
b.) P(4)= 9/50
From the past records, out of the ten matches a school football team has played, it has won seven.How many possible games might the school win in thirty matches ?.
P(winning in one math) = 7/10.
Therefore the number of possible wins in thirty matches = 7/10 x 30 = 21 matches
Range of probability Measure
If P(A) is the probability of an event A happening and P(A’) is the probability of an event A not happening, Then P(A’)= 1 – P(A) and P(A’) + P(A)= 1
Probability are expressed as fractions, decimals or percentages.
Probability space
A list of all possible outcomes is probability space or sample space .The coin is such that the head or tail have equal chances of occurring. The events head or tail are said to be equally likely or
Theoretical probability
This can be calculated without necessarily using any past experience or doing any experiment. The probability of an event happening #number of favorable outcomes /total number of outcomes.
A basket contains 5 red balls, 4green balls and 3 blue balls. If a ball is picked at random from the basket, find:
a.)The probability of picking a blue ball
b.) The probability of not picking a red ball
a.)Total number of balls is 12
The number of blue balls is 3
a.) therefore, P (a blue ball) =3/12
b.)The number of balls which are not red is 7.
Therefore P ( not a red ball)= 7/12
A bag contains 6 black balls and some brown ones. If a ball is picked at random the probability that it is black is 0.25.Find the number of brown balls.
Let the number of balls be x
Then the probability that a black ball is picked at random is 6/x
Therefore 6/x = 0.25
x = 24
The total number of bald is 24
Then the number of brown balls is 24 – 6 =18
When all possible outcomes are count able, they are said to be discrete.
Types of probability
Combined Events
These are probability of two or more events occurring
Mutually Exclusive Events
Occurrence of one excludes the occurrence of the other or the occurrence of one event depend on the occurrence of the other.. If A and B are two mutually exclusive events, then ( A or B) = P (A) + P
(B). For example when a coin is tossed the result will either be a head or a tail.
P(head) + P( tail)
If [OR] is used then we add
Independent Events
Two events A and B are independent if the occurrence of A does not influence the occurrence of B and vice versa. If A and B are two independent events, the probability of them occurring together is
the product of their individual probabilities .That is;
P (A and B) = P (A) x P(B)
When we use [AND] we multiply ,this is the multiplication law of probability.
A coin is tosses twice. What is the probability of getting a tail in both tosses?
The outcome of the 2^nd toss is independ of the outcome of the first .
P (T and T ) = P( T) X P( T)
= =
A boy throws fair coin and a regular tetrahedron with its four faces marked 1,2,3 and 4.Find the probability that he gets a 3 on the tetrahedron and a head on the coin.
These are independent events.
P (H) = P(3) =
P (H and 3) = P (H) x P (3)
= ½ x ¼
= 1/8
A bag contains 8 black balls and 5 white ones.If two balls are drawn from the bag, one at a time,find the probability of drawing a black ball and a white ball.
• Without replacement
• With replacement
• There are only two ways we can get a black and a white ball: either drawing a white then a black,or drawing a black then a white.We need to find the two probabilities;
P( W followed by B) = P (W and B)
• P(B followed by W) = P (B and W)
The two events are mutually exclusive, therefore.
P (W followed by B) or ( B followed by W )= P( W followed by B ) + P ( B followed by W)
= P (W and B) + P( B and W)
Since we are replacing, the number of balls remains 13.
P (W and B) =
P ( B and W) =
P [(W and B) or (B and W)] = P (W and B) + P (B and W)
Kamau ,Njoroge and Kariuki are practicing archery .The probability of Kamau hitting the target is 2/5,that of Njoroge hitting the target is ¼ and that of Kariuki hitting the target is 3/7 ,Find the
probability that in one attempt;
• Only one hits the target
• All three hit the target
• None of them hits the target
• Two hit the target
• At least one hits the target
• P(only one hits the target)
=P (only Kamau hits and other two miss) =2/5 x 3/5 x 4/7
= 6/35
P (only Njoroge hits and other two miss) = 1/4 x 3/5 x 4/7
= 3/35
P (only Kariuki hits and other two miss) = 3/7 x 3/5 x ¾
= 27/140
P (only one hits) = P (Kamau hits or Njoroge hits or Kariuki hits)
= 6/35 + 3/35 +27/140
= 9/20
• P ( all three hit) = 2/5 x 1/4 x 3/7
= 3/70
• P ( none hits) = 3/5 x 3/4 x 4/7
= 9/35
• P ( two hit the target ) is the probability of ;
Kamau and Njoroge hit the target and Kariuki misses = 2/5 x 3/7 x 4/7
Njoroge and Kariuki hit the target and Kamau misses = 1/4 x 3/7 x 3/5
Kamau and Kariuki hit the target and Njoroge misses = 2/5 x 3/7 x 3/4
Therefore P (two hit target) = (2/5 x 1/4 x 4/7) + (1/4 x 3/7 x 3/5) + (2/5 x 3/7 x 3/4)
= 8/140 + 9/140 + 18/140
= ¼
• P (at least one hits the target) = 1 – P ( none hits the target)
= 1 – 9/35
= 26/35
P (at least one hits the target) = 1 – P (none hits the target)
= 26/35
P (one hits the target) is different from P (at least one hits the target)
Tree diagram
Tree diagrams allows us to see all the possible outcomes of an event and calculate their probality.Each branch in a tree diagram represents a possible outcome .A tree diagram which represent a coin
being tossed three times look like this;
From the tree diagram, we can see that there are eight possible outcomes. To find out the probability of a particular outcome, we need to look at all the available paths (set of branches).
The sum of the probabilities for any set of branches is always 1.
Also note that in a tree diagram to find a probability of an outcome we multiply along the branches and add vertically.
The probability of three heads is:
P (H H H) = 1/2 × 1/2 × 1/2 = 1/8
P (2 Heads and a Tail) = P (H H T) + P (H T H) + P (T H H)
= 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2
= 1/8 + 1/8 + 1/8
= 3/8
Bag A contains three red marbles and four blue marbles.Bag B contains 5 red marbles and three blue marbles.A marble is taken from each bag in turn.
• What is the probability of getting a blue bead followed by a red
• What is the probability of getting a bead of each color
• Multiply the probabilities together
P( blue and red) =4/7 x 5/8 = 20/56
• P(blue and red or red and blue) = P( blue and red ) + P (red and blue)
= 4/7 x 5/8 + 3/7 x 3/8
= 20/56 + 9/56
The probability that Omweri goes to Nakuru is ¼ .If he goes to Nakuru, the probability that he will see flamingo is ½ .If he does not go to Nakuru, the probability that he will see flamingo is 1/3
.Find the probability that;
• Omweri will go to Nakuru and see a flamingo.
• Omweri will not go to Nakuru yet he will see a flamingo
• Omweri will see a flamingo
Let N stand for going to Nakuru ,N’ stand for not going to Nakuru, F stand for seeing a flamingo and F’ stand for not seeing a flamingo.
• P (He goes to Nakuru and sees a flamingo) = P(N and F)
= P(N) X P(F)
= ¼ X ½
= 1/8
• P( He does not go to Nakuru and yet sees a flamingo) =P( N’) X P( F)
= P (N’ and F)
= 3/4 X 1/3
= ¼
• P ( He sees a flamingo) = P(N and F) or P ( N’ and F)
= P (N and F) + P (N’ and F)
= 1/8 + 1/4
= 3/8
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic.
1. The probabilities that a husband and wife will be alive 25 years from now are 0.7 and 0.9 respectively.
Find the probability that in 25 years time,
• Both will be alive
• Neither will be alive
• One will be alive
• At least one will be alive
2. A bag contains blue, green and red pens of the same type in the ratio 8:2:5 respectively. A pen is picked at random without replacement and its colour noted
(a) Determine the probability that the first pen picked is
(i) Blue
(ii) Either green or red
(b) Using a tree diagram, determine the probability that
(i) The first two pens picked are both green
(ii) Only one of the first two pens picked is red.
3. A science club is made up of boys and girls. The club has 3 officials. Using a tree diagram or otherwise find the probability that:
(a) The club officials are all boys
(b) Two of the officials are girls
4. Two baskets A and B each contain a mixture of oranges and limes, all of the same size. Basket A contains 26 oranges and 13 limes. Basket B contains 18 oranges and 15 limes. A child selected a
basket at random and picked a fruit at a random from it.
(a) Illustrate this information by a probabilities tree diagram
(b) Find the probability that the fruit picked was an orange.
5. In form 1 class there are 22 girls and boys. The probability of a girl completing the secondary education course is 3 whereas that of a boy is ^2/[3 ]
(a) A student is picked at random from class. Find the possibility that,
• The student picked is a boy and will complete the course
• The student picked will complete the course
(b) Two students are picked at random. Find the possibility that they are a boy
and a girl and that both will not complete the course.
6. Three representatives are to be selected randomly from a group of 7 girls and 8
boys. Calculate the probability of selecting two girls and one boy.
7. A poultry farmer vaccinated 540 of his 720 chickens against a disease. Two months later, 5% of the vaccinated and 80% of the unvaccinated chicken, contracted the disease. Calculate the
probability that a chicken chosen random contacted the disease.
8. The probability of three darts players Akinyi, Kamau, and Juma hitting the bulls eye are 0.2, 0.3 and 1.5 respectively.
(a) Draw a probability tree diagram to show the possible outcomes
(b) Find the probability that:
(i) All hit the bull’s eye
(ii) Only one of them hit the bull’s eye
(iii) At most one missed the bull’s eye
9. (a) An unbiased coin with two faces, head (H) and tail (T), is tossed three
times, list all the possible outcomes.
Hence determine the probability of getting:
(i) At least two heads
(ii) Only one tail
• During a certain motor rally it is predicted that the weather will be either dry (D) or wet (W). The probability that the weather will be dry is estimated to be ^7/[10]. The probability for a
driver to complete (C) the rally during the dry weather is estimated to be ^5/[6]. The probability for a driver to complete the rally during wet weather is estimated to be ^1/[10]. Complete the
probability tree diagram given below.
What is the probability that:
(i) The driver completes the rally?
(ii) The weather was wet and the driver did not complete the rally?
10. There are three cars A, B and C in a race. A is twice as likely to win as B while B is twice as likely to win as c. Find the probability that.
11. a) A wins the race
12. b) Either B or C wins the race.
13. In the year 2003, the population of a certain district was 1.8 million. Thirty per cent of the population was in the age group 15 – 40 years. In the same year, 120,000 people in the district
visited the Voluntary Counseling and Testing (VCT) centre for an HIV test.
If a person was selected at random from the district in this year. Find the probability that the person visited a VCT centre and was in the age group 15 – 40 years.
12. (a) Two integers x and y are selected at random from the integers 1 to 8. If the
same integer may be selected twice, find the probability that
• |x – y| = 2
• |x – y| is 5 or more
(iii) x>y
(b) A die is biased so that when tossed, the probability of a number r showing up, is given by p ® = Kr where K is a constant and r = 1, 2,3,4,5 and 6 (the number on the faces of the die
(i) Find the value of K
(ii) If the die is tossed twice, calculate the probability that the total
score is 11
13. Two bags A and B contain identical balls except for the colours. Bag A contains 4 red balls and 2 yellow balls. Bag B contains 2 red balls and 3 yellow balls.
□ If a ball is drawn at random from each bag, find the probability that both balls are of the same colour.
□ If two balls are drawn at random from each bag, one at a time without replacement, find the probability that:
(i) The two balls drawn from bag A or bag B are red
(ii) All the four balls drawn are red
14. During inter – school competitions, football and volleyball teams from Mokagu high school took part. The probability that their football and volleyball teams would win were ^3/[8] and ^4/[7]
Find the probability that
(a) Both their football and volleyball teams
(b) At least one of their teams won
15. A science club is made up of 5 boys and 7 girls. The club has 3 officials. Using a tree diagram or otherwise find the probability that:
(a) The club officials are all boys
(b) Two of the officials are girls
16. Chicks on Onyango’s farm were noted to have either brown feathers brown or black tail feathers. Of those with black feathers 2/3 were female while ^2/[5] of those with brown feathers were male.
Otieno bought two chicks from Onyango. One had black tail feathers while the other had brown find the probability that Otieno’s chicks were not of the same gender
17. Three representatives are to be selected randomly from a group of 7 girls and 8 boys. Calculate the probability of selecting two girls and one boy
18. The probability that a man wins a game is ¾. He plays the game until he wins. Determine the probability that he wins in the fifth round.
19. The probability that Kamau will be selected for his school’s basketball team is ¼. If he is selected for the basketball team. Then the probability that he will be selected for football is ^1/[3]
if he is not selected for basketball then the probability that he is selected for football is ^4/[5]. What is the probability that Kamau is selected for at least one of the two games?
20. Two baskets A and B each contains a mixture of oranges and lemons. Baskets A contains 26 oranges and 13 lemons. Baskets B contains 18 oranges and 15 lemons. A child selected a basket at random
and picked at random a fruit from it. Determine the probability that the fruit picked an orange.
Specific Objectives
By the end of the topic the learner should be able to:
(a) Locate a point in two and three dimension co-ordinate systems;
(b) Represent vectors as column and position vectors in three dimensions;
(c) Distinguish between column and position vectors;
(d) Represent vectors in terms of i, j , and k;
(e) Calculate the magnitude of a vector in three dimensions;
(f) Use the vector method in dividing a line proportionately;
(g) Use vector method to show parallelism;
(h) Use vector method to show collinearity;
(i) State and use the ratio theorem,
(j) Apply vector methods in geometry.
(a) Coordinates in two and three dimensions
(b) Column and position vectors in three dimensions
(c) Column vectors in terms of unit vectors i, j , and k
(d) Magnitude of a vector
(e) Parallel vectors
(f) Collinearity
(g) Proportional division of a line
(h) Ratio theorem
(i) Vector methods in geometry.
Vectors in 3 dimensions:
3 dimensional vectors can be represented on a set of 3 axes at right angles to each other (orthogonal), as shown in the diagram.
Note that the z axis is the vertical axis.
To get from A to B you would move:
4 units in the x-direction, (x-component)
3 units in the y-direction, (y-component)
2 units in the z-direction. (z-component)
In component form: =
In general: = ,
Column and position vectors
In three dimensions, a displacement is represented b a column vector of the form where p,q and r are the changes in x,y,z directions respectively.
The displacement from A ( 3, 1, 4 ) to B ( 7 ,2,6) is represented b the column vector,
The position vector of A written as OA is where O is the origin
Addition of vectors in three dimensions is done in the same way as that in two dimensions.
If a = then
Column Vectors in terms of unit Vectors
In three dimension the unit vector in the x axis direction is = ,that in the dirction of the y axis is while that in the direction of z – axis is .
Diagrammatic representation of the vectors.
Three unit vectors are written as ; i =
Express vector in terms of the unit vector I , j and k
=5i – 2j +7k
The column vector can be expressed as a i + b j + ck
Magnitude of a 3 dimensional vector.
Given the vector AB = xi + y j + 2 k,then the magnitude of AB is written as |AB| =
This is the length of the vector.
Use Pythagoras’ Theorem in 3 dimensions.
AB^2 = AR^2 + BR^2
= (AP^2 + PR^2) + BR^2
and if u = then the magnitude of u, | u | = length of AB
Distance formula for 3 dimensions
Recall that since: = , then if then
Since x =
1. If A is (1, 3, 2) and B is (5, 6, 4)
2. If Find
Parallel vectors and collinearity
Parallel vectors
Two vectors are parallel if one is scalar multiple of the other.i.e vector a is a scalar multiple of b ,i.e .
a =kb then the two vectors are parallel.
Scalar multiplication is simply multiplication of a regular number by an entry in the vector
Multiplying by a scalar
A vector can be multiplied by a number (scalar).e.g. multiply a by 3 is written as 3 a.Vector 3a has three times the length but is in the same direction as a .In column form, each
component will be multiplied by 3.
We can also take a common factor out of a vector in component form. If a vector is a scalar multiple of another vector, then the two vectors are parallel, and differ only in magnitude. This is a
useful test to see if lines are parallel.
Example if
Collinear Points
Points are collinear if one straight line passes through all the points. For three points A, B, C – if the line AB is parallel to BC, since B is common to both lines, A, B and C are collinear.
Test for collinearity
A is (0, 1, 2), B is (1, 3, –1) and C is (3, 7, –7) Show that A, B and C are collinear.
are scalar multiples, so AB is parallel to BC.Since B is a common point, then A, B and C are collinear.
In general the test of collinearity of three points consists of two parts
• Showing that the column vectors between any two of the points are parallel
• Showing that they have a point in common.
A (0,3), B (1,5) and C ( 4,11) are three given points. Show that they are collinear.
AB and BC are parallel if AB = kBC ,where k is a scalar
AB= BC =
Therefore AB//BC and point B (1,5) is common. Therefore A,B,and C are collinear.
Show that the points A (1,3,5) ,B( 4,12,20) and C are collinear.
Consider vectors AB and AC
AB =
AC =
Hence k =
AC =
Therefore AB//AC and the two vectors share a common point A.The three points are thus collinear.
In the figure above OA = a OB = b and OC = 3OB
• Express AB and AC in terms of a and b
• Given that AM = ¾ AB and AN = , Express OM and O in terms of a and b
• Hence ,show that OM and N are collinear
= – a + b
AC = – a + 3b
= OA +
= a +
= a – b
= b
ON =OA +AN
=OA + AC
a a +
= b
Comparing the coefficients of a;
Thus, OM = ON.
Thus two vectors also share a common point ,O .Hence, the points are collinear.
Proportional Division of a line
In the figure below, the line is divided into 7 equal parts
The point R lies 4/7 of the ways along PQ if we take the direction from P to Q to be positive, we say R divides PQ internally in the ratio 4 : 3..
If Q to P is taken as positive,then R divides QP internally in the ratio 3 : 4 .Hence,QR : RP = 3 : 4 or ,4 QR = 3RP.
External Division
In internal division we look at the point within a given interval while in external division we look at points outside a given interval,
In the figure below point P is produced on AB
The line AB is divided into three equal parts with BP equal to two of these parts. If the direction from A to B is taken as positive, then the direction from P to B is negative.
Thus AP : PB = 5 : -2.In this case we say that P divides AB externally in the ratio 5 : -2 or P divides AB in the ratio 5 : -2.
Points, Ratios and Lines
Find the ratio in which a point divides a line.
The points A(2, –3, 4), B(8, 3, 1) and C(12, 7, –1) form a straight line. Find the ratio in which B divides AC.
B divides AC in ratio of 3 : 2
Points dividing lines in given ratios.
P divides AB in the ratio 4:3. If A is (2, 1, –3) and B is (16, 15, 11), find the co-ordinates of P.
\ 3(p – a) = 4(b – p)
3p – 3a = 4b – 4p
7p = 4b + 3a
Points dividing lines in given ratios externally.
Q divides MN externally in the ratio of 3:2. M is (–3, –2, –1) and N is (0, –5, 2).Find the co-ordinates of Q.
Note that QN is shown as –2 because the two line segments are MQ and QN, and QN is in the opposite direction to MQ.
\ –2(q – m) = 3(n – q)
–2q + 2m = 3n – 3q
q = 3n – 2m
P is P(10, 9, 5)
The Ration Theorem
The figure below shows a point S which divides a line AB in the ratio m : n
Taking any point O as origin, we can express s in terms of a and b the positon vectors of a and b respectively.
OS = OA + AS
But AS =
Therefore, OS = OA +
Thus S = a +
= a –
= (1 –
= +
This is called the ratio theorem. The theorem states that the position vectors s of a point which divides a line AB in the ratio m: n is given by the formula;
S = , where a and b are positon vectors of A and B respectively. Note that the sum of co-ordinates 1
Thus ,in the above example if the ratio m : n = 5 : 3
Then m = 5 and n = 3
OR =
Thus ,r = a +
A point R divides a line QR externally in the ratio 7 : 3 .If q and r are position vectors of point Q and R respectively, find the position vector of p in terms of q and r.
We take any point O as the origin and join it to the points Q, R and P as shown below
QP: PR = 7: -3
Substituting m =7 and n = -3 in the general formulae;
OP =
P =
Vectors can be used to determine the ratio in which a point divides two lines if they intersect
In the below OA = a and OB = B.A point P divides OAin the ratio 3:1 and another point O divides AB in the ratio 2 : 5 .If OQ meets BP at M Determine:
Let OM : MQ = k : ( 1 – k) and BM –MP = n : ( 1 – n )
Using the ratio theorem
OQ =
Also by ratio theorem;
OM = n OP +( 1 – n ) OB
But OP = a
Therefore , OM = n (
Equating the two expressions;
Comparing the co-efficients
= 10: 3
End of topic
Did you understand everything?
If not ask a teacher, friends or anybody and make sure you understand before going to sleep!
Past KCSE Questions on the topic
1. The figure below is a right pyramid with a rectangular base ABCD and VO as the height. The vectors AD= a, AB = b and DV = v
1. Express
(i) AV in terms of a and c
(ii) BV in terms of a, b and c
(b) M is point on OV such that OM: MV=3:4, Express BM in terms of a, b and c.
Simplify your answer as far as possible
2. In triangle OAB, OA = a OB = b and P lies on AB such that AP: BP = 3.5
• Find the terms of a and b the vectors
• AB
• AP
• BP
• OP
• Point Q is on OP such AQ = -5 + 9
8a 40b
Find the ratio OQ: QP
3. The figure below shows triangle OAB in which M divides OA in the ratio 2: 3 and N divides OB in the ratio 4:1 AN and BM intersect at X
(a) Given that OA = a and OB = b, express in terms of a and b:
(i) AN
(ii) BM
(b) If AX = s AN and BX = tBM, where s and t are constants, write two expressions
for OX in terms of a,b s and t
Find the value of s
Hence write OX in terms of a and b
4. The position vectors for points P and Q are 4 I + 3 j + 6 j + 6 k respectively. Express vector PQ in terms of unit vectors I, j and k. Hence find the length of PQ, leaving your answer in
simplified surd form.
5. In the figure below, vector OP = P and OR =r. Vector OS = 2r and OQ = ^3/[2]p.
1. a) Express in terms of p and r (i) QR and (ii) PS
2. b) The lines QR and PS intersect at K such that QK = m QR and PK = n PS, where m and n are scalars. Find two distinct expressions for OK in terms of p,r,m and n. Hence find the values of m and n.
3. c) State the ratio PK: KS
4. Point T is the midpoint of a straight line AB. Given the position vectors of A and T are i-j + k and 2i+ 1½ k respectively, find the position vector of B in terms of i, j and k
5. A point R divides a line PQ internally in the ration 3:4. Another point S, divides the line PR externally in the ration 5:2. Given that PQ = 8 cm, calculate the length of RS, correct to 2 decimal
6. The points P, Q, R and S have position vectors 2p, 3p, r and 3r respectively, relative to an origin O. A point T divides PS internally in the ratio 1:6
(a) Find, in the simplest form, the vectors OT and QT in terms p and r
(b) (i) Show that the points Q, T, and R lie on a straight line
(ii) Determine the ratio in which T divides QR
9. Two points P and Q have coordinates (-2, 3) and (1, 3) respectively. A translation map point P to P’ (10, 10)
• Find the coordinates of Q’ the image of Q under the translation
• The position vector of P and Q in (a) above are p and q respectively given that mp – nq = -12
Find the value of m and n
10. Given that q i + ^1/[3] j + ^2/[3] k is a unit vector, find q
11. In the diagram below, the coordinates of points A and B are (1, 6) and (15, 6) respectively). Point N is on OB such that 3 ON = 2 OB. Line OA is produced to L such that OL = 3 OA
(a) Find vector LN
(b) Given that a point M is on LN such that LM: MN = 3: 4, find the coordinates of M
(c) If line OM is produced to T such that OM: MT = 6:1
(i) Find the position vector of T
(ii) Show that points L, T and B are collinear
12. In the figure below, OQ = q and OR = r. Point X divides OQ in the ratio 1: 2 and Y divides OR in the ratio 3: 4 lines XR and YQ intersect at E.
• Express in terms of q and r
(i) XR
(ii) YQ
(b) If XE = m XR and YE = n YQ, express OE in terms of:
(i) r, q and m
(ii) r, q and n
(c) Using the results in (b) above, find the values of m and n.
13. Vector q has a magnitude of 7 and is parallel to vector p. Given that
p= 3 i –j + 1 ½ k, express vector q in terms of i, j, and k.
14. In the figure below, OA = 3i + 3j ABD OB = 8i – j. C is a point on AB such that AC:CB 3:2, and D is a point such that OB//CD and 2OB = CD (T17)
Determine the vector DA in terms of I and j
15. In the figure below, KLMN is a trapezium in which KL is parallel to NM and KL = 3NM
Given that KN = w, NM = u and ML = v. Show that 2u = v + w
16. The points P, Q and R lie on a straight line. The position vectors of P and R are 2i + 3j + 13k and 5i – 3j + 4k respectively; Q divides SR internally in the ratio 2: 1. Find the
(a) Position vector of Q
(b) Distance of Q from the origin
17. Co-ordinates of points O, P, Q and R are (0, 0), (3, 4), (11, 6) and (8, 2) respectively. A point T is such that the vector OT, QP and QR satisfy the vector equation OT = QP ½ QT. Find the
coordinates of T.
18. In the figure below OA = a, OB = b, AB = BC and OB: BD = 3:1
(a) Determine
(i) AB in terms of a and b
(ii) CD, in terms of a and b
(b) If CD: DE = 1 k and OA: AE = 1m determine
(i) DE in terms of a, b and k
(ii) The values of k and m
19. The figure below shows a grid of equally spaced parallel lines
AB = a and BC = b
(a) Express
(i) AC in terms of a and b
(ii) AD in terms of a and b
(b) Using triangle BEP, express BP in terms of a and b
(c) PR produced meets BA produced at X and PR = ^1/[9]b – ^8/[3]a
By writing PX as kPR and BX as hBA and using the triangle BPX determine the ratio PR: RX
20. The position vectors of points x and y are x = 2i + j – 3k and y = 3i + 2j – 2k respectively. Find XY
2. Given that X = 2i + j -2K, y = -3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures. | {"url":"https://www.citizennewsline.co.ke/form-3-maths-free-notes/","timestamp":"2024-11-07T00:23:30Z","content_type":"text/html","content_length":"358194","record_id":"<urn:uuid:5f22526a-22bc-4b6f-8032-2d9e01fd6117>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00267.warc.gz"} |
One simulation to fit them all - changing the background parameters of a cosmological N-body simulation
We demonstrate that the output of a cosmological N-body simulation can, to remarkable accuracy, be scaled to represent the growth of large-scale structure in a cosmology with parameters similar to
but different from those originally assumed. Our algorithm involves three steps: a reassignment of length, mass and velocity units; a relabelling of the time axis and a rescaling of the amplitudes of
individual large-scale fluctuation modes. We test it using two matched pairs of simulations. Within each pair, one simulation assumes parameters consistent with analyses of the first-year Wilkinson
Microwave Anisotropy Probe (WMAP) data. The other has lower matter and baryon densities and a 15 per cent lower fluctuation amplitude, consistent with analyses of the three-year WMAP data. The pairs
differ by a factor of a thousand in mass resolution, enabling performance tests on both linear and non-linear scales. Our scaling reproduces the mass power spectra of the target cosmology to better
than 0.5 per cent on large scales (k < 0.1hMpc^-1) both in real and in redshift space. In particular, the baryonic acoustic oscillation features of the original cosmology are removed and are
correctly replaced by those of the target cosmology. Errors are still below 3 per cent for k < 1hMpc^-1. Power spectra of the dark halo distribution are even more precisely reproduced, with errors
below 1 per cent on all scales tested. A halo-by-halo comparison shows that centre-of-mass positions and velocities are reproduced to better than 90h^-1kpc and 5 per cent, respectively. Halo masses,
concentrations and spins are also reproduced at about the 10 per cent level, although with small biases. Halo assembly histories are accurately reproduced, leading to central galaxy magnitudes with
errors of about 0.25mag and a bias of about 0.13mag for a representative semi-analytic model. This algorithm will enable a systematic exploration of the coupling between cosmological parameter
estimates and uncertainties in galaxy formation in future large-scale structure surveys.
Monthly Notices of the Royal Astronomical Society
Pub Date:
June 2010
□ cosmology: theory;
□ large-scale structure of Universe;
□ Astrophysics - Cosmology and Nongalactic Astrophysics;
□ Astrophysics - Astrophysics of Galaxies
14 pages, 12 figures. Submitted to MNRAS | {"url":"https://ui.adsabs.harvard.edu/abs/2010MNRAS.405..143A/abstract","timestamp":"2024-11-03T16:37:51Z","content_type":"text/html","content_length":"44389","record_id":"<urn:uuid:cdb3ecdf-2baa-49dd-9ca2-dcb44689bcfd>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00111.warc.gz"} |
1800 - 1810 - Chronology
• Lacroix completes publication of his three volume textbook Traité de Calcul differéntiel et intégral.
• Gauss publishes Disquisitiones Arithmeticae (Discourses on Arithmetic). It contains seven sections, the first six of which are devoted to number theory and the last to the construction of a
regular 17-gon by ruler and compasses.
• The minor planet Ceres is discovered but then lost. Gauss computes its orbit from the few observations that had been made leading to Ceres being rediscovered in almost exactly the position
predicted by Gauss.
• Gauss proves Fermat's conjecture that every number can be written as the sum of three triangular numbers.
• Lazare Carnot publishes Géométrie de position in which sensed magnitudes are first used systematically in geometry.
• Bessel publishes a paper on the orbit of Halley's comet using data from Harriot's observations 200 years earlier.
• Argand introduces the Argand diagram as a way of representing a complex number geometrically in the plane.
• Legendre develops the method of least squares to find best approximations to a set of observed data.
• Fourier discovers his method of representing continuous functions by the sum of a series of trigonometric functions and uses the method in his paper On the Propagation of Heat in Solid Bodies
which he submits to the Paris Academy.
• Poinsot discovers two new regular polyhedra.
• Gauss describes the least-squares method which he uses to find orbits of celestial bodies in Theoria motus corporum coelestium in sectionibus conicis Solem ambientium (Theory of the Movement of
Heavenly Bodies). | {"url":"https://mathshistory.st-andrews.ac.uk/Chronology/18/","timestamp":"2024-11-14T11:19:44Z","content_type":"text/html","content_length":"13167","record_id":"<urn:uuid:714cafe8-9485-49b8-8bad-ce9b230a01a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00093.warc.gz"} |
Peter decides to invest $1,200,000 in a period annuity that earns 4.6% APR compounded monthly for a period of 20 years. How much money will Peter be paid each month?
1. Home
2. General
3. Peter decides to invest $1,200,000 in a period annuity that earns 4.6% APR compounded monthly for a... | {"url":"https://math4finance.com/general/peter-decides-to-invest-1-200-000-in-a-period-annuity-that-earns-4-6-apr-compounded-monthly-for-a-period-of-20-years-how-much-money-will-peter-be-paid-each-month","timestamp":"2024-11-10T16:05:29Z","content_type":"text/html","content_length":"30678","record_id":"<urn:uuid:1046f90e-1e29-485b-8cd7-5f0d9ed7063f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00429.warc.gz"} |
Heads up for a hot new root seeking algorithm!!
01-10-2017, 06:30 PM
Post: #1
Namir Posts: 1,107
Senior Member Joined: Dec 2013
Heads up for a hot new root seeking algorithm!!
The famous Russian mathematician Ostrowski, who taught for many years at the University of Basil, Switzerland, proposedf an enhancement to Newton’s root seeking algorithm. Ostrowski suggested that
each iteration offers two refinements for the root—one of them being intermediate per each iteration. The Ostrowski algorithm matches Halley’s root seeking algorithm in its third order rate of
convergence. Recently, the Ostrowski algorithm inspired many mathematicians to device root-seeking algorithms with two or more refinements to the root per iteration.
I recently applied Ostrowski’s approach to the Illinois algorithm (an improved version of the False Position algorithm) and was able to obtain better rates of convergence than with the Illinois
algorithm. I posted the pseudo-code for this algorithm on this web site. I was a little baffled as to why Ostrowski improved only the Newton’s method and did not become more ambitious to enhance
Halley’s method!
A few days ago, I decided to experiment with applying Ostrowski’s approach to Halley’s algorithm. Since the latter method is a bit more advanced than Newton’s method (requiring the calculations of
approximations for the first AND second derivatives), applying the Ostrowski approach was NOT trivial. I decided, nevertheless, to give it a go. I started with a simple improvement to Halley’s
method, but that did not yield better calculations. After two or three incarnations, I was able to find a satisfactory marriage between Ostrowski and Halley. I plan to publish a report on my website
and include a comparison between the methods of Newton, Halley, Ostrowski, and my new Ostrowski-Halley method. The results will include testing these algorithms with two dozen functions and reporting
the number of function calls AND iterations.
I am happy to report that the new Ostrowski-Halley method competes well with Halley’s method and its cousin the Ostrowski method. The competition is stiff between these three algorithms., but the
Ostrowski-Halley method shows good promise. Stay tuned for my announcement of posting a paper on this subject in my personal web site.
01-10-2017, 08:06 PM
Post: #2
JurgenRo Posts: 205
Member Joined: Jul 2015
RE: Heads up for a hot new root seeking algorithm!!
Hi Namir, very promising! Do you have any stict proof of convergence? Also, interesting woul be an estimate for the achievable convergence rate (and how this depends on type/smoothness etc. of the
function to be "rooted").
Anyways, am curious to read your paper and the comparison with the original algorithms.
01-10-2017, 09:54 PM
Post: #3
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-10-2017 06:30 PM)Namir Wrote: The famous Russian mathematician Ostrowski, who taught for many years at the University of Basil, Switzerland, proposedf an enhancement to Newton’s root seeking
algorithm. Ostrowski suggested that each iteration offers two refinements for the root—one of them being intermediate per each iteration. The Ostrowski algorithm matches Halley’s root seeking
algorithm in its third order rate of convergence. Recently, the Ostrowski algorithm inspired many mathematicians to device root-seeking algorithms with two or more refinements to the root per
I recently applied Ostrowski’s approach to the Illinois algorithm (an improved version of the False Position algorithm) and was able to obtain better rates of convergence than with the Illinois
algorithm. I posted the pseudo-code for this algorithm on this web site. I was a little baffled as to why Ostrowski improved only the Newton’s method and did not become more ambitious to enhance
Halley’s method!
A few days ago, I decided to experiment with applying Ostrowski’s approach to Halley’s algorithm. Since the latter method is a bit more advanced than Newton’s method (requiring the calculations
of approximations for the first AND second derivatives), applying the Ostrowski approach was NOT trivial. I decided, nevertheless, to give it a go. I started with a simple improvement to Halley’s
method, but that did not yield better calculations. After two or three incarnations, I was able to find a satisfactory marriage between Ostrowski and Halley. I plan to publish a report on my
website and include a comparison between the methods of Newton, Halley, Ostrowski, and my new Ostrowski-Halley method. The results will include testing these algorithms with two dozen functions
and reporting the number of function calls AND iterations.
I am happy to report that the new Ostrowski-Halley method competes well with Halley’s method and its cousin the Ostrowski method. The competition is stiff between these three algorithms., but the
Ostrowski-Halley method shows good promise. Stay tuned for my announcement of posting a paper on this subject in my personal web site.
Keep them coming. I hope you don't mind if one or more of your algorithms end up implemented in newRPL (proper credit will be given, of course!). Always interesting to discover something new in a
subject that seemed closed already.
01-10-2017, 10:14 PM
Post: #4
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
Since it's on-topic:
What do you think about more advanced methods like this one:
There's a lot of operations in each iteration, but smaller number of evaluations to find the root. Considering that newRPL has relatively slow CORDIC transcendental functions, evaluation is heavy and
one of these could perhaps be beneficial, but have you ever tested if it's actually worth it? Sometimes they just want to prove a theoretical point, but they are not really better than plain Newton.
01-11-2017, 08:05 PM
(This post was last modified: 01-11-2017 08:19 PM by Namir.)
Post: #5
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-10-2017 10:14 PM)Claudio L. Wrote: Since it's on-topic:
What do you think about more advanced methods like this one:
There's a lot of operations in each iteration, but smaller number of evaluations to find the root. Considering that newRPL has relatively slow CORDIC transcendental functions, evaluation is heavy
and one of these could perhaps be beneficial, but have you ever tested if it's actually worth it? Sometimes they just want to prove a theoretical point, but they are not really better than plain
I looked at the article and my new method does not involve
as much calculations
per iterations! I think my new algorithm has the same order of convergence as Halley, which is third order.
Also the author of teh article is improving on Ostrowski (who improved on Newton). My new algorithms improves on Halley's method using Ostroskwi's approach.
Anyway, thanks for the articles. A few years ago I did a survey of recent algorithms that were inspired by Ostrowski. I found algorithms with three or more intermediate refinements for the root per
iterations. While the number of iterations of these algorithms decreased (compared to Newton or Halley), the number of function calls went up! So I am conscious that the number of function calls, per
iteration, should not get out of hand.
01-17-2017, 09:13 PM
Post: #6
JurgenRo Posts: 205
Member Joined: Jul 2015
RE: Heads up for a hot new root seeking algorithm!!
(01-11-2017 08:05 PM)Namir Wrote:
(01-10-2017 10:14 PM)Claudio L. Wrote: ... but have you ever tested if it's actually worth it? Sometimes they just want to prove a theoretical point, but they are not really better than
plain Newton.
I think my new algorithm has the same order of convergence as Halley, which is third order.
But again Namir, as you wrote "you think it is of 3rd order", but do you know for sure by means of a strict mathematical proof? Mathematics can't go without that ;-) A modification of a 3rd order
algorithm does not necessarily result in an algorithm of the same convergence behavior. Did you try to adopt the proves of Halley/Ostrowski to your new algorithm? This would indeed make your work
01-17-2017, 10:42 PM
Post: #7
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-17-2017 09:13 PM)JurgenRo Wrote: [quote='Namir' pid='66624' dateline='1484165132']
I think my new algorithm has the same order of convergence as Halley, which is third order.
But again Namir, as you wrote "you think it is of 3rd order", but do you know for sure by means of a strict mathematical proof? Mathematics can't go without that ;-) A modification of a 3rd order
algorithm does not necessarily result in an algorithm of the same convergence behavior. Did you try to adopt the proves of Halley/Ostrowski to your new algorithm? This would indeed make your work
The new algorithm matches or slightly improve on Halley. Sicne Halley is third order, the new algorithm could not be second or fourth order. I am saying 3rd order by induction.
01-18-2017, 05:47 AM
(This post was last modified: 01-18-2017 08:39 AM by Ángel Martin.)
Post: #8
Ángel Martin Posts: 1,445
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-17-2017 10:42 PM)Namir Wrote: The new algorithm matches or slightly improve on Halley. Sicne Halley is third order, the new algorithm could not be second or fourth order. I am saying 3rd
order by induction.
And how do you define "it matches"? what criteria is used? What parameters? In what cases? all possible expressions??
IMHO induction is not applicable here...
"To live or die by your own sword one must first learn to wield it aptly."
01-18-2017, 09:47 AM
Post: #9
emece67 Posts: 379
Senior Member Joined: Feb 2015
RE: Heads up for a hot new root seeking algorithm!!
The paper pointed by Claudio shows how the authors use a numerical computation to ascertain the order of convergence of the algorithm. I think that such approach can be used here to estimate the
order of convergence of Namir's algorithm not needing a, perhaps cumbersome, rigorous demonstration. Provided, of course, you have a math software capable to work with high precision numbers (say
Knowing the order of convergence, the efficiency index can be calculated, perhaps a better metric to compare algorithms.
In any way, AFAIK, the basic (Newton based) Ostrowsky's algorithm has a convergence order of 4, not 3.
01-18-2017, 08:22 PM
Post: #10
JurgenRo Posts: 205
Member Joined: Jul 2015
RE: Heads up for a hot new root seeking algorithm!!
(01-18-2017 05:47 AM)Ángel Martin Wrote:
(01-17-2017 10:42 PM)Namir Wrote: The new algorithm matches or slightly improve on Halley. Sicne Halley is third order, the new algorithm could not be second or fourth order. I am saying 3rd
order by induction.
And how do you define "it matches"? what criteria is used? What parameters? In what cases? all possible expressions??
IMHO induction is not applicable here...
I fully agree with Ángels rating. I do not see how Induction could do the work here. Also, the convergence rate might (and likely will) depend on the properties of the function to be "rooted"
(smoothness for example). That is why a "demonstration" of convergence with a couple of functions is not reliable at all. Then again it would be legitimate to say, that you
that the scheme is of 3rd order (for sufficient smooth?) functions but that a rigorous proof is still missing ...
01-18-2017, 11:53 PM
Post: #11
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
My study compared the number of iterations and functions for the method of Newton, Halley, Ostrowski, and my new algorithm. I used two dozen tes functions. The last three method generally had close
number of iterations and function calls. Since Halley has been discussed on several books (and Wikipedia) to be third order, then Ostrowski and my new algorithm should have the same convergence rate.
When you compare Newton's method with the other three, you can see that Newton's method is slower.
I am not sure that Ostrowski's method has an order 4 of convergence. Any reference that confirm this?
01-19-2017, 03:23 AM
Post: #12
Claudio L. Posts: 1,885
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-18-2017 11:53 PM)Namir Wrote: I am not sure that Ostrowski's method has an order 4 of convergence. Any reference that confirm this?
The paper I cited, shows at the very top on page 78 (it's actually the third page) that plain Newton has order 2, Ostrowski has order 4, and another formula has order 8. It says "it's well
established", so I guess it must be true.
01-19-2017, 05:26 AM
(This post was last modified: 01-19-2017 05:36 AM by Namir.)
Post: #13
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-19-2017 03:23 AM)Claudio L. Wrote:
(01-18-2017 11:53 PM)Namir Wrote: I am not sure that Ostrowski's method has an order 4 of convergence. Any reference that confirm this?
The paper I cited, shows at the very top on page 78 (it's actually the third page) that plain Newton has order 2, Ostrowski has order 4, and another formula has order 8. It says "it's well
established", so I guess it must be true.
Yes the paper states, on page 78, that Ostrowski's convergence is of order 4. The
author is mistaken
!! I tested 24 functions and compared Newton, Halley, Ostrowski, and my new Ostrowski-Halley method. The methods of Halley and Ostrowski often found the root in the same number of iterations. If
Ostrowski's method is order 4, I didn't see it in the test results. My new algorithm did, in general a little bit better than Halley and Ostrowski (after all it's a combination of these two methods).
Many mathematicians have, somewhat recently, improved on Ostrowski's method to yield algorithms with high to very high convergence rates. That's nice and dandy, but it comes at the cost of very high
number of function calls, making the simpler (and slower converging) algorithms more practical if one regards the number of function calls as
the cost of doing business
in finding roots.
01-19-2017, 10:44 AM
(This post was last modified: 01-19-2017 11:12 AM by emece67.)
Post: #14
emece67 Posts: 379
Senior Member Joined: Feb 2015
RE: Heads up for a hot new root seeking algorithm!!
(01-19-2017 05:26 AM)Namir Wrote: Yes the paper states, on page 78, that Ostrowski's convergence is of order 4. The author is mistaken!!
states that the order of convergence of Newton's is 2, Halley's 3 and Ostrowsky's 4.
states that the basic Ostrowsky's method has order of convergence 4.
states again that the order of convergence of Ostrowsky method is 4. It also says that the Ostrowsky's method is optimal in the sense that it satisfies the Kung-Traub conjecture (which states that
the maximum attainable order of convergence for an algorithm with \(n\) function evaluations is \(2^{n-1}\), neither Halley's nor Newton's are optimal in such sense.
I'm not sure about the first reference, but the other two are peer reviewed publications, so I assume them to be safe sources.
I cannot find now your algorithms description (didn't they were in this thread or in another one?) so, how many function evaluations do they need on each iteration?
Obtaining the computational order of convergence COC of your algorithm seems not to be a hard task (only 3 iterations are needed for some test function/root combinations). Such COC combined with the
number of function evaluations needed on each iteration will give us all a much better indication of your algorithm performance.
: I located your algorithm in the other thread (sorry I forgot it). It needs four function evaluations on each iteration. Supposing its order of convergence is 4, its efficiency index is \(\sqrt[4]
{4}=1.4142...\). As Ostrowsky's algorithm only requires 3 function evaluations on each iteration and it has an order of convergence of 4, its efficiency index is \(\sqrt[3]{4}=1.5874...\), which is
better. If the order of convergence of your algorithm is not 4, but 3, then the situation is even worse. To compete with the Ostrowsky's algorithm (in terms of the efficiency index) we need your
algorithm to be of order 6-7. For comparison, the efficiency index of Newton's is also 1.4142... and that for Halley is 1.44225...
01-19-2017, 04:09 PM
(This post was last modified: 01-19-2017 06:07 PM by Namir.)
Post: #15
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-19-2017 10:44 AM)emece67 Wrote:
(01-19-2017 05:26 AM)Namir Wrote: Yes the paper states, on page 78, that Ostrowski's convergence is of order 4. The author is mistaken!!
http://folk.uib.no/ssu029/Pdf_file/Varona02.pdf states that the order of convergence of Newton's is 2, Halley's 3 and Ostrowsky's 4.
http://www.sciencedirect.com/science/art...2706006716 states that the basic Ostrowsky's method has order of convergence 4.
http://www.sciencedirect.com/science/art...2111008078 states again that the order of convergence of Ostrowsky method is 4. It also says that the Ostrowsky's method is optimal in the sense that it
satisfies the Kung-Traub conjecture (which states that the maximum attainable order of convergence for an algorithm with \(n\) function evaluations is \(2^{n-1}\), neither Halley's nor Newton's
are optimal in such sense.
I'm not sure about the first reference, but the other two are peer reviewed publications, so I assume them to be safe sources.
I cannot find now your algorithms description (didn't they were in this thread or in another one?) so, how many function evaluations do they need on each iteration?
Obtaining the computational order of convergence COC of your algorithm seems not to be a hard task (only 3 iterations are needed for some test function/root combinations). Such COC combined with
the number of function evaluations needed on each iteration will give us all a much better indication of your algorithm performance.
Edit: I located your algorithm in the other thread (sorry I forgot it). It needs four function evaluations on each iteration. Supposing its order of convergence is 4, its efficiency index is \(\
sqrt[4]{4}=1.4142...\). As Ostrowsky's algorithm only requires 3 function evaluations on each iteration and it has an order of convergence of 4, its efficiency index is \(\sqrt[3]{4}=1.5874...\),
which is better. If the order of convergence of your algorithm is not 4, but 3, then the situation is even worse. To compete with the Ostrowsky's algorithm (in terms of the efficiency index) we
need your algorithm to be of order 6-7. For comparison, the efficiency index of Newton's is also 1.4142... and that for Halley is 1.44225...
Thank you for the interesting information. I will look at the various articles you mentioned in your links.
I have a question for you. Given a function f(x) with an initial guess x0 and a specified tolerance. If I apply root-seeking algorithm AL1, which has an efficient index EI1, and it takes N1
iterations to refine the root. Can I predict the number of iterations N2 for another algorithm AL2 with an efficiency index EI2? If we can do that, then we can estimate the efficiency index EI3 of
algorithm AL3, compared with algorithm AL1, given the number of iterations N1 and N3 for both methods (which the initial guess and tolerance being the same) and an Efficiency Index EF1 for AL1.
01-19-2017, 08:13 PM
(This post was last modified: 01-19-2017 08:42 PM by JurgenRo.)
Post: #16
JurgenRo Posts: 205
Member Joined: Jul 2015
RE: Heads up for a hot new root seeking algorithm!!
(01-19-2017 03:23 AM)Claudio L. Wrote:
(01-18-2017 11:53 PM)Namir Wrote: I am not sure that Ostrowski's method has an order 4 of convergence. Any reference that confirm this?
The paper I cited, shows at the very top on page 78 (it's actually the third page) that plain Newton has order 2, Ostrowski has order 4, and another formula has order 8. It says "it's well
established", so I guess it must be true.
You have to distinguish between
- and
error. Plain Newton is simply Tayler series Expansion of 1st order:
f(x_(n+1)) = f(x_n) + hf'(x_n) + (1/2)h^2f''(ceta),
where h= x_(n+1)-x_n, ceta in (x_n+1,x_n) and R:=(1/2)h^2f''(ceta) is the term defining the local error (of each step). That is, locally plain Newton is o(h^2), i.e. quadratic. The global error
though (the accumulated error of all N steps ) is o(Nh^2)=o(h), i.e. linear behavior. Normally you are interested in the total accumulated error, i.e. in the global error. So to my understanding
plain Newton is a linear scheme.
01-20-2017, 01:24 AM
Post: #17
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
Using Taylor series can easily explain the order of convergence for Newton, Halley, and Halston. But as the algorithms becomes more complication, generating one or more additional intermediate
guesses per iteration, this get complicated.
I was asking in a previous message whether the number of iterations can be estimated using the efficiency index, if we know the number of iterations achieved by another algorithm with a known
efficiency index--- for solving the root of the same function, using the same initial guess and tolerance value for the refined root.
My own perception is that the efficiency index is a qualitative indicator that gives you a general idea about the convergence rate. I doubt there exists a general formula for what I am asking above.
01-20-2017, 08:59 AM
Post: #18
emece67 Posts: 379
Senior Member Joined: Feb 2015
RE: Heads up for a hot new root seeking algorithm!!
(01-20-2017 01:24 AM)Namir Wrote: My own perception is that the efficiency index is a qualitative indicator that gives you a general idea about the convergence rate. I doubt there exists a
general formula for what I am asking above.
That's also my perception.
Regarding your question, seeing that algorithms can behave very differently (even with different orders of convergence) upon different functions and even around different roots of the same function,
I also doubt that there's a way to ascertain the required number of iterations for a given algorithm to converge to a given root of a given function.
I think that what we can do is to build tables such that given here
in order to "have a feeling" about the behaviour of an algorithm compared to others.
01-20-2017, 09:27 AM
(This post was last modified: 01-20-2017 09:38 AM by Namir.)
Post: #19
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
If you download the ZIP file from my web site, you get the report and an Excel file that has worksheets for the dozen test functions. What I observed is that in the majority of the cases, the methods
of Halley and Ostrowski give the same results or don't differ by more than one iteration (both require three function calls per iteration). The results DO SHOW a consistent improvement over Newton's
method. You can say that the methods of Halley and Ostrowski have a higher convergence rate than that of Newton. My new algorithm (which uses the Ostrowski's approach to enhance Halley) shows several
cases where either the number of iterations or both the number of iterations AND number of function calls is less than that of Halley and Ostrowski. Since I developed the new algorithm using an
empirical/heuristic approach I did not yield a long set of mathematical derivations that indicate the convergence rate. It may well be at least one order more than that of Ostrowski's method. It's
hard to measure ... compared to, say determining the order of array sorting methods where you can widely vary the array size and calculate the number of array element comparisons and number of
element swaps. See my article about
enhancing the CombSort method
where I was able to calculate the sort order of this method in Tables 3 and 4 of the article.
01-21-2017, 02:35 PM
(This post was last modified: 01-21-2017 02:37 PM by Namir.)
Post: #20
Namir Posts: 1,107
Senior Member Joined: Dec 2013
RE: Heads up for a hot new root seeking algorithm!!
(01-20-2017 08:59 AM)emece67 Wrote:
(01-20-2017 01:24 AM)Namir Wrote: My own perception is that the efficiency index is a qualitative indicator that gives you a general idea about the convergence rate. I doubt there exists a
general formula for what I am asking above.
That's also my perception.
Regarding your question, seeing that algorithms can behave very differently (even with different orders of convergence) upon different functions and even around different roots of the same
function, I also doubt that there's a way to ascertain the required number of iterations for a given algorithm to converge to a given root of a given function.
I think that what we can do is to build tables such that given here http://article.sapub.org/10.5923.j.ajcam...01.03.html in order to "have a feeling" about the behaviour of an algorithm compared
to others.
Regarding the link to the article you mentioned in your message. Can you check the new algorithm by the author Thukral. I implemented equation 7 in the article and got bizarre results! Perhaps I am
doing something wrong? When I replaced the second subtraction in equation 7 with a multiplication, the algorithm worked but was painfully slow to converge!
I suspect typos om the article since the title has one " Nonlinear Equations of Type f(0)=0" instead of " Nonlinear Equations of Type f(x)=0".
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Leetcode Daily | 12–09–22 | Bag Of Tokens
Hello everyone,
So from today onwards I have decided to post this blog for the Leetcode daily challenge with its explanation from brute force to optimized one. Since many people face problems while understanding the
code, I decided to give a proper explanation for every statement in the code.
Here is the question that we will be looking at today.
You have an initial power of power, an initial score of 0, and a bag of tokens where tokens[i] is the value of the ith token (0-indexed).
Your goal is to maximize your total score by potentially playing each token in one of two ways:
• If your current power is at least tokens[i], you may play the ith token face up, losing tokens[i] power and gaining 1 score.
• If your current score is at least 1, you may play the ith token face down, gaining tokens[i] power and losing 1 score.
Each token may be played at most once and in any order. You do not have to play all the tokens.
Return the largest possible score you can achieve after playing any number of tokens.
• 0 <= tokens.length <= 1000
• 0 <= tokens[i], power < 104
You can find the link to the question here, https://leetcode.com/problems/bag-of-tokens/
For the first approach, we will use two pointers and greedy method. This can only be applied to the array if it is sorted in acending order.
We will place on pointer at the staring and another one at the end. After that, we will teaversing the array till the starting pointer becomes equal to the ending pointer. While traversing, we will
keep a track of elements.
If the element is greater than the power, we will look the score and gain the power. Else, if the element is less than the power, we will gain a score and loose the power.
Here is the algorithm that we will be using in this code-
1. sort the token in increasing order
2. create two pointer i, j, where i = 0, j = n ; initially
3. 0 <= x <= i represent the tokens face up i.e, [Score ⬆, Power ⬇]
4. j <= x <= n represent the tokens face down i.e, [Score ⬇, Power⬆]
5. if we have sufficient power to face up ith token, power — token[i], i++.
6. else if we have some score [i - (n - j)] face down jth token, j--, power + token[j].
i - (n - j) represents the score since (face up - face down) = score, face up = i and face down = n - j
Note :we also need to check if we face down j we should have some unplayed token to face up. That's why we are checking j > i + 1
7. if we have no move to make break the loop.
C++ Code
Java Code
Python Code
The time complexity of the above code is O(nlogn) and the space complexity is O(1). | {"url":"https://aditijha201200.medium.com/leetcode-daily-12-09-22-bag-of-tokens-3766cbb61662?source=author_recirc-----5473f61a2511----0---------------------03f5a753_59b3_43ff_bfa1_67c1deb8dd91-------","timestamp":"2024-11-08T10:32:02Z","content_type":"text/html","content_length":"116335","record_id":"<urn:uuid:40101f5f-9e28-4832-9747-9f4f6b2f7d21>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00579.warc.gz"} |