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Radians to Degrees - Conversion, Formula, Examples
Radians and degrees conversion is a very essential skill for higher math learners to grasp.
Initially, we need to explain what radians are thereby you can understand how this formula is used in practice. After that we’ll take it one step further by exhibiting some examples of going from
radians to degrees quickly!
What Is a Radian?
Radians are units of measurement for angles. It is originated from the Latin word "radix," which suggests nostril or ray, and is a critical theory in mathematics and geometry.
A radian is the SI (standard international) unit for angles, while a degree is a more generally used unit in math.
That being said, radians and degrees are simply two distinct units of measure used for measuring the identical thing: angles.
Note: a radian is not to be mixed with a radius. They are two completely different things. A radius is the distance from the center of a circle to the edge, whereas a radian is a unit of measure for
Association Between Radian and Degrees
There are two manners to think about this question. The initial method is to think about how many radians are present in a full circle. A full circle is equivalent to 360 degrees or two pi radians
(precisely). Hence, we can say:
2π radians = 360 degrees
Or easily:
π radians = 180 degrees
The second way to think regarding this question is to think about how many degrees exists in a radian. We know that there are 360 degrees in a complete circle, and we also recognize that there are
two pi radians in a complete circle.
If we divide each side by π radians, we’ll see that 1 radian is approximately 57.296 degrees.
π radiansπ radians = 180 degreesπ radians = 57.296 degrees
Both of these conversion factors are useful relying on what you're trying to get.
How to Go From Radians to Degrees?
Now that we've gone through what degrees and radians are, let's practice how to convert them!
The Formula for Converting Radians to Degrees
Proportions are a helpful tool for converting a radian value into degrees.
π radiansx radians = 180 degreesy degrees
Simply plug in your given values to obtain your unknown values. For example, if you wanted to convert .7854 radians into degrees, your proportion will be:
π radians.7854 radians = 180 degreesz degrees
To solve for z, multiply 180 by .7854 and divide by 3.14 (pi): 45 degrees.
This formula works both ways. Let’s recheck our workings by changing 45 degrees back to radians.
π radiansy radians = 180 degrees45 degrees
To work out the value of y, multiply 45 with 3.14 (pi) and divide by 180: .785 radians.
Now that we've changed one type, it will always work out with another simple calculation. In this scenario, afterwards converting .785 from its original form back again, following these steps
produced precisely what was expected -45°.
The formulas work out like this:
Degrees = (180 * z radians) / π
Radians = (π * z degrees) / 180
Examples of Converting Radians to Degrees
Let's go through a handful of examples, so these theorems become easier to digest.
At the moment, we will change pi/12 rad to degrees. Much like before, we will plug this number into the radians slot of the formula and calculate it like this:
Degrees = (180 * (π/12)) / π
Now, let divide and multiply as you generally would:
Degrees = (180 * (π/12)) / π = 15 degrees.
There you have it! pi/12 radians equals 15 degrees.
Let's try one more general conversion and transform 1.047 rad to degrees. One more time, use the formula to get started:
Degrees = (180 * 1.047) / π
One more time, you multiply and divide as appropriate, and you will wind up with 60 degrees! (59.988 degrees to be exact).
Right away, what happens if you have to convert degrees to radians?
By employing the very same formula, you can do the contrary in a pinch by solving it considering radians as the unknown.
For example, if you want to transform 60 degrees to radians, put in the knowns and work out with the unknowns:
60 degrees = (180 * z radians) / π
(60 * π)/180 = 1.047 radians
If you recollect the formula to find radians, you will get identical answer:
Radians = (π * z degrees) / 180
Radians = (π * 60 degrees) / 180
And there it is! These are just some of the examples of how to change radians to degrees and vice versa. Bear in mind the equation and try solving for yourself the next time you are required to make
a transformation between radians and degrees.
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Time for Spring Cleaning? Here’s How You Can Unclutter Classroom Culture - Solution Tree Blog
It’s that time of year again: spring! For many, this means it is time to unclutter the home and “spring-clean” as winter gives way to summer. I know for me, it is this time of year when I suddenly
feel the need to purge unused items and scrub windows and floors to make the house feel like new.
Lately, as I work with mathematics teachers and teams, the urge to unclutter makes me wonder what spring-cleaning might look like in classrooms. I wonder what the clutter is that hampers student
learning and might need to be “cleaned” or freshened as students work to learn remaining standards.
Clutter can be literal when there are too many anchor charts on the walls or an overwhelming amount of manipulatives or papers strewn around the room without clear organization. However, clutter is
also found in the culture of a classroom and ambiguity of expectations. Consider the following types of cultural clutter that can hamper student learning:
• Some students are left to struggle in confusion, rather than productive struggle, as they work on a high-level task because they are unsure where to start.
• Students and/or teachers talk louder and louder to be heard while few actually listen.
• Students repeatedly ask for the directions because they were not listening and the directions or expectations for student work are not visually available.
• Students are unsure of how to behave or function as an effective group when working independently or in groups.
• Students are unsure of how to engage in mathematical discourse or how to approach a task.
• Students complete activities without understanding how it connects to the essential learning standard.
• Students are assigned lengthy homework assignments and then passively listen as the problems are reviewed in class the following day.
• Students are only given low-level tasks to complete with many finishing early or working in isolation on a computer.
• Students dutifully take notes without checking their understanding and getting formative feedback from peers.
• Weeks are spent on state test preparation rather than on a conceptual understanding of standards with procedural fluency and application.
• Only a few students raise their hands, answering all of the questions asked.
• Students know that expectations for learning mathematics in one classroom differ from another.
How does one unclutter a classroom so each student is engaged in learning mathematics? What might need to be scrubbed or polished in these last few weeks of learning?
Consider some of the following ideas.
Lose the Consider…
clutter of…
Student • Provide students with sentence frames to use, example questions to ask one another, and vocabulary words to practice when working with one another in pairs or groups.
disengagement • Teach students to be resources for one another in the learning process and ask questions of one another.
Confusion Clarify expectations for working in groups and make sure all students are accountable to learning from the task.
Too many • Stop and teach students to listen and paraphrase when working with one another.
voices • Gain the attention of students before giving directions to avoid talking over students. Consider how to make the directions visible.
Passive Determine how all students can answer questions posed in class. This may mean each student shares the answer with a shoulder partner and popsicle sticks are used to randomly call on a
engagement student to answer in front of the whole group.
Learning Determine how to help students make connections from previous learning to the current standard(s) and develop a visual model or strategy to use to make sense of any algebraic
content in representation (e.g., graph, tape diagram, picture).
Consider using a timer to create urgency around trying a task as well as knowledge that in a short amount of time some sort of check will occur to clarify the task if needed (e.g., 1
Lost time minute to read the task, 2 minutes to start the task, 2 minutes to discuss the task with a partner and choose one of the starts while continuing to work toward a solution, 2 minutes to
check with another partner group, and 2 minutes to complete the solution prior to having students share with the class.
Inequity Work with your colleagues on your collaborative team to clarify proficiency expectations, scoring agreements, and calibrate grading of assessments.
Too often clutter in the classroom interferes with student learning of mathematics whether it be chaotic noise, ambiguity of expectations, auditory confusion without a visual representation, or an
overabundance of chart paper posters, to name a few. It is time for some classroom spring cleaning? What needs to be polished?
Recently I visited a middle school where teachers are working to unclutter their classrooms by analyzing instructional practices. As a result, I observed students solving a higher-level cognitive
demand task in groups of four, sharing their solutions, and learning from one another. All the while, the teacher masterfully worked to scaffold the task and extend the task, as needed, by walking
around the room and engaging with students while requiring them to learn from one another.
It turns out that I will begin spring-cleaning at home this week. And though it promises to be challenging and aggravating at times, I can’t wait for the result!
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Variations in Water Demand
There are wide variations in the use of water in different seasons, in different months of the year, in different days of the month and in different hours of the day.
Seasonal or monthly variation are prominent in tropical countries like India rate of consumption reaches maximum in summer season due to greater use of water for street and lawn sprinkling etc… It
goes down in winter months. The fluctuation in the rate of consumption may be as much as 150% of the average annual consumption.
Avg. Demand on a maxm. Day
TIME IN HOURS.
Daily and hourly Fluctuations depend on the general habits of people, climatic conditions etc… more water demand will be on Sundays and holidays due to more comfortable working, etc…. as compared to
other working days. Peak hours may be 6 a.m to 10a.m and 10a.m to 4.p.m minimum flow and between 10.p.m to 4.a.m it is very less. The above graph shows the hourly variation in demand of water or rate
of consumption 20% of average hourly demand
The maximum demands (monthly, daily or hourly) are generally expressed as ratios of their means. The following figures are generally adopted.
1. MAXIMUM DAILY CONSUMPTION is generally taken as 180% of the average, therefore
Maximum daily demand(MDD) = 1.8 Average daily demand (^DD)
= 1.8 q
2. MAXIMUM HOURLY CONSUMPTION is generally taken as 150% of its average hooray consumption of maximum day, there fore
Maximum hourly consumption of the maximum day or peak demand
= 1.5 (150%) x Average hourly consumption of the maxm. Day.(Litres/day)
= 1.5 [MDD / 24 ] (Litres/hr)
= 1.5 [1.8 q / 24 ] (Litres/hr)
= 2.7 / 24 (Litres/hr)
Maximum hourly consumption of the maximum day = (2.7 Annual Average hourly demand)
The formula given by GOODRICH is also used for finding out the rather of peak demand rates to their corresponding average values.
GOODRICH FORMULA P=180 t ^0.10
Where, P = % of annual average draft for the time„t? in days
T = Time in days from 1/ 24 to 365
When t= 1 day (For daily variations)
P = 180 x [1] 0.10
P = 180%
MDD/ADD = 180%
When t = 7 days (For weekly variations)
P = 180 x [(7)]-0.10
P = 148%
MWD/ AWD = 148%
T = 30 days (For monthly variations)
P = 180 (30)^-0.10
P =128%
MMD / AMD = 128%
Maxm monthly Demand / Avg .MonthlyDemand = 128%
1. The design population of a town is 15000 Determine the Average daily, Maximum hourly demand under suitable assumptions
Soln: Assuming Average percapita demand as 270 Lpcd
i. ADD = design population Avg. per capita demand
= 1500 x 270 = 4050000 Litres/day
ADD = 4050 m^3/day
ii. MDD = 1`.8 x Average daily demand
= 1.8 x 4050
= 7290 m^3/day
iii. Maximum hourly demand of maximum day
= 2.7 x q/24
= 2.7 x (4050/24)
= 455.625 m^3/hr or
= 10935 m^3/day
1. The sources of supplies such as wells etc… may be designed for MDD
2. The pipe mains taking water from the source upto the service reservoirs may be designed for MDD.
3. The filter and OTHER UNITS at water treatment plant may also be designed for MDD. Sometimes an additional provision for reserve is also made for break down and repairs therefore they may be
designed for twice the ADD instead of MDD.
4. The pumps may be designed for MDD plus some additional reserve (say twice the ADD)
When the pumps do not work for all the 24 hrs such as in small town supplies, the design draft should be multiplied by
24/ Number of hours in the day foe which the pumps are running
5 The distribution system is generally designed for the maximum hourly demand of the maximum day or coincident draft whichever is more.
6 Service reservoirs are generally designed for 8 days consumption. | {"url":"https://www.brainkart.com/article/Variations-in-Water-Demand_3265/","timestamp":"2024-11-02T01:17:37Z","content_type":"text/html","content_length":"65220","record_id":"<urn:uuid:bc879b92-780d-4899-8db6-19dd7ae3ed7e>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00340.warc.gz"} |
Question #7a0d4 | Socratic
Question #7a0d4
1 Answer
10,000 L (if you are talking about concentration)
the question can be a little more clear. Is that 5% a concentration or a molarity percent.
Anyway if you were thinking about concentration, if you are given a percent, you have to divide your persent by 100 due to the fact that you want to get rid of your percentage.
Then since you are given a solute for your solution, you have to divide the grams given by the number of percentage.
$\frac{500 g}{0.05} = 10 , 000 L$
This seems like a large number but thinking about what was given to us, if you are mixing 500g (which is half a kg) in water to get a small concentration, you will need to have lots and lots of
Let me know in the comments if I answered the question correctly. I enjoy comments that people ask me.
Impact of this question
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1.(adapted from Prob. 2, p. 127, Brown's Introductory Physics II)Supp
1.(adapted from Prob. 2, p. 127, Brown's Introductory Physics II)Suppose you have a charge q at position y a and a charge -g at y -d, both on the y
axis (an electric dipole as studied in chapter 1.see Figure I below). (a)Write an expression for the electric potential of the dipole at an arbitrary position (x, y (c)Evaluate your result in part
(b) for a position (x, y) = (x, 0) on the x axis and compare it to the result of HW2B #1(b). We usually try to simplify early in a problem before doing the tough math. Why is it important in this
problem to differentiate the potential before setting y= 0? (b)Find expressions for the x and y components of the electric field at an arbitrary position (x, y) by carefully differentiating your
result in part (a).
Fig: 1
Fig: 2
Fig: 3
Fig: 4
Fig: 5 | {"url":"https://tutorbin.com/questions-and-answers/1adapted-from-prob-2-p-127-browns-introductory-physics-iisuppose-you-h","timestamp":"2024-11-09T02:36:59Z","content_type":"text/html","content_length":"70910","record_id":"<urn:uuid:b25f7418-141f-4356-9000-958f4f80ae59>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00783.warc.gz"} |
What Is The Unit Rate Of Scoops Of High-protein Food To Scoops Of High-fiber Food If The Food Is Made
10. 31.14
Step-by-step explanation:
Answer an essay on nothing
Step-by-step explanation:
In philosophy there is a lot of emphasis on what exists. We call this ontology, which means, the study of being. What is less often examined is what does not exist.
It is understandable that we focus on what exists, as its effects are perhaps more visible. However, gaps or non-existence can also quite clearly have an impact on us in a number of ways. After all,
death, often dreaded and feared, is merely the lack of existence in this world (unless you believe in ghosts). We are affected also by living people who are not there, objects that are not in our
lives, and knowledge we never grasp.
Upon further contemplation, this seems quite odd and raises many questions. How can things that do not exist have such bearing upon our lives? Does nothing have a type of existence all of its own?
And how do we start our inquiry into things we can’t interact with directly because they’re not there? When one opens a box, and exclaims “There is nothing inside it!”, is that different from a real
emptiness or nothingness? Why is nothingness such a hard concept for philosophy to conceptualize?
Let us delve into our proposed box, and think inside it a little. When someone opens an empty box, they do not literally find it devoid of any sort of being at all, since there is still air, light,
and possibly dust present. So the box is not truly empty. Rather, the word ‘empty’ here is used in conjunction with a prior assumption. Boxes were meant to hold things, not to just exist on their
own. Inside they might have a present; an old family relic; a pizza; or maybe even another box. Since boxes have this purpose of containing things ascribed to them, there is always an expectation
there will be something in a box. Therefore, this situation of nothingness arises from our expectations, or from our being accustomed. The same is true of statements such as “There is no one on this
chair.” But if someone said, “There is no one on this blender”, they might get some odd looks. This is because a chair is understood as something that holds people, whereas a blender most likely not.
The same effect of expectation and corresponding absence arises with death. We do not often mourn people we only might have met; but we do mourn those we have known. This pain stems from expecting a
presence and having none. Even people who have not experienced the presence of someone themselves can still feel their absence due to an expectation being confounded. Children who lose one or both of
their parents early in life often feel that lack of being through the influence of the culturally usual idea of a family. Just as we have cultural notions about the box or chair, there is a standard
idea of a nuclear family, containing two parents, and an absence can be noted even by those who have never known their parents.
This first type of nothingness I call ‘perceptive nothingness’. This nothingness is a negation of expectation: expecting something and being denied that expectation by reality. It is constructed by
the individual human mind, frequently through comparison with a socially constructed concept.
Pure nothingness, on the other hand, does not contain anything at all: no air, no light, no dust. We cannot experience it with our senses, but we can conceive it with the mind. Possibly, this sort of
absolute nothing might have existed before our universe sprang into being. Or can something not arise from nothing? In which case, pure nothing can never have existed.
If we can for a moment talk in terms of a place devoid of all being, this would contain nothing in its pure form. But that raises the question, Can a space contain nothing; or, if there is space, is
that not a form of existence in itself?
This question brings to mind what’s so baffling about nothing: it cannot exist. If nothing existed, it would be something. So nothing, by definition, is not able to ‘be’.
Is absolute nothing possible, then? Perhaps not. Perhaps for example we need something to define nothing; and if there is something, then there is not absolutely nothing. What’s more, if there were
truly nothing, it would be impossible to define it. The world would not be conscious of this nothingness. Only because there is a world filled with Being can we imagine a dull and empty one.
Nothingness arises from Somethingness, then: without being to compare it to, nothingness has no existence. Once again, pure nothingness has shown itself to be negation. | {"url":"https://community.carbonfields.net/question-handbook/what-is-the-unit-rate-of-scoops-of-high-protein-food-to-scoo-cn2p","timestamp":"2024-11-13T01:57:45Z","content_type":"text/html","content_length":"75361","record_id":"<urn:uuid:5f0cb736-21ac-4737-8e57-d01887f69dbd>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00405.warc.gz"} |
Pandora's Box Gittins Index
Gittins index $\alpha^\star$: optimal for cost-per-sample problem
• What about expected budget-constrained problem?
Theorem. Assume the expected budget constraint is feasible and active. Then there exists a $\lambda \gt 0$ and a tie-breaking rule such that the policy defined by maximizing the Gittins index
acquisition function $\alpha^\star(\.)$, defined using costs $\lambda c(x)$, is Bayesian-optimal.
Proof idea: Lagrangian duality
Our work: extends special case of a result of Aminian et al. (2024) to non-discrete reward distributions | {"url":"https://presentations.avt.im/2024-08-01-Pandoras-Box-BayesOpt/","timestamp":"2024-11-06T18:33:41Z","content_type":"text/html","content_length":"623386","record_id":"<urn:uuid:438b5240-8b20-4211-a048-9b6bf1efec5c>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00822.warc.gz"} |
The Chain Rule | mathhints.com
The Chain Rule Basics
The Equation of the Tangent Line with the Chain Rule
More Practice
The chain rule says when we’re taking the derivative, if there’s something other than $ \boldsymbol {x}$, like in the parenthesis or under a radical sign, for example, we have to multiply what we get
by the derivative of what’s inside the parentheses. It all has to do with Composite Functions, since $ \displaystyle \frac{{dy}}{{dx}}=\frac{{dy}}{{du}}\cdot \frac{{du}}{{dx}}$. Note that we’ll learn
how to “undo” the chain rule here in the U-Substitution Integration section.
Think of it this way when we’re thinking of rates of change, or derivatives: if we are running twice as fast as Person A, and then Person B is running three times as fast as us, Person B is running
six times as fast as Person A. It’s all about relativity! Here is what it looks like in Theorem form:
If $ \displaystyle y=f\left( u \right)$ and $ u=g\left( x \right)$ are differentiable and $ y=f\left( {g\left( x \right)} \right)$, then:
$ \displaystyle \frac{{dy}}{{dx}}=\frac{{dy}}{{du}}\cdot \frac{{du}}{{dx}}$, or
$ \displaystyle \frac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right]={f}’\left( {g\left( x \right)} \right){g}’\left( x \right)$ (more simplified): $ \displaystyle \frac{d}{{dx}}\left[
{f\left( u \right)} \right]={f}’\left( u \right){u}’$
We’ve actually been using the chain rule all along, since the derivative of an expression with just an $ \boldsymbol {x}$ in it is just 1, so we are multiplying by 1.
For example, if $ \displaystyle y={{x}^{2}},\,\,\,\,\,{y}’=2x\cdot \frac{{d\left( x \right)}}{{dx}}=2x\cdot 1=2x$.
Do you see how when we take the derivative of the “outside function” and there’s something other than just $ \boldsymbol {x}$ in the argument (for example, in parentheses, under a radical sign, or in
a trigonometric function), we have to take the derivative again of this “inside function”? In a nutshell, we are taking the derivative of the “outside function” and multiplying this by the derivative
of the “inside” function(s). That’s pretty much it!
In the problems below, see how we take the derivative again of what’s in red? And sometimes, again, what’s in blue? Yes, sometimes we have to use the chain rule twice, in the cases where we have a
function inside a function inside another function. We could theoretically take the chain rule a very large number of times, with one derivative!
Here’s one more problem, where we have to think about how the chain rule works:
The Equation of the Tangent Line with the Chain Rule
Here are a few problems where we use the chain rule to find an equation of the tangent line to the graph $ f$ at the given point. Note that we saw more of these problems here in the Equation of the
Tangent Line, Tangent Line Approximation, and Rates of Change Section.
Understand these problems, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Implicit Differentiation and Related Rates – you’re ready! | {"url":"https://mathhints.com/differential-calculus/the-chain-rule/","timestamp":"2024-11-12T06:18:37Z","content_type":"text/html","content_length":"576479","record_id":"<urn:uuid:5b9c0ac2-99ac-496f-aa48-086d59783db7>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00764.warc.gz"} |
Laser Divergence Calculator - 405nm.com
Laser Divergence Calculator
This laser divergence calculator can quickly tell you how large a laser’s dot will become for a given distance and divergence. Enter the specifications of your laser in the fields below to see how
the beam will diverge.
Enter laser specifications to calculate the divergence of a laser over a given distance.
Laser divergence is a measure of how much a laser beam spreads out as it travels over distance. It is typically measured in milliradians (mrad), which is a unit of angle. A milliradian is equal to
0.001 radians. A small divergence angle results in a more collimated beam which stays narrow over a greater distance, while a larger divergence angle results in a beam that spreads out more quickly
over distance.
Looking to convert between wavelength and frequency?
Try the Wavelength to Frequency Calculator. | {"url":"https://405nm.com/laser-divergence-calculator/","timestamp":"2024-11-11T20:50:45Z","content_type":"text/html","content_length":"79330","record_id":"<urn:uuid:bab7ae5d-5120-4652-806c-a07eb0cc78a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00234.warc.gz"} |
Overview and Objective
In this lesson, students explore a game to build the background knowledge needed to classify and define the terms factor, multiple, prime, and composite. The game introduces the idea as to why 1 is
neither prime or composite. The game is played by taking turns selecting a number on a hundreds chart, and then collecting points. A point is scored for each unique rectangular array that they can
create with that number of square tiles.
This activity can be used with younger students who can identify numbers up to 100, and who have the dexterity to use a mouse. This activity can also be used with high school and upper level
mathematics as it explores number systems such as highly composite, abundant numbers, superior numbers and their intersections.
Invite students to explore how many unique rectangles they can make with 100 square tiles. Share this Polypad with students so they can explore this question. This video shows how to “push and pull”
the merged squares to find rectangles and how to duplicate and recolor rectangles so that students can see all of their findings. Click here to learn more about using Number Tiles on Polypad.
Be sure to include a 1 x 100 and a 100 x 1 as different rectangles. This will set the groundwork for why 1 is not prime or composite. Also, 100 is a perfect square, so remind students that squares
are a type of rectangle, and should be included.
There are a total of 9 unique rectangles: 1x100, 2x50, 4x25, 5x20, 10x10, 20x5, 25x4, 50x2, and 100x1.
Main Activity
Play the Rectangle Game. Make a copy of this gameboard and then share a copy with students.
Each player clicks on a number on the gameboard to open a Polypad with that number of square tiles. They find as many unique rectangles as possible (note: 1 x 3 is different than 3 x 1). Then score a
point for each unique rectangle.
3 has two unique rectangles (3x1 and 1x3), so this would score 2 points.
4 has three unique rectangles (4x1, 2x2, 1x4), so this would score 3 points.
Once points are scored, players can block that number using a colored chip. After four rounds, the player with the most points is the winner. Students can play additional games as time allow.
Gather as a class to discuss the following.
Worst Gameboard Number: 1. It only has 1 rectangle (1x1). Because it only has 1 factor, it is not prime, but it is also not composite.
Bad Gameboard Numbers: 2, 3, 7, 11, and other prime numbers because you can only make 2 rectangles.
Good Gameboard Numbers: 4, 6, 8, 9, 10, 12 and other composite numbers because you can make multiple rectangles.
Best Gameboard Numbers: 60, 72 and 96 because you can make the most (12 rectangles) with these 3 numbers. These numbers have the most factors of any number under 100.
Amount of different rectangles: factors
Flipped rectangles: (example: 3 x 2 and 2x 3) use the same factors and are an example of the commutative property of multiplication.
Square rectangles: (example: 2 x 2) these are square numbers because the two sides use the same factor.
• *Smallest Gameboard number with the most rectangles: 60 because it is a highly composite number. Highly composite numbers is a positive integer with more divisors than any smaller positive
integer has.
• *Upper level mathematics number theory terms: Highly composite number, smooth number, abundant number, superior number
Support and Extension
To help support students, ask them the following: Sarah picked 100, because it was the biggest number. She scored 9 points because she found 9 different rectangles (one of which was a square). Zarah
found another number on the gameboard that would give her more points than Sarah. What number could it be?
For an extension question, ask students the following: What if the gameboard extended to 1,000. What are the best gameboard numbers?
Polypad and Gameboard For This Lesson
To assign these to your classes in Mathigon, save a copy to your Mathigon account. Click here to learn how to share Polypads with students and how to view their work.
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Extreme spontaneous deformations of active crystals
Statistical Physics and Complexity Group meeting
Extreme spontaneous deformations of active crystals
• Event time: 3:00pm until 4:00pm
• Event date: 9th January 2024
• Speaker: Hugues Chaté (CEA-Saclay, Paris)
• Location: Online - see email.
Event details
The Kosterlitz-Thouless-Halperin-Nelson-Young (KTHNY) theory of melting is a much-admired result that revealed that equilibrium 2D crystals can melt into in two successive KT-like transitions leading
to liquid state. A well-known result of KTHNY is to predict an upper bound $\frac{1}{3}$ to the spin wave exponent $\eta$ governing the algebraic decay of the two-point correlation function of
positional order in the crystal phase. Whereas this result is often used in equilibrium to decide to decide whether a crystal has melted, it has been claimed (or assumed) to apply also for crystals
made of active units, which are intrinsically out of equilibrium.
I will show that such active crystals are not subjected to the KTHNY bound $\eta< \frac{1}{3}$. They can be either more resilient to fluctuations without melting ($\eta> \frac{1}{3}$), or more
fragile and break for $\eta< \frac{1}{3}$. These results are rationalized within linear elastic theory in terms of two well-defined effective temperatures governing respectively large scale
positional deformations and short scale bond-order fluctuations.
Event resources
This is a weekly series of webinars on theoretical aspects of Condensed Matter, Biological, and Statistical Physics. It is open to anyone interested in research in these areas..
Find out more about Statistical Physics and Complexity Group meetings.
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How to Randomly Select Cells Based on Criteria in Excel
To randomly select cells based on criteria in Excel, you can use a combination of the RAND, IF, and INDEX functions. Here's a step-by-step guide on how to do this:
1. Organize your data: Make sure your data is organized in columns, and your criteria are defined in separate columns.
2. Create a helper column: Add a helper column to your dataset. This column will be used to generate random numbers.
3. Use the RAND function: In the first cell of the helper column, enter the following formula:
4. Copy the formula: Copy the RAND formula to all the cells in the helper column corresponding to the rows in your dataset.
5. Create a new column for the randomly selected cells: Add a new column to your dataset where you want the randomly selected cells to be displayed.
6. Use the INDEX, IF, and RANK functions: In the first cell of the new column, enter the following formula (replace "A" with the column where your data is, "B" with the criteria column, and "C" with
the helper column):
=INDEX($A$1:$A$10, RANK(C1, $C$1:$C$10) * (IF($B$1:$B$10 = "criteria", 1, 0)))
7. Replace "criteria" with your actual criteria: Replace the "criteria" in the formula with the actual criteria you want to use for selecting the cells.
8. Copy the formula: Copy the formula to all the cells in the new column corresponding to the rows in your dataset.
9. Hide the helper column (optional): If you don't want to display the helper column, you can hide it by right-clicking on the column header and selecting "Hide".
Now, you will have a list of randomly selected cells from your dataset that meet the specified criteria.
Let's say you have a list of students and their scores, and you want to randomly select 3 students who scored above 80. Here's how you can do that in Excel:
1. Organize your data: Assume the student names are in Column A (A2:A11) and their scores are in Column B (B2:B11).
2. Create a helper column: Add a helper column in Column C (C2:C11).
3. Use the RAND function: In cell C2, enter the formula:
4. Copy the formula: Copy the formula to all the cells in the helper column (C2:C11).
5. Create a new column for the randomly selected students: Add a new column in Column D.
6. Use the INDEX, IF, and RANK functions: In cell D2, enter the following formula:
=INDEX($A$2:$A$11, RANK(C2, $C$2:$C$11) * (IF($B$2:$B$11 > 80, 1, 0)))
7. Copy the formula: Copy the formula to cells D3 and D4.
You will now have a list of 3 randomly selected students who scored above 80 in Column D. If you want to refresh the random selection, press F9 to recalculate the RAND function.
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Exporting Small Language Models to ONNX with KV-cache Support - Esperanto Technologies
At Esperanto Technologies, we are interested in running generative AI models in ONNX format for three reasons:
1. ONNX is an interoperable format that allows two-way conversion to most other formats.
2. ONNX is supported by the ONNXRuntime framework, which is open-source and one of today’s main frameworks.
3. The ONNX format presents a static view of the models that is easier to load and optimize for Esperanto’s ML compiler, which leads to faster inferencing.
An ONNX file stores an implicit graph of operations. Vertices and edges of the graph are identified by operation and tensor names. The precise locations (i.e., names) can be found by inspection in
the visualization tool Netron.
The figure showing the ONNX interoperability.
In this blog article, we will review the pipeline that we follow for converting a PyTorch Small Language Model (foundational, or fine-tuned) into a fully functional ONNX model using fp16 precision
with Key-Value cache (KV-cache) support.
First of all, we assume that we have a Small Language Model (SLM: Transformer-based language models with less than 10 billion parameters) in the torch format, the format employed by PyTorch, the most
widespread format for ML. Our first step will be to convert our torch model into a preliminary ONNX model using the torch.onnx.export command. For this, we need to provide inputs to the model so that
we can trace the different operations that form the network.
A non-KV-cache-enabled SLM only requires two independent inputs: the input_ids, or the list of tokens that our prompt is converted to after tokenization, and the attention_mask, a mask that provides
1’s to the actual tokens that are being consumed and 0’s to padding. Only one output is needed for text generation, which is the “logits”. This output provides for each token in the input, one score
for each token in the tokenizer’s dictionary that represent the predicted token likelihood distribution of the next position in the sequence. In a regular use case, we would ignore all but the last
of these lists to predict one new token. When providing these inputs, we can also define “onnx_symbols” or dynamic_shapes, such as the batch or the sequence_length, which are placeholders that can
adapt to the needs for the inference. These are particularly relevant to get different configurations from the same model (with different sequence length, batch, etc…). The last relevant flag for
this conversion is the ONNX opset. As of September 2024, we can only guarantee that model with opset <=14 will be fully supported since some nodes’ definitions can vary drastically from an opset to
the next.
So we finally get a command similar to the following:
dummy_input = torch.ones((1, 128), dtype = torch.int64)
symbolic_names = {0: "batch", 1: "sequence"}
model, #torch model
(dummy_input, dummy_input), #inputs
"model.onnx", #path of the output onnx model
input_names = ["input_ids", "attention_mask"],
output_names = ["logits"],
dynamic_axes = {"input_ids": symbolic_names, "attention_mask": symbolic_names, "logits": symbolic_names},
opset_version = 14
Once we obtain our preliminary model out of this command, we might want to tune it:
• Save the weights as external files.
• Convert the model to fp16, if the PyTorch model wasn’t already in fp16 precision. Note that when converting to fp16, we still want LayerNormalization performed using fp32 precision to give a
higher dynamic range in those calculations.
• Change the output’s precision: if the PyTorch model was in fp16, the logits might still be converted to fp32. This can easily be modified by removing the Cast to fp32 in the last step of the
Once all these steps are run, we can run inferences easily thanks to ONNXRuntime and verify that we have a fully functional ONNX model without KV-cache.
The language models under consideration all use the decoder-only Transformer architecture. In particular, these models operate with the embedded vectors representing different text tokens in parallel
everywhere except at the self-attention layers. In those blocks, the auto-regressive nature of the models (in the sense that a new token is predicted taking into account only the past tokens) makes
it possible to store intermediate results to reduce inference complexity. The cached tensors are known as keys and values.
We implement KV-caching using a sliding window approach. That is, only a range of a few tokens is processed by each inference of the model and this range moves as new tokens are generated. All the
internal tensors of the model are cut accordingly, except at the self-attention layers where the embedded vectors representing past tokens are retrieved from the KV-cache.
Thus, the sequence length is replaced with several magnitudes:
1. The window size is the number of tokens processed by one inference.
2. The sequence length is the total number of tokens that the model can take into account in an inference. This is the length of an implicit sequence (possibly with padding at the beginning) that is
made of the past tokens (implicitly stored in the caches) plus the window tokens.
3. The maximum context length is the maximum number of tokens that the caches can store. This is useful in case one needs to increase the sequence length without moving the cached data.
Simplified diagram of the KV-cache. In grey, padding. In yellow, the window filled by an inference.
The figure above shows the shape of a KV-cache tensor. As a matter of fact, this figure is a simplification because the lines corresponding to a token index are fragmented. The new model computes the
yellow part, corresponding to the current window. To complete the picture, we add an input tensor for the white and grey parts (the past), an output tensor that represents the whole cache, an
operation to concatenate all parts and another operation to extract the white and yellow parts (i.e., without the padding represented in grey). Esperanto Technologies’ ML compiler keeps the KV-cache
tensor as a device-resident tensor, thus avoiding unnecessary data movement and copies.
Operations added to form the keys/values tensors.
The input_ids tensor’s dimension corresponding to the sequence length is simply replaced with the window size. By ONNX’s shape inference, this change propagates to the rest of the model.
Change applied to the tensor of input tokens.
As for the attention mask, there are two tensors to consider. The mask applied to the self-attention blocks of the models is originally a square matrix of dimension equal to the sequence length. In
the caching version of the model, only the last few rows (as many as the window size) of this matrix are used. In the simplest usage, auto-regressive language models are provided always with a
triangular matrix known as causal mask, as it encodes that every token prediction depends on all the previous tokens. Then the input attention mask tensor is just a vector that the model internally
expands as a matrix. That vector remains unchanged, but the part of the model that expands it has to be modified to use the right magnitude between the window size and the sequence length.
Change applied to the attention mask.
Rotary positional embeddings
Some SLMs include another piece that requires some careful modifications in our implementation: the rotary positional embeddings. This is a transformation applied to the queries and keys tensors of
the self-attention layers before comparing them via dot products. Essentially, tokens are indexed starting from 0 and the embedded vectors corresponding to the k-th token are rotated by angles of the
form kθ (for the queries) and -kθ (for the keys). When combining queries and keys in the self-attention, these rotations partially cancel each other. All in all, these operations capture how far
apart two tokens are from each other inside the sequence.
The tokens inside the window cannot be indexed starting from 0 at every inference as the window slides. Rather, we modified the model so that the window is seen as the final part of the sequence
(i.e., its first index is the sequence length minus the window size). Then it is crucial that the keys tensors are cached before applying the rotary embeddings transformations. In this way, even
though the absolute index of a given token changes from inference to inference, the relative differences between indices are preserved and the model behaves as expected.
All the work described in this section is generic and can be applied to any SLM based on a transformers decoder-only architecture. At Esperanto, we advocate for the use of optimized ONNX models and
to that end, we are committed to open-source as many SLM as possible, using different precisions, and extended to implement different optimization techniques. Note that these models can be used for
any ONNX provider but they are specially modified to target the strengths of Esperanto’s products and achieve faster inference.
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Multiplication Chart 1-144 2024 - Multiplication Chart Printable
Multiplication Chart 1-144
Multiplication Chart 1-144 – If you are looking for a fun way to teach your child the multiplication facts, you can get a blank Multiplication Chart. This may let your kid to complete the information
independently. You will discover empty multiplication maps for various product ranges, which includes 1-9, 10-12, and 15 goods. You can add a Game to it if you want to make your chart more exciting.
Below are a few tips to get your youngster started: Multiplication Chart 1-144.
Multiplication Maps
You may use multiplication maps as part of your child’s pupil binder to help them remember math facts. While many youngsters can remember their mathematics information naturally, it takes many others
time to do this. Multiplication charts are an effective way to strengthen their boost and learning their self confidence. As well as being educational, these graphs can be laminated for added
durability. Listed here are some valuable approaches to use multiplication maps. You can even take a look at these websites for helpful multiplication fact sources.
This lesson addresses the essentials from the multiplication desk. Together with discovering the guidelines for multiplying, individuals will comprehend the idea of elements and patterning. By
understanding how the factors work, students will be able to recall basic facts like five times four. They will also be able to use the property of zero and one to fix more complicated items.
Students should be able to recognize patterns in multiplication chart 1, by the end of the lesson.
Different versions
As well as the common multiplication graph or chart, pupils should develop a graph with additional elements or a lot fewer elements. To create a multiplication graph or chart with additional
elements, individuals must produce 12 furniture, every with a dozen series and a few posts. All 12 furniture should fit on a single page of document. Collections should be drawn having a ruler. Graph
paper is the best for this project. If graph paper is not an option, students can use spreadsheet programs to make their own tables.
Activity tips
Whether you are teaching a beginner multiplication training or focusing on the competence in the multiplication table, you can develop entertaining and fascinating online game suggestions for
Multiplication Graph or chart 1. A couple of exciting ideas are the following. This game requires the students to remain work and pairs on the same issue. Then, they may all hold up their cards and
explore the answer for a min. They win if they get it right!
When you’re instructing youngsters about multiplication, one of the better resources you may provide them with is a computer multiplication graph or chart. These printable linens come in a variety of
patterns and may be published in one page or a number of. Youngsters can understand their multiplication facts by copying them from the chart and memorizing them. A multiplication graph or chart will
be helpful for a lot of good reasons, from aiding them learn their arithmetic details to instructing them the way you use a calculator.
Gallery of Multiplication Chart 1-144
Multiplication Chart UDL Strategies Multiplication Chart Printable
Table Of 144 Learn 144 Times Table Multiplication Table Of 144
FREE Printable Multiplication Chart Printable Multiplication Table
Leave a Comment | {"url":"https://www.multiplicationchartprintable.com/multiplication-chart-1-144-2/","timestamp":"2024-11-12T05:42:11Z","content_type":"text/html","content_length":"52879","record_id":"<urn:uuid:62f7017d-76ab-4a3b-97d4-2db03f37194f>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00285.warc.gz"} |
- Teaching
Lecture Course: Topology 1
Time and Venue: Tuesday 12-14 ct (A 027) and Friday 12-14 ct (A 027)
Here is the moodle website which contains more information. The subscription key is the first name of a famous german topologist whose last name is the german word for ``doll'' (it is also the first
name of a colleague in the topology group here at LMU).
Oberseminar Arithmetic and Algebraic Geometry
Time and Venue: Wednesday 16-18 (TBD)
Workgroup Seminar Homotopy theory
Time and Venue: TBD
Lecture Course: Commutative Algebra
Time and Venue: Monday 14-16 ct (B 005) and Thursday 12-14 ct (B 005)
Here is the moodle website which will contain more information. The subscription key is the last name (first letter capitalized) of a famous algebraist whose first name is David.
Workgroup Seminar Homotopy theory
Time and Venue: Wednesday 13-15 st B 416, next meeting: 26.04.
Lecture Course: Algebra
Time and Venue: Wednesday 12-14 ct (B 006) and Friday 12-14 ct (B 006)
Exercise Session: Monday 16-18 ct (B 006)
Here is the website for the course with more information.
Lecture Course: Condensed Mathematics
Time and Venue: Monday 12-14 (B 251) and Thursday 10-12 (B 041)
Exercise Session: Monday 14-16 (B 046)
There will soon be a moodle website which contains the relevant course material.
Lectures on Condensed Mathematics
These are the lecture notes of the course. They are unfinished and not thoroughly proofread. Use at own risk. Comments more than welcome!
Verdichtete Mathematik.pdf
PDF-Dokument [635.2 KB]
Lecture course: Hermitian K-theory of Poincaré categories
Time and venue: Tuesday and Thursday 13-15, via Zoom.
Exercises sessions (biweekly): Thursday 15-17, via Zoom.
There is a webpage at Absalon which will contain lecture notes.
Lecture course: An Introduction to Infinity-Categories
Time and venue: Monday 10-12, Auditorium 6 and Friday 10-12, Auditorium 6
Exercises (by Piotr Pstragowski): Tuesday 13-16, Bio Center
There is a webpage at Absalon which will contain exercise sheets and lecture notes.
Lecture course: Infinity-categories
Time and venue: Monday 8-10, M103 and Wednesday 16-18, M101
Exercises (by Georg Tamme): Friday 10--12 H32
There is now a webpage at GRIPS -- if you plan to attend, please sign up for it there.
The GRIPS page will contain the exercise sheets. They will be uploaded on each monday. You will not have to hand in solutions, but they will be discussed in the exercise sessions.
K-theory seminar: Galois descent in telescopically localized K-theory, with Georg Tamme
Time and venue: Tuesday, 14-16, M311
Seminar webpage with further information.
CONFERENCE: Bavarian Geometry & Topology Meeting V -- Regensburg, July 4-5 2019.
The Homepage for the conference is here.
Autumn School: Computations in motivic homotopy theory -- Regensburg, September 16--20, 2019, organized with Cisinski, Strunk, and Tamme.
The Homepage for the conference is here.
Lecture course: Introduction to higher categories
Time and venue: Monday 16-18, M311 and Thursday 8-10, M311
Exercises (by Adeel Khan): Friday 10-12 in M009
There is now a webpage at GRIPS -- if you plan to attend the course, please sign up for it there.
The GRIPS page will contain the exercise sheets. They will be uploaded on each monday and will have to be handed in on the forthcoming monday, preferably in the break.
K-theory Seminar: p-adic K-theory of p-adic rings
Time and venue: Tuesday, 14-16, M311
Seminar webpage with further information.
On Sabbatical granted by the Academic Research Sabbatical Program of the University of Regensburg.
CONFERENCE: Bavarian Geometry & Topology Meeting III -- Regensburg, July 11-12 2018, organised jointly with G. Raptis
The Homepage for the conference is here.
CONFERENCE: Bavarian Geometry & Topology Meeting II -- Augsburg, December 14-15 2017, organised jointly with F. Hebestreit
The Homepage for the conference is here.
Research Seminar of the SFB: HIOB 6 -- On Topological Cyclic Homology
Time and venue: Mondays, 12-14, Seminar Room SFB. Seminar Homepage
Block Seminar on Representation theory of finite groups
Time and venue: Tuesday 10.10. -- Thursday 12.10, M009.
Seminar "Stable homotopy theory II" (with Georgios Raptis)
Time and venue: Wednesdays, 16-18, Seminar Room SFB. The Seminar webpage is hosted by GRIPS.
Lecture course on homotopy theoretic methods in the topology of manifolds
Time and venue: Mondays, 14-16, M102
Seminar on representation theory of finite groups
Time and venue: Fridays, 10-12, M101
---- CANCELLED, but will probably take place next semester ----
Seminar on Automorphisms of manifolds and algebraic K-theory (along Weiss-Williams)
Seminar on Homological and Representation Stability
Time and venue: Tuesdays, 14-16 in the SFB room: Bio 1.1.34.
Exercise Session for Topology 1
Time and venue: Wednesdays, 8:30-10, in room M101.
Winter School on Bordism, L-theory and real algebraic K-theory
Time and venue: December 5-9, 2016 at Kastell Windsor outside of Regensburg. Here is the link to the webpage of the Winter School.
Summer Term 15: Teaching Assistant for Topology 2
Organization of the exercise classes for the lecture of Thomas Nikolaus.
Winter Term 14/15: Teaching Assistant for Topology 1
Organization of the exercise classes for the lecture of Thomas Nikolaus.
Summer Term 14: Organization of the GRK 1150 Phd student seminar about Browder's result on the Kervaire invariant one problem.
Summer Term 13: Organization of the GRK 1150 Phd student seminar about C*-algebras and K-theory.
Sumer Term 12: Exercise Sessions in Linear Algebra 2 (lectured by Prof. Rapoport)
Winter Term 11/12: Exercise Sessions in Linear Algebra 1 (lectured by Prof. Rapoport)
Summer Term 11: Exercise Sessions in Global Analysis 2 (lectured by Prof. Lesch)
Winter Term 10/11: Exercise Sessions in Global Analysis 1 (lectured by Prof. Lesch)
Summer Term 10: Exercise Sessions in Analysis 1 (lectured by Prof. Conti)
Winter Term 09/10: Exercise Sessions in Analysis 2 (lectured by Prof. Conti) | {"url":"https://www.markus-land.de/teaching/","timestamp":"2024-11-05T23:19:42Z","content_type":"text/html","content_length":"28567","record_id":"<urn:uuid:ee193f64-ba3c-4a96-806a-56d1b0d46d20>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00468.warc.gz"} |
Expand Your Academic Horizons with Our Comprehensive Linguix Dictionary
How To Use Unit of measurement In A Sentence
• A nit is a unit of measurement used to describe screen brightness; one nit equals one candela per square meter and the more nits, the brighter the display. Post-gazette.com - News
• The OHM, as a unit of measurement, equals a unit of _resistance_ that is equivalent to the resistance of a hundred feet of copper wire the size of a pin. Steam, Steel and Electricity
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• The usual modern method is to use the vertical length of the head as a unit of measurement.
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• Ampère also devised a rule governing the mutual interaction of current-carrying wires (known as Ampère's law) and defined the unit of measurement for current flow, now known as the ampère.
Ampère, André Marie
• This is the antilog in terms of the original unit of measurement that defines the 50th percentile.
• From there, sizing descendant elements with ems (a relative unit of measurement) becomes much more logical: 1em is 10 px, 1.6em is 16 px, .9em is 9px, and so forth.
• A megapixel is a unit of measurement for the resolution of digital cameras. Everything2 New Writeups
• A number and its unit of measurement are hyphenated if they modify another noun.
• Likewise one may call the price index a ‘statistical illusion’ based on the chimera of a fixed basket of products as the unit of measurement.
• In astronomy, the preferred unit of measurement for such distances is the parsec, which is defined as the distance at which an object will appear to move one arcsecond of parallax when the
observer moves one astronomical unit perpendicular to the line of sight to the observer. Ann Aguirre » Blog Archive » A day in the life – blog Jeopardy
• The usual modern method is to use the vertical length of the head as a unit of measurement.
• Simply put, a bitcoin is an algorithm-based mathematical construct - a unit of measurement invented to quantify value.
• The unit of measurement is actually millimeters of mercury, and that figure of 120 just means the pressure is high enough to hold up a column of mercury 120 mm high.
• The unit of measurement was 8 'reales' (royals), later known as a peso. Safehaven
• According to Priscilla, the genius waitress, an alobar is a unit of measurement that describes the rate at which Old Spice after-shave lotion is absorbed by the lace on crotchless underpants,
although at other times she has defined it as the time it takes Chanel No. 5 to evaporate from the wing tips of a wild duck flying backward. La insistencia de Jürgen Fauth
• A diopter is a unit of measurement that indicates the amount of correction needed to change your vision to be as close to 20/20 as possible. Eyeglasses 101
• This is the antilog in terms of the original unit of measurement that defines the 50th percentile.
• Beloved by Scrabble fans, em (/εm/) can mean the letter M or a unit of measurement in typography – hence em dash (one of which appeared just there, before hence). May « 2009 « Sentence first
• The 42-gallon barrel is still a standard unit of measurement in the oil industry, though.
• The @ symbol was also used as an abbreviation for "amphora", the unit of measurement used to determine the amount held by the large terra cotta jars that were used to ship grain, spices and wine.
discovered this use of the @ symbol in a letter written in 1536 by a Florentine trader named Francesco Lapi. TheNextWeb.com
• Every time, the unit of measurement modifies the view.
• Commodity uses value commonly with the society necessary production time is unit of measurement.
• Whereas exchange is the concept of marketing, A transaction is marketing's unit of measurement.
• In astronomy, the preferred unit of measurement for such distances is the parsec, which is defined as the distance at which an object will appear to move one arcsecond of parallax when the
observer moves one astronomical unit perpendicular to the line of sight to the observer. Ann Aguirre » Blog Archive » A day in the life – blog Jeopardy
• However, Member States may require that measuring instruments bear indications of quantity in a single legal unit of measurement. | {"url":"https://linguix.com/english/word/unit%20of%20measurement/examples","timestamp":"2024-11-08T21:46:35Z","content_type":"text/html","content_length":"76580","record_id":"<urn:uuid:0a8df65f-bce2-4f80-b593-6fba80d5ec45>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00225.warc.gz"} |
WRITING PIECEWISE FUNCTIONS From a Real World Scenario
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WRITING PIECEWISE FUNCTIONS From a Real World Scenario
Creating Piecewise Functions from Real World Scenarios – Day 3 Today’s objective: I can write a piecewise function from multiple representations. (From a Real World Scenario)
4 -Step Problem Solving Process: STEP 1: Understand the problem: a) Read the entire problem. b) Can you restate the problem?
4 -Step Problem Solving Process: STEP 2: Devise a Plan: a) Highlight any given information. b) Eliminate any unnecessary info. c) Define the variable using the unknown info. d) Relate the given info
to the unknown info with a formula or equation.
4 -Step Problem Solving Process: STEP 3: Execute the Plan: a) Model the problem with the equation. b) Solve the equation for the unknown. c) Be sure to label the units.
4 -Step Problem Solving Process: STEP 4: Check Your Work: a) Check the solution in the original equation. b) Does your answer make sense in the context of the problem?
Example 1: Create a Piecewise Function You go to Wal-Mart to buy some candy. You decide to buy snickers because they have a special deal on snickers. A bag of snickers costs $3. 45, but if you buy 4
or more bags, they only cost $3. 00 per bag. Create a piecewise function to represent the cost of the bags of snickers.
Solution: Example 1 • b = number of bags of snickers • C(b) = Cost of all bags of snickers purchased • $3. 45 = cost of a bag if less than 4 bags are purchased • $3. 00 = cost of a bag if 4 or more
bags are purchased.
Example 2: Create a Piecewise Function The Mad Hatter is ordering cups from Teacups, Limited, for his tea party. The Teacups, Limited catalog prices cups according to the number of cups ordered. For
orders of 20 or fewer cups, the price is $1. 40 per cup plus $12 shipping and handling on the order. For orders of more than 20 cups, the price is $1. 10 per cup plus $15 shipping and handling. a)
Create a function to describe the price of cups.
Example 2: Create a Piecewise Function b) How much will it cost the Mad Hatter to order 16 cups? c) If the Mad Hatter wants to spend at most $45, what is the maximum number of cups he can order?
Solution: Example 2 • c = the number of cups ordered • P(c) = the price of all cups ordered • $1. 40 cost of each cup if 20 or less are ordered • $12. 00 shipping if 20 or less are ordered • $1. 10
cost of each cup if more than 20 are ordered • $15. 00 shipping if more than 20 are ordered b) $34. 40 c) 27 cups
Example 3: Create a Piecewise Function Every month your cell phone plan costs $75 and gives you unlimited talk, 500 text messages, and no data plan. It costs $0. 10 per text message sent in excess of
the 500 that you are originally allotted. a) Write a piecewise function to determine the amount of your cell phone bill. b) How much will it cost you if you send 750 text messages?
Solution: Example 3 • $75 = monthly cost of cell phone bill • $0. 10 = cost of each text in excess of the plan’s allotted 500. • t = number of text messages sent • T-500 = number of texts in excess
of 500 allotted • C(t) = Cost of your cell phone bill b) $100
Example 4: Create a Piecewise Function A construction worker earns $17 per hour for the first 40 hours of work and $25. 50 per hour for work in excess of 40 hours. a) Create a function to represent
the amount of her paycheck. b) One week she earned $896. 75. How much overtime did she work?
Solution: Example 4 • $17 = hourly pay: hours up to 40. • $25. 50 = hourly pay; hours in excess of 40 • h = # of hours worked • h-40 =# hours worked in excess of 40 • P(h) = total amount of paycheck
b) 8. 5 overtime hours
Example 5: Create a Piecewise Function Southeast Electric charges $0. 09 per kilowatt-hour for the first 200 k. Wh. The company charges $0. 11 per kilowatt-hour for all electrical usage in excess of
200 k. Wh. a) Create a function to model this scenario. b) How many kilowatt-hours were used if a monthly electric bill was $57. 06?
Solution: Example 5 • h = # kilowatt-hours • C(h) = Cost of total kilowatt-hours b) 555. 09 Kwh
YOU TRY: On Your Whiteboard! Ex. 1: A city parking lot uses the following rules to calculate parking fees: ① A flat rate of $5. 00 for any amount of time up to and including the first hour. ① A flat
rate of $12. 50 for any amount of time over 1 hour and up to and including 2 hours. ① A flat rate of $13 plus $3 per hour for each hour after 2 hours. Create a piecewise function to model the parking
YOU TRY: On Your Whiteboard! Ex. 2: Your job pays you $8. 50 an hour for a normal 40 hour work week. If you work over 40 hours then you get paid overtime, at time and half (1. 5 times your normal
rate) for anything over 40 hours, but you are not allowed to work more than 20 overtime hours. a) Create a function that models this scenario. b) How much would your paycheck be if you worked 45
YOU TRY: On Your Whiteboard! Ex. 3: Every month your cell phone plan costs $230 and gives you unlimited talk, text messages, and 2 GB of data. Extra data for each month is $15 per GB you go over your
allotted 2 GB. a)Write a piecewise function to represent this situation.
YOU TRY: On Your Whiteboard! Ex. 4: You are teaching tomorrow with an activity and you want to use candy to motivate your students. You go to Fred’s Food Club and they have a special going on for
blow pops. For 2 or less bags it cost $4. 35 for each bag, but if you buy more than 2 bags you get the special price of $3. 25 per bag. a) Write a piecewise function to represent this situation.
YOU TRY: On Your Whiteboard! Ex. 5: Greenville Utilities charges a basic customer charge of $10. 99 and $0. 1260 per k. Wh for 400 k. Wh or less. If you go over 400 k. Wh you will then pay $0. 1151
per k. Wh for all k. Wh in excess of the original 400 KWh. a) Write a piecewise function to represent this situation. | {"url":"https://slidetodoc.com/writing-piecewise-functions-from-a-real-world-scenario/","timestamp":"2024-11-12T16:51:28Z","content_type":"text/html","content_length":"115963","record_id":"<urn:uuid:c2907361-4069-493d-8dc0-c100e2596867>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00734.warc.gz"} |
George Whitesides: Lorenz Effect and Curiosity | ChemTalk
Lorenz Effect
Dr. George Whitesides is curious about the Lorenz effect and force. Lorenz force (also known as the Lorentz force) describes the force experienced by a charged particle due to electric and magnetic
fields. First described by the Dutch physicist Hendrik Lorenz, this force can be described by this equation: F = qE + qv x B [1]. The bolded parts represent vector quantities. A vector is an object
with both a magnitude and a direction.
Let’s now dive into what each term refers to! F is the entire electromagnetic force on the charged particle—and this is our Lorenz force. We can then break down the equation into two parts consisting
of an electric force and a magnetic force. The first term is qE, often called the electric force. E is the electric field while q is the electric charge of the particle. Generated by electric charge,
electric fields provide information about force per unit charge at every point in space surrounding a distribution of charges. Multiplying this electric field with the electric charge of the particle
will give us the electric force on the particle.
The second term of the Lorenz equation is the magnetic force: F = qv x B. q represents the charge of the particle; v represents the velocity of the charged particle; B represents the magnetic field
vector. The x in the equation represents a cross-product between the velocity and the magnetic field vector.
Cross-product between two vectors (let’s say a and b) produce another vector that has special properties. Namely, the resultant vector is perpendicular to both a and b. Here is an image that
represents this relationship:
Figure 1. Cross product parallelogram
The formula for a cross-product between two vectors is a x b = ||a|| ||b|| sin (θ) where θ is the angle between a and b. We can apply this to the magnetic force part of the Lorenz force equation: F =
qv x B = q ||v|| ||B|| sin (θ).
The Lorenz effect has important applications in various areas of physics and engineering. It is the principle behind the operation of many electric motors and generators, where the interaction
between the magnetic field and the moving charged particles produces mechanical work or electrical energy. Dr. George Whitesides, a remarkable chemist, a pioneer in nanotechnology, and a visionary in
the field of materials science, finds the Lorenz effects exciting and useful [3]. His undergraduate thesis concerned electrochemistry, and he continues to ponder how magnetic fields affect chemistry
as well as the bigger question of the origin of life. Find more on his lab’s work with the Lorenz effect here.
Dr. Whitesides has had an illustrious career as a scientist and inventor. His work extends beyond the realms of chemistry and has left long-lasting impacts in various fields. For example, these
fields include but are not limited to medical diagnostics, nanofabrication techniques, and public health. He believes in doing work that benefits society and sees chemistry as a vehicle to accomplish
such work [3]. It is incredibly hard to capture all that Dr. Whitesides has worked on and accomplished, but we would like to leave one message from him for our readers: be and stay curious [3]. Dr.
Whitesides finds many topics interesting. He tells future scientists that it is up to them to decide what is important out of those curiosity-inducing phenomena, questions, and problems.
Learn More
If you’d like to hear more about Dr. Winhitesides’ journey as well as his current passions and previous projects, visit us on Spotify, Apple Podcasts, and many other streaming services to listen to
our ChemTalk podcast with Dr. George Whitesides, Woodford L. and Ann A. Flowers University Research Professor at Harvard University’s Department of Chemistry and Chemical Biology.
Find the ChemTalk podcast here.
Works Cited
[1] Encyclopaedia Britannica, inc. (n.d.). Lorentz force. Encyclopaedia Britannica. https://www.britannica.com/science/Lorentz-force
[2] Williams, M. (2016, October 23). What is a magnetic field? Universe Today. https://www.universetoday.com/76515/magnetic-field/
[3] Whitesides, George. Personal Interview. Conducted by Olivia Lambertson and Riya Jain. 6 April 2023. | {"url":"https://chemistrytalk.org/spotlight-george-whitesides/","timestamp":"2024-11-14T01:57:50Z","content_type":"text/html","content_length":"230718","record_id":"<urn:uuid:20ac7ebb-74b5-49d9-a0e1-ec49d63d3020>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00228.warc.gz"} |
Higher-Order Spectral Clustering of Directed Graphs
NeurIPS 2020
Higher-Order Spectral Clustering of Directed Graphs
Review 1
Summary and Contributions: The paper considers a graph clustering on directed graphs. The authors introduce a new notion of clustering objective denoted by flow ratio. For any ordered partition of
vertex set V into k pairwise disjoint subset (S0, ..., Sk-1), the flow ratio of the partition is sum of the average flow (i.e. w(i, i+1)/Vol(Si)+Vol(Si+1)) along the path from S0 to Sk-1. The optimal
clustering is the partitioning of V that maximizes the flow ratio among all possible partitions. The authors represent the directed graph using the Hermitian adjacency matrix. The main contribution
of this work is proving that the optimal clustering is well-embedded with the first eigenvector of the corresponding Laplacian matrix. Equipped with this observation, authors design an algorithm to
find clustering. The algorithm is very similar to the standard spectral clustering algorithms for undirected graph; It first estimates the embedding and then runs k-means on the corresponding
embedding. They show that for graphs with few clusters (i.e., poly(log n)) the algorithm runs in nearly linear time. Moreover, they prove that the clustering returned by the algorithm has a small
symmetric difference with optimal clustering where the recovery quality depends on the gap between the first and second eigenvalues of the Laplacian. Finally, the authors show how to improve the
runtime of the algorithm by subsampling edges with probabilities proportional to the weight by degree. In addition, they evaluate the performance of their algorithm on the stochastic block model and
UN Comtrade Dataset.
Strengths: This paper has a conceptual contribution in considering the flow ratio notion for clustering directed graphs. It is an initial step towards applying known spectral techniques for
undirected graphs to the directed graphs with the help of complex Hermitian adjacency matrix. Although the algorithm and most of the theoretical analysis are very similar to the standard spectral
clustering literature such as work by Peng, Sun and Zanetti, the novel observation of this work is embedding the points into the space of dimension 2 with the help of complex Hermitian matrix. This
approach shows how to differentiate points from different clusters using angles, although to find the clustering for undirected graphs it is crucial to look at the subspace of dimension k that is
spanned by the first k eigenvectors.
Weaknesses: The authors succeed in showing that the standard spectral techniques can be applied to the directed graph clustering using flow ratio formulation. However, the notion of flow ratio is not
well-motivated. It is not clear why one should consider a specific path between clusters. In other words, the relationship between pairwise different clusters is not captured in the objective
function and it only takes care of consecutive terms which is a restrictive definition in my opinion. Moreover, since this is a new objective function for directed graph clustering the authors should
justify the hardness of finding the optimal clustering, also they should compare it with other clustering objectives for directed graphs. Regarding the experiments on stochastic block model, first,
it's not clear how large the input graph is, also, the number of clusters (i.e, k) is a small constant, so it's not clear from the experiments if the sublinear algorithm scales very well on large
graphs. Second, the number of different runs in Figure 1 seems to be very small which might not be enough to get a good concentration in the accuracy of the results. Finally, the authors evaluate the
quality of their methods using the Adjusted Rand Index. It might be useful to report the quality of the generated output using other known evaluation methods for spectral clusterings such as inner
and outer conductance, precision and recall.
Correctness: Yes both theoretical claims and empirical methodology are correct. In line 185 of the main file it's not clear for me why do you have (1+APT)(yk-1)? It shouldn't be APT*(yk-1)?
Clarity: The paper is well-written, and the algorithms and proofs are well-explained.
Relation to Prior Work: The authors clearly explain their theoretical contribution comparing to the previous works.
Reproducibility: Yes
Additional Feedback:
Review 2
Summary and Contributions: After reading the authors' rebuttal, my evaluation remains the same. The work studies clustering of directed graphs by spectral means. Traditional spectral clustering on
undirected graphs seek to find dense clusters that induce small cuts. Instead here the goal is to find a directed structure, for example, a cluster that has mostly outgoing edges and one that has
mostly incoming edges (so sparse clusters inducing large cuts are welcome). The authors introduce a notion of flow ratio and discuss algorithms to find a clustering that approximately maximize that
ratio. The paper contains experiments where the aforementioned algorithms find patterns in international trade markets (e.g. clusters of crude oil exporting countries).
Strengths: The topic is interesting and matches the NeurIPS audience. The techniques are nontrivial and yield interesting insights. The whole analysis and the algorithms seem theoretically well
grounded. The spectral analysis is of interest on its own (I remark that in many points it bears similarities with previous work). Thus I believe the work connects nicely the original goal of
clustering with the spectral analysis for directed graphs and with the resulting approximation algorithms. The experiments are interesting as well, and suggest that the proposed algorithm is indeed
capable of finding the cluter structure that one would expect. In particular the authors recover the crude oil international market structure (say, exporters vs importers) using data involving 246
different countries and regions over several years.
Weaknesses: The main weakness of the paper is that here and there the exposition is poor. In particular, it is hard to make the necessary connections and understand where the discussion is going.
There are several technical results stated in a "standalone" form although they are clearly meant to be connected/used somewhere else. This makes it quite hard to understand the key ideas of the
paper and in what they differ from previous work. In particular, many concepts and results are similar to [22] (Peng et al @ SICOMP 17).
Correctness: I could not verify the proofs of the claims although the results make sense. There is at least one mathematical expression that I found suspicious since it is undefined (division by 0)
when k=2 and G has a single edge or is bipartite (see detailed comments below). I do not know if this is only a corner case.
Clarity: The paper is fairly well written, but as I said it is generally hard to understand where the discussion is heading. There are also a few sentences that are confusing or incorrect (for
example, at the very beginning of the introduction the paper presents a complicate expression and none of the symbols involved were defined before). I think the presentation can be improved.
Relation to Prior Work: Yes and no. The authors acknowledge previous work in the sense that they correctly point to the relevant papers. What leaves me in doubt is that it is not clear what is
similar to what. For example, the authors introduce in Equation (3) a kind of indicator vector for the clusters. Similar things are standard, so this is an (interesting) variation, but this it is not
written. (Specifically, the equation just above (3) is the standard normalized indicator vector, and (3) does a linear combination which is the novelty part). This leaves the non-specialist reader
like me in doubt; what is conceptually new, and what is just an application of standard techniques?
Reproducibility: Yes
Additional Feedback: Detailed comments: - line 42: "any set of vertices" is confusing, here there is a partition - lines 43-35: I could not understand this expression as none of the involved symbols
has been introduced except S_j - lines 73-74: "in particular we have vol(G) = vol(V) = 2m" this is not true since you define d() as weights instead of degrees - lines 98-99: "the fraction of directed
edges with endpoints in S_j or S_{j-1} which cross the cut from S_j and S_{j-1}" this sounds like the direction of the edge is irrelevant (it is also oddly phrased) - line 122: this expression seems
to have a 0 denominator, for example when k=2 and G is a single edge; this is strange - lines 138-139: "we propose to apply k-means on the embedded points from f_1" I do not understand this sentence,
in particular "from f_1" and what points the sentence is talking about - S 4.2: I cannot understand how the three lemmata imply the theorem; I see they are related, but they are just listed without
explaining the connection
Review 3
Summary and Contributions: This paper studies k-way partitions of directed graphs using k (complex) eigenvectors of the directed Laplacian. It gives bounds similar to those for k-way partitions of
undirected graphs, and demonstrates good performances of the algorithms experimentally on small synthetic and real world graphs.
Strengths: Extending spectral clustering into directed graphs is a difficult task: it's quite surprising to me that spectral techniques and their guarantees can be readily extended to directed
graphs. The experimental verifications of the algorithms are fairly through.
Weaknesses: Aside from the extension to the complex domain, it's not clear what are the new ideas in this paper compared to spectral methods for (k-way) partitioning undirected graphs. The
experiments were mostly for small sized graphs, and it's not clear how effective the algorithm would scale to larger instances. Even the largest dataset for oil trading only involves < 300 vertices
(but with about 100 commodities): I suspect the data becomes much smaller after the input is converted to graph form. If the data size is indeed large, it would be useful to directly mention the
sizes of the intermediate graphs generated.
Correctness: I believe the algorithm and the bounds claimed. The experimental set up is sound, although I'd like to have seen some comparisons with other methods such as pagerank or matrix
factorization based methods. While those methods are not geared toward the flow ratios studied here, I believe their output clusters can still be compared directly.
Clarity: The paper introduces the problem and methods clearly, but I find some of the formal components a bit lacking: for example, k was assumed to be O(log^{c}n) on line 144 and omitted from
subsequent bounds involving running times. A pseudo-code of the overall algorithm is also not present in sec 4: this made it difficult to figure out the overall algorithmic picture.
Relation to Prior Work: Prior works, especially for spectral partitioning of directed graphs, are discussed systematically.
Reproducibility: Yes
Additional Feedback: I believe the paper has some interesting ideas, but have difficulty figuring out its improvements over previous results. I feel there are enough theoretical new ideas for the
paper to be accepted, although also feel that the ideas are still an iteration away from being directly used on large networks. | {"url":"https://papers.neurips.cc/paper_files/paper/2020/file/0a5052334511e344f15ae0bfafd47a67-Review.html","timestamp":"2024-11-11T10:22:03Z","content_type":"text/html","content_length":"13627","record_id":"<urn:uuid:042afd4d-78d0-41ba-8b46-253a1f42599d>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00138.warc.gz"} |
What are Binary Numbers: Understanding the Foundation of Modern ComputingWhat are Binary Numbers: Understanding the Foundation of Modern Computing
What are Binary Numbers: Understanding the Foundation of Modern Computing
Binary numbers are fundamental to modern computing and play a pivotal role in digital communication data storage and computation. In this article we will delve into the intricacies of binary numbers
exploring their significance applications and conversion methods.
Introduction to Binary Numbers
In the realm of mathematics and computer science binary numbers serve as the backbone of digital technology. Unlike the decimal system which utilizes ten digits (0-9) binary numbers rely on only two
digits: 0 and 1. This binary system forms the basis of all digital data representation.
Understanding the Binary Number System
What are Binary Numbers?
Binary numbers often referred to as base-2 numbers represent numeric values using only two symbols: 0 and 1. Each digit in a binary number holds a specific place value similar to the decimal system.
However in binary each digit's value doubles as you move from right to left.
Why are Binary Numbers Used?
The utilization of binary numbers is primarily due to their compatibility with digital electronic systems. Since computers operate using binary switches (on/off) representing data in binary form
aligns seamlessly with the underlying hardware architecture.
Binary Digits (Bits) and Place Value
In binary each digit is called a bit short for binary digit. The position of a bit within a binary number determines its place value following a pattern of powers of 2. For instance the rightmost bit
represents 2^0 (1) the next bit to the left represents 2^1 (2) then 2^2 (4) and so forth.
Conversion between Binary and Decimal
Binary to Decimal Conversion
Converting binary numbers to decimal involves multiplying each digit by its corresponding power of 2 and summing the results. For example the binary number 1011 is equivalent to (1 × 2^3) + (0 × 2^2)
+ (1 × 2^1) + (1 × 2^0) = 11 in decimal.
Decimal to Binary Conversion
Converting decimal numbers to binary requires dividing the decimal number by 2 repeatedly and noting the remainders. The binary equivalent is obtained by reading the remainders in reverse order. For
example converting decimal 13 to binary yields 1101.
Applications of Binary Numbers
Computing and Digital Electronics
Binary numbers form the foundation of computing systems facilitating data processing storage and transmission. From microprocessors to memory units digital devices rely on binary logic to perform
Binary Representation in Computers
Inside computers all data is stored and processed in binary format. Text images audio and video are encoded into binary digits for manipulation by electronic circuits. Binary representation enables
the binary arithmetic operations that computers perform.
Advantages of Binary Numbers
Binary numbers offer several advantages in digital systems including simplicity efficiency and compatibility with electronic hardware. Their concise representation and straightforward arithmetic
operations make them indispensable in computing.
Disadvantages of Binary Numbers
Despite their utility binary numbers can pose challenges in human comprehension due to their lengthy representation for large values. Additionally manual binary arithmetic can be cumbersome compared
to decimal arithmetic.
Binary Numbers in Daily Life
While binary numbers may seem abstract they influence various aspects of modern life. From digital clocks and electronic gadgets to internet protocols and encryption algorithms binary principles
underpin countless technological innovations.
FAQs (Frequently Asked Questions)
1. Why are binary numbers used in computers? Binary numbers align with the on/off nature of electronic switches in computers making them ideal for digital representation.
2. Can you give an example of binary numbers in everyday life? Yes digital clocks which represent time using binary digits are a common example of binary numbers in daily use.
3. What are the advantages of binary numbers over decimal numbers in computing? Binary numbers offer simplicity efficiency and compatibility with electronic hardware making them well-suited for
digital systems.
4. How do computers perform arithmetic operations with binary numbers? Computers utilize electronic circuits designed to perform binary arithmetic operations including addition subtraction
multiplication and division.
5. Are there any drawbacks to using binary numbers? While binary numbers are efficient for computers they can be challenging for humans to work with due to their lengthy representation and complex
arithmetic operations.
Binary numbers serve as the cornerstone of modern computing enabling the digital revolution that has transformed society. Understanding binary concepts is essential for anyone seeking to grasp the
inner workings of computers and digital technology.
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A bidirectional circulator with value type boost::graph_traits<Graph>::halfedge_descriptor over all halfedges having the same vertex as target.
Let h be a halfedge of graph g. For a Halfedge_around_target_circulator havc with h = *havc; the following holds: opposite(next(h,g),g) == *++havc.
Template Parameters
Graph must be a model of the concept HalfedgeGraph | {"url":"https://doc.cgal.org/5.5.1/BGL/classCGAL_1_1Halfedge__around__target__circulator.html","timestamp":"2024-11-07T01:02:48Z","content_type":"application/xhtml+xml","content_length":"9902","record_id":"<urn:uuid:ad922733-a18c-420f-8622-5fe83268d15a>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00336.warc.gz"} |
Fundamental Theorem of Genera
Consider proper equivalence classes of forms with discriminant equal to the field discriminant, then they can be subdivided equally into genera of forms which form a subgroup of the proper
equivalence class group under composition (Cohn 1980, p. 224), where is the number of distinct prime divisors of . This theorem was proved by Gauß
Arno, S.; Robinson, M. L.; and Wheeler, F. S. ``Imaginary Quadratic Fields with Small Odd Class Number.'' http://www.math.uiuc.edu/Algebraic-Number-Theory/0009/.
Cohn, H. Advanced Number Theory. New York: Dover, 1980.
Gauss, C. F. Disquisitiones Arithmeticae. New Haven, CT: Yale University Press, 1966.
© 1996-9 Eric W. Weisstein | {"url":"http://drhuang.com/science/mathematics/math%20word/math/f/f371.htm","timestamp":"2024-11-14T20:13:56Z","content_type":"text/html","content_length":"4314","record_id":"<urn:uuid:cdbae206-3ff5-4009-85e3-786db1055304>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00166.warc.gz"} |
On the Sufficiency of Using Degree Sequence of the Vertices to Generate Random Networks Corresponding to Real-World Networks
The focus of research in this paper is to investigate whether a random network whose degree sequence of the vertices is the same as the degree sequence of the vertices in a real-world network would
exhibit values for other analysis metrics similar to those of the real-world network. We use the well-known Configuration Model to generate a random network on the basis of the degree sequence of the
vertices in a real-world network wherein the degree sequence need not be Poisson-style. The extent of similarity between the vertices of the random network and real-world network with respect to a
particular metric is evaluated in the form of the correlation coefficient of the values of the vertices for the metric. We involve a total of 24 real-world networks in this study, with the spectral
radius ratio for node degree (measure of variation in node degree) ranging from 1.04 to 3.0 (i.e., from random networks to scale-free networks). We consider a suite of seven node-level metrics and
three network-level metrics for our analysis and identify the metrics for which the degree sequence would be just sufficient to generate random networks that have a very strong correlation
(correlation coefficient of 0.8 or above) with that of the vertices in the corresponding real-world networks.
Configuration model; degree sequence; correlation; random network; real-world network
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How to Use the 3D Convolution In Tensorflow?
To use 3D convolution in TensorFlow, you first need to import the necessary modules such as tensorflow and tensorflow.keras.layers. Next, you can create a 3D convolutional layer by using the Conv3D
class provided by TensorFlow's Keras API. You can specify the number of filters, kernel size, strides, padding, and activation function for the convolutional layer.
For example, you can create a 3D convolutional layer with 64 filters, a kernel size of (3, 3, 3), and a ReLU activation function by using the following code:
1 import tensorflow as tf
2 from tensorflow.keras.layers import Conv3D
4 model = tf.keras.Sequential()
5 model.add(Conv3D(64, (3, 3, 3), activation='relu', input_shape=(height, width, depth, channels)))
You can then compile and train your 3D convolutional neural network using your data. 3D convolution is commonly used in tasks such as video recognition, medical image analysis, and volumetric data
processing where the spatial and temporal dimensions need to be considered.
How to save and load a trained 3D convolutional model in TensorFlow?
To save and load a trained 3D convolutional model in TensorFlow, you can use the save and load_model functions from the TensorFlow library. Here is a step-by-step guide on how to do this:
1. Save the trained model:
1 model.save('3d_conv_model.h5')
1. Load the saved model:
1 from tensorflow.keras.models import load_model
3 # Load the model
4 model = load_model('3d_conv_model.h5')
1. To use the loaded model for making predictions or further training:
1 # Make predictions
2 predictions = model.predict(test_data)
4 # Continue training
5 model.fit(train_data, train_labels, epochs=5)
By following these steps, you can save and load your trained 3D convolutional model in TensorFlow. Remember to replace 3d_conv_model.h5 with the desired filename for your saved model.
How to perform data augmentation for 3D convolution in TensorFlow?
Data augmentation for 3D convolution in TensorFlow can be performed using the following steps:
1. Load and preprocess your 3D data: Start by loading your 3D data and preprocessing it as needed. This may include data normalization, resizing, and any other necessary preprocessing steps.
2. Define data augmentation techniques: 3D data augmentation techniques can include rotations, translations, scaling, flipping, adding noise, and more. Define the data augmentation techniques you
want to apply to your 3D data.
3. Create TensorFlow data augmentation functions: Write functions in TensorFlow that apply the defined data augmentation techniques to your 3D data. These functions should take the input data and
apply the specified augmentation techniques to generate augmented data.
4. Incorporate data augmentation into the training pipeline: Integrate the data augmentation functions into your training pipeline in TensorFlow. This can be done using TensorFlow's data input
pipelines or data augmentation layers.
5. Train your model with augmented data: Train your 3D convolutional neural network using the augmented data generated in the training pipeline. Monitor the training process and evaluate the
performance of your model on the augmented data.
By following these steps, you can perform data augmentation for 3D convolution in TensorFlow to improve the robustness and generalization of your 3D convolutional neural networks.
How to perform 3D pooling after convolution in TensorFlow?
To perform 3D pooling after convolution in TensorFlow, you can use the tf.nn.max_pool3d function. Here is an example code snippet demonstrating how to perform 3D pooling after convolution in
1 import tensorflow as tf
3 # Perform 3D convolution
4 input_data = tf.placeholder(tf.float32, [None, height, width, depth, channels])
5 conv_filter = tf.Variable(tf.random_normal([filter_size, filter_size, filter_size, input_channels, output_channels]))
6 conv_output = tf.nn.conv3d(input_data, conv_filter, strides=[1, 1, 1, 1, 1], padding='SAME')
8 # Perform 3D pooling
9 pool_output = tf.nn.max_pool3d(conv_output, ksize=[1, pool_size, pool_size, pool_size, 1], strides=[1, pool_strides, pool_strides, pool_strides, 1], padding='SAME')
In the code snippet above, we first define a 3D convolution operation using tf.nn.conv3d with the input data, convolution filter, and specified strides and padding. Then, we perform 3D pooling using
tf.nn.max_pool3d with the convolution output, pooling size, pooling strides, and padding type.
You can adjust the filter_size, pool_size, and pool_strides parameters to customize the size of the convolution filter and pooling operation.
How to apply 3D convolution to a video input in TensorFlow?
To apply 3D convolution to a video input in TensorFlow, you can use the tf.keras.layers.Conv3D layer. Here is an example code snippet on how to do this:
1 import tensorflow as tf
3 # Define your video input shape (num_frames, height, width, channels)
4 input_shape = (10, 128, 128, 3)
6 # Create a Conv3D layer with desired parameters
7 conv3d_layer = tf.keras.layers.Conv3D(filters=32, kernel_size=(3, 3, 3), activation='relu', input_shape=input_shape)
9 # Define your video input tensor
10 input_tensor = tf.random.normal(input_shape)
12 # Apply convolution to the input tensor
13 output = conv3d_layer(input_tensor)
15 # Print the output shape
16 print(output.shape)
In this code snippet, we first define the input shape of our video input (num_frames, height, width, channels). We then create a Conv3D layer with specified parameters such as the number of filters,
kernel size, and activation function. Next, we define our video input tensor using tf.random.normal and apply the convolution operation using the Conv3D layer. Finally, we print the shape of the
output tensor after applying the convolution operation.
You can customize the parameters of the Conv3D layer such as the number of filters, kernel size, padding, etc., based on your requirements. | {"url":"https://stlplaces.com/blog/how-to-use-the-3d-convolution-in-tensorflow","timestamp":"2024-11-14T23:58:11Z","content_type":"text/html","content_length":"362013","record_id":"<urn:uuid:1fb47fbe-730f-49d0-aab5-3b08b4d421d3>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00685.warc.gz"} |
Conditional Quasi-Exact Solvability of the Quantum Planar Pendulum and its Anti-Isospectral Hyperbolic Counterpart
Becker, S. and Mirahmadi, M. and Schmidt, B. and Schatz, K. and Friedrich, B. (2017) Conditional Quasi-Exact Solvability of the Quantum Planar Pendulum and its Anti-Isospectral Hyperbolic
Counterpart. Eur. J. Phys. D, 71 (6). p. 149.
Full text not available from this repository.
Official URL: https://dx.doi.org/10.1140/epjd/e2017-80134-6
We have subjected the planar pendulum eigenproblem to a symmetry analysis with the goal of explaining the relationship between its conditional quasi-exact solvability (C-QES) and the topology of its
eigenenergy surfaces, established in our earlier work [Frontiers in Physical Chemistry and Chemical Physics 2, 1-16, (2014)]. The present analysis revealed that this relationship can be traced to the
structure of the tridiagonal matrices representing the symmetry-adapted pendular Hamiltonian, as well as enabled us to identify many more - forty in total to be exact - analytic solutions.
Furthermore, an analogous analysis of the hyperbolic counterpart of the planar pendulum, the Razavy problem, which was shown to be also C-QES [American Journal of Physics 48, 285 (1980)], confirmed
that it is anti-isospectral with the pendular eigenproblem. Of key importance for both eigenproblems proved to be the topological index κ, as it determines the loci of the intersections (genuine and
avoided) of the eigenenergy surfaces spanned by the dimensionless interaction parameters η and ζ. It also encapsulates the conditions under which analytic solutions to the two eigenproblems obtain
and provides the number of analytic solutions. At a given κ, the anti-isospectrality occurs for single states only (i.e., not for doublets), like C-QES holds solely for integer values of κ, and only
occurs for the lowest eigenvalues of the pendular and Razavy Hamiltonians, with the order of the eigenvalues reversed for the latter. For all other states, the pendular and Razavy spectra become in
fact qualitatively different, as higher pendular states appear as doublets whereas all higher Razavy states are singlets.
Repository Staff Only: item control page | {"url":"http://publications.imp.fu-berlin.de/2037/","timestamp":"2024-11-09T03:32:06Z","content_type":"application/xhtml+xml","content_length":"21373","record_id":"<urn:uuid:dbecef9f-ba3d-47e7-ad5b-0aebccf3ef8b>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00045.warc.gz"} |
Barrel (Petroleum) to Cubic Inch Converter
Enter Barrel (Petroleum)
Cubic Inch
⇅ Switch toCubic Inch to Barrel (Petroleum) Converter
How to use this Barrel (Petroleum) to Cubic Inch Converter 🤔
Follow these steps to convert given volume from the units of Barrel (Petroleum) to the units of Cubic Inch.
1. Enter the input Barrel (Petroleum) value in the text field.
2. The calculator converts the given Barrel (Petroleum) into Cubic Inch in realtime ⌚ using the conversion formula, and displays under the Cubic Inch label. You do not need to click any button. If
the input changes, Cubic Inch value is re-calculated, just like that.
3. You may copy the resulting Cubic Inch value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Barrel (Petroleum) to Cubic Inch?
The formula to convert given volume from Barrel (Petroleum) to Cubic Inch is:
Volume[(Cubic Inch)] = Volume[(Barrel (Petroleum))] × 9702
Substitute the given value of volume in barrel (petroleum), i.e., Volume[(Barrel (Petroleum))] in the above formula and simplify the right-hand side value. The resulting value is the volume in cubic
inch, i.e., Volume[(Cubic Inch)].
Calculation will be done after you enter a valid input.
Consider that a tanker truck transports 200 barrels (petroleum) of crude oil.
Convert this volume from barrels (petroleum) to Cubic Inch.
The volume in barrel (petroleum) is:
Volume[(Barrel (Petroleum))] = 200
The formula to convert volume from barrel (petroleum) to cubic inch is:
Volume[(Cubic Inch)] = Volume[(Barrel (Petroleum))] × 9702
Substitute given weight Volume[(Barrel (Petroleum))] = 200 in the above formula.
Volume[(Cubic Inch)] = 200 × 9702
Volume[(Cubic Inch)] = 1940400
Final Answer:
Therefore, 200 bl is equal to 1940400 in^3.
The volume is 1940400 in^3, in cubic inch.
Consider that a pipeline transfers 500 barrels (petroleum) of refined oil per day.
Convert this flow rate from barrels (petroleum) to Cubic Inch.
The volume in barrel (petroleum) is:
Volume[(Barrel (Petroleum))] = 500
The formula to convert volume from barrel (petroleum) to cubic inch is:
Volume[(Cubic Inch)] = Volume[(Barrel (Petroleum))] × 9702
Substitute given weight Volume[(Barrel (Petroleum))] = 500 in the above formula.
Volume[(Cubic Inch)] = 500 × 9702
Volume[(Cubic Inch)] = 4851000
Final Answer:
Therefore, 500 bl is equal to 4851000 in^3.
The volume is 4851000 in^3, in cubic inch.
Barrel (Petroleum) to Cubic Inch Conversion Table
The following table gives some of the most used conversions from Barrel (Petroleum) to Cubic Inch.
Barrel (Petroleum) (bl) Cubic Inch (in^3)
0.01 bl 97.02 in^3
0.1 bl 970.2 in^3
1 bl 9702 in^3
2 bl 19404 in^3
3 bl 29106 in^3
4 bl 38808 in^3
5 bl 48510 in^3
6 bl 58212 in^3
7 bl 67914 in^3
8 bl 77616 in^3
9 bl 87318 in^3
10 bl 97020 in^3
20 bl 194040 in^3
50 bl 485100 in^3
100 bl 970200 in^3
1000 bl 9702000 in^3
Barrel (Petroleum)
The petroleum barrel is a standard unit of measurement for crude oil and other petroleum products. Originating in the early oil industry of the 19th century, it has become the globally accepted unit
for quantifying oil volumes. Historically, the use of the petroleum barrel facilitated trade and transport, allowing for standardized transactions and efficient handling. Today, it remains a
fundamental measure in the oil industry, used extensively in production, shipping, and trading.
Cubic Inch
The cubic inch is a unit of measurement used to quantify three-dimensional volumes, particularly in engineering, manufacturing, and real estate. It is defined as the volume of a cube with sides each
measuring one inch in length. Historically, the cubic inch has been used for precise measurements in industries such as automotive and aerospace, where detailed volume calculations are essential.
Today, it remains relevant in various fields, including product design, packaging, and spatial analysis, especially in contexts where detailed and small-scale volume measurements are required.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Barrel (Petroleum) to Cubic Inch in Volume?
The formula to convert Barrel (Petroleum) to Cubic Inch in Volume is:
Barrel (Petroleum) * 9702
2. Is this tool free or paid?
This Volume conversion tool, which converts Barrel (Petroleum) to Cubic Inch, is completely free to use.
3. How do I convert Volume from Barrel (Petroleum) to Cubic Inch?
To convert Volume from Barrel (Petroleum) to Cubic Inch, you can use the following formula:
Barrel (Petroleum) * 9702
For example, if you have a value in Barrel (Petroleum), you substitute that value in place of Barrel (Petroleum) in the above formula, and solve the mathematical expression to get the equivalent
value in Cubic Inch. | {"url":"https://convertonline.org/unit/?convert=barrel_petroleum-cubic_inch","timestamp":"2024-11-04T16:41:28Z","content_type":"text/html","content_length":"93767","record_id":"<urn:uuid:56eaea36-1312-4b9b-af9f-3022c72f631b>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00615.warc.gz"} |
Calcium - AlladiRamakrishnanHall
Alladi Ramakrishnan Hall
Duality made Simple: From Ising model to SU(N) lattice gauge theory
Manu Mathur
Retired Professor, SN Bose National Center for Basic Sciences
We use simple canonical transformations to obtain exact dualities in Ising Spin Model (Kramers-Wannier duality) as well as in U(1) & SU(N) Lattice Gauge Theories (LGT) in the Hamiltonian framework.
We show that the SU(N) LGT duality can be directly obtained
from the simple U(1) LGT duality through the well known minimal coupling procedure. Like in the Ising model, the dual construction naturally leads to the SU(N) LGT Disorder Operators which are
analogous to the Z_2 Kink operators in the Ising Model. These
SU(N) disorder operators create SU(N) magnetic vortices and are dual to the SU(N) Wilson loop operators. The SU(N) order-disorder algebra and the plausible role of the SU(N) magnetic vortices in
electric color confinement, topological quantum computing will be
briefly discussed. | {"url":"https://www.imsc.res.in/cgi-bin/CalciumShyam/Calcium40.pl?CalendarName=InstituteEvents&ID=4763&Date=2023%2F6%2F23&Source=AlladiRamakrishnanHall&Op=PopupWindow&DoneURL=Calcium40.pl%3FCalendarName%3DInstituteEvents%26Op%3DShowIt%26Amount%3DWeek%26NavType%3DNeither%26Type%3DBlock%26Date%3D2023%252F6%252F19&Amount=Week&NavType=Neither&Type=Block","timestamp":"2024-11-06T01:34:29Z","content_type":"application/xhtml+xml","content_length":"3781","record_id":"<urn:uuid:11915844-6951-41d3-a4ac-60617b52b2d0>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00422.warc.gz"} |
Home page of András Gilyén
András Gilyén
Alfréd Rényi Institute of Mathematics
Reáltanoda street 13-15, H-1053, Budapest, Hungary
Room: R.3. (on the second floor)
liam-e: uh[tod]iyner[ta]neylig
Links to my arXiv papers and my Google scholar page.
I am a Marie Curie fellow at the Rényi Institute. My main research topic is quantum algorithms and complexity, with a recent focus on stochastic quantum processes related to Glauber and Metropolis
dynamics, and more generally quantum walks and quantum linear algebra methods (quantum singular value transformation and the block-encoding framework) with application in optimization and related
fields. I received my PhD in 2019 from the University of Amsterdam, where I was supervised by Ronald de Wolf and co-supervised by Harry Buhrman at CWI/QuSoft. Between 2019 and 2021 I was an IQIM
postdoctoral fellow at Caltech, meanwhile I received the ERCIM Cor Baayen Young Researcher Award in 2019 and was a Google Research Fellow at the Simons Institute for the Theory of Computing in
Berkeley during the "The Quantum Wave in Computing" program in the spring of 2020.
Quantum Computer Science Seminar, Budapest
- I am organizing the Quantum Computer Science Seminar Series in Budapest. If you would like to join the mailing list please send me an e-mail.
Supervising BSc, MSc and PhD students (including voluntary research projects [TDK in Hungarian]): Please write me an e-mail if you are interested!
Introduction to Quantum Computing [in Hungarian] (2024 Fall @Eötvös Loránd University)
Theory of Computation Exercise Class [in Hungarian] (2024 Fall @Eötvös Loránd University)
- Autumn school (Arbeitsgemeinschaft) on
"Quantum Signal Processing and Nonlinear Fourier Analysis"
, Oberwolfach, Germany (6-11 October 2024)
Theory of Computation Exercise Class [in Hungarian] (2024 Spring @Eötvös Loránd University)
Quantum Computing (2023 Fall @Eötvös Loránd University)
IAS / PCMI 2023 Quantum Computing Graduate Summer School
, lecture series on
Quantum Fourier transform beyond Shor’s algorithm
-- Slides from day
and Exercise Sheet
Theory of Computation Exercise Class [in Hungarian] (2023 Spring @Eötvös Loránd University)
Quantum Computing (2022 Fall @Eötvös Loránd University)
Bad Honnef 2022 Quantum Computing Summer School
Slides for Quantum Machine Learning
Slides for Grand Unification of Quantum Algorithms
Summer School in Post-Quantum Cryptography
, Eötvös Loránd University, Budapest, Hungary (August 2022) --
Theory of Computation Exercise Class [in Hungarian] (2022 Spring @Eötvös Loránd University)
Quantum Computing (2021 Fall @Eötvös Loránd University)
- Quantum Computing TA (
Spring @University of Amsterdam)
- Quantum Computing (2014 Spring @Eötvös Loránd University) --
notes for the introductory lecture in Hungarian | {"url":"https://gilyen.hu/index.html","timestamp":"2024-11-11T07:29:58Z","content_type":"text/html","content_length":"9124","record_id":"<urn:uuid:411aae09-a32d-46d7-a883-6fe898cdf810>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00435.warc.gz"} |
45-45-90 Right Triangles (solutions, examples, videos)
Related Pages
30-60-90 Right Triangle
Other Special Right Triangles
More Geometry Lessons
Recognizing special right triangles in geometry can provide a shortcut when answering some questions. A special right triangle is a right triangle whose sides are in a particular ratio. You can also
use the Pythagorean theorem formula, but if you can see that it is a special triangle it can save you some calculations.
What is a 45-45-90 Triangle? {#45-45-90triangles}
A 45-45-90 triangle is a special right triangle whose angles are 45°, 45° and 90°. The lengths of the sides of a 45-45-90 triangle are in the ratio of 1:1:√2.
The following diagram shows a 45-45-90 triangle and the ratio of its sides. Scroll down the page for more examples and solutions using the 45-45-90 triangle.
Note that a 45-45-90 triangle is an isosceles right triangle. It is also sometimes called a 45-45 right triangle.
A right triangle with two sides of equal lengths is a 45-45-90 triangle.
You can also recognize a 45-45-90 triangle by the angles. As long as you know that one of the angles in the right-angle triangle is 45° then it must be a 45-45-90 special right triangle.
A right triangle with a 45° angle must be a 45-45-90 special right triangle.
How to solve problems with 45-45-90 triangles?
Example 1:
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches.
Step 1: This is a right triangle with two equal sides so it must be a 45-45-90 triangle.
Step 2: You are given that the both the sides are 3. If the first and second value of the ratio n:n:n√2 is 3 then the length of the third side is 3√2
Answer: The length of the hypotenuse is 3√2 inches.
Example 2:
Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 4√2 inches and one of the angles is 45°.
Step 1: This is a right triangle with a 45° so it must be a 45-45-90 triangle.
Step 2: You are given that the hypotenuse is 4√2. If the third value of the ratio n:n:n√2 is 4√2 then the lengths of the other two sides must 4.
Answer: The lengths of the two sides are both 4 inches.
The following videos show more examples of 45-45-90 triangles.
How to find the length of a leg or hypotenuse in a 45-45-90 triangle using the Pythagorean Theorem and then derive the ratio between the length of a leg and the hypotenuse?
This video gives an introduction to the 45-45-90 triangles and shows how to derive the ratio between the lengths of legs and the hypotenuse.
How to solve a 45-45-90 triangle given the length of one side by using the ratio?
Special Right Triangles in Geometry: 45-45-90 and 30-60-90
This video discusses two special right triangles, how to derive the formulas to find the lengths of the sides of the triangles by knowing the length of one side, and then does a few examples using
Example problems of finding the sides of a 45-45-90 triangle with answer in simplest radical form.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. | {"url":"https://www.onlinemathlearning.com/45-45-90-right-triangle.html","timestamp":"2024-11-08T05:15:26Z","content_type":"text/html","content_length":"41429","record_id":"<urn:uuid:bdadbb91-85ce-453c-9087-f452de2cc2b7>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00210.warc.gz"} |
In mathematical analysis, the maximum and minimum^[a] of a function are, respectively, the greatest and least value taken by the function. Known generically as extremum,^[b] they may be defined
either within a given range (the local or relative extrema) or on the entire domain (the global or absolute extrema) of a function.^[1]^[2]^[3] Pierre de Fermat was one of the first mathematicians to
propose a general technique, adequality, for finding the maxima and minima of functions.
Local and global maxima and minima for cos(3πx)/x, 0.1≤ x ≤1.1
As defined in set theory, the maximum and minimum of a set are the greatest and least elements in the set, respectively. Unbounded infinite sets, such as the set of real numbers, have no minimum or
In statistics, the corresponding concept is the sample maximum and minimum.
A real-valued function f defined on a domain X has a global (or absolute) maximum point at x^∗, if f(x^∗) ≥ f(x) for all x in X. Similarly, the function has a global (or absolute) minimum point at x^
∗, if f(x^∗) ≤ f(x) for all x in X. The value of the function at a maximum point is called the maximum value of the function, denoted ${\displaystyle \max(f(x))}$ , and the value of the function at a
minimum point is called the minimum value of the function, (denoted ${\displaystyle \min(f(x))}$ for clarity). Symbolically, this can be written as follows:
${\displaystyle x_{0}\in X}$ is a global maximum point of function ${\displaystyle f:X\to \mathbb {R} ,}$ if ${\displaystyle (\forall x\in X)\,f(x_{0})\geq f(x).}$
The definition of global minimum point also proceeds similarly.
If the domain X is a metric space, then f is said to have a local (or relative) maximum point at the point x^∗, if there exists some ε > 0 such that f(x^∗) ≥ f(x) for all x in X within distance ε of
x^∗. Similarly, the function has a local minimum point at x^∗, if f(x^∗) ≤ f(x) for all x in X within distance ε of x^∗. A similar definition can be used when X is a topological space, since the
definition just given can be rephrased in terms of neighbourhoods. Mathematically, the given definition is written as follows:
Let ${\displaystyle (X,d_{X})}$ be a metric space and function ${\displaystyle f:X\to \mathbb {R} }$ . Then ${\displaystyle x_{0}\in X}$ is a local maximum point of function ${\displaystyle f}$
if ${\displaystyle (\exists \varepsilon >0)}$ such that ${\displaystyle (\forall x\in X)\,d_{X}(x,x_{0})<\varepsilon \implies f(x_{0})\geq f(x).}$
The definition of local minimum point can also proceed similarly.
In both the global and local cases, the concept of a strict extremum can be defined. For example, x^∗ is a strict global maximum point if for all x in X with x ≠ x^∗, we have f(x^∗) > f(x), and x^∗
is a strict local maximum point if there exists some ε > 0 such that, for all x in X within distance ε of x^∗ with x ≠ x^∗, we have f(x^∗) > f(x). Note that a point is a strict global maximum point
if and only if it is the unique global maximum point, and similarly for minimum points.
A continuous real-valued function with a compact domain always has a maximum point and a minimum point. An important example is a function whose domain is a closed and bounded interval of real
numbers (see the graph above).
Finding global maxima and minima is the goal of mathematical optimization. If a function is continuous on a closed interval, then by the extreme value theorem, global maxima and minima exist.
Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the domain. So a method of finding a global
maximum (or minimum) is to look at all the local maxima (or minima) in the interior, and also look at the maxima (or minima) of the points on the boundary, and take the greatest (or least) one.Minima
For differentiable functions, Fermat's theorem states that local extrema in the interior of a domain must occur at critical points (or points where the derivative equals zero).^[4] However, not all
critical points are extrema. One can often distinguish whether a critical point is a local maximum, a local minimum, or neither by using the first derivative test, second derivative test, or
higher-order derivative test, given sufficient differentiability.^[5]
For any function that is defined piecewise, one finds a maximum (or minimum) by finding the maximum (or minimum) of each piece separately, and then seeing which one is greatest (or least).
The global maximum of ^x√x occurs at x = e.
Function Maxima and minima
x^2 Unique global minimum at x = 0.
x^3 No global minima or maxima. Although the first derivative (3x^2) is 0 at x = 0, this is an inflection point. (2nd derivative is 0 at that point.)
${\displaystyle {\sqrt[{x}]{x}}}$ Unique global maximum at x = e. (See figure at right)
x^−x Unique global maximum over the positive real numbers at x = 1/e.
x^3/3 − x First derivative x^2 − 1 and second derivative 2x. Setting the first derivative to 0 and solving for x gives stationary points at −1 and +1. From the sign of
the second derivative, we can see that −1 is a local maximum and +1 is a local minimum. This function has no global maximum or minimum.
|x| Global minimum at x = 0 that cannot be found by taking derivatives, because the derivative does not exist at x = 0.
cos(x) Infinitely many global maxima at 0, ±2π, ±4π, ..., and infinitely many global minima at ±π, ±3π, ±5π, ....
2 cos(x) − x Infinitely many local maxima and minima, but no global maximum or minimum.
cos(3πx)/x with 0.1 ≤ x ≤ 1.1 Global maximum at x = 0.1 (a boundary), a global minimum near x = 0.3, a local maximum near x = 0.6, and a local minimum near x = 1.0. (See figure at top of
x^3 + 3x^2 − 2x + 1 defined over the Local maximum at x = −1−√15/3, local minimum at x = −1+√15/3, global maximum at x = 2 and global minimum at x = −4.
closed interval (segment) [−4,2]
For a practical example,^[6] assume a situation where someone has ${\displaystyle 200}$ feet of fencing and is trying to maximize the square footage of a rectangular enclosure, where ${\displaystyle
x}$ is the length, ${\displaystyle y}$ is the width, and ${\displaystyle xy}$ is the area:
${\displaystyle 2x+2y=200}$
${\displaystyle 2y=200-2x}$
${\displaystyle {\frac {2y}{2}}={\frac {200-2x}{2}}}$
${\displaystyle y=100-x}$
${\displaystyle xy=x(100-x)}$
The derivative with respect to ${\displaystyle x}$ is:
{\displaystyle {\begin{aligned}{\frac {d}{dx}}xy&={\frac {d}{dx}}x(100-x)\\&={\frac {d}{dx}}\left(100x-x^{2}\right)\\&=100-2x\end{aligned}}}
Setting this equal to ${\displaystyle 0}$
${\displaystyle 0=100-2x}$
${\displaystyle 2x=100}$
${\displaystyle x=50}$
reveals that ${\displaystyle x=50}$ is our only critical point. Now retrieve the endpoints by determining the interval to which ${\displaystyle x}$ is restricted. Since width is positive, then ${\
displaystyle x>0}$ , and since ${\displaystyle x=100-y}$ , that implies that ${\displaystyle x<100}$ . Plug in critical point ${\displaystyle 50}$ , as well as endpoints ${\displaystyle 0}$ and ${\
displaystyle 100}$ , into ${\displaystyle xy=x(100-x)}$ , and the results are ${\displaystyle 2500,0,}$ and ${\displaystyle 0}$ respectively.
Therefore, the greatest area attainable with a rectangle of ${\displaystyle 200}$ feet of fencing is ${\displaystyle 50\times 50=2500}$ .^[6]
Functions of more than one variable
Peano surface, a counterexample to some criteria of local maxima of the 19th century
The global maximum is the point at the top
Counterexample: The red dot shows a local minimum that is not a global minimum
For functions of more than one variable, similar conditions apply. For example, in the (enlargeable) figure on the right, the necessary conditions for a local maximum are similar to those of a
function with only one variable. The first partial derivatives as to z (the variable to be maximized) are zero at the maximum (the glowing dot on top in the figure). The second partial derivatives
are negative. These are only necessary, not sufficient, conditions for a local maximum, because of the possibility of a saddle point. For use of these conditions to solve for a maximum, the function
z must also be differentiable throughout. The second partial derivative test can help classify the point as a relative maximum or relative minimum. In contrast, there are substantial differences
between functions of one variable and functions of more than one variable in the identification of global extrema. For example, if a bounded differentiable function f defined on a closed interval in
the real line has a single critical point, which is a local minimum, then it is also a global minimum (use the intermediate value theorem and Rolle's theorem to prove this by contradiction). In two
and more dimensions, this argument fails. This is illustrated by the function
${\displaystyle f(x,y)=x^{2}+y^{2}(1-x)^{3},\qquad x,y\in \mathbb {R} ,}$
whose only critical point is at (0,0), which is a local minimum with f(0,0) = 0. However, it cannot be a global one, because f(2,3) = −5.
Maxima or minima of a functional
If the domain of a function for which an extremum is to be found consists itself of functions (i.e. if an extremum is to be found of a functional), then the extremum is found using the calculus of
In relation to sets
Maxima and minima can also be defined for sets. In general, if an ordered set S has a greatest element m, then m is a maximal element of the set, also denoted as ${\displaystyle \max(S)}$ .
Furthermore, if S is a subset of an ordered set T and m is the greatest element of S with (respect to order induced by T), then m is a least upper bound of S in T. Similar results hold for least
element, minimal element and greatest lower bound. The maximum and minimum function for sets are used in databases, and can be computed rapidly, since the maximum (or minimum) of a set can be
computed from the maxima of a partition; formally, they are self-decomposable aggregation functions.
In the case of a general partial order, the least element (i.e., one that is less than all others) should not be confused with a minimal element (nothing is lesser). Likewise, a greatest element of a
partially ordered set (poset) is an upper bound of the set which is contained within the set, whereas a maximal element m of a poset A is an element of A such that if m ≤ b (for any b in A), then m =
b. Any least element or greatest element of a poset is unique, but a poset can have several minimal or maximal elements. If a poset has more than one maximal element, then these elements will not be
mutually comparable.
In a totally ordered set, or chain, all elements are mutually comparable, so such a set can have at most one minimal element and at most one maximal element. Then, due to mutual comparability, the
minimal element will also be the least element, and the maximal element will also be the greatest element. Thus in a totally ordered set, we can simply use the terms minimum and maximum.
If a chain is finite, then it will always have a maximum and a minimum. If a chain is infinite, then it need not have a maximum or a minimum. For example, the set of natural numbers has no maximum,
though it has a minimum. If an infinite chain S is bounded, then the closure Cl(S) of the set occasionally has a minimum and a maximum, in which case they are called the greatest lower bound and the
least upper bound of the set S, respectively.
Argument of the maximum
As an example, both unnormalised and normalised sinc functions above have ${\displaystyle \operatorname {argmax} }$ of {0} because both attain their global maximum value of 1 at x = 0.
The unnormalised sinc function (red) has arg min of {−4.49, 4.49}, approximately, because it has 2 global minimum values of approximately −0.217 at x = ±4.49. However, the normalised sinc function
(blue) has arg min of {−1.43, 1.43}, approximately, because their global minima occur at x = ±1.43, even though the minimum value is the same.^[7]
, the arguments of the maxima (abbreviated
arg max
or argmax) and arguments of the minima (abbreviated arg min or argmin) are the input points at which a
output value is
maximized and minimized
, respectively.
While the
are defined over the
domain of a function
, the output is part of its
See also
1. ^ PL: maxima and minima (or maximums and minimums).
2. ^ PL: extrema.
External links
Wikimedia Commons has media related to Extrema (calculus).
Look up maxima, minima, or extremum in Wiktionary, the free dictionary.
• Thomas Simpson's work on Maxima and Minima at Convergence
• Application of Maxima and Minima with sub pages of solved problems
• Jolliffe, Arthur Ernest (1911). . Encyclopædia Britannica. Vol. 17 (11th ed.). pp. 918–920. | {"url":"https://www.knowpia.com/knowpedia/Maximum_and_minimum","timestamp":"2024-11-09T03:42:12Z","content_type":"text/html","content_length":"171186","record_id":"<urn:uuid:960c7e10-459d-4006-8326-32bec18735d4>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00022.warc.gz"} |
#what is the name of each part of the plane formed by these two lines?
write the answer of each of the following questions.
#what is the name of each part of the plane formed by these two lines?
The name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively. The name of each part of the plane formed by these two
lines x-axis and the y-axis is quadrants. The point where these two lines intersect is called the origin.
Without knowing the equations of the two lines, it's difficult to give you the exact names of the different parts of the plane they form. However, in general, when two lines intersect in a plane,
they divide the plane into four regions or parts, which are commonly referred to as quadrants. These quadrants are typically labeled using Roman numerals I, II, III, and IV, as shown below:
II | I
III| IV
The point of intersection of the two lines is called the origin and is typically denoted by the coordinates (0, 0). The region above and to the right of the origin (quadrant I) is where both x and y
are positive. The region above and to the left of the origin (quadrant II) is where x is negative and y is positive. The region below and to the left of the origin (quadrant III) is where both x and
y are negative. Finally, the region below and to the right of the origin (quadrant IV) is where x is positive and y is negative.
Again, without the equations of the two lines, it's difficult to determine which quadrant(s) each line occupies or which region(s) of the plane are enclosed by the lines.
Step-by-step explanation: | {"url":"https://alumniagri.in/task/write-the-answer-of-each-of-the-following-questions-what-is-42346029","timestamp":"2024-11-06T09:16:04Z","content_type":"text/html","content_length":"26558","record_id":"<urn:uuid:ee9a9801-e8bf-4561-b6ba-249952b088ca>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00540.warc.gz"} |
2022 Audi Q3 S line quattro MPG and Fuel Economy
On this page, you will find a complete guide to the miles per gallon and fuel efficiency data for the 2022 Audi Q3 S line quattro.
The 2022 Audi Q3 S line quattro runs on regular gasoline and is in the Small Sport Utility Vehicle 4WD car class.
Whether you are considering a purchase of the car, or just wanting to find out how economical and environmentally friendly (or un-friendly) your 2022 Audi Q3 S line quattro is, we have the
information you need.
🛣 How Many Miles per Gallon (MPG) Does a 2022 Audi Q3 S line quattro Get?
First off, the most commonly asked question. A 2022 Audi Q3 S line quattro gets up to 21 miles per gallon in the city, and 28 miles per gallon on the highway.
The combined average MPG for the 2022 Audi Q3 S line quattro is 24 miles per gallon.
💵 What is the Average Yearly Fuel Cost for a 2022 Audi Q3 S line quattro?
The estimated fuel costs for the 2022 Audi Q3 S line quattro is $2,600 per year.
That value has been estimated by government regulators based on 15,000 miles driven per year, using regular gasoline, and a split of 55% city driving and 45% highway driving.
If you were to compare a 2022 Audi Q3 S line quattro to an average vehicle over 5 years, you will spend $1,500 more on fuel.
🛢 How Many Barrels of Petroleum Does a 2022 Audi Q3 S line quattro Consume?
The 2022 Audi Q3 S line quattro will consume roughly 12 barrels of petroleum per year, using the standard estimate of 15,000 miles per year.
The majority of the world's petroleum is sourced from countries like Saudi Arabia, Russia, Iraq, and the United States.
💨 How Many Grams of CO[2] Does a 2022 Audi Q3 S line quattro Emit?
For every mile driven, the 2022 Audi Q3 S line quattro will emit 375 grams of CO[2], which is about 5,625,000 grams of CO[2] per year.
🌳 To help you understand the impact of this, a normal tree will absorb around 21,000 grams of CO[2] per year. This means that roughly 192 trees would be needed to offset emissions from the 2022
Audi Q3 S line quattro.
The table below covers all of the miles per gallon, engine specifications, and emission details we have for the 2022 Audi Q3 S line quattro.
City MPG 21
Highway MPG 28
Combined MPG 24
Save/Spend vs Average Car Spend $1,500
Fuel Cost $2,600
Fuel Type 1 Regular
Barrels of Petroleum 12
Grams of CO[2] 5,625,000
Trees Needed to Offset 268
Drive All-Wheel Drive
Cylinders 4
Transmission Automatic (S8)
Vehicle Class Small Sport Utility Vehicle 4WD
Start/Stop? Yes
Supercharged? No
Turbocharged? Yes
Related Audi Models from 2022
If you're interested in other Audi models from 2022, you can find them in the table below. You can click the model link to find miles per gallon and emission information for that model.
Ratings were provided by the manufacturer to the U.S. Department of Energy, which is where we sourced the data. Be advised that manufacturers may have upgraded, downgraded, or changed these ratings
following the compilation of this data.
MPG Buddy does not guarantee the accuracy of this data, nor are we liable for any decisions made by referencing this data. Make sure to contact the specific car manufacturer to confirm accuracy.
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More Student-Created Geogebras - and some pushback
In this post:
• The ggb assignments so far
• The benefits of student-created geogebras (and the evidence)
• Student feedback
The assignments so far:
I've spent so much time trying to write this post without making it ridiculously long, so I gave up and did part of it in a video! This shows you two things at once - what they had to do, and what
they actually did:
The benefits, and the evidence:
First the benefits associated with the specific tasks, in other words, the benefits I foresaw. I've also included the kids' actual word-for-word reflections (in colour), which I think provide
evidence of great learning:
• Formulas:
□ Finding them: Benefit: To find those formulas, they had to move up a step on the ladder of abstraction. They were manipulating equations with no actual numbers in them. Up until now, someone
else has done this for them.
☆ Today I figured out the coordinates for my zero. I also edited my y-int and zeros conditions.
☆ I feel so proud of myself for figuring out the rule of the zero on my own. I dont know why but that was something i was having a hard time with and i got it!
□ Entering them properly: Benefit: To get them to work, they had to be very careful about where to put brackets, about using only variables that were already defined, and of course to not make
any typos.
☆ I had trouble with the y-intercept and the zero. Turns out, in both cases, I wasn't putting the brackets at the good place!
☆ OH MY GOSH!!!! It works!!! FINALLY!! Okay, so my mistake was sillly, I had written my P like this: P = (t, a sqrt(b(x-h)))+k I had put my last bracket in the wrong place. It is now: P =
(t, asqrt(b(x-h))+k).
☆ With today's class though, I was able to know what to do and come with this product! What bothers me with this one though, is that the y-intercept doesn't seem right. The rule is what we
usually use I guess, but the number geogebra gives me in the text doesn't seem correct! I'll try to find what the simplified rule is and write that instead of the big thing ( y= a*sqrt(b
(x-h))+k )
• Conditions:
□ Finding them: Benefit: This was very challenging for everyone. They've never had to do this before - systematically list all the possibilities for a certain math situation. I used to give out
all this information for free, but no more. This stretched their algebra minds, no doubt.
☆ .I have to say one of the toughest part was figuring out when my text box should appear or not appear. It can get very complicated. I noticed though that when the theres a y intercept
text box need's to show up it's the same as when a theres a zero textbox; the only difference being the parameters used.
☆ I made domain, range, function is increasing/decreasing, y int, and now I am working on the zeros but I am having trouble figuring out what to write as conditions for zeros to appear.
☆ The longest part about the texts was actually finding WHAT the conditions were. This is why making students use geogebra helps them understand how the function actually works and it's
like having animated notes you can use to study with.
☆ I am SO SO SO SO exited with this last version:) I have my increasing and decreasing in however i also made the function change colour according to whether or not its decreasing! i did
look at the vt for help but i must say that i did learn a few things on my own! like at first i accidentally made the line only appear when it was decreasing so i needed to play around
with it until i was able to make it change for both:) so exited!
□ Entering them properly: Benefit: Logic! Boolean operators! Truth! I never even got to teach this before!
☆ I was really happy to have figured out how to get my conditions right for my text that I did a happy dance in my head. So I figured that a>0, b>0 and a<0,b<0 means that it will increase
and if a<0,b>0 and a>0, b<0 it will decrease. I was really happy.
☆ So for the y textbox in this example it was (h≥0&&b≤0)||(h≤0&&b≥0). The zero text box was the same thing except with k and a.
• Expression for the moving point P
□ Finding it: Another step up the ladder of abstraction. That t-slider may well be their first opportunity to see a letter as a number whose value is varying, because that's literally what's
happening as they move that beautiful little dot along. And that's not the same thing as a variable. Some letters are more variable than others. Not to mention that that moving point P was
really a preparation for the next big project for physics about projectiles.
☆ Because of our last ggb assig nment i was able to figure out that point P was x as t and then the entire rule represented y (t as x).
☆ i had no problem with entering the rule and the 4 parameters but when i entered point P with t-slider the point P does not always stay on that functions line segment, im not sure what
could be the problem. So far everything seems to be simple since we have done similar to this in the linear function except slider-t.
□ Entering it properly: Benefit: Again, being careful with the brackets, and also getting the big picture after 3 or 4 functions.
☆ like i had said i have no problems with the parameters but i did have a problem with correctly adding point P with slider t and making it follow the line but with help from you we have
noticed that it was tiny mistake in the way i had written my rule when i was entering the coordinates of the point P
Other benefits that I honestly wasn't even expecting:
• Doing one function would have been good, but doing several has allowed them to get a bigger picture. Some used their older assignments to figure out what to do in the new one, and saw patterns
• Engagement: There are some students who are definitely more engaged doing this than anything else. Some students had 10 or 11 versions before they finished. It's addictive. Even when it's not
working. Especially when it's not working.
• Each time they do something they can check it right away. And that involves action, ie moving a slider, which then causes another action ie a colour change - it's like watching a movie!
• Opportunities now to talk about why different formulas work ie |t - h| and |h - t| give same result
• Opportunities for eloquence - eg they all input this for their formula for the x intercept of the square root function:$\left&space;(&space;\left&space;(&space;\frac{-k}{a}&space;\right&space;)^
{2}\div&space;b+h,0&space;\right&space;)$, which is completely correct, but not very pretty. I get to ask them which formula they'd rather type in, the one above, or this one:$\left&space;(&
space;\frac{k^{2}}{a^{2}b}+h,0&space;\right&space;)$, and by the way this is why we simplify algebraic expressions.
• Opportunities to talk about presentation - lining up your sliders so they're nice and neat, colour coding, using checkboxes to not crowd the screen
And here's more general feedback:
This was a good practice to learn about square root functions. When using geogebra, it gives a more indepth explination of how the function works by letting me explore its movement.
This was a great way to learn. I was able to see how the parameters affect each other and how it affects the function.
I feel that these assignments are really helpful for making sure we understand the concept. It allows us to put what we learned in class and in voice threads to use in a very creative way. I can't
wait to see what else we will be doing with geogebra.
I am happy to be done and i have realised that this geogebra was similar to the other one except that there were more conditions to show objects and more rules about certain intercepts or zeros that
we needed to enter. But by doing all these things i have learned more about the square root function and my understanding has been increased about what happens to the function when certain parameters
are either negative or positive and so on. i'm really happy i finished it on time ( i have an english response that has been due for a week, so you should be quite happy, i'm not very good with
Finally: I had a bit of pushback. A group of students asked to meet with me to talk about all the geogebra they've been doing. Their concerns were that they were getting too dependent on the software
and were not developing their algebra skills. They pointed out that they can't use it on their tests, so they feel unprepared for them. Really important input! Then another student approached me and
mentioned that they would rather have more notes and less geogebra. Part of me says "Listen to your students, they're your eyes on the ground" and the other part says "They're not used to working
this hard or doing this much independent thinking, let them get used to it."
Well, for the time being, they've got one more geogebra project to do, and that's the big one, the one I've been thinking about since last spring - the physics/math virtual manipulative project.
After that, I'll give them a geoge-break. But till then, I'm full steam ahead, because there's just too much great stuff happening that's telling me this is all worth it.
5 comments:
1. Audrey
Thanks for an inspiring post! I've been in love with GeoGebra as a demonstration tool for a few years now. I have not made the commitment that you have to using it as an exploration tool for my
kids. I now feel an extra charge to do so. I shared this with a colleague who is interested in taking the GeoGebra plunge and I hope that this gives her the energy to do so.
1. Thanks so much Mr. Dardy! You have no idea how much it means to me that you are inspired by this. One reason it took me so long to write is that I wanted to inspire and not overwhelm any
teacher who reads it. I really feel like this is something big, that it's getting closer to the R at the end of the SAMR model.
My next post will be about their projectile projects. They're due Monday, and I've already seen some of them. They. Are. Spectacular. Stay tuned!
2. "Then another student approached me and mentioned that they would rather have more notes and less geogebra."
I got that comment a lot during my last years of teaching, when I really placed a lot of responsibility on students for learning (like you're doing!). I think you're right in thinking that
they're not used to working that hard. My response was to teach them how to take their own notes from the research they did or the videos I published. While that didn't stop them from from
wanting me to give the notes, the feedback I got was always that they were better prepared when doing work on their own. Just my two cents....
1. Terie, I remember reading about the same kind of push back on your blog. I so value your advice on this, and I will definitely be trying that out. I've been scaffolding their notes for them,
and it's time I let that go.
The funny thing is that I am still providing them with notes in which all they have to do is fill in blanks etc, so I was really surprised to hear there weren't enough notes! I felt like
saying - geez kid, if you think there aren't enough of them now....just wait, less is on the way!
I also think that it's probably not accurate to say that "they're not used to working this hard", because some of them do work hard, but on the traditional kind of stuff. Working hard on this
kind of assignment probably uses the more creative side of the brain, and that's what they're not used to - at least in math class.
Thanks for your comment and great thoughts!
3. I'm totally frustrated w/ figuring out things on Geogebra. SO many important things (like construction protocol) that take too long to trip over. I'm typing this comment to avoid going back to
beating my head against its walls... | {"url":"http://audrey-mcsquared.blogspot.com/2013/11/more-student-created-geogebras-and-some.html","timestamp":"2024-11-04T15:17:55Z","content_type":"text/html","content_length":"126180","record_id":"<urn:uuid:99beeb1f-15e8-4938-bd7c-ab35ffd42cee>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00581.warc.gz"} |
Help with alias analysis framework
berpast@hotmail.com (BerPast)
13 Aug 2004 17:25:16 -0400
From comp.compilers
| List of all articles for this month |
From: berpast@hotmail.com (BerPast)
Newsgroups: comp.compilers
Date: 13 Aug 2004 17:25:16 -0400
Organization: http://groups.google.com
Keywords: analysis, question
Posted-Date: 13 Aug 2004 17:25:16 EDT
In the paper "Interprocedural may-alias analysis for pointers: beyond
k-limiting", the author presents a parametric framework for the
analysis of pointer aliases.
The framework is parametrised by a numeric lattice V#.
The lattice has to have some abstract operators; the 4th are 5th are
for me the most complex:
4) resolution of a linear system: given a system S, I need to compute
a member of the lattice which is an upper approximation of the integer
solutions to S;
5) intersection with a linear system: if S is a system of linear
equations and K is a member of the lattice, I need to compute an upper
approximation of the solutions to S that are also in K (no empty
intersection with K I guess).
I need to implement the lattice operators, but I have no idea of how
to do this. Probably using Fourier-Motzkin projection I can solve the
problem for the lattice obtained by the smash product of the interval
lattice (any other, more efficient method?), but how can this problem
be solved for a generic lattice?
Can someone point out to me some documentation and (if available code
implemenation) explaining how to solve this problem for a generic
Thanks in advance for your help.
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How Do You Check Continuity Of A Function?
Continuity of a function can be checked by examining the limits of the function as x approaches a certain value and ensuring the function's output is equal to the value of the function at that point.
Continuity of a function is an important concept in calculus and other branches of mathematics. In order to check the continuity of a function, the limits of the function must be examined as x
approaches a certain value. This means that the left-hand limit and the right-hand limit must be equal. If the limits are equal, then the function is continuous at that point. If the limits are not
equal, then the function is not continuous at that point. Additionally, the value of the function at that point must also be equal to the limits in order for the function to be continuous. If the
function is discontinuous at any point, then it is not considered continuous. Continuity is necessary for certain properties of functions, such as derivatives and integrals, to be defined.
Learn more about function here
Number of pizza deliver on Saturday = 14 pizza
Step-by-step explanation:
Number of delivery day (Friday, Saturday) = 2
Number of pizza deliver on Friday = 8
Average of delivery (Mean) = 11 pizza
Number of pizza deliver on Saturday
Mean = Sum of pizza / Number of delivery day
11 = [8 + Number of pizza deliver on Saturday] / 2
22 = 8 + Number of pizza deliver on Saturday
Number of pizza deliver on Saturday = 22 - 8
Number of pizza deliver on Saturday = 14 pizza | {"url":"https://www.cairokee.com/homework-solutions/how-do-you-check-continuity-of-a-function-vjy0","timestamp":"2024-11-07T09:16:37Z","content_type":"text/html","content_length":"67507","record_id":"<urn:uuid:efbe17d9-4bf9-44a2-b9cc-228b1eab0660>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00583.warc.gz"} |
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The break-even point 2 is the level of sales at which a company's total revenues equal its total costs, resulting in neither profit nor loss. It is a key financial metric used to assess the
viability of a business and its ability to cover its fixed and variable costs. By reaching the break-even point 2, a company can start generating profits beyond that level of sales. It is an
important milestone for businesses to achieve in order to ensure long-term sustainability and growth.
• Where is the break-even point located?
The break-even point is located at the intersection of the total revenue and total cost curves on a graph. It represents the level of output or sales at which a company's total revenues equal its
total costs, resulting in neither profit nor loss. At this point, the company has covered all its expenses and has reached a point of financial equilibrium. Beyond the break-even point, the
company starts to generate profit, while below the break-even point, it incurs losses.
• What is the break-even point at 5?
The break-even point is the level of sales at which total revenue equals total costs, resulting in neither profit nor loss. At a sales level of 5, the break-even point can be calculated by
determining the total costs and total revenue at that level of sales. If the total revenue equals the total costs at a sales level of 5, then that would be the break-even point. It is important
to consider both fixed and variable costs when calculating the break-even point.
• How do you determine the break-even point?
The break-even point is determined by finding the level of sales at which total revenue equals total costs, resulting in zero profit or loss. To calculate the break-even point, you can use the
formula: Break-even point (in units) = Fixed costs / (Selling price per unit - Variable cost per unit). This formula helps you determine the number of units you need to sell in order to cover all
your fixed and variable costs. By knowing the break-even point, you can make informed decisions about pricing, production levels, and overall business strategy.
• How do you calculate the break-even point?
To calculate the break-even point, you need to determine the fixed costs and the contribution margin per unit. The break-even point is reached when total revenue equals total costs, which can be
expressed as: Break-even point (in units) = Fixed costs / Contribution margin per unit. This calculation helps businesses understand the level of sales needed to cover all costs and start making
a profit.
• What is the break-even point at 6?
The break-even point is the level of sales at which total revenue equals total costs, resulting in neither profit nor loss. At a sales level of 6, the break-even point can be calculated by
determining the total costs and total revenue at that level of sales. If the total costs at a sales level of 6 are $600 and the total revenue is also $600, then the break-even point is 6. This
means that at a sales level of 6, the company is neither making a profit nor incurring a loss.
• What is the equation for the break-even point?
The equation for the break-even point is: Break-even point = Fixed costs / (Selling price per unit - Variable cost per unit) This equation calculates the number of units that need to be sold in
order to cover all fixed and variable costs, resulting in a net profit of zero. The fixed costs are the expenses that do not change regardless of the level of production, while the variable costs
are the expenses that vary with the level of production. The selling price per unit represents the revenue generated from each unit sold.
• What is the break-even point in business administration?
The break-even point in business administration is the level of sales at which a company's total revenues equal its total costs, resulting in neither profit nor loss. It is the point at which a
company covers all its expenses and begins to make a profit. By calculating the break-even point, businesses can determine the minimum level of sales needed to cover costs and make informed
decisions about pricing, production, and overall business strategy.
• How do I calculate the break-even point P?
To calculate the break-even point P, you need to know the fixed costs (F), the variable cost per unit (V), and the selling price per unit (S). The break-even point P can be calculated using the
formula P = F / (S - V), where F is the fixed costs, S is the selling price per unit, and V is the variable cost per unit. This formula helps you determine the level of sales needed to cover all
fixed and variable costs, resulting in zero profit or loss. By calculating the break-even point, you can understand the minimum level of sales required to cover all costs and start making a
• What is an ROI and what is a break-even?
ROI stands for Return on Investment, which is a measure used to evaluate the efficiency or profitability of an investment. It is calculated by dividing the net profit of the investment by the
initial cost of the investment, and is usually expressed as a percentage. On the other hand, break-even refers to the point at which total revenues equal total costs, resulting in neither profit
nor loss. It is a crucial milestone for businesses as it marks the point where they start making a profit from their operations.
• How can one have an even more intense voice break?
To have an even more intense voice break, one can practice vocal exercises that focus on expanding vocal range and control. Working with a vocal coach can also help in developing techniques to
strengthen the voice and create a more powerful break. Additionally, incorporating proper breathing techniques and warm-up exercises before singing can help in achieving a more intense voice
break. Experimenting with different styles of music and emoting through the lyrics can also add depth and intensity to the voice break.
• Can you break your ribs or even die from sneezing?
It is extremely rare to break a rib or die from sneezing. While it is possible for a forceful sneeze to cause a rib fracture, it is uncommon and usually only occurs in individuals with weakened
or fragile bones. As for the possibility of dying from sneezing, it is highly unlikely. Sneezing is a reflexive response to irritants in the nasal passages and is a normal bodily function that
does not typically pose a serious risk to one's health.
* All prices are inclusive of VAT and, if applicable, plus shipping costs. The offer information is based on the details provided by the respective shop and is updated through automated processes.
Real-time updates do not occur, so deviations can occur in individual cases. | {"url":"https://www.vsayt.com/%20Break-even","timestamp":"2024-11-05T06:46:41Z","content_type":"text/html","content_length":"56158","record_id":"<urn:uuid:f4aa2a88-319f-4647-8c4a-e8c539a43f54>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00523.warc.gz"} |
MAX + COLLECT Formula Help
I am trying to create a formula that will find the largest % from the % cells in each row and then label it correctly. For example, if the largest % is Novice in the first row, I want the blank box
column to note that the Overall Expertise Level is Novice. Please advise how to set up this formula. I think it is a combination of max and collect, but I am not sure how to format it.
Thank you.
• I actually think this is a MATCH/MAX formula
• What if there is more than one that have the same percentage?
• Paul, I am not sure. That's a great question.
• Either way, you are going to need a series of IF statements. The order and exact syntax depends on how to handle duplicates though.
If there will never be a duplicate in the same row, then it is a basic nested IF in any order.
If there are duplicates:
Displaying all would be "adding" IF statements together.
Displaying the lowest (novice) would be a nested IF moving from left to right.
Displaying the highest (expert) would be a nested IF moving from right to left.
• Could you help me with a formula? I have this formula already to find the max number, but now I need it to tell me where that number came from. Will I need a helper row since it won't read the
column header?
That formula reads the 4 cells on that row (Novice, Intermediate, Skilled, Expert) and tells me the highest number. But now, I want it to tell me which column that came from and give me the name
instead of the number itself.
Thank you!
• You would need either a helper row or a series of IF statements. Whether you use a helper row or not, we would still need to know how to deal with the possibility of two of them being the same on
a single row (if that is even a possibility).
I would be happy to help you with a formula, but we are looking at a possibility of at least 8 different formulas depending on helper row or not and duplicates or not. I just need to know which
one to help with.
• Let's go with there will be no duplicates.
I can create a helper row if that will be more helpful, but currently it is not set up that way.
• IN that case you would use a series of nested IFs to compare each cell in turn to the MAX.
=IF(Novice@row = MAX(Novice@row:Expert@row), "Novice", IF(Intermediate@row = MAX(Novice@row:Expert@row), "Intermediate", IF(Skilled@row = MAX(Novice@row:Expert@row), "Skilled", "Expert")))
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/104218/max-collect-formula-help","timestamp":"2024-11-12T17:25:23Z","content_type":"text/html","content_length":"424548","record_id":"<urn:uuid:a949f850-f8c1-45ea-a181-166ca1321a37>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00797.warc.gz"} |
Notion VIP: Convert Property Values for Formulas
This is a free version of a lesson from Notion A-to-Z. For all lessons in an intuitive sequence, along with functional demos, practical exercises and certification prep, join this unrivaled learning
Before converting property values, you'll want to be familiar with the fundamentals of formulas.
The operations your perform with formulas require particular data types. To accommodate these requirements, you'll often need to convert properties from one data type to another. For example, you're
unable to merge a date with a text string to form a phrase like "Birthday: January 14, 1987". The date must first become a string.
Rollups can be particularly confusing because their data types can differ from the properties they retrieve.
In this lesson, we'll explore how to convert data types and configure Rollups for use in formulas.
Data Type Refresher
As I cover in Formula Fundamentals, every property value in a Notion database is one of four data types:
• Number
• String (Text)
• Bolean (true or false)
• Date
Formula operations require particular data types for their inputs:
• You can only calculate the sum of numbers.
• You concatenate, or merge, one or more text strings.
• You find the time between dates.
• You compare a boolean (true or false) with another boolean.
Therefore, when you reference other properties as inputs for your formula, you need to remain mindful of their data types.
If you attempt an operation on incompatible data types, Notion will will throw a type mismatch error, which typically means you need to convert the data type of one or more of the input values.
Conversion Functions
Convert values to strings with format().
The format() function accepts as its argument a number, date or boolean, which it returns as a string. The converted value can then be concatenated, or merged, with other strings.
Demo: Contextualize Gallery Properties
I often use format() when adding context to properties in the Gallery format. In the example below, each card includes the term "Age: " before the person's age, which would otherwise be a standalone
number out of context.
To achieve this, we create a new Formula property called "Age: Contextualized." In the formula, we reference the Age property within the format() function, and prepend that with the string "Age: ":
"Age: " + format(prop("Age"))
Convert values to numbers with toNumber().
Just as format() converts a value to a string, the toNumber() function converts its sole argument to a number, which can be used for mathematical calculations.
• Strings like "2" to 2.
• For booleans, true and false convert to 1 and 0, respectively.
• Dates convert to their Unix timestamp, or the number of milliseconds since January 1, 1970 12:00 AM (GMT) (Unix epoch).
I use toNumber() most often after extracting a number from a string with replaceAll(), which you'll learn in other lessons.
Demo: Total Checkboxes
Another useful example is calculating progress from checked Checkbox properties. The example below imagines a set of requirements, where Progress calculates the percent checked.
The keys to the formula are:
• converting each Checkbox to a number;
• adding those numbers; then
• dividing that sum by 3 (the number of checkboxes).
toNumber(prop("Req. 1"))
+ toNumber(prop("Req. 2"))
+ toNumber(prop("Req. 3")),
However, that returns an egregious decimal. Thus, we need to:
• multiply by 100;
• round that product to an integer using the round() function; then
• divide by 100.
toNumber(prop("Req. 1"))
+ toNumber(prop("Req. 2"))
+ toNumber(prop("Req. 3")),
Of course, this can be accomplished more simply by using arithmetic operators in place of functions:
round(((toNumber(prop("Req. 1")) + toNumber(prop("Req. 2")) + toNumber(prop("Req. 3"))) / 3) * 100) / 100
Rollup Configuration
A Rollup property retrieves a specified property from related items. Typically, those items are in another database.
Consider Transactions and Invoices databases, for example, where each invoice relates to its payments. Rollup properties in the Invoices database can retrieve the Date and Money In values from the
corresponding transactions to populate Payment Date and Total Paid:
Formula properties can then reference Payment Date and Total Paid to calculate Days Late and Balance. When we compose these formulas, however, we Notion throws a type mismatch, reporting that Payment
Date is not a date and Total Paid is not a number:
That's because Rollups, by default, are strings, regardless of the property type they retrieve.
To convert the value to the original data type, Notion requires you to choose a Calculation other than Show original when configuring your Rollup.
In the case of our Payment Date, we can choose Latest date. Upon doing so, the value aligns to the right, as dates do.
It also assumes "relative" formatting, which I find far less useful than a traditional variation of MM/DD/YYYY)
We can then add our Days Late formula, which utilizes dateBetween() from Essential Date Functions:
dateBetween(prop("Payment Date"), prop("Due Date"), "days")
For Total Paid, we can change the Calculation to Sum, which right-aligns the values, thus indicating numbers. That allows us to subtract Total Paid from Amount Due to calculate Balance:
prop("Amount Due") - prop("Total Paid")
The "Unique Values" Caveat
As of this writing, an unintended behavior persists in Rollups: If you choose Show unique values as your calculation, then reference the Rollup in a formula, the input value will be the count of
unique values, not the values themselves.
If you hit any snags as you convert property values, feel free to tweet @WilliamNutt. | {"url":"https://www.notion.vip/insights/convert-property-values-for-formulas","timestamp":"2024-11-14T20:31:05Z","content_type":"text/html","content_length":"41872","record_id":"<urn:uuid:d1eb1aa0-7c75-4dc0-9581-b598c4112cce>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00738.warc.gz"} |
etworks and
Neural Networks and Deep Learning is a free online book. The book will teach you about:
• Neural networks, a beautiful biologically-inspired programming paradigm which enables a computer to learn from observational data
• Deep learning, a powerful set of techniques for learning in neural networks
Neural networks and deep learning currently provide the best solutions to many problems in image recognition, speech recognition, and natural language processing. This book will teach you many of the
core concepts behind neural networks and deep learning.
For more details about the approach taken in the book, see here. Or you can jump directly to Chapter 1 and get started. | {"url":"http://neuralnetworksanddeeplearning.com/?ref=producthunt","timestamp":"2024-11-05T18:20:13Z","content_type":"text/html","content_length":"22738","record_id":"<urn:uuid:6516fa60-cfe4-4d20-8565-371f9f0b0e24>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00005.warc.gz"} |
Solving for an Unknown Involving Triangles - Workshop49Solving for an Unknown Involving Triangles, Workshop49
Solving for an Unknown Involving Triangles
In triangles the total sum of the degrees is 180. However, it is also true that any exterior angle of a triangle, will equal the sum of the two interior angles to which it isn’t adjacent to. We can
use this rule to help solve adjacent angles much quicker than solving the third angle of a triangle and then using a straight angle rule.
First try to solve for an unknown, x, where you are only working with solving a single interior angle.
Now we can solve for an unknown, x, where you are only working with interior angles. Remember that even if multiple angles are variables that we can still set the sum of all of the angles to 180 and
then solve the equation for x.
Next try to solve for an unknown, x, where we have introduced an exterior angle.
This style of problem is still solving for an unknown, x, where we have introduced an exterior angle. However, in this situation you will have to do a little more algebra in order to get your final
If you are looking for additional practice beyond this, there is a printable worksheet. | {"url":"https://workshop49.com/solving-for-an-unknown-inside-triangles/","timestamp":"2024-11-03T02:56:05Z","content_type":"text/html","content_length":"49590","record_id":"<urn:uuid:dd9e4cb4-b8b3-4e42-ae58-e0f8a3e0bbcb>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00531.warc.gz"} |
EXCEL FUNCTION-FIND - GyanKosh | Learning Made Easy
FIND FUNCTION is one of the very important and useful functions in Excel.
As FIND FUNCTION is concerned with the text, it is present under the TEXT FUNCTION CATEGORY of the functions.
FIND FUNCTION counts the characters whether it is single byte or double byte. [Byte is the size of the characters].
It finds out the exact same copy of the characters or text to be searched. It compares content as well as case of the text.
FIND FUNCTION is quite useful in manipulating text to get the desired results easily.
In this article we would learn about the purpose, syntax, formula of the FIND FUNCTION and get a better understanding with the help of the examples.
FIND FUNCTION is used to find any character or text fragment within a longer text and return its starting position as the character number.
For examples,
Suppose we need to find the location of any character or text pattern in the given text.
Let us take a text snippet ” What a beautiful day!“
Now if we want to find the location of the first ‘a’ or the word ‘beautiful’, we can do so with the use of the FIND FUNCTION.
• Basic understanding of how to use a formula or function.
• Basic understanding of rows and columns in Excel.
• Some information about the TEXT HANDLING IN EXCEL is an advantage for the use of such formulas.
• Of course, Excel software.
Helpful links for the prerequisites mentioned above What Excel does? How to use formula in Excel?
The Syntax for the FIND function is
=FIND ( TEXT TO BE FIND, TEXT IN WHICH THE TEXT IS TO BE FOUND, STARTING CHARACTER POSITION FOR THE SEARCH)
TEXT TO BE FIND is the character, text, word or pattern which we want to find
TEXT IN WHICH THE TEXT IS TO BE FOUND is the longer text in which we want the first argument to be found.
STARTING CHARACTER POSITION FOR THE SEARCH is the starting position from where we’d start our search.
All characters before this position will be ignored.
FIND FUNCTION IS CASE SENSITIVE. IT’LL FIND EXACTLY WHAT IS IS ASKED FOR INCLUDING THE CONTENT AND THE CASE OF THE TEXT.
Suppose we want to find ‘ap’ in the word ‘Happy’.
TEXT TO BE FIND is “ap”
TEXT IN WHICH THE TEXT IS TO BE FOUND is “Happy”
STARTING CHARACTER POSITION FOR THE SEARCH is 1
as we want to start the search from the first letter.
EXAMPLE TYPE 1: FINDING A LETTER [FIRST OCCURRENCE]
Let us take a word say “GOOD MORNING”.
Let us try to find out the position of ‘G’, space, ‘I’.
• Select the cell where we want the result.
• Enter the following formulas for G , space and I in the given word.
• For finding the G, put the formula as =FIND(“G”,E6,1)
• For finding the ” ” SPACE, put the formula as =FIND(” “,E7,1)
• For finding the I, put the formula as =FIND(“I”,E8,1)
EXAMPLE TYPE 2: FINDING A TEXT PATTERN IN THE GIVEN TEXT
Let us take a sentence ” Humpty Dumpty sat on a wall”
Let us try to find out the pattern ‘pty’ in the given sentence.
• Select the cell where we want the result.
• Enter the following formulas for finding “pty”.
• For finding the G, put the formula as =FIND(“pty”,E20,1)
• Click Enter.
• The result would appear.
• The picture below shows the result and other details.
Let us understand the working of the formula.
We used the formula =FIND(“pty”,E20,1)
Where the first argument “pty” is the one which we want to find.
E20 contains the complete sentence in which we want to find the occurrence position of this pattern
Third argument is 1 which tells us that we want to start the search from the position 1.
Now, we might have a thinking that there are two occurrences of “pty” in the sentence. What if we need the second occurrence.
Well, there is no direct function for finding out the second occurrence but we can use a trick to find one.
The next example is the extension to this one and finds out the second occurrence of “pty”
Let us take a sentence ” Humpty Dumpty sat on a wall”
Let us try to find out the second occurrence of the pattern ‘pty’ in the given sentence.
• Select the cell where we want the result.
• Enter the following formulas for finding the second occurrence of the pattern “pty”.
• For finding the G, put the formula as =FIND(“pty”,E29,FIND(“pty”,E29,1)+1)
• Click Enter.
• The result would appear.
• The picture below shows the result and other details.
Let us understand the working of the formula.
We used the formula =FIND(“pty”,E29,FIND(“pty”,E29,1)+1)In the previous example, we found the “pty” text pattern simply and started the search from the letter 1, which returned us the starting
position of first occurrence of “pty” text pattern.But this time, we want to find the second occurrence. For that we planned to start the search from the position , after the first occurrence of the
text pattern.In the formula, the first argument is the text which we want to find.The second argument is the complete text in which we want to search for the text pattern.The third argument, which is
the starting position of the search is positioned by adding one to the first occurrence of the text pattern “pty”.The portion of the function as the third argument FIND(“pty”,E29,1) returns the
position of the first “pty” in the given text. Now if we start the search after this position,it’ll return us the second occurrence.
SEARCH FUNCTION also helps to search for a particular character, or patter or text within another longer text.
The only difference between the FIND and SEARCH FUNCTION is that the
FIND FUNCTION IS CASE SENSITIVE and SEARCH FUNCTION IS NOT CASE SENSITIVE.
For example.If we want to find “ARE” in a sentence say ” How are you today” using both the functions.
We’ll find the following results.
FIND:=FIND(“ARE”,”How are you today”,1). The result will be a VALUE ERROR as it won’t be able to find our ARE as the sentence contains “ARE” and not “are”.SEARCH=SEARCH(“ARE”,”How are you today”,1).
The result will be 5 as “are” starts at the position 5. It’ll ignore the case and match “are” with “ARE”. | {"url":"https://gyankosh.net/msexcel/how-to-insert-text-box-in-excel/excel-function-find/","timestamp":"2024-11-03T20:16:00Z","content_type":"text/html","content_length":"164933","record_id":"<urn:uuid:f321a36c-10bb-4b58-8073-3fc15b3a4929>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00158.warc.gz"} |
Degree/Square Minute to Turn/Square Month
Degree/Square Minute [deg/min2] Output
1 degree/square minute in degree/square second is equal to 0.00027777777777778
1 degree/square minute in degree/square millisecond is equal to 2.7777777777778e-10
1 degree/square minute in degree/square microsecond is equal to 2.7777777777778e-16
1 degree/square minute in degree/square nanosecond is equal to 2.7777777777778e-22
1 degree/square minute in degree/square hour is equal to 3600
1 degree/square minute in degree/square day is equal to 2073600
1 degree/square minute in degree/square week is equal to 101606400
1 degree/square minute in degree/square month is equal to 1921068900
1 degree/square minute in degree/square year is equal to 276633921600
1 degree/square minute in radian/square second is equal to 0.0000048481368110954
1 degree/square minute in radian/square millisecond is equal to 4.8481368110954e-12
1 degree/square minute in radian/square microsecond is equal to 4.8481368110954e-18
1 degree/square minute in radian/square nanosecond is equal to 4.8481368110954e-24
1 degree/square minute in radian/square minute is equal to 0.017453292519943
1 degree/square minute in radian/square hour is equal to 62.83
1 degree/square minute in radian/square day is equal to 36191.15
1 degree/square minute in radian/square week is equal to 1773366.22
1 degree/square minute in radian/square month is equal to 33528977.46
1 degree/square minute in radian/square year is equal to 4828172754.62
1 degree/square minute in gradian/square second is equal to 0.00030864197530864
1 degree/square minute in gradian/square millisecond is equal to 3.0864197530864e-10
1 degree/square minute in gradian/square microsecond is equal to 3.0864197530864e-16
1 degree/square minute in gradian/square nanosecond is equal to 3.0864197530864e-22
1 degree/square minute in gradian/square minute is equal to 1.11
1 degree/square minute in gradian/square hour is equal to 4000
1 degree/square minute in gradian/square day is equal to 2304000
1 degree/square minute in gradian/square week is equal to 112896000
1 degree/square minute in gradian/square month is equal to 2134521000
1 degree/square minute in gradian/square year is equal to 307371024000
1 degree/square minute in arcmin/square second is equal to 0.016666666666667
1 degree/square minute in arcmin/square millisecond is equal to 1.6666666666667e-8
1 degree/square minute in arcmin/square microsecond is equal to 1.6666666666667e-14
1 degree/square minute in arcmin/square nanosecond is equal to 1.6666666666667e-20
1 degree/square minute in arcmin/square minute is equal to 60
1 degree/square minute in arcmin/square hour is equal to 216000
1 degree/square minute in arcmin/square day is equal to 124416000
1 degree/square minute in arcmin/square week is equal to 6096384000
1 degree/square minute in arcmin/square month is equal to 115264134000
1 degree/square minute in arcmin/square year is equal to 16598035296000
1 degree/square minute in arcsec/square second is equal to 1
1 degree/square minute in arcsec/square millisecond is equal to 0.000001
1 degree/square minute in arcsec/square microsecond is equal to 1e-12
1 degree/square minute in arcsec/square nanosecond is equal to 1e-18
1 degree/square minute in arcsec/square minute is equal to 3600
1 degree/square minute in arcsec/square hour is equal to 12960000
1 degree/square minute in arcsec/square day is equal to 7464960000
1 degree/square minute in arcsec/square week is equal to 365783040000
1 degree/square minute in arcsec/square month is equal to 6915848040000
1 degree/square minute in arcsec/square year is equal to 995882117760000
1 degree/square minute in sign/square second is equal to 0.0000092592592592593
1 degree/square minute in sign/square millisecond is equal to 9.2592592592593e-12
1 degree/square minute in sign/square microsecond is equal to 9.2592592592593e-18
1 degree/square minute in sign/square nanosecond is equal to 9.2592592592593e-24
1 degree/square minute in sign/square minute is equal to 0.033333333333333
1 degree/square minute in sign/square hour is equal to 120
1 degree/square minute in sign/square day is equal to 69120
1 degree/square minute in sign/square week is equal to 3386880
1 degree/square minute in sign/square month is equal to 64035630
1 degree/square minute in sign/square year is equal to 9221130720
1 degree/square minute in turn/square second is equal to 7.7160493827161e-7
1 degree/square minute in turn/square millisecond is equal to 7.7160493827161e-13
1 degree/square minute in turn/square microsecond is equal to 7.716049382716e-19
1 degree/square minute in turn/square nanosecond is equal to 7.7160493827161e-25
1 degree/square minute in turn/square minute is equal to 0.0027777777777778
1 degree/square minute in turn/square hour is equal to 10
1 degree/square minute in turn/square day is equal to 5760
1 degree/square minute in turn/square week is equal to 282240
1 degree/square minute in turn/square month is equal to 5336302.5
1 degree/square minute in turn/square year is equal to 768427560
1 degree/square minute in circle/square second is equal to 7.7160493827161e-7
1 degree/square minute in circle/square millisecond is equal to 7.7160493827161e-13
1 degree/square minute in circle/square microsecond is equal to 7.716049382716e-19
1 degree/square minute in circle/square nanosecond is equal to 7.7160493827161e-25
1 degree/square minute in circle/square minute is equal to 0.0027777777777778
1 degree/square minute in circle/square hour is equal to 10
1 degree/square minute in circle/square day is equal to 5760
1 degree/square minute in circle/square week is equal to 282240
1 degree/square minute in circle/square month is equal to 5336302.5
1 degree/square minute in circle/square year is equal to 768427560
1 degree/square minute in mil/square second is equal to 0.0049382716049383
1 degree/square minute in mil/square millisecond is equal to 4.9382716049383e-9
1 degree/square minute in mil/square microsecond is equal to 4.9382716049383e-15
1 degree/square minute in mil/square nanosecond is equal to 4.9382716049383e-21
1 degree/square minute in mil/square minute is equal to 17.78
1 degree/square minute in mil/square hour is equal to 64000
1 degree/square minute in mil/square day is equal to 36864000
1 degree/square minute in mil/square week is equal to 1806336000
1 degree/square minute in mil/square month is equal to 34152336000
1 degree/square minute in mil/square year is equal to 4917936384000
1 degree/square minute in revolution/square second is equal to 7.7160493827161e-7
1 degree/square minute in revolution/square millisecond is equal to 7.7160493827161e-13
1 degree/square minute in revolution/square microsecond is equal to 7.716049382716e-19
1 degree/square minute in revolution/square nanosecond is equal to 7.7160493827161e-25
1 degree/square minute in revolution/square minute is equal to 0.0027777777777778
1 degree/square minute in revolution/square hour is equal to 10
1 degree/square minute in revolution/square day is equal to 5760
1 degree/square minute in revolution/square week is equal to 282240
1 degree/square minute in revolution/square month is equal to 5336302.5
1 degree/square minute in revolution/square year is equal to 768427560 | {"url":"https://hextobinary.com/unit/angularacc/from/degpmin2/to/turnpm2","timestamp":"2024-11-11T06:42:42Z","content_type":"text/html","content_length":"113369","record_id":"<urn:uuid:db87de20-7e5b-449b-a300-3fe6532e02c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00034.warc.gz"} |
Non – Linear Chemotactic Hydromagnetic Bioconvection
Volume 02, Issue 01 (January 2013)
Non – Linear Chemotactic Hydromagnetic Bioconvection
DOI : 10.17577/IJERTV2IS1130
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Prof. Dr. P. K. Srimani, Mrs. Radha. D, 2013, Non – Linear Chemotactic Hydromagnetic Bioconvection, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 02, Issue 01 (January
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• Authors : Prof. Dr. P. K. Srimani, Mrs. Radha. D
• Paper ID : IJERTV2IS1130
• Volume & Issue : Volume 02, Issue 01 (January 2013)
• Published (First Online): 02-02-2013
• ISSN (Online) : 2278-0181
• Publisher Name : IJERT
• License: This work is licensed under a Creative Commons Attribution 4.0 International License
Text Only Version
Non – Linear Chemotactic Hydromagnetic Bioconvection
Prof. Dr. P. K. Srimani Mrs. Radha.D
R&D,Director (DSI) Assistant Professor, Dept of Mathematics,
Bangalore BMS College of Engineering, Bangalore
The effect of external magnetic field on the chemotactic bacterial bioconvection by considering a Continuum model is investigated. Chemotaxis causes cells to swim out of the plume because the high
concentration of the cells constituting the plumes leads to a lower concentration of oxygen in the surrounding fluid. Further worlds major portion consists of bio-mass; therefore it is of immense
interest and at most importance to study bioconvection under different types of constraints. A similarity solution is found for the plume in which the cell flux and the volume flux could be matched
to those in the boundary layer and also outside the suspension regions. Axisymmetric plumes is formed by applying two scales one with respect to the radial co-ordinate and the other with respect to
the similarity variable. The effects of magnetic field are remarkable and encouraging and the computed results are in excellent agreement with those of hydrodynamic case in the limiting case.
1. Introduction
The spontaneous formation of patterns in suspensions of swimming microorganisms due to their tactic nature viz. oxytactic, gyrotactic, chemotactic etc., is termed as bioconvection. The
microorganisms exhibiting bioconvection have the following key features: (i) They are denser than water (ii) They swim upwards due to their tactic nature. This leads to an unstable situation in
the system and thus an overturning instability develops leading to pattern formation [1][2][3]. Also Magnetic field has a strong influence on the system in many real time situations. Experiments
on bioconvection containing suspensions of bacteria (Bacillus Subtilis) have revealed the formation of Falling plumes (Figure 1.) when the system becomes unstable.
Cell rich upper boundary layer
Outer region
Figure 1. Formation of Bioconvection Plume
Some literatures pertaining to bioconvection in deep chambers are [4][5]. The study of such a phenomenon has a variety of applications in biological and physiological problems. Further,
chemotaxis and oxygen consumption are important in setting up the basic state and soon after, the resulting plumes are entirely buoyancy driven and the cells are merely advected. In such cases,
the velocity would vary across the plume [6][7] . The present work investigated the nonlinear Hydromagnetic bioconvection in order to study the effect of magnetic field on the formation of
falling plumes (Axisymmetric) where the oxygen consumption and chemotaxis are important. The model constituted the quasi steady situation in which an upper boundary layer containing a high
concentration of bacteria feeds a falling plume of cell-rich fluid. The suspension was divided into three separate regions as shown in Figure 1, a cell-rich upper boundary layer of known
thickness , a falling plume of unknown width
which also contained a high concentration of
bacteria and the fluid outside the plume which had to circulate in order to conserve mass. Here, the assumption of the axisymmetric nature of the plume reduced the 3D-problem to 2D-problem [8].
No much literature is available in this direction. The solutions were obtained by a Fast Computational Technique.
2. Mathematical Formulation
The bacterial suspension (Bacillus Subtilis) contained in a deep chamber reveal the development of a thin upper boundary layer of cell-rich saturated fluids which becomes unstable, leading to the
formation of falling plumes which is a complex phenomenon. This was used as a basis for our mathematical model. The whole suspension was under the influence of uniform magnetic field .
The dimensionless governing equations are: The equation of cell conservation
: Strength of Oxygen consumption relative to i ts diffusion
: Measures the relative strengths of
directional and random swimming
: Ratio of Oxygen diffusivity to cell diffusivity
: Bio-Rayleigh number
Sc : Schmidth number
Pm : Magnetic Prandtl number
: Kinematic viscosity of the fluid
c ,w : Densities of cell and water g : Acceleration due to gravity
: Volume of a cell
N .HN UN HN
B : Modified Hotmann n umber
: Magnetic Pr eamibility
The equation of oxygen concentration
. U HN
m : Magnetic viscosity of the fluid
2.2 Boundary Conditions
The Navier Stokes equation (with Boussinesq approximation)
Sc1 U U.U P 2U N B H.H
1. No slip condition at Z = 1(bottom of the chamber).
2. Stress free condition at the upper surface of the
The conservation of mass
.U 0
The magnetic induction equation
H U.H H.U P 2H
chamber, i.e., at Z = 0.
3. The vertical components of velocity vanish at both the boundaries.
4. Zero cell flux at both the boundaries.
t m
The variables are non dimensionalized as :
5. Zero oxygen flux at the bottom surface and C = Co at the free surface.
1 , N
N ,
C Cmin , D
D H,
6. H = 0 at both the boundaries
h N0 C0 Cmin
N N0
7. The vertical components of velocity vanish at both
the boundaries.
V b VsH, K K0H.b Vs , t N0 ,
U U h , H H
D H
Mathematically, At Z = 0,
N0 0
U.Z 0, z2 U.Z 0, 1
Where h is the Depth of the chamber
Vs : It has dimensions of velocity
H Z NHZ 0, HZ 0 .
N0 : Initial cell concentration
U : Saturated fluid velocity DN0, K0 : Constants
At Z = 1,
DN : The cell diffusivity
: The oxygen concentration C0 : Initial Concentration
H : The step function
T : The time
H0 : The constant magnetic field
U.Z 0, U Z 0, Z 0, Z 0, H.Z 0
3. Axisymmetric Plumes using Radial Co-ordinates
In the plume, the radial co-ordinate is scaled as
2.1 Dimensionless Parameters
R r, 0 1
K N p
b V
N NAn, 1 CAC, W WA w,
Dc C0 Cmin
s , DN0
c , B DN0
U WA u, P PAp
N gp
where NA , C A, WA , and PA are scale factors. Then the
0 c w
,Sc , P
axisymmetric governing equations (Neglecting O ( 2 )
D D m D
N0 w N0 N0
terms) are,
1 n 2n 2
1 u w
2w w w
r R r2
WA u r
w z
Sc r r u
r2 r
Z w rZ
n C n C
n 3w 1 2w 1 w
CA r
r r
r n r2
3 r
r r
hz hz
WA u C w C NAn C 1 C>
B r r hz 2 r Z hz rZ
r Z CA r r r
u u w 0
Differentiating (14) w.r.t r and substituting for HA,WA
r r Z
u w 2w w w 2w
2 1 u
P p
r u
• r
Z w rZ
1 u 2 B HA h
1 u 2 B HA h
W Sc
u w A
r r
hx w
hz w
x hx hz
r hx
r2 r
Z hz rZ
r r W r Z
w w P p N
Pm 1 hz 1
2W Sc1 u
w 2 A 2 A n
r2 r r
A r
W Z W
A A (12)
2w 1 w
2 H 2 h
B A hx z hz z
Now, imposing the boundary conditions:
r r WA r
n C w
2 W
u u
• w u
2H h
u h
r 0, r
0, u 0, r 0
x r
z Z
Also, r , n 0, C
0, w 0
P A
x x
m W r r
3.1 Similarity solution for axisymmetric case
2W u w w w 2H h
w h
A r
x r
z Z
In order to obtain a similarity solution [9][10] for
HA 1hz
(16), (17), (18), (19) the solution was posed of the form
Pm W r
r 2
h Za , w Zb,n Zc,C Zd,u Zab1
(h: width of the plume, a = ½, b = 0, c = -1, d = 0) Since h :: Z1/2 , the similarity variable is defined as
WA = 2
(to retain advection terms)
r Z1/ 2
C A = 2/ (to retain chemotaxis term)
NA = / 2 (to retain the oxygen consumption term in 9)
Assuming the solution in the form
n Z 1H, C G , ZF,
= O ( 1 2 ) (to retain the buoyancy term in 12)
1 F F
PA = 4 (to retain the pressure term in 12)
u Z 2
, w F
H 2 (to retain the induction term in 14)
This leads to p = 0 hence p = p (Z) in (11). (15)
Substituting for CA , WA and NA in (8) and (9) :
( : Stream functions, Primes denote differentiation
w.r.t ). Substituting these into (16) and integrating once w.r.t with the boundary conditions at = 0, we
n n
n C n C
get the following: HF 2HG (23)
u r
w Z 2 r
r 2 r
Substituting into (17)
2n 2C 1 n 2n 0
G G 1 GF H 0
r2 r r r2
1 C C 1 C 2C
Substituting into (18)
u r w Z n r r r2 0
F F F Sc FF FF
Differentiating (12) w.r.t. r and substituting for NA WA ,
HA and :
H B* 1 MM 1 MM 0
Substituting into (19)
M F
The boundary conditions are ,
5. Results and Discussion
At 0
H G F F F F 0
H 0,G 0, F 0, M 0
1. Solution
For 0 , CFD technique is employed. However for
0 (i.e., when the chemotaxis is unimportant in the plume) analytical solutions are possible with = O (1),
In this study, the deep chamber experiments of Figure.1 have been modelled in three separate regions:
1. an upper boundary layer of depth R
2. a falling plume of width
3. the region outside the plume.
In the sections 3 and 4, solutions for the cell and the oxygen concentration and the fluid velocity in the upper boundary layer are determined under the influence of a uniform vertical magnetic
field. The solutions are found to depend on the parameters like, Sc (Schmidth
number), Q ( the cell flux ), (Bio-Rayleigh number),
Co = 1, No
= 2
and = 2 . Following [4] [8] the
B* (Magnetic Parameter) and (diffusivity ratio). The
computations are performed using the MATLAB tool;
solutions for the equations (23, 24, 25, 26) are found to be ( see table 1)
B 12
1 Sc1 B
the computed results are presented through graphs in Figures 2 to 14.
The following observations are made: In Figures 2,3,4,5. the effect of variation in the magnetic parameter
192 A2 1
B* on the profiles of velocity w F1 , the cell
C 1 Sc 1 B
Table 1. Solutions for F, H, G'
At Sc 1, 1 At Sc 2, 1
12A2 24A2
F 2 B 1 A2 F 3 2B 1 A2
192A2 384A2
H 6 H 12
2 B 1 A2 2B 3 2B 1 A2 32B
G 96A 192A
6 G 12
2 B 1 A2 2B 3 2B 1 A2 32B
At Sc 1, 1 At Sc 2, 1
12A2 24A2
F 2 B 1 A2 F 3 2B 1 A2
192A2 384A2
H 6 H 12
2 B 1 A2 2B 3 2B 1 A2 32B
G 96A 192A
6 G 12
2 B 1 A2 2B 3 2B 1 A2 32B
concentration H and the oxygen concentration G is studied for the values, Sc = = = Q = 1.0. Here, the oxygen concentration is considered as
1 2 G where is assumed to be always
positive so that, all the bacteria are active. In Figure.2 the effect of similarity variable on the F profile are
shown. F decreases enormously and remains constant as
for all values of B* . Further, F is negative and
its value is highest in the hydrodynamic case. In other
words, F increases in absolute value as B* increases,
Also solutions satisfy the boundary conditions at 0
and .
Q 2 B 8 B
and the vertical fluid velocity w, at the center of the plume increases indicating that the horizontal fluid flow into the plume increases. From Figure.5, w0 as
as expected. In Figure.3 the effect of similarity
variable on H profile is shown, it reveals that, as B* decreases the cell concentration in the plume increases as expected. Physically it means that, the higher the concentration of the cells, the
greater is the consumption of oxygen which means that the oxygen
for Sc 1
concentration at the center of the plume decreases. In Figure.4 the effect of similarity variable on G profile is
Q 3 2B 15 2B
shown. Clearly the width of th plume decreases as
for Sc 2
B* increases. The oxygen concentration is more in the
hydrodynamic case ( B* = 0) when compared to the hydro magnetic case ( B* 0).
Since HF d Q
Figure 2. F vs
Figure 5. F
Figure 3. H vs
Figure 4. G vs
Figures 6,7, 8. reveal the effect of variation of Q (= 0.5, 3, 5) on the profiles of F, H, G'. As the cell flux increases, F slightly decreases and increases in absolute value. The values of F are
considerably very large inside the plume and drastically decrease and become constant
for large values of and accordingly w0 for large .
Fig.7 reveals that the width of the plume drastically increases as the value of Q decreases and the plume becomes narrower for large values of Q. Thus, the high concentration of the cells leads to a
greater consumption of oxygen which in turn reduces the oxygen concentration at the center of the plume. Thus, as the cell flux Q increases, the vertical fluid velocity w, at the center of the plume
increases and the values of M increases, indicating the increase in the horizontal fluid flow into the plume.
Figure 6. F vs
Figure 7. H vs
Figure 8. G vs
From Figure 8., it is found that the oxygen concentration in the plume is high for large Q(= 5) and the width of the plume drastically increases as Q decreases. This clearly indicates that the oxygen
concentration at the center of the plume is less since there is a greater consumption of oxygen in the plume for large Q.
Figures 9,10,11. represent the effect of variation of
on the profiles F, H, G' for fixed values of the parameters Sc B* 1. The values of
considered are 0.2, 2 and 5.
The effect of buoyancy becomes important when
is large. The cell concentration is more for small values of and the plume becomes narrower for large .
When the cell concentration in the centre of the plume increases, the plume becomes narrower, accordingly the oxygen profile becomes narrower and the oxygen concentration at the center of the plume
increases. The consumption of oxygen will be more. Therefore, the velocity of the fluid in the center of the plume will be
larger when the buoyancy force is dominant. But w0 more rapidly than for small values of . The decrease in M for the increase in indicates that less fluid is
entrained by the plume.
Figure 9. F vs
Figure 10. H vs
Figure 11. G vs
Figures 12,13,14. present the graph of the profiles F, H, G' when the values of Sc (= 1, 2) are varied. The other
parameters have fixed values viz., = 1, Q = 1 B* = 1.
It is observed that as in the hydrodynamic case ( B* = 0), the variation in Sc has a significant effect on the behaviour of the profiles. There is a drastic change in the values of F for Sc = 1 and
2. As Sc increased, F decreases rapidly and F a constant value as
as expected. The cell concentration will be
more for Sc = 2 and accordingly the oxygen consumption in the plume will be more and there will be a reduction in the oxygen concentration in the plume.
Figure 12. F vs
Figure 13. H vs
Figure 14. G vs
Finally, it is concluded that (i) the governing dimensionless parameters have a remarkable effect in the hydrodynamic as well as in the hydromagnetic cases
(ii) the qualitative nature of the profiles is almost the same in both the cases but there is a drastic difference in the quantitative nature of the profiles. Figure 1. clearly indicates the strong
influence of the magnetic parameter on the present bioconvective system, these clearly suggest that the plume convection could be suppressed or enhanced through the proper choice of the magnetic
parameter. The results are in excellent agreement with the hydrodynamic case.
1. J.O.Kessler, Cooperative and concentrative phenomena of swimming microorganisms,Contemp.Phys.26, 1985 147 166.
2. A.V.Kuznetsov, A.A. Avramenko, and P. Geng, A similarity solution for a falling plume in bioconvection of oxytactic bacteria in a porous medium, Int Commun Heat Mass Transfer,30,2003, 37 – 46
3. T.J.Pedley, N.A.Hill, and J.O. Kessler, The growth of bioconvection patterns in a uniform suspension of gyrotactic microorganisms, J.Fluid Mech, 195,1988, 223 238.
4. A.J. Hillesdon, T.J. Pedley, and J.O. Kessler, The development of concentration gradient in a suspension of chemotactic bacteria, Bull. Math. Biol, 57: 2,1995, 99-344
5. A.J. Hillesdon, and T.J. Pedley, Bioconvection in suspensions of oxytactic bacteria: linear theory, J. F.M, 1996, 324: 223 259.
6. A.M. Metcalfe, et al., Bacterial bioconvection: weakly nonlinear theory for pattern selection, J. Fluid Mech, 1998, 370:249 270.
7. S. Ghorai, and N.A. Hill, Gyrotactic Bioconvection in Three dimensions, Phys. Fluid, 2007, 19: 054107.
8. S. Ghorai, and N.A. Hill, Gyrotactic Bioconvection in Three dimensions, Phys. Fluid, 2007, 19: 054107.
9. C.S. Yih, Free convection due to a point source of heat, In Proc. 1st US Nat. Cong. Appl. Mech, (ed. E. Sternberg), 1951, pp. 941 947.
10. A.M. Metcalfe, T.J. Pedley, Falling plumes in bacterial bioconvection,J. Fluid Mech. 445, 2001, 121 – 149.
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Approximate control of parabolic equations by spectral decomposition
Friday, June 16th, 2017, 12:00h.,
Central Meeting Room at DeustoTech
Martin Lazar
University of Dubrovnik, Dubrovnik, Croatia
Cesare Molinari
Universidad Técnica Federico Santa María, Valparaíso, Chile.
We consider the constrained minimisation problem
where $x^T$ is some given target state, J is a given cost functional and x is the solution of
If the cost functional J is given by $J(u) = {\|u\|}_{L^2}$ the problem (P) is reduced to a classical minimal norm control problem which can be solved by Hilbert uniqueness method (HUM). In this
paper we allow for a more general cost functional and analyse examples in which, apart from the target state and the control norm, one considers a desired trajectory and penalise a distance of the
state from it. Such problem
requires a more general approach, and it has been addressed by dierent methods throughout last decades.
In this paper we suggest another method based on the spectral decomposition in terms of eigenfunctions of the operator A . Surprisingly, the problem reduces to an algebraic equation for a scalar
unknown, representing a Lagrangian multiplier. The same approach has been recently introduced in [1] for an optimal control problem of the heat equation in which the control was given through the
initial datum.
This paper generalises the method to the distributed control problems. As can be expected, in this case one has to consider the associated dual problem which makes the calculation more complicated,
although the algorithm steps follow a similar structure as in [1]. In the talk basic steps of the method will be explained, followed by numerical examples demonstrating its efficiency.
[1] Lazar, M, Molinari C, J. Peypouquet Optimal control by spectral decomposition of parabolic equations , Optimisation, (2017) | {"url":"https://cmc.deusto.eus/parabolic-eq-control/","timestamp":"2024-11-05T04:09:18Z","content_type":"text/html","content_length":"83584","record_id":"<urn:uuid:a2b22dab-2695-48e4-b949-a5690be80c7c>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00856.warc.gz"} |
Asparagus - Encoding Class Sokoban
Description Sokoban is a game puzzle developed by the Japanese company Thinking
Rabbit, Inc. in 1982. 'Sokoban' means 'warehouse-keeper' in Japanese.
Each puzzle consists of a room layout (a number of square fields
representing walls or parts of the floor, some of which are marked as
storage space) and a starting situation (one sokoban and a number of
boxes, all of which must reside on some floor location, where one box
occupies precisely one location and each location can hold at most one
box). The goal is to move all boxes onto storage locations. To this end,
the sokoban can walk on floor locations (unless occupied by some box),
and push single boxes onto unoccupied floor locations.
In our setting, an instance contains the warehouse layout, representing
the floor locations, and in particular their horizontal and vertical
relationships, and storage locations, where the boxes should eventually
go to:
right(L1,L2) : location L2 is immediately right of location L1
top(L1,L2): location L2 is immediately on top of location L1
solution(L): location L is a storage location
An instance also consists of a description of the initial configuration:
box(L): location L initially holds a box
sokoban(L): the sokoban is at location L
It can be assumed that each instance has exactly one sokoban.
In order to keep the search space small, we consider the sokoban walking
to a box and pushing it any number of fields in one direction as an
atomic action. This is motivated that in any minimal solution, the
sokoban will walk without pushing only in order to push a box, so making
walking an action on its own is superfluous. Moreover, pushing a box
several fields in one direction does not involve any walking (in a
minimal solution), and thus it makes sense to collapse it into one
action. An instance also contains a fixed number of labels ('steps') for
configurations, between which these atomic actions occur, and their
successorship relation:
step(S): S is a step
next(S1,S2): step S2 is the successor of step S1
Please note that for n steps, exactly n-1 actions should be performed
if not minimizing the number of pushes, while the goal for the optimization
problem is to find the minimum amount of pushes as described next.
Any answer set should contain a sequence of push actions (as defined
above, in the syntax described next), such that between each pair of
successive configurations exactly one push action is performed and such
that in the final configuration all target locations contain a box. The
sequence of push actions should be represented by atoms of the form
push(Bbefore,D,Bafter,S), where Bbefore is the location of the pushed
box at step S, D is a direction (one of the constants right, left, up,
down), Bafter is the location on which the box ends (where it will be
in the next step), and S is the step in which the push is initiated.
A push action is feasible if the sokoban can reach the field, from which
the pushed box is in the adjacent location in the pushed direction (i.e.
o the location adjacent to the pushed box in opposite push direction),
on a box-free path of locations at step S (going any direction).
Furthermore the location on which the box ends must obviously be in the
correct direction and all fields in the pushed direction up to and
including the end location must not contain any box at step S. In the
successive step, the pushed box will be on the new location, and the
sokoban will be adjacent to the pushed box in opposite pushing
direction. All other boxes are on the same locations as in the previous
Author: Wolfgang Faber ()
Level-Author: Jacques Duthen | {"url":"https://asparagus.cs.uni-potsdam.de/encodingclass/show/id/30","timestamp":"2024-11-13T02:44:00Z","content_type":"application/xhtml+xml","content_length":"14134","record_id":"<urn:uuid:5001047a-21e7-4064-93b0-a2aed44aec0e>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00139.warc.gz"} |
Stick to the Content
Posted on February 27, 2009 by Frank Morgan
By Daniel Erman
There is a wealth of information on the internet about how to give a good Mathematics talk, including:
(*) Gian-Carlo Rota’s essay “Ten lessons I wish I had been taught”
(*) John Baez’s “Advice for the young scientist”
(*) Terence Tao’s “Talks are not the same as papers”
In this post, I’d like to consider an aspect of lecturing which is not emphasized in the above sources, but which I think is extremely important. Lecturing necessarily involves choosing to emphasize
some aspects of the material over others. A common pitfall I’ve seen among speakers—especially student speakers—is to apologize during the talk for such choices, or to make self-deprecating jokes.
This is nearly always a bad idea, as it distracts from the point of your talk.
Let’s take an example. Imagine you are giving an expository talk on a topic which involves a theorem with a complicated proof, and you don’t understand the proof very well. However, you have a
clear view of why this theorem is important, and you know how to apply the theorem. So you decide that, in your talk, you will avoid the proof of the theorem and focus instead on an application
which illustrates the big picture. You prepare your talk along these lines.
As the date for the talk approaches, you start getting nervous about the fact that you don’t understand some of the details of the proof. A common reaction to such nervousness is, when skipping the
proof, to apologize for this choice; e.g. “Now I’m going to skip the proof, since I don’t understand it.” Offering up this personal detail doesn’t help the audience understand your talk any better,
and it generally makes the speaker look uncomfortable. After presenting the statement of theorem, it would be better to state: “We will focus on applications of the theorem rather than the proof.”
This is not to say that you should try to fool the audience into thinking you know more than you do. However, you should only include such details when relevant. For instance, if an audience member
asks, “Could you tell us about the proof?”, then this is the time to say, “Honestly, I didn’t understand the proof well enough to give a sketch.” However, preempting such questions by discussing
your own struggles tends to distract from the message of your lecture.
Your lecture should focus on the material you have chosen to present. Though you may feel the desire to apologize for/joke about some aspect you are omitting, such apologies/jokes are usually
1 Response to Stick to the Content
1. Brie Finegold says:
Very helpful! Nice to see all of those references in one spot.
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Graphing systems of equations worksheets
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Author Message
Varthantel Posted: Friday 15th of Mar 07:10
Can someone please assist me? I simply need a quick way out of my problem with my algebra . I have this exam coming up in the next few days. I have a problem with graphing systems of
equations worksheets. Finding a good tutor these days quickly is difficult. Would value greatly any tips .
Buenos Aires
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AllejHat Posted: Saturday 16th of Mar 08:32
Hi! I guess I can give you ideas on how to solve your assignment. But for that I need more details. Can you give details about what exactly is the graphing systems of equations
worksheets assignment that you have to work out. I am quite good at solving these kind of things. Plus I have this great software Algebrator that I downloaded from the internet which is
soooo good at solving math homework. Give me the details and perhaps we can work something out...
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fveingal Posted: Saturday 16th of Mar 15:46
I tried out each one of them myself and that was when I came across Algebrator. I found it really apt for long division, multiplying fractions and factoring polynomials. It was actually
also effortless to run this. Once you key in the problem, the program carries you all the way to the answer clearing up each step on its way. That’s what makes it superb . By the time
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Algebra in math. I am also sure that you too will appreciate this program just as I did. Wouldn’t you want to check this out?
From: Earth
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Homuck Posted: Sunday 17th of Mar 13:44
I remember having problems with radical expressions, exponent rules and fractional exponents. Algebrator is a really great piece of algebra software. I have used it through several math
classes - Basic Math, Intermediate algebra and College Algebra. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly
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NejhdLimks Posted: Tuesday 19th of Mar 08:44
Hi all , Thanks a ton for all your answers. I shall definitely give Algebrator at https://softmath.com/news.html a try and would keep you updated with my experience. The only thing I am
very specific about is the fact that the tool should give sufficient aid on Algebra 1 which in turn would help me to complete my homework on time.
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Koem Posted: Wednesday 20th of Mar 09:09
I think you will get the details here: https://softmath.com/algebra-software-guarantee.html. They also claim to provide an unconditional money back guarantee, so you have nothing to
lose. Try this and Good Luck!
From: Sweden
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Paper Abstract: 2009a
Title: A geometric approach to energy shaping (71 pages)
Author(s): Bahman Gharesifard
Detail: PhD Thesis, Queen's University
Original manuscript: 2009/07/09
In this thesis is initiated a more systematic geometric exploration of energy shaping. Most of the previous results have been dealt with particular cases and neither the existence nor the space of
solutions has been discussed with any degree of generality. The geometric theory of partial differential equations originated by Goldschmidt and Spencer in late 1960s is utilized to analyze the
partial differential equations in energy shaping. The energy shaping partial differential equations are described as a fibered submanifold of a k-jet bundle of a fibered manifold. By revealing the
nature of kinetic energy shaping, similarities are noticed between the problem of kinetic energy shaping and some well-known problems in Riemannian geometry. In particular, there is a strong
similarity between kinetic energy shaping and the problem of finding a metric connection initiated by Eisenhart and Veblen. We notice that the necessary conditions for the set of so-called
lambda-equation restricted to the control distribution are related to the Ricci identity, similarly to the Eisenhart and Veblen metric connection problem. Finally, the set of lambda-equations for
kinetic energy shaping are coupled with the integrability results of potential energy shaping. This gives new insights for answering some key questions in energy shaping that have not been addressed
to this point. The procedure shows how a poor design of closed-loop metric feedback can make it impossible to achieve any flexibility in the character of the possible closed-loop potential function.
The integrability results of this thesis have been used to answer some interesting questions about the energy shaping method. In particular, a geometric proof is provided which shows that linear
controllability is sufficient for energy shaping of linear simple mechanical systems. Furthermore, it is shown that all linearly controllable simple mechanical control systems with one degree of
underactuation can be stabilized using energy shaping feedback. The result is geometric and completely characterizes the energy shaping problem for these systems. Using the geometric approach of this
thesis, some new open problems in energy shaping are formulated. In particular, we give ideas for relating the kinetic energy shaping problem to a problem on holonomy groups. Moreover, we suggest
that the so-called Fakras lemma might be used for investigating the stabilization condition of energy shaping.
650K pdf
Last Updated: Fri Mar 15 07:55:59 2024
Andrew D. Lewis (andrew at mast.queensu.ca) | {"url":"https://mast.queensu.ca/~andrew/papers/abstracts/2009a.html","timestamp":"2024-11-10T11:23:08Z","content_type":"text/html","content_length":"3561","record_id":"<urn:uuid:60ffd73b-6f30-4c88-a214-7eddff28dff7>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00439.warc.gz"} |
In Text Question Answer
In Text Question Answer
Question 15: What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer: The quantity of distance is measured by the area occupied below the velocity time graph.
Question 16: A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.
Answer: Here we have, Initial velocity (u) = 0
Acceleration (a) = 0.1ms-2
Time (t) = 2 minute = 120 second
(a) The speed acquired:
We know that, v = u + at
`⇒ v = 0 + 0.1m//s^2 xx 120 s`
`⇒ v = 120 m//s`
Thus, the bus will acquire a speed of 120 m/s after 2 minute with the given acceleration.
(b) The distance travelled:
We know that, `s=ut+1/2at^2`
`=>s=0xx120s+1/2xx0.1\ m//s^2xx(120s)^2`
`= 1/2 xx 1440m = 720 m`
Thus, bus will travel a distance of 720 m in the given time of 2 minute.
Question 17: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s^-2. Find how far the train will go before it is brought to rest.
Answer: Here,we have,
Initial velocity, `u=90\ km//h`
`=(90xx1000m)/(60xx60s)=25\ m//s`
Final velocity `v=0`
Acceleration, `a = -0.5m//s^2`
Thus, distance travelled =?
We know that, `v^2=u^2+2as`
`=> 0 = (25\ m//s)^2 +2 xx-0.5\ m//s^2 xx s`
`=>0=625\ m^2s^(-2) - 1\ m\ s^(-2)s`
`=>1\ ms^(-2)s = 625\ m^2 s^(-2)`
`=>s=(625\ m^2\ s^(-2))/(1\ m\ s^(-2))=625m`
Therefore, train will go 625 m before it brought to rest.
Question 18: A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?
Answer: Here we have,
Initial velocity, u = 0
Acceleration (a) = 2cm/s^2 = 0.02m/s^2
Time (t) = 3s
Therefore, Final velocity, v =?
We know that, `v=u+at`
`:. v=0+0.02\ m//s^2 xx 3s`
`=>v=0.06\ m//s`
Therefore, the final velocity of trolley will be 0.06m/s after start
Question 19: A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?
Answer: Here we have,
Acceleration, a = 4m/s2
Initial velocity, u =0
Time, t = 10s
Therefore, Distance (s) covered =?
We know that, `s=ut+1/2 at^2`
`=>s= 0xx10s+1/2xx4\ m//s^2 xx(10s)^2`
`=>s=1/2xx4\ m//s^2 xx 100s^2`
`=>s = 2xx100m = 200m`
Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.
Question 20: A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be
the height attained by the stone and how much time will it take to reach there?
Answer: Here we have,
Initial velocity (u) = 5m/s
Final velocity (v) =0 (Since from where stone starts falling its velocity will become zero)
Acceleration (a) = -10m/s^2
(Since given acceleration is in downward direction, i.e. the velocity of the stone is decreasing, thus acceleration is taken as negative)
Height, i.e. Distance, s =?
Time (t) taken to reach the height =?
We know that, `v^2=u^2+2as`
`=>0= (5\ m//s)^2+2xx-10\ m//s^2 xxs`
`=>0= 25\ m^2s^2 - 20\ m//s^2 xxs`
`=>20\ m//s^2 xxs = 25\ m^2s^2`
`=>s=(25\ m^2s^2)/(20\ m//s^2)`
`=>s = 1.25\ m`
Now, we know that, `v=u+at`
`=>0= 5\ ms^(-1) +(-10\ ms^(-2))xxt`
`=>0=5\ ms^(-1) - 10\ ms^(-2) xxt`
`=>10\ ms^(-2)xxt = 5\ ms^(-1)`
`=>t=(5\ ms^(-1))/(10\ ms^(-2))`
`=>t=1/2s = 0.5\ s`
Thus, stone will attain a height of 1.25m. And time taken to attain this height is 0.5s | {"url":"https://www.excellup.com/classnine/sciencenine/motionnineIntextQ2.aspx","timestamp":"2024-11-14T03:59:38Z","content_type":"text/html","content_length":"11490","record_id":"<urn:uuid:a1b7df62-95e6-47f6-a557-e7ea0bb383d3>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00230.warc.gz"} |
What's My Area?
Karen L. Mickel William Carter Elementary School
5740 S. Michigan Avenue
Chicago IL 60637
(312) 535-0860
To cover an area with uniform tiles and count the number of tiles needed.
To recognize rectangles, triangles and squares.
To review basic colors.
Materials needed:
One inch multi-color square tiles
Rectangular tiles
Triangular tiles
Worksheet with different-sized shapes
Each student will be given some square tiles. The students will find a
space on the floor to play with their tiles. After the children have played
with the tiles, tell them it is time to learn about area. Distribute papers
with different-sized shapes. Tell children to cover the shapes completely with
their tiles. After the shapes are covered, ask how many tiles did it take to
cover the shapes. Repeat the procedure with rectangular and triangular tiles.
After the first concept has been taught you may introduce a new way of
finding area. You may give the students different square patterns. Some
patterns may require the students to use rectangular and triangular shapes.
As a follow-up activity, the students can make up their own worksheet with
different-sized shapes. They would trade worksheets with one of their group
members. The group member would use square tile, rectangular tiles and
triangular tiles to cover the shapes on the worksheet.
Repeat this lesson until concept is mastered.
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Coulombs to Picocoulombs Conversion (C to pC) - Inch Calculator
Coulombs to Picocoulombs Converter
Enter the electric charge in coulombs below to convert it to picocoulombs.
Result in Picocoulombs:
1 C = 1,000,000,000,000 pC
Do you want to convert picocoulombs to coulombs?
How to Convert Coulombs to Picocoulombs
To convert a measurement in coulombs to a measurement in picocoulombs, multiply the electric charge by the following conversion ratio: 1,000,000,000,000 picocoulombs/coulomb.
Since one coulomb is equal to 1,000,000,000,000 picocoulombs, you can use this simple formula to convert:
picocoulombs = coulombs × 1,000,000,000,000
The electric charge in picocoulombs is equal to the electric charge in coulombs multiplied by 1,000,000,000,000.
For example,
here's how to convert 5 coulombs to picocoulombs using the formula above.
picocoulombs = (5 C × 1,000,000,000,000) = 5,000,000,000,000 pC
How Many Picocoulombs Are in a Coulomb?
There are 1,000,000,000,000 picocoulombs in a coulomb, which is why we use this value in the formula above.
1 C = 1,000,000,000,000 pC
Coulombs and picocoulombs are both units used to measure electric charge. Keep reading to learn more about each unit of measure.
What Is a Coulomb?
One coulomb is the electric charge equal to one ampere of current over one second.^[1]
The coulomb can be expressed as Q[C] = I[A] × t[s]
The charge in coulombs is equal to the current in amperes times the time in seconds.
The coulomb is the SI derived unit for electric charge in the metric system. Coulombs can be abbreviated as C; for example, 1 coulomb can be written as 1 C.
Learn more about coulombs.
What Is a Picocoulomb?
The picocoulomb is 1/1,000,000,000,000 of a coulomb, which is the electric charge equal to one ampere of current over one second.
The picocoulomb is a multiple of the coulomb, which is the SI derived unit for electric charge. In the metric system, "pico" is the prefix for 10^-12. Picocoulombs can be abbreviated as pC; for
example, 1 picocoulomb can be written as 1 pC.
Learn more about picocoulombs.
Coulomb to Picocoulomb Conversion Table
Table showing various coulomb
measurements converted to picocoulombs.
Coulombs Picocoulombs
0.000000000001 C 1 pC
0.000000000002 C 2 pC
0.000000000003 C 3 pC
0.000000000004 C 4 pC
0.000000000005 C 5 pC
0.000000000006 C 6 pC
0.000000000007 C 7 pC
0.000000000008 C 8 pC
0.000000000009 C 9 pC
0.0000000000001 C 0.1 pC
0.000000000001 C 1 pC
0.00000000001 C 10 pC
0.0000000001 C 100 pC
0.000000001 C 1,000 pC
0.00000001 C 10,000 pC
0.0000001 C 100,000 pC
0.000001 C 1,000,000 pC
0.00001 C 10,000,000 pC
0.0001 C 100,000,000 pC
0.001 C 1,000,000,000 pC
0.01 C 10,000,000,000 pC
0.1 C 100,000,000,000 pC
1 C 1,000,000,000,000 pC
1. International Bureau of Weights and Measures, The International System of Units, 9th Edition, 2019, https://www.bipm.org/documents/20126/41483022/SI-Brochure-9-EN.pdf
More Coulomb & Picocoulomb Conversions | {"url":"https://www.inchcalculator.com/convert/coulomb-to-picocoulomb/","timestamp":"2024-11-11T18:11:11Z","content_type":"text/html","content_length":"67766","record_id":"<urn:uuid:24046c7b-ef0a-4711-ae42-cfee4e03c7eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00356.warc.gz"} |
ncl_vvudmv - Linux Manuals (3)
ncl_vvudmv (3) - Linux Manuals
VVUDMV - This routine is the user-definable external subroutine used to draw masked vectors. The default version of the routine draws any polyline all of whose area identifiers are greater than or
equal to zero.
CALL VVUDMV (XCS,YCS,NCS,IAI,IAG,NAI)
XCS (REAL array, assumed size NCS, input): Array of X coordinates of the points defining the polyline with the given set of area identifiers.
YCS (REAL array, assumed size NCS, input): Array of Y coordinates of the points defining the polyline with the given set of area identifiers.
NCS (INTEGER, input): Number of points; assumed size of the X and Y coordinate arrays, XCS and YCS.
IAI (INTEGER array, assumed size NAI, input): Array of area identifier values. Each value represents the area identifier with respect to the area group in the area group array with the same array
IAG (INTEGER array, assumed size NAI, input): Array of area-identifier groups.
NAI (INTEGER, input): Number of area identifier groups. The current version of Vectors supports up to 64 area groups.
'VVUDMV' is the name given to the default version of the masked vector drawing routine, and it is also the name given to the argument through which the external subroutine is passed to VVECTR.
However, you may choose any acceptable FORTRAN identifier as the name of a user-defined version of the routine. The substitute routine must have an argument list equivalent to the default version of
VVUDMV. Also, whether or not the default version is used, the subroutine that calls VVECTR should contain an external declaration of the routine similar to the following:
If the MSK parameter is set to the value 1, specifying high precision masking, Vectors sends one set of X and Y polyline coordinate arrays to the area masking routine, ARDRLN, for each vector arrow.
ARDRLN subdivides the polyline into pieces such that each smaller polyline has a single area identifier with respect to each area identifier group, then makes a call to VVUDMV for each polyline
piece. While the default version of VVUDMV only checks to see that none of the area identifiers are negative, a user-defined version could perform more complicated decision processing based on
knowledge of the meaning of specific area identifier groups and/or area identifier values. Note that, before invoking VVUDMV, ARDRLN modifies the user coordinate space by making the following calls:
CALL GETSET(VPL,VPR,VPB,VPT,WDL,WDR,WDB,WDT,LLG) CALL SET(VPL,VPR,VPB,VPT,VPL,VPR,VPB,VPT,1)
These calls temporarily turn the user to NDC mapping into an identity, allowing the user to call any of the routines, CURVE, CURVED, or the GKS routine GPL, to render the polygon piece , without
worrying about a possible non-identity mapping between user and world coordinates.
If MSK has a value greater than 1, specifying low precision masking, Vectors calls the routine ARGTAI to get the area identifiers with respect to the area identifier groups for a single point that
locates the base position of the vector. Vectors then calls the VVUDMV routine itself, passing the coordinate arrays for a complete vector arrow. Thus, a vector arrow whose base position is within an
area to be masked can be eliminated, but an arrow whose base position is nearby, but outside, a masked area may intrude into the area. Also, in this case, since faster rendering is the goal, Vectors
does not convert the coordinate arrays into normalized device coordinates and do the identity SET call. Therefore, the user should use only CURVE or CURVED to render the polyline, unless there is no
possibility of a non-identity user to world coordinate mapping.
The current version of Vectors supports masked drawing with up to 64 area groups. Vectors will exit with an error message if an area map with more than 64 groups is passed to it.
Use the ncargex command to see the following relevant examples: ffex05, vvex01.
To use VVUDMV, load the NCAR Graphics libraries ncarg, ncarg_gks, and ncarg_c, preferably in that order.
Copyright (C) 1987-2009
University Corporation for Atmospheric Research
The use of this Software is governed by a License Agreement.
Online: vectors, vectors_params, vvectr, vvgetc, vvgeti, vvgetr, vvinit, vvrset, vvsetc, vvseti, vvsetr, vvumxy, ncarg_cbind. | {"url":"https://www.systutorials.com/docs/linux/man/3-ncl_vvudmv/","timestamp":"2024-11-13T02:57:14Z","content_type":"text/html","content_length":"11222","record_id":"<urn:uuid:7816072c-712c-4392-a96f-e850bada9eae>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00778.warc.gz"} |
Proactive Caching at the Edge Leveraging Influential User Detection in Cellular D2D Networks
Department of Computer Science and Electrical Engineering, Information Technology University (ITU), Lahore 54000, Pakistan
BSON Lab, ECE, University of Oklahoma, Norman, OK 73019, USA
Computer Lab, University of Cambridge, Cambridge CB2 1TN, UK
Author to whom correspondence should be addressed.
Submission received: 28 July 2018 / Revised: 28 August 2018 / Accepted: 28 August 2018 / Published: 21 September 2018
Caching close to users in a radio access network (RAN) has been identified as a promising method to reduce a backhaul traffic load and minimize latency in 5G and beyond. In this paper, we investigate
a novel community detection inspired by a proactive caching scheme for device-to-device (D2D) enabled networks. The proposed scheme builds on the idea that content generated/accessed by influential
users is more probable to become popular and thus can be exploited for pro-caching. We use a Clustering Coefficient based Genetic Algorithm (CC-GA) for community detection to discover a group of
cellular users present in close vicinity. We then use an Eigenvector Centrality measure to identify the influential users with respect to the community structure, and the content associated to it is
then used for pro-active caching using D2D communications. The numerical results show that, compared to reactive caching, where historically popular content is cached, depending on cache size, load
and number of requests, up to 30% more users can be satisfied using a proposed scheme while achieving significant reduction in backhaul traffic load.
1. Introduction
Generational shifts in the world of mobile networks have been driven by the unprecedented growth in mobile data traffic. The exponentially growing number of connected devices and the unquenchable
thirst for better mobile broadband experience is also a driving force behind this evolution. The story of 5G is no more different as it is envisioned to provide a gigabit experience and virtually
zero latency while withstanding an expected 500-fold increase in mobile video traffic over the next ten years. It is predicted that 75% of global mobile data will be video content in which 6.7% will
be machine-to-machine (M2M) communication. Such massive growth of multimedia traffic was further fueled by social media feeds, such as Facebook and Twitter (representing 15% of the traffic [
]). This trend is undoubtedly going to stress the capacity of core networks, wireless links and mobile backhauls to their limits eventually leading to poor quality-of-experience (QoE).
In order to cope with this issue, a 5G network is gearing up with various technologies including device-to-device (D2D) communication [
], Massive MIMO (Multiple-input multiple-output) [
], Device-centric architecture [
], Small Cells, Caching, mmWave communication [
] all orchestrated by self organizing networks (SON) [
]. Among them, caching through D2D mode of communication is being considered a promising technology aimed for reducing backhaul load and minimizing latency [
]. The key idea of D2D enabled caching is to store the popular content of the network at the user end. The key benefit here is that it brings likely to be accessed content as close as possible to the
users. This approach can reduce latency substantially, a key challenging requirement in 5G [
]. It can also increase the effective throughput and reduce the backhaul load [
]. The three key research questions in designing an optimal caching enabled next generation Radio Access Network (RAN) are:
Where to cache? How to cache?
What to cache?
Where to cache focuses on the problem of caching the content over the network in such a way that it reduces the network traffic. How to cache focuses on multiple ways of content caching, and what to
cache focuses on the problem of finding content having the highest probabilities to be used in the near future. In order to address these questions, various attempts have been made by the researchers
such as [
]. For a detailed review of caching in RAN, the reader is referred to [
]. However, to the best of the authors’ knowledge, this is a first study that investigates the potential of the leveraging detection of influential users via centrality measure in cellular networks
and then uses that detection for identifying what to cache and pro-actively caching those likely to be accessing content using D2D communications.
Online Social Networks, one of the key sources of mobile data explosion, are known to be scale-free networks, which follow power-law distribution. For example, the Twitter network has millions of
users and an enormous flow of information is published daily. There are small number of Twitter users who have a large number of followers while there is a huge number of users who have a small
number of followers. In this paper, our key idea to design and evaluate a novel proactive yet low complexity caching solution is as follows. It is highly probable that content accessed or generated
by influential users will become popular due to their relatively higher following and connectivity. Furthermore, popular content is expected to be in demand in the same community i.e., the community
of the influential user first and then spread slowly in the remaining network. The mapping structure, information diffusion and influence phenomena of the social networks can thus be exploited for
proactively caching those content which are expected to be highly demanded in the near future, for reducing the backhaul traffic load [
]. Therefore, we represent a mobile cellular network in the form of a graph and use graph theory and Social Network Analysis (SNA) techniques to improve QoE and reduce the backhaul traffic load.
This study considers social networks aware D2D caching by focusing on influential users, their communities, and content. Initially, we use a preferential attachment model [
] for representing the mobile cellular network. We then perform clustering using a CC-GA algorithm [
], and find influential users using an eigenvector centrality measure [
]. Finally, based on the users’ previous history, we proactively cache the content of the influential users with respect to the available cache size. We perform extensive system level experiments to
evaluate the gain of the proposed approach. The results reveal that 48% of backhaul traffic load can be reduced (48% satisfied requests) when users generating requests are 100% using the proposed
The rest of the paper is organized as follows.
Section 2
presents a brief background to caching methodologies in future cellular networks.
Section 3
gives the most important related work.
Section 4
discusses the methodology of the proposed approach.
Section 5
presents the experimental setup and numerical results. Finally,
Section 6
summarizes and concludes the paper.
2. Caching in Future Cellular Networks
We have seen a tremendous growth in cellular networks in the last decade. This growth is mainly driven by the exponentially growing demands of a high data rate by the end users. Various techniques
have been employed to achieve a high data rate as well as spectral efficiency. Spectral efficiency can be achieved by efficiently managing cellular spectrum. These techniques range from control data
split architecture (CDSA) [
], a cloud-radio access network (CRAN) to ultra-dense cellular networks (UDCN) [
] and D2D communication [
]. Almost all of these technologies are considered as enabling technologies for an upcoming fifth-generation (5G) cellular network. Among these, multi-tier architecture and D2D communication are well
thought out as key enablers for providing access to a massive number of users as well as high data rate connectivity. In multi-tier architecture, there is an overlaid macro-cell (MC) comprising
multiple micro-cells (mCs) operating in its coverage area. MC provides an overall coverage service and mobility management, whereas mCs can provide local area services and high data rate
On the other hand, caching in cellular networks has been introduced to exploit storage capacity of diverse network devices. These devices range from user side (user equipment (UE), edge devices) to
the network side (mCs, MC and extended packet core). In the following, we briefly describe how caching can be done on these diverse network devices.
Edge Caching: When an UE generates a request for specific content, first it will search for that content in its own memory and if that content is cached locally (edge), the UE will access that
without any delay. This kind of caching can drastically reduce the backhaul traffic as well as access delay at the cost of large UE memory requirements.
Cluster-Head Caching
: If the requested content is not available at UE itself, it will then ask its peers for that content through D2D communication. One way is to search for the content sequentially in every peer
device. The other way could be to search for an influential user where it is highly probable that content accessed or generated by this influential user will become popular due to its relativity
higher connectivity [
]. D2D communication will enable this kind of caching in the network. This mechanism also reduces backhaul traffic and access delay.
Micro-Cell Caching: If the requested content is not found using D2D communication, then mC will provide that content to the user if it is cached there. mC will utilize a radio access network (RAN)
for delivering that content and not affect the backhaul channel.
Macro-Cell Caching: If the content can not be accessed using the above-mentioned ways, then MC will provide that content by downloading it from EPC (Evolved Packet Core) or from the cloud itself. It
will ensure that content is delivered successfully. This mechanism can cause large access delays and an increase in backhaul traffic.
In the above, we answer the question where to cache?; however, in order to understand how to cache?, we might have two different approaches such as reactive and proactive. In the following,
we describe these approaches in the context of cellular communication.
Reactive Caching: In the reactive caching approach, files are stored in cache if they were repeatedly requested in the past based on the history of the files accessed. In this caching, any content
that remains in high demand for a specific period of time can be cached—for example, viral videos, popular tweets and other highly requested social media content.
Proactive Caching:
It is a mechanism to predict future content that might be requested by the users. Based on previous content and user histories, certain content might become popular in the near future, which will be
cached before they are actually requested, e.g., popular entities in various societies might trigger similar kinds of trends such as top trends on Twitter, etc. Due to the exponential increase of the
demand of cellular data, it is very difficult to maintain the user satisfaction rate. Therefore, the importance of backhaul is increasing dramatically. Various studies [
] have shown that proactive caching in D2D communication can reduce a significant amount of backhaul traffic load, and is a smarter way to exploit D2D cellular networks. Using this approach in D2D
communication in small cellular networks, the popular content is cached at the UE.
In the next section, we focus on the related work that deals with proactive caching using D2D communication.
3. Related Work
In recent years, a large number of studies have investigated caching on the edge in small cell networks. In [
], a proactive caching approach has been proposed. The authors studied two cases and utilized the spatial and social structure of the network. First, based on correlations among users and file
popularity, files are cached during off-peak time. Secondly, influential users are detected and strategic content is cached using D2D communication and social networks. The authors used an
eigenvector centrality measure for detecting influential users of the social network and model the content dissemination as a Dirichlet Process (CRP). In the experimental setup, a preferential
attachment model is used to map the D2D small cellular network. The numerical results show a significant improvement in gain ratio; however, the proposed technique does not consider content of
influential users in a specific community, which makes it less suitable for D2D communication in future cellular networks. The study in [
] introduced a new QoE metric for satisfying a given file request using proactive caching and then they proposed an optimization algorithm (
) to maximize QoE. This algorithm is based on the popularity statistics of the requested files and cache files with highest popularity. However, unlike our proposed work, the algorithm has a
limitation in that it chooses the popular files for caching based on their statistics or previous history. The work in [
] considers UE devices as cooperating nodes on which distributed cache is implemented for efficient downloading. However, this approach has certain drawbacks. Firstly, it does not consider the
distance between D2D devices for cooperation and, secondly, it stores random content without prioritizing high in demand content. Golrezaei et al. [
] introduces a collaborative architecture which uses the distributed storage of popular content. In order to choose and cache random files at the user end, authors compute an average number of D2D
links that can coexist without interference. However, this work also does not consider content of influential users to cache.
A probabilistic approach for optimizing scheduling policies and D2D caching is considered in [
]. The authors first derived approximated uploading probability for uploading gain and then they optimize the scheduling factor and caching distribution to optimize the successful offloading
probability. In [
], authors investigate an optimal caching strategy by formulating an optimization problem and showing the relationship between D2D caching distribution and demand distribution for homogeneous Poisson
Point Process models with different noise levels. With the exponential increase in multimedia traffic, content caching at every edge node might become a challenging task. To overcome this problem,
authors in [
] proposed an idea of peer-to-peer content delivery networks. The authors have exploited the benefits of distrusted fog nodes for content delivery among the networks. In order to efficiently deploy
enabling 5G technologies, authors in [
] proposed an algorithm to reduce total installation costs. The proposed algorithm considers both the hardware and cloud enabled softwarized components (reusable functional blocks) to ensure the
users’ good performance as well as reduction in computation times. Kennington et al. [
] gives a comprehensive overview of modelling and solving optimization problems arising in wireless networks. Moreover, for a detailed review of applications of big data and caching techniques in
mobile computing, the reader is referred to [
Table 1
summarizes caching strategies in comparison with the one proposed in this paper.
Finding social hubs in social networks remains an active research area of Social Network Analysis (SNA). Various measures including Eigenvector Centrality measure, Betweenness measure, Closeness
measure, Degree centrality measure and PageRank algorithm have been used extensively for finding the nodes importance in social graphs. Among these measures, Eigenvector centrality measure is the
most widely used measure for detecting social hubs in social graphs. This measure has been used by three popular methods: PageRank [
], HITS (Hypertext Induced Topic Search) [
], and SALSA (Stochastic Approach for Link Structure Analysis) [
] for retrieving web information.
It is worth mentioning that several prior studies exist that propose various design approaches for proactive caching schemes [
]. However, the novelty of the proposed work stems from the simple yet under-explored idea of considering the content of influential users as popular content. Furthermore, our work uses a new
community detection algorithm for detecting communities in the network and then
influential users are found, one for each community. In the next section, we will explain our proposed methodology for proactive caching in socially aware D2D networks.
4. Proposed Methodology
We exploit the social and spatial structure of social networks and use SNA approaches to find the content of the network that is expected to become viral in the future and proactively cache them in
order to reduce the backhaul load. As the number of active users increases in the network, load on Small Base Station (SBS) increases. Therefore, D2D communication can play a vital role in reducing
the load and increasing the user satisfaction rate. The D2D communication paradigm can be exploited to perform proactive caching and store highly probable content in the users’ caches. In this paper,
we introduce a new proactive caching approach, which uses SNA techniques to address this problem. In the following subsection, after describing the system model and the essentials, we shall present
our proposed algorithm.
4.1. System Model
Assume that there are
number of users who generate number of requests
$R = { r 1 , r 2 , … r n }$
, number of files
$F = { f 1 , f 2 , … f n }$
with lengths
$L = { l 1 , l 2 , … l n }$
, and bit rate
$B r = { b 1 , b 2 , … b n }$
. Note that bit rate of the file represents the transferring rate of the file per unit time. The requests are generated by the users in different time frames
$T = { t 1 , t 2 , … t n }$
. A request
$r 1 ∈ R$
generated by user
is said to be satisfied iff D2D link capacity is higher than the bit rate of the file. The user satisfaction can be formulated as follows:
$B r ≤ l i t i ′ − t i ,$
$l i$
is the size(length) of the file,
$t i ′$
$t i$
is the end and starting time of delivery of file
$B r$
represents the bit rate of file
. Based on Equation (
), the user satisfaction rate
$S r$
can be defined as:
$S r = 1 R ∑ r ∈ R x r ,$
$x r = { 1 , if L t z ′ − t z ≥ B r , 0 , otherwise .$
In order to reduce backhaul traffic load, our main goal is to maximize $S r$.
The users are considered to be connected with other users of their community in a D2D mode of communication. Thus, the request r of the user i for a file f is directed to its nearest user to search
the requested file and will be served if the file exists in the nearby cache. Thus, a good proactive caching strategy can satisfy more requests at the user end, which results in reducing the backhaul
traffic load.
The architecture considered in this work assumes the D2D communication enabled small cellular network. Every community in the network finds a central node that can serve a maximum number of D2D users
with a minimum distance as shown in
Figure 1
. There are a total of
number of users that are connected with each other via D2D communication with a link capacity
$C b$
. The users are also connected with the small cell base station with a link capacity
$C w$
. Assume that the
$∑ C w ≤ ∑ C b$
, then the average backhaul traffic load due to each file of length
$l i$
can be formulated as follows [
$B l ( r ) = 1 R ∑ r ∈ R 1 l i ∑ t = t r t = t ′ λ r ( t ) ,$
$λ r ( t )$
is the backhaul data rate during the content delivery for request
We consider the network as a graph
$G = ( V , E )$
represents set of users
$V = { v 1 , v 2 , v 3 , … v n }$
is set of social links
$( E ⊆ v 2 )$
between the users. For generating the social network for our experimental setup, we used a well known model called preferential attachment proposed by Albert Barabasi [
] to form the small cellular network. This model basically represents degree distribution of the random generated network i.e., how users connect to each other. The core idea behind the preferential
attachment model is that new users prefer to connect to well-connected users over less-connected users proportional to the probability of their degree. The generated network using this model follows
power law distribution with exponent
$α = 3$
. This process is also known as rich-get-richer or cumulative advantage or Matthew effect. The process starts with some initial subgraph. Each new user comes with some initial
links. The probability
$Π ( u )$
of connecting user
with other users
in the network is:
$Π ( u ) = m k u ∑ j k j ,$
$Π ( u )$
represents the probability of connecting user
with other users
in the network.
represents the number of links to which the new user joins the network and
represents the degree of each user in the network. This process results in a network with a power-law distribution with exponent = 3.
Figure 2
shows a network of 256 users generated using a preferential attachment model where nodes represent users while links represent the D2D communication. We see that the network contains very few users
having a large number of links while the rest of the users have very few links. The degree distribution of the generated network on a log–log scale is plotted in
Figure 3
. We can see that the network degree distribution follows power-law distribution.
4.2. Cluster Formation in a Social Network
The first step of our proposed approach is to detect the cluster of each user in the social network as discussed in the previous subsection. For a given graph $G = ( V , E )$, there exist partitions
$C = { c 1 , c 2 , … . c k }$ of V, where $| c i | >$ 0. Our goal is to identify a set of clusters that are densely intra-connected and sparsely interconnected. The process of identifying clusters in
social networks is known as community detection.
Various kinds of community detection algorithms have been proposed. In this paper, we use the CC-GA algorithm [
]. The main reason for using CC-GA is its promising results when compared to other state-of-the-art algorithms such as the information theoretic algorithm (Infomap) [
] and the label prorogation algorithm (LPA) [
] on various kinds of networks including power-law and Forest fire models. This algorithm uses a well-known SNA measure called a Clustering Coefficient (CC) for generating an initial population. CC
gives a better population for a genetic algorithm and results in better clustering of the network. Additionally, the algorithm using CC has shown that it converges very quickly [
]. In the following equation,
$C v i$
represents the value of clustering coefficient of node
$v i$
$C v i = 2 L v i k v i ( k v i − 1 ) ,$
$L v i$
represents the total number of links between the
$k v i$
neighbors. The higher value of
$C v i$
represents the dense neighborhood of a node. The main intuition behind using CC is that if the neighborhood of a node is densely populated, then it is more likely that all of the neighborhood nodes
belong to the same cluster. Furthermore, CC-GA uses Modularity measure [
] as an objective function that is a quality index for clustering of social networks. It can be formally defined as follows:
$Q = ∑ C ∈ c | E ( C ) | m − | E ( C ) | + ∑ C ′ ∈ c | E ( C , C ′ ) | | 2 m | 2 ,$
$| E ( C ) |$
represents the number internal links within the cluster while
$E ( C , C ′ )$
represents the links between clusters.
represents the total number of links in the whole network. Equation (
) can be rewritten into a precise mathematical formulation as given below:
$Q = ∑ C ∈ c | E ( C ) | m − ∑ v ∈ c d e g ( v ) | 2 m | 2 .$
In Equation (
), the first fraction represents the true intra-connectivity of the users, while the second fraction represents the expected number of links within the cluster. The value of modularity ranges between
$− 1$
and 1. Greater value of modularity indicates better clustering of the network while a negative value represents an absence of true clustering in the network.
CC-GA tries to maximize the value of modularity in order to get nearly optimal clustering of the network. It is to be noted that modularity optimization is an NP-hard problem, which can not be solved
with a guaranteed optimal solution. Algorithm 1 presents the pseudocode of CC-GA. Initially, the algorithm requires a graph consisting of vertices (V) and edges (E) as inputs. It also requires a set
of input parameters including: the termination criteria
, population size (
$P n$
), crossover rate
$( P s )$
, mutation rate
$( P m )$
, mutation extension rate
$( α )$
and percentage of population
$( T c )$
to be selected for the next iteration. Initially, the algorithm computes CC of all nodes (steps 2–4) and generates the initial population (step 5). Afterwards, the genetic operators (crossover and
mutation) are performed iteratively until the termination criterion is satisfied (steps 5-25). The algorithm also uses extension of mutation to merge small communities (step 19), and uses modularity
measure as an objective function (steps 9, 15, 21). In the end, the algorithm returns the community structure of the input network. For more details about this algorithm, readers are referred to [
]. The interested readers are referred to [
] for more about genetic algorithms.
Algorithm 1 CC-GA (G)
$P s =$ Crossover rate;
$P m =$ Mutation rate;
$P n =$ Population size;
$α =$ Rate of mutation extension;
$r =$ Termination criteria;
$T c =$ Selection criteria;
$C ∗$ = ${ c 1 , c 2 , … . . c k }$;
procedure CC-GA($G , P n , T c , P s , P m , r ,$$α$)
for $i ← 1 , V$do
$C v i = 2 L v i K v i ( K v i − 1 )$
end for
$P = P 1 , P 2 , P 3 … … P n ←$ Initialize population $( g , P n , C v ) ;$
$O ←$ Apply Crossover ($P , T c , P s ,$);
Q = Compute fitness (P);
$Q ′ ←$ Compute modularity (O);
if $Q ′ > Q$then
P ← update population (O);
end if
O← Apply mutation (O, $P m$);
Q = Compute fitness (P);
$Q ″ ←$ Compute modularity (O);
if $Q ″ > Q$then
P ← update population (O);
end if
O ← Apply mutation extension (O, $α$);
Q = Compute fitness (P);
$Q ′ ′ ′ ←$ Compute modularity (O);
if $Q ′ ′ ′ > Q$then
P ← update population (O);
end if
until (termination-criterion(r))
$C ∗ ←$ Chromosome having highest fitness value
Return $C ∗$
end procedure
4.3. Finding Influential Users and Popular Content
After clustering of the social network, our task is to find the Influential Users (IUs) of the network and then find their content. An influential user can be defined as the most connected and active
user of the network [
]. In order to find the IUs, we use the notion of centrality [
], which is a well-known concept in SNA. The centrality measure provides a way to quantify the importance of nodes with respect to their positions and neighborhood connectivity within a social
network. While various centrality measures exist, eigenvalue centrality is the most successful measure for detecting the social hubs or IUs within a social network. It is a widely used measure for
finding the nodes centrality in the networks [
In order to find the IUs, we exploit the EigenVector Centrality measure. This measure uses the notion of eigenvector and eigenvalue of the adjacency matrix. It quantifies centrality of nodes based on
centrality of its neighbors. Based on this measure, a node having more central neighbors will have greater centrality value. It can be formally defined as follows:
$x v i = 1 λ ∑ N ∈ G a i , j x v i ,$
is a normalization constant and
represents neighbors of node
$v i$
$a i , j$
is 1 if nodes
are directly connected to each other, and 0, otherwise. The greatest value of
$x v i$
represents the highest influence of the user in the network. Equation (
) can be rewritten in a vector form as follows:
where A represents the adjacency matrix of the graph, and
is the vector of the eigenvalue scores. Based on eigenvector centrality measure, if our clustering algorithm returns
clusters of the network, then we find
influential users, one for each cluster in our social network. As we are considering D2D communication, it is necessary to identify at least one influential user in each cluster in order to
communicate with other users directly. In our social network model shown in
Figure 2
, different clusters of the network have been shown in different colors after applying our clustering algorithm. There is one central influential user (nodes with greater sizes) found by using an
eigenvector centrality measure. For example, user 11 is the influential user of its community represented with the green color, and there are 16 members of its community. Overall, the network is
divided into 15 different communities and each community has one influential user. Furthermore, users who join the network earlier have more of a chance to become more connected and are hence more
popular and influential (users 0–20).
Once we know the influential users and their communities, our next step is to determine the popular content of each community. For this purpose, inspired by the pervasive use of online social media,
we assume that the content requested, generated, or accessed by the influential users is going to be the popular content for the same community, and we proactively cache this content based on the
available cache size.
Algorithm 2 shows a pseudocode for computing the centrality values. We initialize an empty matrix $M N$ for storing the centrality value of each node. Then, the algorithm recomputes the centrality
score of each node as a weighted sum for centralities of all nodes in a node neighborhood. In order to normalize the centrality score, $λ$ is used. We set the value of $λ$ to the largest value of $x
u$. The procedure repeats until the values of each node converge.
After describing all the essentials in the following, we shall now present the proposed algorithm.
Algorithm 2 Compute Centrality
procedureCompute Centrality(A)
Initialize matrix $M N$ by 1
for $v ∈ V$do
$x u = 1 λ ∑ u ∈ | A | A u , v x u$
Update $M u$
end for
until values of $x u$ converge
end procedure
Return M
4.4. Proposed Algorithm
Algorithm 3 presents our proposed approach for centrality based D2D proactive caching. Initially, we model the system based on the procedure presented in
Section 4.1
. After designing the model (step 2), we apply the CC-GA algorithm (step 3) for identifying clusters in the network that returns a set
depicting the communities of the network (communities and clustering have the same meaning and we use them interchangeably). Afterwards, we initialize a matrix
(step 4) having the dimensions
$( C × U )$
, where
represents the total number of clusters (
$C i$
row contains the centrality values of cluster
$C i$
) and
$m a x ( | C i | )$
. This matrix stores the centrality value of each node with respect to its community. Note that
$C i$
$| C i |$
values and the remaining indexes which are
$m a x ( | C i | ) − ( | C i | )$
will have 0 entries. We also initialize an
$I U$
matrix to find and store the influential users. The dimensions of this matrix are
$( C × 2 )$
. Then, we call the compute centrality algorithm (step 5), presented in Algorithm 2, which returns a vector of centrality values (
$M ′$
) indexes by node labels that are in the form of integers. As we need matrix
for computing the influential users, we therefore compute it from
$M ′$
(step 6). We compute the primary influential users and secondary influential users from the matrix in each cluster and store them in the matrix (steps 7–10). The reason for finding secondary
influential users is to ensure reliability of the system. In the end, we compute the popular content of influential users as described in
Section 4.3
and load it into the cache (step 12).
We see that the proposed Algorithm 3 relies on one primary influential user from one cluster, which raises important questions such as how many clusters the network has and how many users each
cluster contains. To answer these questions, we present the distribution of community sizes in power-law networks in
Figure 4
. The figure shows the community sizes distribution of the network shown in
Figure 2
. We can see that the community size also follows a power law distribution that implies that there are no such communities who have either a large number of users that cover a great part of the
network and neither there are such communities who have very few numbers of nodes. Taking this into consideration, we also follow network community size distribution to load the content in the cache
of influential users. The user that has a greater community size has more content in the cache and vice versa.
The cellular networks suffer from a network outage problem, which can be in the form of power failure, signal loss or link break down. For maintaining the reliability of the network, we also maintain
a list of secondary influential users as shown in Algorithm 3. In case of network outage problems, the secondary influential users will be used for D2D communication.
Algorithm 3 Proactive Caching
procedure D2D Proactive Caching
$G =$ Model the system
$C =$ CC-GA (G)
initialize $M C × U$ and $I U C × 2$ matrix by 0
$M ′ =$ Compute Centrality(G)
compute matrix M from $M ′$
for $i ← 1 , C$do
$I U 0 = m a x ( M i )$
remove($m a x ( M i )$)
$I U 1$ = $m a x ( M i )$
end for
Compute the popular content of influential users and load into cache
end procedure
5. Experimental Setup and Results
In this section, we present simulation setup and experimental results for evaluating the effectiveness of the proposed approach.
5.1. Simulation Setup
Currently, simulations are performed in Python 3. For community detection, CC-GA is implemented using Python’s library NetworkX to interact with the system model. Initial parameters for the genetic
algorithm are initialized with the values shown in
Table 2
a. Typically, for network clustering problems, GA parameters like population size, termination criteria and mutation probability are set as 200, 50, and 0.15, respectively [
]. However, we performed experiments with different parameter settings to select values that perform best in our case. We set population size, termination criteria and mutation probability values as
500, 50 and 0.2, respectively. Note that the higher values of population size and mutation probability may increase the running time of the algorithm but give better results.
Table 2
b shows a parameter setup for performing simulations. Two parameters,
and D2D cache size
, are considered for performance evaluation as we are interested in evaluating backhaul traffic load with respect to cache size and the number of requests. For simulation purposes, we selected values
of bitrate and size of each file, link capacity, and total number of requests from well-studied and well-cited papers on proactive caching [
]. Moreover, for a total number of files, number of users, and cache size, we selected higher values, in order to evaluate our proposed algorithm in more scalable scenarios.
Table 2
b presents simulation parameter values.
5.2. Simulation Results
In order to evaluate the performance of proposed approach, results are compared with a reactive caching approach, in which files are stored in the cache if they were repeatedly requested in the past
based on the history of files accessed. Initially, a system model is generated using a preferential attachment model [
], and uniform distribution is considered for user requests’ generation. Based on the parameters setup shown in
Table 2
, the proposed algorithm is evaluated for a different number of requests and cache size.
Figure 5
, we plot satisfaction rate with respect to the number of requests generated by the users and the user’s cache size. Our experimental tests are based on number of active users in the network. We
considered two main scenarios, first high load scenario when 100% users are active i.e., 256 users and second, and a low load scenario when 50% of the users are active i.e., 128 users. In
Figure 5
(left), we measured satisfied requests of the proposed approach based on proactive caching with low and high load against reactive caching with respect to the number of requests. We see that the
performance of reactive caching decreases more as compared with proactive caching when the number of requests increases. We see that with 100% requests, the number of satisfied requests is about 48%.
We also note that, with 50% of requests, the proposed approach satisfies 52% requests on average with low load and satisfies 45% requests with high load, whereas these are 36% and 31%, respectively,
in the case of the reactive approach. In
Figure 5
(right), the effect of cache size on satisfaction rate is shown. We can see that with caching 20% of files, 65% of requests are satisfied by the proposed approach while caching 80% files increases
the satisfied requests to 100% with low load, whereas these are 30% and 100%, respectively, in the case of reactive caching.
Figure 6
, we plot backhaul traffic load, with respect to the number of requests generated by the users and also with respect to the user’s cache size. It can be seen from
Figure 6
(left) that, as we increase the number of requests, the backhaul traffic also increases. High load scenarios cause more backhaul traffic to increase as compared to low load scenarios. With the
proposed approach, we see that on 100% requests, the backhaul load is 52% on low load and 65% on high load, whereas these are 75% and 79%, respectively, in the reactive approach. The proposed
approach thus obtains significant reduction in backhaul traffic load as compared with the reactive caching approach. The effect of cache size on backhaul traffic is shown in
Figure 6
(right). We see that with low values of cache size, the backhaul traffic load is significantly lower in the case of the proposed approach as compared to the reactive approach. For instance, with
cache size of 10%, the backhaul traffic load is more than 90% in the case of reactive caching, whereas it is half of this in the proposed approach. As we increase cache size of the influential user,
backhaul traffic decreases. For a low load scenario, using the proposed approach with 60% of the files cached, backhaul traffic load drops down to less than 10% and, when 80% of the files are cached,
there is no backhaul traffic.
Figure 5
Figure 6
also reveal the effect of increasing the number of users. Specifically from plots in
Figure 5
(left) and
Figure 6
(left), we see that as the number of active users increases, the number of satisfied requests decreases, leading to an increase in backhaul traffic. On the other hand, reducing the number of active
users increases the number of satisfied requests and decreases the backhaul traffic load.
6. Conclusions and Future Directions
In this paper, we present a proactive caching approach for D2D in small cellular communication networks. We exploit the social and spatial structure of social networks and represent the D2D cellular
network as a social graph. In order to find the influential users and clusters formation, we use an eigenvector centrality measure and CC-GA clustering algorithms, and cache the content of
influential users of each community. Based on the experimental results, we conclude that caching content of influential users at the user cache can reduce a significant amount of backhaul load and
increase the user satisfaction rate, compared to reactive caching or no caching. The key advantage of the proposed scheme is the simplicity of its popular content identification process, which is a
key high complexity challenge in both reactive and proactive caching.
This work can be extended in the following directions:
• Event-driven influential users detection: When there is a change in the network being detected, it is possible that current influential users may not be available or as influential as they were
before the change. In such scenarios, the system should be able to dynamically detect the influential users through re-computing the centrality scores.
• Joint optimization techniques for D2D enabled caching: Another interesting future work would be to investigate joint optimization of proactive content caching, scheduling techniques, and
interference management.
• Machine learning for network optimization: For optimizing D2D caching, effective machine learning approaches can be exploited for clustering the network users and detection of influential users.
• Mobility aware proactive caching: Most of the work discussed in the literature focuses on cache placement and delivery strategy while the user is stationary. Caching becomes challenging when a
user is moving. For example, when a user with a live streaming session moves from the connectivity of one influential user to another, the user will not be able to stream if the content is not
available at the new influential user cache. In such scenarios, deep learning and big data analytics are essential tools that can be leveraged.
Author Contributions
A.S., S.W.H.S. and A.I. discussed and confirmed the idea. A.S. and S.W.H.S. carried out the experiments and wrote the main paper. H.F., A.N.M., A.I. and J.C. provided the technical guidance for
proactive caching in D2D small cellular networks with proofreading the paper. All authors provided comments on the manuscript at all stages.
The authors would like to thank Saeed Ul Hassan for the insights provided and Tayyaba Liaqat for her help in reviewing the manuscript.
Conflicts of Interest
The authors declare that there is no conflict of interests regarding the publication of this paper.
Figure 5. Proactive vs. Reactive caching showing satisfied requests vs. no. of requests (left) and cache size (right).
Figure 6. Proactive vs. Reactive caching showing backhaul load vs. no. of requests (left) and cache size (right).
Objective References Cache Location Methodology
[9] UE Using SNA and CRP
[10,11] UE Files popularity statistics
[23] UE Random content
[24] UE Probabilistic approach
Cache/Contents Placement [25] UE Optimization
[30] Small Cell Files instantaneous popularity learning and multi-armed bandit (MAB) technique
[31,32] Small Cell Coded placement strategy
[33] Small Cell Problem formulation using cost of files retrieval and bandwidth
This paper UE SNA and influential users content as file popularity index
(a) Genetic Algorithm Parameters
Parameter Description Value
$P n$ Population size 500
$P m$ Mutation rate 0.2
$P s$ Crossover rate 0.2
$T c$ Selection criteria 0.1
$α$ Mutation extension rate 0.02
r Termination criteria 50
(b) Proposed Approach Parameters
Parameter Description Value
K Total of communities 15
N Users 256
$| F |$ Total number of files 1024
$l i$ Length of each file 1 Mbits
$b i$ Bitrate of each file 1 Mbits/s
$C w$ Total small base link capacity 64 Mbits/s
$C b$ Total D2D link capacity 128 Mbits/s
R Maximum number of requests 10,000
D Total D2D cache size 256 Mbits
© 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/
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Said, A.; Shah, S.W.H.; Farooq, H.; Mian, A.N.; Imran, A.; Crowcroft, J. Proactive Caching at the Edge Leveraging Influential User Detection in Cellular D2D Networks. Future Internet 2018, 10, 93.
AMA Style
Said A, Shah SWH, Farooq H, Mian AN, Imran A, Crowcroft J. Proactive Caching at the Edge Leveraging Influential User Detection in Cellular D2D Networks. Future Internet. 2018; 10(10):93. https://
Chicago/Turabian Style
Said, Anwar, Syed Waqas Haider Shah, Hasan Farooq, Adnan Noor Mian, Ali Imran, and Jon Crowcroft. 2018. "Proactive Caching at the Edge Leveraging Influential User Detection in Cellular D2D Networks"
Future Internet 10, no. 10: 93. https://doi.org/10.3390/fi10100093
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details
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Regression VI - Polynomial Features - The Stubborn Coder
Regression VI – Polynomial Features
It’s time to talk about feature engineering. Using just the tools of linear regression, we can greatly increase the complexity of our models by creating features from our underlying data. A feature
is a new predictor that we create by applying some function to our existing set of predictors. We can think of it as manipulating the data to dig out underlying patterns.
The best example are polynomial features. Suppose we have a single predictor X which is spaced evenly between 0 and 10, and a response Y. Say we plot Y against X and it looks like this
Lets try and do a simple linear regression against Y.
import numpy as np
from sklearn.linear_model import LinearRegression
from matplotlib import pyplot as plt
linear_model = LinearRegression().fit([[x] for x in X],Y)
linear_predictions = linear_model.predict([[x] for x in X])
plt.plot(X, linear_predictions)
This will create something like this
Not very good at all. So, let’s try adding a new feature. We will define this feature as the square of X, like so
square_model = LinearRegression().fit([[x, x**2] for x in X], Y)
square_predictions = square_model.predict([[x, x**2]])
plt.plot(X, square_predictions)
This will give you something that looks a bit better
Although our regression is still linear, now we are fitting the closest quadratic rather than the closest straight line. This is also called polynomial regression. But I think it makes more sense to
think of it as a form of feature engineering.
Now, let’s add another feature, the cube of X.
cubic_model = LinearRegression().fit([[x, x**2, x**3] for x in X], Y)
cubic_predictions = cubic_model.predict([[x, x**2, x**3]])
plt.plot(X, cubic_predictions)
Now we’re getting there. If we add the fourth powers of X as a feature, we will get an even better looking fit
We are not just limited to polynomial features, as we shall see. However we shouldn’t bother with linear features, as any linear term is already captured in the regression itself. When we fit, we
already find the best possible linear combination of the predictors . So our features need to be non-linear in our existing predictors. A big part of building and fitting models is finding useful
features in our data like this. | {"url":"https://www.stubbornprogrammer.com/2021/07/04/regression-vi-ipolynomials/","timestamp":"2024-11-03T23:02:28Z","content_type":"text/html","content_length":"86448","record_id":"<urn:uuid:30d426c5-398d-4d01-82bf-e8b842793e6a>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00458.warc.gz"} |
NumPy: Create a 3x3x3 array filled with arbitrary values - w3resource
NumPy: Create a 3x3x3 array filled with arbitrary values
NumPy: Basic Exercise-31 with Solution
Write a NumPy program to create a 3x3x3 array filled with arbitrary values.
This problem involves writing a NumPy program to generate a 3x3x3 array filled with arbitrary values. The task requires utilizing NumPy's array creation functionalities to efficiently construct the
multi-dimensional array with arbitrary values. By specifying the desired shape of the array and providing arbitrary values, the program generates a 3x3x3 array suitable for various computational and
analytical tasks.
Sample Solution :
Python Code :
# Importing the NumPy library with an alias 'np'
import numpy as np
# Creating a 3x3x3 array filled with random numbers between 0 and 1 using np.random.random()
x = np.random.random((3, 3, 3))
# Printing the randomly generated 3x3x3 array 'x'
[[[ 0.51919099 0.31268732 0.58506582]
[ 0.12730206 0.30556451 0.55981097]
[ 0.92910493 0.73947119 0.14252086]]
[[ 0.96159407 0.21341612 0.6814465 ]
[ 0.71884351 0.01271011 0.98812225]
[ 0.47553515 0.04584955 0.59425412]]
[[ 0.44064227 0.26612916 0.58861619]
[ 0.37935141 0.0640071 0.53717733]
[ 0.77986385 0.0771148 0.92183522]]]
The above code creates a 3D array of random numbers between 0 and 1 with the shape (3, 3, 3) and prints the resulting array.
Here ‘np.random.random((3, 3, 3))’ creates a 3D array with the shape (3, 3, 3), filled with random numbers between 0 and 1. The input to np.random.random() is a tuple representing the desired shape
of the output array.
Finally print(x) prints the generated 3D array to the console.
Visual Presentation:
Python-Numpy Code Editor:
Previous: NumPy program to create an 4x4 matrix in which 0 and 1 are staggered, with zeros on the main diagonal.
Next: NumPy program to compute sum of all elements, sum of each column and sum of each row of an given array.
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Test your Programming skills with w3resource's quiz.
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For what values of a do the roots x1 and x2 of the equation x... | Filo
Question asked by Filo student
For what values of a do the roots and of the equation satisfy the relation ? Find the roots.
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Question Text For what values of a do the roots and of the equation satisfy the relation ? Find the roots.
Updated On May 20, 2023
Topic Algebra
Subject Mathematics
Class Class 12
Answer Type Video solution: 3
Upvotes 263
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Math Games
• KS1
Key Stage 1 Standards
Top Mathematicians
□ KS1.NS.1.1
Understanding Number and Number Notation
• Pupils should be able to:
- count, read, write and order whole numbers, initially to 10, progressing to at least 1,000;
- understand the empty set and the conservation of number;
- understand that the place of the digit indicates its value;
- make a sensible estimate of a small number of objects and begin to approximate to the nearest 10 or 100;
- recognise and use simple everyday fractions.
□ KS1.NS.1.2
Patterns, Relationships and Sequences in Number
• Pupils should be able to:
- copy, continue and devise repeating patterns;
- explore patterns in number tables;
- understand the commutative property of addition and the relationship between addition and subtraction;
- understand the use of a symbol to stand for an unknown number;
- understand and use simple function machines.
□ KS1.NS.1.3
Operations and their Applications
• Pupils should be able to:
- understand the operations of addition, subtraction, multiplication and division (without remainders) and use them to solve problems;
- know addition and subtraction facts to 20 and the majority of multiplication facts up to 10 x 10;
- develop strategies for adding and subtracting mentally up to the addition of two two-digit numbers within 100.
□ KS1.NS.1.4
• Pupils should be able to:
- recognise coins and use them in simple contexts;
- add and subtract money up to £10, use the conventional way of recording money, and use these skills to solve problems;
- talk about the value of money and ways in which it could be spent, saved and kept safe;
- talk about what money is and alternatives for paying;
- decide how to spend money.
□ KS1.GMD.1.1
Exploration of Shape
• Pupils should be able to:
- sort 2-D and 3-D shapes in different ways;
- make constructions, pictures and patterns using 2-D and 3-D shapes;
- name and describe 2-D and 3-D shapes; recognise reflective symmetry;
- explore simple tessellation through practical activities.
□ KS1.GMD.1.2
Position, Movement and Direction
• Pupils should be able to:
- use prepositions to state position;
- understand angle as a measure of turn; understand and give instructions for turning through right angles;
- recognise right-angled corners in 2-D and 3-D shapes;
- know the four points of the compass;
- use programmable devices to explore movement and direction.
□ KS1.ID.1.1
Collecting, Representing and Interpreting Data
• Pupils should be able to:
- sort and classify objects for one or two criteria and represent results using Venn, Carroll and Tree diagrams;
- collect data, record and present it using real objects, drawings, tables, mapping diagrams, simple graphs and ICT software;
- discuss and interpret the data;
- extract information from a range of charts, diagrams and tables;
- enter and access information using a database.
□ KS1.N.1.1
• Pupils should be able to:
- understand and use the language associated with length, 'weight', capacity, area and time;
- use non-standard units to measure and recognise the need for standard units;
- know and use the most commonly used units to measure in purposeful contexts;
- make estimates using arbitrary and standard units;
- choose and use simple measuring instruments, reading and interpreting them with reasonable accuracy;
- sequence everyday events; know the days of the week, months of the year and seasons; explore calendar patterns;
- recognise times on the analogue clock and digital displays;
- understand the conservation of measures. | {"url":"https://nir.mathgames.com/standards/keystage1","timestamp":"2024-11-08T18:54:54Z","content_type":"text/html","content_length":"532961","record_id":"<urn:uuid:2cca7929-f714-482b-aaba-2113951b190d>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00447.warc.gz"} |
Mastering the Python Square Function – A Comprehensive Guide
Introduction to the Python Square Function
In mathematics, the square function is a fundamental operation that calculates the square of a given number. This function is widely used in various fields such as physics, engineering, and data
analysis. Similarly, in Python, the square function allows us to calculate the square of a given value with ease.
Basic Usage of the Square Function
The syntax for the square function in Python is simple and straightforward. To square a number, we use the exponentiation operator, “**”, with the number we want to square as the base and 2 as the
exponent. Here’s an example:
result = number ** 2
Here, “number” represents the value we want to square, and “result” stores the squared value.
When using the square function, we have multiple ways to pass arguments:
Using a Single Integer Argument
The square function in Python can accept a single integer argument. For example:
# Using a single integer argument result1 = square(5) result2 = square(10)
result1 will store the square of 5, which is 25, while result2 will store the square of 10, which is 100.
Using Variables as Arguments
Instead of using a single integer argument, we can also pass variables to the square function. This allows us to calculate the square value dynamically. Here’s an example:
# Using variables as arguments num1 = 7 result = square(num1)
In this example, num1 is a variable that holds the value 7. The square function is called with num1 as the argument, which calculates the square of num1 and stores it in the result variable.
Using Expressions as Arguments
In addition to using single integer arguments or variables, we can also use expressions as arguments for the square function. This gives us the flexibility to perform various calculations. For
# Using expressions as arguments result = square(2 + 3)
In this example, the square function is called with the expression 2 + 3 as the argument. The result will be the square of the sum of 2 and 3, which is 25.
Common Mistakes and Errors
When using the square function in Python, it’s important to be aware of common mistakes and errors that can occur. Let’s explore a couple of them:
Understanding Common Errors when Using the Square Function
1. Syntax errors:
Syntax errors occur when there is a mistake in the structure or format of the code. These errors can prevent the square function from running correctly. For example:
result = square(# number ** 2)
In this case, the opening parenthesis is missing the corresponding closing parenthesis, resulting in a syntax error.
2. Type errors:
Type errors occur when the square function is called with an argument of an incompatible data type. For instance:
result = square("5")
Here, the square function is called with a string argument instead of an integer. Since the square function operates on numeric values, it will raise a type error in this case.
Tips to Avoid Mistakes when Using the Square Function
To prevent common mistakes and errors when using the square function, consider the following tips:
1. Double-check the syntax: Ensure that you have correctly formatted the square function calls by verifying the placement of parentheses, commas, and other syntax elements.
2. Validate the data type: Make sure to pass arguments of the appropriate data type to the square function. If necessary, convert the input to the appropriate data type using casting.
Square Function with Negative Numbers
When passing negative numbers to the square function in Python, it’s important to understand the result and handle it properly. Here’s what you need to know:
Explanation of the Result when Passing Negative Numbers to the Square Function
The square function in Python calculates the square of a given number regardless of its sign. When a negative number is squared, the result is always positive. For example:
result = square(-5)
In this case, the result will be 25, which is the square of -5.
Handling the Square Function with Negative Numbers
While the square function treats negative numbers as expected, it’s crucial to handle the results appropriately based on the context of your program. Here are a few scenarios you may encounter:
1. Calculating the square root of a negative number: Since the square root of a negative number is not a real number, an error will occur. It’s essential to handle such cases by utilizing conditional
statements or exception handling.
2. Working with absolute values: In some scenarios, you may need to work with the magnitude of a number without considering its sign. Using the “abs” function before squaring the number can help
achieve this. For example:
result = square(abs(-5))
In this case, abs(-5) evaluates to 5, and the square function calculates the square of 5, resulting in 25.
Advanced Usage of the Square Function
Besides basic usage, the square function in Python can also handle more complex scenarios. Let’s explore a few:
Using the Square Function with Floating-Point Numbers
The square function can work with floating-point numbers just as effectively as it does with integers. Consider the following:
Using the Square Function with Float Arguments
With the square function, we can pass floating-point arguments and obtain accurate results. For example:
result = square(2.5)
In this case, the square function will calculate the square of 2.5, resulting in 6.25.
Casting Float Values to Integers before Using the Square Function
When necessary, we can cast a floating-point value to an integer before using the square function. Here’s an example:
result = square(int(2.5))
In this case, the square function will convert the floating-point value 2.5 to an integer, resulting in 2. The square of 2 is 4.
Employing the Square Function in Complex Mathematical Operations
The square function can be utilized as part of more complex mathematical operations. For example:
result = square(x + y)
In this case, the square function is called with the expression x + y as the argument. The result will be the square of the sum of x and y.
Performance Considerations
While the square function is simple and efficient, there are a few performance considerations to keep in mind:
Understanding the Time Complexity of the Square Function
The time complexity of the square function is constant, denoted as O(1). This means that the execution time does not depend on the size of the input value. Regardless of whether the input is small or
large, the square function will take the same amount of time to calculate the result.
Identifying Potential Performance Issues when Using the Square Function
Although the square function itself has excellent performance, there are certain scenarios where performance may become a concern:
1. Large input values: When working with extremely large input values, the computational resources required to calculate the square may increase. It’s important to consider the limitations of the
system and optimize the code accordingly.
2. Iterating through a large collection: If you need to square multiple elements in a large collection, the iteration process may introduce performance issues. In such cases, consider using
vectorized operations or alternative approaches that can process the data more efficiently.
In conclusion, the Python square function provides a straightforward way to calculate the square of a given value. It can be used with both integers and floating-point numbers, and it handles
negative numbers as expected. By understanding the common mistakes, handling negative numbers properly, and exploring advanced usage, you can leverage the square function to efficiently perform
various mathematical operations. Remember to consider performance considerations when working with large input values or iterating through collections. Now it’s time to practice and explore further! | {"url":"https://skillapp.co/blog/mastering-the-python-square-function-a-comprehensive-guide/","timestamp":"2024-11-11T03:43:50Z","content_type":"text/html","content_length":"113673","record_id":"<urn:uuid:b80f12d2-db8e-43d2-9434-b0520ab5ff29>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00073.warc.gz"} |
Missing Multipliers
What is the smallest number of answers you need to reveal in order to work out the missing headers?
The multiplication square below has had all its headings and answers hidden. All of the headings are numbers from 2 to 12.
By clicking on some of the cells to reveal the answers, can you work out what the headings must be?
Once you've worked out what each heading must be, drag the purple numbers to the appropriate spaces. When you think you have cracked it, click "Show the solution" to see if you are right.
What is the smallest number of answers you need to reveal in order to work out the missing headers?
Here are some more Missing Multiplier challenges you might like to try:
If you enjoyed this problem you may also enjoy Gabriel's Problem
Getting Started
If you are having difficulty getting started, take a look at
Mystery Matrix
Student Solutions
Joey from St Theresa's Catholic College in Australia and Aditya from St Columbas College in the UK sent in completed grids. This is Aditya's work (click here here to see a bigger version):
Janeen from Westridge School
Rhihito from the American School in
in America only needed to
Japan sent in this result:
reveal six answers:
Keshav from Colchester Royal Grammar School and Anh Minh from the British Vietnamese International School, Hanoi in Vietnam described a general method revealing seven squares. This is Keshav's work:
You have to reveal at least seven squares to find all the headings.
Firstly, reveal any square you want. Then, reveal any other square next to it (horizontally or vertically). Using your knowledge of factors, you can then find the heading of that row.
Then, reveal a square next to either of the two that are already revealed, and make sure that square doesn't belong to the heading that you've already figured out.
Now you have two headings. You can use these two headings to find all the others, still using your knowledge of factors, using just the row and column you've formed.
Could Janeen's method be described in this way?
Tuấn Vé from the British Vietnamese International School in Hanoi described a method for larger grids. How could you use what we've seen here to make this method more efficient?
You need to know at least two values of each row to solve the problem, you will need to open it in two vertical lines. So you will need 2$y$ if the table is $y \times y$.
So for 4$\times$4 you need to open 8 values; for 6$\times$6 you need 12;for 8$\times$8 you need 16 and so on.
That's not quite right, since Keshav's method for a 4 by 4 grid uses 7 values, and Janeen's uses 6.
In fact, one in each row and one in each column means $y\times y-1$ because the row and column will cross.
As Janeen noticed, you don't actually need the sqaure in the row and column, because you can find both of those headings from the other rows. So in fact you need to reveal $y\times y-2$ reveals.
Teachers' Resources
Why do this problem?
This problem offers an opportunity for students to consider common factors while gaining fluency in multiplication facts. The interactivity engages students' curiosity and perseverance by challenging
them to complete the grid using a minimum number of 'reveals'.
Multiplication tables are often presented with row and column headings filled in, with students challenged to fill in the products. This task inverts that concept, as students can reveal chosen
products and work out possibilities for the headings.
Possible approach
If computers or tablets are available, students could work in pairs using the interactivity. Students could try a few examples to get the idea, and then work on the challenge of trying to find the
grid headings by revealing as few cells as possible. Once they have developed some strategies, they could try the larger grids that include bigger numbers.
If computers are not available, the task can be recreated by asking each student to create a multiplication grid of their own, and then draw a blank grid for their partner. As in the interactivity,
the challenge is to ask for as few entries as possible from the grid in order to work out what the headers are.
Once students have had plenty of time to develop strategies, the key questions below provide a good basis for a plenary discussion, after which students could revisit the interactivity to test out
each other's suggestions.
Key questions
Which numbers, when revealed, make it straightforward to work out the row and column headings?
Which numbers give lots of possibilities for row and column headings?
Is there a strategy for working out the row and column headers in fewer than 10 reveals?
Can you find a way to work out the row and column headers using only 6 reveals?
Possible support
Mystery Matrix works in the same way, but some helpful cells have already been revealed.
Possible extension
There are natural extensions within the problem - working on the 10 by 10 grid provides a real mental workout!
Gabriel's Problem
Product Sudoku
would make nice follow-up activities.
Finding Factors
has a very similar interactivity but the context is factorisation of quadratic expressions. | {"url":"http://nrich.maths.org/problems/missing-multipliers","timestamp":"2024-11-07T04:20:46Z","content_type":"text/html","content_length":"46105","record_id":"<urn:uuid:32efa8ab-b38e-403c-a25a-d41e18fca06b>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00279.warc.gz"} |
How to Calculate Standardized Residuals in Excel | Online Tutorials Library List | Tutoraspire.com
How to Calculate Standardized Residuals in Excel
by Tutor Aspire
A residual is the difference between an observed value and a predicted value in a regression model.
It is calculated as:
Residual = Observed value – Predicted value
If we plot the observed values and overlay the fitted regression line, the residuals for each observation would be the vertical distance between the observation and the regression line:
One type of residual we often use to identify outliers in a regression model is known as a standardized residual.
It is calculated as:
r[i] = e[i] / s(e[i]) = e[i] / RSE√1-h[ii]
• e[i]: The i^th residual
• RSE: The residual standard error of the model
• h[ii]: The leverage of the i^th observation
In practice, we often consider any standardized residual with an absolute value greater than 3 to be an outlier.
This tutorial provides a step-by-step example of how to calculate standardized residuals in Excel.
Step 1: Enter the Data
First, we’ll enter the values for a small dataset into Excel:
Step 2: Calculate the Residuals
Next, we’ll go to the Data tab along the top ribbon and click Data Analysis within the Analysis group:
If you haven’t installed this Add-in already, check out this tutorial on how to do so. It’s easy to install and completely free.
Once you’ve clicked Data Analysis, click the option that says Regression and then click OK. In the new window that pops up, fill in the following information and click OK:
The residual for each observation will appear in the output:
Copy and paste these residuals in a new column next to the original data:
Step 3: Calculate the Leverage
Next, we need to calculate the leverage of each observation.
The following image shows how to do so:
Here are the formulas used in the various cells:
• B14: =COUNT(B2:B13)
• B15: =AVERAGE(B2:B13)
• B16: =DEVSQ(B2:B13)
• E2: =1/$B$14+(B2-$B$15)^2/$B$16
Step 4: Calculate the Standardized Residuals
Lastly, we can calculate the standardized residuals using the formula:
r[i] = e[i] / RSE√1-h[ii]
The RSE for the model can be found in the model output from earlier. It turns out to be 4.44:
Thus, we can use the following formula to calculate the standardized residual for each observation:
From the results we can see that none of the standardized residuals exceed an absolute value of 3. Thus, none of the observations appear to be outliers.
It’s worth noting in some cases that researchers consider observations with standardized residuals that exceed an absolute value of 2 to be considered outliers.
It’s up to you to decide whether to use an absolute value of 2 or 3 as the threshold for outliers, depending on the specific problem you’re working on.
Additional Resources
What Are Residuals?
What Are Standardized Residuals?
Introduction to Multiple Linear Regression
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Do z scores follow a standard normal distribution?
Do z scores follow a standard normal distribution?
The histogram below illustrates this: if a variable is roughly normally distributed, z-scores will roughly follow a standard normal distribution. For z-scores, it always holds (by definition) that a
score of 1.5 means “1.5 standard deviations higher than average”.
What is z-score chart?
A Z-Score chart, often called a Z-Table, is used to find the area under a normal curve, or bell curve, for a binomial distribution. The Z score itself is a statistical measurement of the number of
standard variations from the mean of a normal distribution.
What is the z-score for a 95% confidence interval?
The value of z* for a confidence level of 95% is 1.96. After putting the value of z*, the population standard deviation, and the sample size into the equation, a margin of error of 3.92 is found. The
formulas for the confidence interval and margin of error can be combined into one formula.
How do you calculate the normal curve?
Set your cursor to find the range of where you want to find the area under the normal curved graph. Press the “Left Arrow” button on your calculator until you reach the left limit. Press the “Enter”
button to set the marker for the left limit. Scroll to the right limit using the “Right Arrow” on your calculator until you reach the right limit.
How do you calculate the area of a standard normal curve?
The area under the curve is 1.00 or 100 percent. The easiest way to determine that your data are normally distributed is to use a statistical software program such as SAS or Minitab and conduct the
Anderson Darling Test of Normality. Given that your data is normal, you can calculate z-score.
What is the formula for standard normal distribution?
Standard Normal Distribution is calculated using the formula given below. Z = (X – μ) / σ. Standard Normal Distribution (Z) = (75.8 – 60.2) / 15.95. Standard Normal Distribution (Z) = 15.6 / 15.95.
What is z score in normal distribution?
A z-score is also known as a standard score and it can be placed on a normal distribution curve. Z-scores range from -3 standard deviations (which would fall to the far left of the normal
distribution curve) up to +3 standard deviations (which would fall to the far right of the normal distribution curve). | {"url":"https://www.handlebar-online.com/articles/do-z-scores-follow-a-standard-normal-distribution/","timestamp":"2024-11-13T23:02:34Z","content_type":"text/html","content_length":"43608","record_id":"<urn:uuid:4b13bcc3-7d4b-43ba-a2dc-f7dbb138260e>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00590.warc.gz"} |
Non-Abelian Berry transport, spin coherent states and Majorana points
We consider the adiabatic evolution of Kramers-degenerate pairs of spin states in a half-integer spin molecular magnet as the molecule is slowly rotated. To reveal the full details of the quantum
evolution, we use Majoranas parametrization of a general state in the (2j + 1)-dimensional Hilbert space in terms of 2j Majorana points. We show that the intricate motion of the Majorana points may
be described by a classical Hamiltonian which is of the same form, but of quite different origin, as that which appears in the spin-coherent-state path integral. As an illustration, we consider
molecular magnets of the j = 9/2 Mn4 family and compute the frequency with which the magnetization varies. This frequency is generally different from the frequency of the rotation.
ASJC Scopus subject areas
• Statistical and Nonlinear Physics
• Statistics and Probability
• Modeling and Simulation
• Mathematical Physics
• General Physics and Astronomy
Dive into the research topics of 'Non-Abelian Berry transport, spin coherent states and Majorana points'. Together they form a unique fingerprint. | {"url":"https://experts.illinois.edu/en/publications/non-abelian-berry-transport-spin-coherent-states-and-majorana-poi","timestamp":"2024-11-03T01:27:20Z","content_type":"text/html","content_length":"55508","record_id":"<urn:uuid:d9539519-8b9a-4fd7-9ae3-9f6209889541>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00038.warc.gz"} |
Procedure 1 – Using Geometric Dimensions to Determine Carrying Capacity (One 45-minute class period)
In this part of the lesson, students will work together to determine if specific populations would be able to survive in a given area.
• Ask the students to brainstorm about which factors may contribute to the carrying capacity of a population. Acceptable answers may include: available resources (food, water, etc.), competition
for resources, # of predators, physical space, etc. Explain that the students will have to find a sustainable habitat for specific groups of animals, based primarily on the geometric dimensions
of a given area.
• Distribute a copy of Handout 1 to each student.
• Refer the students to the groups of species (from the handout) that will require a new sustainable habitat.
• Given that the students will be working with the dimension of acres, demonstrate on the board how to convert the measurements of a given area to acreage (as outlined on Handout #1).
• Divide the class into 4-6 groups (of about 6 students each)
• Have students follow the directions on Handout 1 to calculate the range requirements for each species (given 25 breeding pairs for each species, as per the directions) and complete the chart.
□ Note: There are six different animal species which require a sustainable habitat. Depending upon the amount of time that you would like to devote to this part of the lesson, you may choose to
shorten the activity by having each group do the calculations for only one or two of the given species. When complete, each group can provide the acreage requirements for each assigned
species to the rest of the class.
• Provide each group of students with either:
□ A map of a local or national park
□ The chance to choose a local park area that they believe would be a sufficiently-sized environment for the provided populations. Then, look up the dimensions of the area/acreage online. (Ex/
The Jamaica Bay Wildlife Refuge in Queens, NY is 9155 acres)
Note: If it is not feasible to obtain or locate a local map, then any park map (or even the dimensions of an imaginary park) will suffice.
• Have each group calculate or look up the acreage for their given/chosen area.
• Each group should then compare the calculated acreage of their park to range requirements for each species. Challenge the students to choose sub-areas within the park that may be best suited and
sufficiently sized for each species (based upon the terrain and other needs, as indicated on Handout #1). For a more thorough analysis of habitat sustainability, any information obtained about
the park can be used to complement the students’ geometrical data calculations.
• On a separate piece of paper, have the students complete Questions 1-6, based upon each groups’ park dimensions, calculations & overall analysis.
• When there are 10 minutes remaining in class, have each group share the details of their findings with the rest of the class. Encourage students to elaborate on other factors that will influence
the carrying capacity of the park that they have chosen.
• In the interest of time, some questions may be omitted or assigned for homework.
Procedure 2- Using Technology to Graphically Demonstrate & Analyze Population Mechanics
(Two – Three 45-Minute Class Periods)
Note: Although the following two graphing activities (a & b) are related, each may be completed independently and can be done in any order (or one may be omitted).
Procedure 2a – Graphing Population Growth
In this part of the lesson, students will be presented with an opportunity to graphically demonstrate population growth, based upon the Logistic Model (see Background section for full details). Given
a basic understanding of carrying capacity, students will analyze and make predictions through the graphical depiction of a rabbit population’s growth, as it approaches carrying capacity.
Differentiated (Technology) Learning Options:
Option 1– (Length= Two 45-minute class periods)
(For Students to design, implement & manipulate a Population Growth Model- Excel Spreadsheet)
• Each group of students (2-3 students per group) should have a computer setup with MS Excel
• First, ensure that students have a basic understanding of carrying capacity and the Logistic Model of Population growth (see Background). Then provide an overview of the activity’s goal: to
design & create a working model of population growth of a rabbit population using the discussed Logistic Model.
• Provide each group with a copy of “Handout A” and read through the guidelines for the proper completion of their Excel Model of Population Growth.
• Provide each group with either a digital or a “hard-copy” of Handout 2 (Excel Tutorial), which will provide them with detailed assistance with some Excel tools that will be useful for creating
their population growth model (especially Scroll Bars).
• Instruct the students to complete the Prediction Section of “Handout A”, before beginning the design for the population growth model.
• The design of the model should take at least the entirety of the 1^st lesson period. “Finishing touches” may be applied during the beginning of the 2^nd lesson period.
• Once the students have created their Population Growth Model, instruct them to continue with the Analysis Section of “Handout A” (allowing them to manipulate and analyze their model) and complete
Questions 1-6.
• Students should be able to submit their findings on Handout A by the end of class
Option 2– (Length= Two 45-Minute Class Periods)
(For Students to follow directions to implement & manipulate a Population Growth Model- Excel Spreadsheet)
• The instruction of Option 2 is almost identical to Option 1, with one exception:
□ You can opt to provide step-by-step support to your students (perhaps through interdisciplinary collaboration with the computer/technology department) guiding them through the spreadsheet
creation process.
□ Within Excel, the formula that will model the Logistic equation can be expressed as:
☆ =B2+B2*Birthrate*((Capacity-B2)/Capacity)
○ where B2 represents the population size of the preceding month (to be sequentially compounded within each subsequent month)
○ Birthrate & Capacity represent the cell “name” for each respective parameter
□ Once the students have completed their spreadsheet models, they will complete the rest of the lesson using “Handout A” as described above within Option 1.
Option 3(Length= One 45-Minute Class Period)
(For students to (only) manipulate a Population Growth Model- Excel Spreadsheet)
• The instruction of Option 3 omits the spreadsheet design section of the lesson.
• Instead, instruct the students to open the supplied “Logistic Model” Spreadsheet using MS Excel. They will then manipulate the supplied model in order to complete the questions on “Handout A3”.
Procedure 2b- Using Technology to Graphically Analyze the Predator-Prey Relationship (Length= One 45-Minute Class Period)
In this part of the lesson, students will be presented an opportunity to graphically analyze a model of the predator-prey relationship. Since the Lotka-Volterra Model is based upon a higher-order
pair of inter-dependent equations, advanced programming knowledge is required in order to execute proper spreadsheet design, the students will instead manipulate and analyze a supplied spreadsheet
model of the Lotka-Volterra Equations.
• Each group of students (2-3 students per group) should have a computer setup with MS Excel.
• First, ensure that students have a basic understanding of the predator/prey relationship, as exemplified by the Lotka-Volterra Equations (see Background). Then provide an overview of the
activity’s goal: to manipulate and analyze a working model of the predator/prey relationship .
• Provide each group with a copy of “Handout B” and read through the activity directions, explaining how to manipulate and analyze the Predator-Prey Relationship Model.
• Open Spreadsheet 2 (Predator-Prey Relationship) using MS Excel.
• Instruct the students to complete the Prediction Section of “Handout B”, before manipulating the Lotka-Volterra model.
• Once predictions have been made, the students must complete Questions 1-6 on “Handout B” as indicated.
• Students should be able to submit their findings on “Handout B” by the end of class | {"url":"https://www.teacherstryscience.org/ko/node/1147","timestamp":"2024-11-01T20:33:14Z","content_type":"application/xhtml+xml","content_length":"72258","record_id":"<urn:uuid:54fb4f83-d634-4cad-8ae5-c66a02435a5f>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00865.warc.gz"} |
Brain Teaser Geography Puzzle: Find The Name Of The Country Based On The Outline Given In This Picture - Minh Khang Cente
Brain Teaser Geography Puzzle: Find The Name Of The Country Based On The Outline Given In This Picture
The impact of brain teasers
Brain games effectively stimulate the brain and improve cognitive function, making them a valuable tool for those seeking to keep their minds sharp and active. Participating in brain games improves
thinking skills, resulting in improved cognitive abilities, faster thinking speed, and higher concentration levels.
Additionally, playing brain games can help you boost your self-confidence and reduce stress levels as you solve challenging problems and achieve success. By training your brain regularly, you can
also improve your overall brain health and reduce your risk of cognitive decline later in life. So whether you’re looking to improve your mental performance or are just looking for a fun and engaging
way to exercise your mind, brain games are a great choice.
You are watching: Brain Teaser Geography Puzzle: Find The Name Of The Country Based On The Outline Given In This Picture
Brain teasers using pictures
Picture Brainteaser is a visual puzzle that can be used for a variety of purposes, including:
• Entertainment: Picture brainteasers are a fun and engaging activity for people of all ages. They can be used at parties, social gatherings, or even as a stand-alone activity to pass the time.
• Educational purposes: Picture brainteasers can be used in schools or other educational settings to help students develop critical thinking skills, visual processing and problem-solving skills.
• Cognitive Development: Picture brainteasers can be used to stimulate children’s cognitive development and help them improve their observation skills, memory and attention to detail.
• Therapeutic Purpose: Picture brainteasers can be used as a form of therapy in the rehabilitation of people with brain injuries, strokes, or other cognitive impairments. They can help retrain the
brain and improve cognitive function.
• Recruiting Tool: Employers can use picture brainteasers during the hiring process to assess candidates’ problem-solving abilities, attention to detail, and critical thinking skills.
Overall, picture brainteasers are a versatile tool that can be used for a variety of purposes, including entertainment, education, cognitive development, therapy, and recruitment.
Brainteaser Geography Puzzle: Find the name of the country based on the outline given in the picture
Brain teasers are generally considered healthy because they trigger your cognitive thinking and allow your brain to think outside the box. But be sure not to tire your eyes! We’ve attached the
question, followed by this text, in the image below. Scroll down for a second and see if you can solve the puzzle.
We know that adding a time limit might add more excitement to your challenge, but it does. Take a closer look at the image we proved below; now try to guess the answer in a few seconds. Time starts
now: 1, 2, 3…
See more : Brain Teaser: If you have Eagle Eyes Find the Word Crane in 13 Secs
Tick tock, tick tock, tick tock, it’s time!
If you can find the right answer, give yourself a pat on the back.
Wait, how do you know if this is the right answer?
The visuals you glean from the information mentioned here may create different perceptions in your brain. This may happen if you have a different answer in mind. After all, it’s common for us to
perceive different meanings of simple images. This means you have a higher IQ level.
Brainteaser Geography Puzzle: Find the name of the country based on the outline given in the picture
Brainteasers have unexpected benefits for those who participate in the challenge, so you can swipe down to participate in the challenge to find out. Netizens are increasingly wondering what the
correct answer to this picture is. You may definitely be confused. But take a deep breath and try again to find the right answer.
Since solving this brainteaser is quite challenging, we are here to reveal the solution to you! Scroll down to reveal the answer! Don’t be discouraged if you can’t come up with the right answer.
Regular practice with our brainteasers blog will improve your observation and problem-solving skills.
Brain Teasers IQ Test Math Test: 89-49÷7+2+33÷3=?
Enter the fun world of brain teasers IQ test math quizzes using the following equation: 89 – 49 ÷ 7 + 2 + 33 ÷ 3. Your challenge is to carefully apply the order of operations and find the final
See more : Brain Test: If you have Hawk Eyes Find the word Shoe In 15 Secs
To solve this equation, follow the order of operations. First, divide 49 by 7 to get 7. Then, add 2 and 7 to get 9. Next, divide 33 by 3 to get 11. Finally, add 9 and 11 to get 20. Subtract 20 from
89 to get the final answer: 89 – 49 ÷ 7 + 2 + 33 ÷ 3=69.
Brain Teasers Math IQ Test: Solve 56÷4×8+3-2
Immerse yourself in this brainteaser math IQ test using the following equation: 56 ÷ 4 x 8 + 3 – 2. Your challenge is to carefully follow the order of operations and determine the final result.
First, perform division: 56 ÷ 4 equals 14. Then, multiply: 14 x 8 equals 112. Add 3 to 112 to get 115, and finally subtract 2 from 115 to get 113. Therefore, equation 56 ÷ 4 x 8 + 3 – 2 equals 113.
Brain teaser math test: equals 760÷40×5+8
Get into fun territory with brainteaser math tests using the following equation: 760 ÷ 40 x 5 + 8. Your task is to carefully follow the order of operations and calculate the final result.
To solve this equation, follow the order of operations. First, perform division: 760 ÷ 40 equals 19. Then, multiply: 19 x 5 equals 95. Add 8 and 95 to get the final answer of 103. Therefore, the
equation 760 ÷ 40 x 5 + 8=103.
Brainteaser math puzzle: 11+11=5, 22+22=11, 33+33=?
Dive deeper into exciting brainteaser math puzzles. The pattern starts with: 11+11 equals 5 and 22+22 equals 11. Now the mystery deepens: What happens when 33+33 goes through this interesting
The sequence uses multiplication and addition to create unexpected but logical progressions. In the first equation, 11+11 equals 5, calculated as (1×1) + (1×1) + 3. Likewise, for 33+33, we have (3×3)
+ (3×3) + 3, which results in 21.
Brainteaser math puzzle: 11+11=5, 22+22=11, 33+33=?
Embark on a fun journey of brainteaser math puzzles. The pattern starts with 11+11 equaling 5 and 22+22 equaling 11. The mystery deepens: when we deal with 33+33 in this sequence, what is the result?
In the first equation, 11+11 equals 5, which is elaborated by (1×1) + (1×1) + 3. Following the same logic, for 33+33, it evolves into (3×3) + (3×3) + 3=21.
Disclaimer: The above information is for general information purposes only. All information on this website is provided in good faith, but we make no representations or warranties, express or
implied, as to the accuracy, adequacy, validity, reliability, availability or completeness of any information on this website.
Source: https://anhngunewlight.edu.vn
Category: Brain Teaser
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How to Calculate Oscillation Frequency
••• noise image by Nicemonkey from Fotolia.com
Oscillation is a type of periodic motion. A motion is said to be periodic if it repeats itself after regular intervals of time, like the motion of a sewing machine needle, motion of the prongs of a
tuning fork, and a body suspended from a spring. If a particle moves back and forth along the same path, its motion is said to be oscillatory or vibratory, and the frequency of this motion is one
of its most important physical characteristics.
The displacement of a particle performing a periodic motion can be expressed in terms of sine and cosine functions. As these functions are called harmonic functions, periodic motion is also known as
harmonic motion.
What is Simple Harmonic Motion?
Among all types of oscillations, the simple harmonic motion (SHM) is the most important type. In SHM, a force of varying magnitude and direction acts on particle. It is important to note that SHM
has important applications not just in mechanics, but also in optics, sound, and atomic physics.
A body is said to perform a linear simple harmonic motion if
1. It moves to and fro periodically along a straight line.
2. Its acceleration is always directed towards its mean position.
3. The magnitude of its acceleration is proportional to the magnitude of its displacement from the mean position.
is used to define a linear simple harmonic motion (SHM), wherein F is the magnitude of the restoring force; x is the small displacement from the mean position; and K is the force constant. The
negative sign indicates that the direction of force is opposite to the direction of displacement.
Some examples of simple harmonic motion are the motion of a simple pendulum for small swings and a vibrating magnet in a uniform magnetic induction.
What is the Oscillation Amplitude?
Consider a particle performing an oscillation along the path QOR with O as the mean position and Q and R as its extreme positions on either side of O. Suppose that at a given instant of the
oscillation, the particle is at P. The distance traveled by the particle from its mean position is called its displacement (x) i.e. OP = x.
The displacement is always measured from the mean position, whatever may be the starting point. For example, even if the particle travels from R to P, the displacement still remains x.
The amplitude (A) of the oscillation is defined as the maximum displacement (x[max]) of the particle on either side of its mean position, i.e., A = OQ = OR. A is always taken as positive,
and so the amplitude of oscillation formula is just the magnitude of the displacement from the mean position. The distance QR = 2A is called the path length or extent of oscillation or total path
of the oscillating particle.
Formula of the Frequency of Oscillation
The period (T) of the oscillation is defined as the time taken by the particle to complete one oscillation. After time T, the particle passes through the same position in the same direction.
The frequency of oscillation definition is simply the number of oscillations performed by the particle in one second.
In T seconds, the particle completes one oscillation.
Therefore, the number of oscillations in one second, i.e. it's frequency f, is:
The oscillation frequency is measured in cycles per second or Hertz.
Type of Oscillation Frequency
The human ear is sensitive to frequencies lying between 20 Hz and 20,000 Hz, and frequencies in this range are called sonic or audible frequencies. The frequencies above the range of human hearing
are called ultrasonic frequencies, while the frequencies which are below the audible range are called infrasonic frequencies. Another very familiar term in this context is “supersonic.” If a body
travels faster than the speed of sound, it is said to travel at supersonic speeds.
Frequencies of radiowaves (an oscillating electromagnetic wave) are expressed in kilohertz or megahertz, while visible light has frequencies in the range of hundreds of terrahertz.
• No matter what type of oscillating system you are working with, the frequency of oscillation is always the speed that the waves are traveling divided by the wavelength, but determining a system's
speed and wavelength may be more difficult depending on the type and complexity of the system.
About the Author
She is a science editor of research papers written by Chinese and Korean scientists. She is a science writer of educational content, meant for publication by American companies. She has a master's
degree in analytical chemistry. She has been a freelancer for many companies in the US and China.
Photo Credits
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Reflections of a High School Math Teacher
In my Introduction to Algebra Class we did a lesson covering the area of a triangle, rectangle, and a parallelogram. We used a differentiated lesson idea of making a cube. The students wrote
questions on 5 sides of the cube and then their answers on the 6th side of the cube. On the five sides students had a choice of what types of question s to put on it. They could choose easy, medium
or difficult questions. The students would then use scissors to cut out and tape together their cube. Once the students were done with making the cube, they found another student to roll the cube
with. They would then they would do the problem and then check with the answer side of the cube. It turned out to be a great activity of choosing problems and checking their answers. I have attached
a link to the the activity that I did.
The activity
A PDF Cube
The note read "I miss you Dad". It was a note that a son had written to honor his Dad at the Naperville 2009 Flag Memorial. The memorial has 2009 flags set up in a park in Naperville. It is amazing.
It really struck me that this man gave his life for his nation. His son knows the meaning of that full well.
I took a 1 minute video of all the flags today. Hopefully you will just get a glimpse of it's power. It has made me reflect on this Veterans Day that so many people have died for my freedom. We are
certainly blessed. I think it is worth while to discuss the importance of honoring our veterans this week with our students.
I have my students change seats every chapter. I often have my students work in pairs throughout the class. So this change of seats is really a change of class partners. This change of seats and
change of partners typically occurs on the first day after a test. We change seats and then I have the new partners interview each other. I just make up random questions for the students to ask each
other. After they interview each other, I pick students at random to and ask them to introduce their partner, and then answer one of the questions from the interview. It does take a little time out
of my class, however, I believe it is well worth while to create this type of community team spirit.
INTERVIEW QUESTIONS and Changing Seats Directions (I give my students these)
We will change seats every chapter.
The purpose of Interview Questions is to get you to know your partner a little bit before you work with them on Math.
You do not need to write them out. However, you might be asked to share part of your interview with the rest of the class.
Students will be picked at random to share their interview with the rest of the class. Remember, to introduce your partner first, and then answer the question that we are on.
Interview Questions
1. What is your name?
2. What are your activities and interests?
3. What extreme sport or activity are you most afraid of? (like 1/2 pipe snowboarding) Why?
4. What former president (living or not) would you like to have a conversation with?
5. Rank the following restaurants from best to worst: Outback, On the Border, Mongolian BBQ and Maggiano's.
6. What insect do you like the most, and hate the most? Why?
7. Without giving a name of a person, give a few positive traits that you admire in someone.
8. Tell about your pets and their names.
We used the class set of Flip Videos last week. It worked out well. The students worked in pairs and solved a word problem from the section we were on. It was actually a homework problem. We have
desks that can be written on. So the pair of students wrote some key information about the problem on the desks with their whiteboard makers. Then, they each explained how to do part of the problem.
They watched what they did and if it wasn't good, they redid the problem. If they liked what they did, they gave me back the Flip Video and I downloaded the video to my computer. I then put the
videos on YouTube and linked them to my site. Students then could watch the different problems from home with the added benefit of stopping and rewinding the video if they needed to.
The rules were simple. 1. Make sure your problem is correct before you explain it. 2. All people in the group must take a turn in explaining how to do the problem. 3. It must be 3 minutes or less in
length. 4. You must do the video in one take, so plan out what you want to say. 5. Lastly, have fun.
See the video for an example.
Here is the link to other Flip Video Problems
How have you used Flip Video's in your class?
This site is a great example of differentiated learning. It catches the student where they feel most comfortable starting. The site gives a math problem and then develops the problem into different
stages A, B, C, D, and so on. I got this from my twitter friend johntaig.
This site has caused me to think about the way I present problems. I'm wondering if I should have a extention whenever I give a problem. For instance, if I give a problem to the class, I should
consider posing a thought provoking question at the bottom of my queston to extend their thinking. Wouldn't this be great for that student who is always done early to chew on something that is a
little more difficult, yet helps promote understanding?
See the example below
What do you think?
Welcome back to a new school year! Homework Assignment Number 1 was to have my students email me a few things about themselves. First, they had to write their name and class in the subject line.
Second, they wrote some activities or interests that they have. Thirdly, they had to go to my website and find a quote that was posted there and explain it's meaning. Not a huge deal for the
students, and a great amount of information for me. First of all, they realize day 1 that they can email me to get some information. I think this is a great thing. Secondly, I have found out some
information about them that really helps me to get to know them. They feel a little more comfortable sharing about themselves in an email compared to talking in class or writing in class. Right after
the emails started coming in, I really felt as if I knew them better. Lastly, I made a distribution list of the whole class off of these emails so that I can send information out to them when I need
My brother in law gave me this idea. It is a uno-stacko review game. Uno-stacko is like Jenga. Here are the rules
1. Groups of 4.
2. Give a problem to the class.
3. Give time for them to solve the problem.
4. Call on someone at random.
5. If they talk you through the problem correctly, they pick two tiles.
If they don't get you through correctly, they pick four tiles.
6. If the tower falls when you are working on it, EVERYONE in the class gets extra credit, EXCEPT that persons GROUP of FOUR.
7. The students really get into it.
8. And yes, I have had students that want to knock it over and give the rest of the class extra credit. It never has happened yet.
Give it a try. Here is the video for it.
I thought I would show a video of a couple of students singing the quadratic formula to the tune of Pop Goes the Weasel. The students really like this. Give it a try.
Ben Grey has a great blog post that has prompted me to think about the question of housing our student and teacher work in or out of district. http://bengrey.com/blog/2009/04/technology-guidelines/
Thanks for the great post Ben.
Should we keep our students' work and our teachers' work within district? Why or Why Not. Is their a compromise on this issue?
My thoughts keep waffling on safety and collaboration. What do you think?
Wow. Today I approached Precalculus class a little different than last year. I asked my students to work in groups and guess the average high temperature in Naperville for the month of January. Then
I gave them the answer which happened to be 32 degrees F. Then I asked them to find the average of all the months of the year. When they had finished this they graphed it with the month being the
x-axis and the temperature being the y-axis. Then, in their groups I asked them to make as many observations about the graph as they could. We spent about 5 minutes going over many ideas such as
there is a max point, and it is periodic, and then someone said it. They said it looks like a sin graph! Awesome! This made my day.
Then I had them draw in the midline and tell where exactly it hit the y axis and it's meaning to the problem. Quickly the students found that this was the average high temperature for the year.
We covered other things like the midline and the amplitude and the phase shift. The period seemed boring to them, but was an important fact.
Then, we looked the actual graph that www.weather.com has on it. We went to the site and typed in our zip code 60565. Then we scrolled down to find AVERAGES . This gave this terrific graph.
Then I asked them to predict what the graph would look like from Honalulu HI. They used all kinds of precalculus vocabulary with each other and described how it would be different.
We looked up other cities like Barrow AK, and Death Valley CA. We also looked up Sydney Austrailia and found that the graph was shifted over significantly.
This lesson was different than last year because they were very interested in their own predictions. I think they were much more engaged.
Thanks to my colleague Kevin Bell for introducing me to the www.weather.com site.
My father-in-law gave me this awesome quote and I thought I'd share it.
I'd rather be a could-be if I cannot be an are;
because a could-be is a maybe who is reaching for a star.
I'd rather be a has-been than a might-have-been, by far;
for a might have-been has never been,
but a has-been was once an are.
Milton Berle
I'm going to give this quote to my students when I get back from spring break.
PLEASE NOTE: THE CREATORS OF THIS SOFTWARE ASKED THAT ONLY OWNERS OF TI 83'S AND 84'S USE THIS PROGRAM. (I ACTUALLY THINK THE MORE THIS PROGRAM IS DISTRIBUTED, THE MORE BUSINESS TI WILL MAKE) THE
LINK BELOW IS NOT MY WEBSITE.
Do you have a virtual calculator on your computer? It is easy to download. Follow the instructions on the website and you will have yourself a TI 83 on your computer and for the data projector in a
snap. The best part of this is if you have a IWB, then you can have a student model the keystrokes that you should use when using the calculator. If you are not too techie, then ask someone who can
help you to take these steps. It will not take too long. With a restart about 10 minutes.
Here is what I did to get the calculator on my laptop:
1. Make a folder called "Virtual TI83" in the Programs folder
2. Click the "vti.zip" file and all of the files should show (if you don't have a zip program, there is one at the bottom of the website "unzip32-312.exe" that you can install to be able to unzip the
3. Put the unzipped files into the folder "Virtual TI83" from step 1.
4. Download "ti83 Plus v1.03.rom" to the folder "Virtual TI83" from step 1.
5. Go to the folder "Virtual TI83" and click on the "vti.exe" file to start the program. It will ask you to set the ROM calculator version and then you will be ready to go.
6. When using the calculator and you would like to turn it off, right click while you are pointed on the calculator, and click on Exit without saving state or Exit and save state.
7. Now make a shortcut for your desktop. I do this by right clicking on the "vit.exe" file and sending it to make a shortcut on the desktop.
Today was something new for me. I put a masking tape angle on the floor. It was an acute angle. Then I had a tall person and a short person volunteer for the activity.
I had the tall person on the ground in line with the acute angle already on the floor. I had the short person forming a triangle with the tall person and the tape on the ground. It was quite a site.
The students were really interested in what was happening. Keeping the same vertex, I asked the students to make another triangle if possible. The students found it quickly. I had a third student put
masking tape on the triangle positions that were formed. I had the students talk with their partner and write down any observations they had about the resulting figure. We had a rich discussion
sharing out their ideas.
Then I formed groups of three with the class. One of the three was to cut out a longer strip of paper, another was to cut out a short strip of paper, and the third was to cut out an acute angle. The
were to physically make the SSA example with their paper Side, Side and Angle. All groups worked well with this and some interesting results occurred. The usual two case scenerio occurred. One group
found that theirs did not reach. Another group found that theirs only formed one triangle. Each group of three drew their results on the board.
It gave me a way to describe something mathematical in an easy way. I used language like Side (Anthony) Side (Michelle) and Angle (Acute). After I felt like the students were much more engaged in the
process. It took most of the period to do it. We did one problem at the end of the class and that problem went extremely well. I think that moving while learning is important for the brain to
remember what is happening.
First, this activity will take about 20 minutes.
Show the students the problem and make sure they know what is being asked. Don't give any clues as to how to solve it. Have them work in pairs to come up with a plan to solve it.
After about 3-5 minutes pick a pair at random to explain their plan. A class discussion will take place on the plan. Then that pair will start executing the plan and all in the class will do the math
for that plan. Plans have differed from class to class. But, they will definitely need to use the Pythagorean Theorem to get the answer.
The students solution: Usually it will be to measure two sides with the ruler, and then find the third side using the Pythagorean Theorem. They now have the 3 sides in centimeters. Then they measure
the key in centimeters. They sometimes use a proportion to convert the centimeters to miles.
It is great fun to watch the student trying to use the ruler on the IWB to measure. The students are using collaboration, problem solving, and math skills with this activity.
Hope you can try something like this in your class,
CBS's Debbie Turner Bell came out to our school in December and she made a very nice 3 minute story called "Pumping up the Brain" It aired nationally in January. Debbie and her crew made us all feel
very comfortable. Hope you enjoy it.
CBS Early Show Link Video
CBS Early Show Written Story
My colleague coined it scaffolding for our students. He is right. We need to build some structure help to our weaker students. Here is what I did. Instead of just solving an equation, I solved the
equation myself and then took out key parts. I made labeled these "blanks" with letters so that the students could communicate with each other which "blank" they were referring to. The students then
were given time to work this out with their partner. Then I randomly picked a student to go up to the board and fill in a missing spot. Interestingly the student who started picked the middle to
start with. See video. Students came up one by one until they hardest spot was left. All of the explanations were very good. Things like, "you have to add it to both sides" came out. I like when that
happens. I'm trying to build a little challenge for my stronger students as well as a little guided help for my weaker students.
The second video looks at a different class in a pre-calculus example. This uses the same idea as above, except it has some diagrams involved.
What do you think?
Remember to Take Time to Enjoy Teaching,
Hope to hear from you,
I am amazed at how Brad Cohen overcame his Tourette Syndrome and became a wonderful teacher. Tourette Syndrome is a neurological condition that is exhibited through twitches, yelps, verbal outbursts,
and more. The problem is that with TS you have no control over these behaviors. It is like blinking for most of us.
Brad's main theme in the book is that he would educate people about TS and they would generally accept it. He would honestly explain his condition to any audience that would listen. Most importantly
he would educate his students. His students would then in turn ask questions which Brad answered frankly. As kids are, they accepted Mr. Cohen quickly. It is really a great story of how positive
attitude and fortitude will get you to realize your dreams. In this case Brad's dream was being a teacher.
Brad's story reminds me as a teacher how I treat all my students. It makes me think of how sometimes I assume that students have complete control of themselves at all times. I always need to
investigate before I make assumptions. It has been a great lesson for me.
Thanks Brad for your inspiring story. I hope you get a chance to read it.
Here is Brad's Website http://www.classperformance.com/ | {"url":"https://teachhighschoolmath.blogspot.com/2009/","timestamp":"2024-11-03T18:38:41Z","content_type":"application/xhtml+xml","content_length":"151433","record_id":"<urn:uuid:0966e49d-1471-476c-94c2-7067533b00d2>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00683.warc.gz"} |
A/Prof N J Wildberger Personal Pages
Rational Trigonometry Site
These pages will attempt to provide an overview of Rational Trigonometry and how it allows us to reformulate spherical and elliptic geometries, hyperbolic geometry, and inversive geometry, and leads
to the new theory of chromogeometry, along with many practical applications.
Reference: The main reference is the book
`Divine Proportions: Rational Trigonometry to Universal Geometry'
by N J Wildberger, ISBN 0-9757492-0-X (hardcover), Wild Egg Books, Sydney 2005 available at http://wildegg.com and also at amazon.com.
Here are free downloads of sections of this book (all in pdf format)
YouTube Videos: I have posted a series of YouTube videos on rational trigonometry and geometry (about 50 so far) in the WildTrig series under user name njwildberger. These videos are meant for a
general audience with an interest in geometry, and are packed with information about rational trigonometry and its applications, and about geometry more generally.
Here is a list of the WildTrig videos with brief descriptions.
My YouTube channel is called `Insights into Mathematics', and also has another series called MathFoundations, which attempts to lay proper foundations for mathematics, and criticizes the current
sloppy approach to logical difficulties. This series is meant for a general audience.
There is a third series which will be devoted to mathematics seminars I have given, these are at a more advanced level (be warned!)
Hyperbolic Geometry:
Here you can find an explanation of how rational trigonometry gives a new approach to hyperbolic geometry, as well as a series of The Geometer's Sketchpad worksheets that illustrate various
constructions and theorems of universal hyperbolic geometry.
The main reference paper is the paper `Universal Hyperbolic Geometry I: Trigonometry'.
Additional Elementary Papers (by N J Wildberger)
The following papers are meant for a high school audience. They provide a resource for teachers and students. You may download as pdf.
Additional Advanced Papers (by N J Wildberger)
The following papers are meant for an academic audience, meaning mathematicians, university math majors, and enthusiastic high school teachers. You may download as pdf.
• N J Wildberger, Universal Hyperbolic Geometry I: Trigonometry (http://arxiv.org/abs/0909.1377)
ABSTRACT: Hyperbolic geometry is developed in a purely algebraic fashion from first principles, without a prior development of differential geometry. The natural connection with the geometry of
Lorentz, Einstein and Minkowski comes from a projective point of view, with trigonometric laws that extend to `points at infinity', here called `null points', and beyond to `ideal points'
associated to a hyperboloid of one sheet. The theory works over a general field not of characteristic two, and the main laws can be viewed as deformations of those from planar rational
trigonometry. There are many new features.
• N J Wildberger, Affine and projective universal geometry (pdf) (http://arxiv.org/abs/math/0612499)
This paper establishes the basics of universal geometry, a completely algebraic formulation of metrical geometry valid over a general field and an arbitrary quadratic form. The fundamental laws
of rational trigonometry are here shown to extend to the more general affine case. Also a corresponding projective version, which has laws which are deformations of the affine case is
established. This unifies both elliptic and hyperbolic geometries, in that the main trigonometry laws are identical in both!
• N J Wildberger, One dimensional metrical geometry (pdf) (http://arxiv.org/abs/math/0701338)
The basics of universal geometry are already visible in the one dimensional situation, the great blind spot of modern geometry. There is both an affine and a projective version. The affine
version is interesting especially with regard to the quadruple quad formula, the relation between quadrances from four points, which anticipates the formula of Brahmagupta for cyclic
quadrilaterals. The projective version depends on the one dimensional analog of a quadratic form. Chromogeometry already makes an appearance. The spread polynomials, which are rational
equivalents of the Chebyshev polynomials of the first kind, contain already the seeds of two dimensional symmetry.
• N J Wildberger, Chromogeometry (pdf) (http://arxiv.org/abs/0806.3617)
My favourite discovery. A three fold symmetry in planar metrical geometry, that ends up transforming almost every aspect of the subject. Euclidean geometry meets two hyperbolic or relativistic
geometries, and all three interact in a lovely way. This introductory paper illustrates applications to triangle geometry, in particular it `explains' in a new way the Euler line and introduces a
new triangle of coloured orthocenters which controls all three Euler lines, along with circumcenters and nine point centers. There is much more to be said about this subject!
• N J Wildberger Chromogeometry and Relativistic Conics (pdf) (http://arxiv.org/abs/0806.2789)
This paper discusses how chromogeometry sheds new light on conics. It gives a novel formulation of an ellipse involving two canonical lines called the diagonals of the ellipse, together with an
associated corner rectangle, and shows how this concept applies both in blue (Euclidean), red and green geometries. Red and green foci come in two pairs, determined geometrically in a simpler
fashion than the usual foci. Hyperbolas are also discussed. The parabola plays a special role, as metrically it is the same in all three geometries, and the interaction between the three colours
is particularly striking.
• N J Wildberger, Neuberg cubics over finite fields (pdf) (http://arxiv.org/abs/0806.2495)
The Neuberg cubic is the most famous triangle cubic, organizing more than a hundred known triangle points and many lines. This paper uses the framework of Universal Geometry to extend triangle
geometry to the finite field setting, and studies an example of a Neuberg cubic over the field of 23 elements. Many, but not all, of the usual real number properties hold. The paper also
discusses tangent conics to elliptic curves in Weierstrasse form, showing them to be disjoint or identical if -3 is not a square.
Additional Papers (by others)
To post your comments/papers/developments involving Rational trigonometry here, send them to me at n.wildberger (at) unsw (dot) edu (dot) au
Links, news and reviews
Here are links to various places on the internet where you can read about Rational Trigonometry. | {"url":"https://web.maths.unsw.edu.au/~norman/Rational1.htm","timestamp":"2024-11-02T04:57:29Z","content_type":"text/html","content_length":"27755","record_id":"<urn:uuid:3a2a032b-c01e-4f0a-929a-15cc1e88f1a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00135.warc.gz"} |
Method and apparatus for generating numbers
Method and apparatus for generating numbers
A non-repeating sequence of numbers having a substantially uniform distribution is obtained from a shift register whose contents are shifted, with the shifted digits being replaced by digits from a
continuous sequence. The contents of the shift register are also replaced by their complement in dependence for the value of the most significant bit, so that the operations performed on the contents
are represented by a tent map, thereby providing uniform distribution of the numbers. A randomising subsystem can be used to convert the output sequence into a random sequence.
[0001] This invention relates to a method and apparatus for generating a non-repeating sequence of numbers, and particularly but not exclusively uncorrelated random numbers distributed
uniformly over a specified interval.
[0002] Random numbers uniformly distributed over a set of integers {0, 1, . . . , 2N−1} are required for stochastic simulation and also for spread-spectrum communications and
radar. It is particularly desirable to provide truly random numbers in contrast to pseudorandom numbers that can be generated in accordance with the many existing techniques.
[0003] It is known that the use of random numbers as frequency-hopping patterns (codes) in radar results in low probability of intercept and enhanced resistance to intelligent jamming.
Furthermore, when random numbers are subjected to suitable digital-to-analogue conversion, the resulting signals can be utilized for modulating radar transmissions, thereby providing radar waveforms
with maximum unpredictability. Random waveforms are also useful for applications in multiuser environments where many similar or disparate systems operate in the same geographical region and those
systems share, at least partly, the same wide frequency band.
[0004] Several classes of methods are known to generate truly random numbers by exploiting various physical phenomena such as thermal or avalanche noise, gaseous discharge, particle-induced
scintillation, phase fluctuation of harmonic oscillators, etc.
[0005] The best known method for generating truly random numbers is based on converting a random signal produced by a physical noise source into a random binary waveform with two
equiprobable values. Next the binary waveform is suitably sampled to yield a sequence of independent random bits occurring with equal probabilities. Random N-bit numbers with uniform distribution are
then formed from this sequence by using different nonoverlapping sub-sequences of length N. In order to obtain a fast random number generator, N independent sequences of random bits may be utilized
in a parallel scheme.
[0006] Because of inherent instabilities of physical noise sources and associated electronic circuits, it is not possible in practice to produce random binary sequences with equiprobable
bits. Accordingly, numbers formed from such sequences will not be uniformly distributed. This results in inefficient use of frequency bands in communications and radar systems, increased
predictability and other disadvantages.
[0007] To overcome this problem, various solutions have been proposed: either by incorporating a suitable stabilising feedback loop or by exploiting a ‘divide-by-two’ operation to obtain
equiprobable bits. Although the technique employing a ‘divide-by-two’ operation can produce equiprobable bits, in general those bits may be correlated.
[0008] Another example of an implementation of a random number generator is the T7001 Random Number Generator chip, manufactured by AT&T. The principle of operation is based on the phase
jitter of a free-running oscillator. As a result, the output bit stream is truly random, not pseudorandom. However, the output data rate is not sufficient for high-frequency operation.
[0009] It would, accordingly, be desirable to provide an apparatus and method for the generation of truly random numbers with uniform distribution suitable for spread-spectrum radar and
other applications.
[0010] It would also be desirable to provide an apparatus and method for the generation of truly random numbers to yield frequency-hopping patterns and other suitable waveforms which are
intended for application in multiuser environments, and/or are resistant to deliberate intelligent jamming, and/or have a low probability of detection and intercept.
[0011] It would also be desirable to provide a method and apparatus for generating a non-repeating sequence of uniformly-distributed numbers. Such numbers would be useful, for example, in
the production of uncorrelated random numbers distributed uniformly over a specified interval, intended for example for simulation or other purposes.
[0012] Aspects of the present invention are set out in the accompanying claims.
[0013] According to a preferred embodiment of the invention, a truly random primary binary sequence, not necessarily with equiprobable bits, is generated. The bits of this sequence are
manipulated in such a way as to obtain a plurality of uncorrelated binary sets with equiprobable bits. Uncorrelated chaotic N-bit binary numbers with uniform distribution are formed from those sets
by suitably selecting subsets comprising N bits. The operational procedure employed for manipulating bits of a primary binary sequence is based on the chaotic behaviour of the so-called “tent” map,
widely explored in chaos theory, resulting from a procedure known as “stretching and folding” to produce successive values. This operational procedure is implemented in an embodiment of the invention
by a hybrid chaos generator.
[0014] Preferably, a truly random auxiliary binary sequence, not necessarily with equiprobable bits, is also generated. Each of the generated N-bit uncorrelated chaotic numbers and N bits
suitably selected from this auxiliary sequence are XOR'ed bit by bit to obtain a resulting truly random N-bit binary number with uniform distribution and also with maximum unpredictability. Thus all
the functions and operations required to achieve maximum unpredictability of generated numbers are implemented by a randomisation subsystem which operates on the chaotic numbers.
[0015] In an embodiment of the present invention, both the primary binary sequence and the auxiliary binary sequence are obtained from a single physical noise source.
[0016] As indicated above, the operational procedure for manipulating the bits of the preliminary binary sequence to obtain sets of equiprobable bits may be based on the tent map, the
simplest form of which, shown in FIG. 1, is given by 1 T ⁡ ( v ) = { 2 ⁢ v , 0 < v < 1 / 2 2 ⁢ ( 1 - v ) , 1 / 2 < v < 1 ( 1 )
[0017] It is well known that a sequence of numbers, generated according to
vk+1=T(vk), k=0, 1, . . . (2)
[0018] will be infinite and nonrepeating with uniform distribution over the unit (0, 1)-interval, if the initial value v0 has been suitably selected. Furthermore, the autocorrelation
function of this sequence will be zero for all non-zero shifts. For example, an infinite and nonrepeating sequence {vk; k=0, 1, . . .} can be obtained when the initial value v0 is an
irrational (or a transcendental) number.
[0019] Although the chaotic behaviour of sequences generated by the tent map T(v) can be observed experimentally in analogue electronic circuits, any attempt to generate such infinite and
nonrepeating sequences digitally will fail because irrational numbers cannot be represented by finite binary numbers. As a result, a sequence of numbers generated by a digital implementation of
equation (2) will be either periodic or it will quickly iterate to zero.
[0020] However, according to a preferred aspect of the present invention an infinite sequence of bits (the primary binary sequence) required to represent a suitable initial value v0 is
supplied, bit by bit, by a generator producing random, but not necessarily equiprobable, bits. Those bits are utilized sequentially by a finite-length shift register with suitable feedback arranged
in such a way as to implement equation (2) governing the generation of a chaotic sequence. This technique is also applicable to other schemes employed to produce infinite and non-repeating chaotic
[0021] Arrangements embodying the present invention will now be described by way of example with reference to the accompanying drawings, in which:
[0022] FIG. 1 shows a tent map having characteristics which can be exploited for the generation of chaotic numbers;
[0023] FIG. 2 is a functional block diagram of a hybrid chaotic number generator according to the invention;
[0024] FIG. 3 is a block diagram of a generator of random binary sequences;
[0025] FIG. 4 shows an exemplary noise signal s(t), a random binary waveform b(t) derived from that signal, and a sequence of random bits obtained by sampling the waveform b(t) at time
instants determined by clock pulses (CLK);
[0026] FIG. 5 is a block diagram of a specific embodiment of a hybrid chaos number generator according to the invention;
[0027] FIG. 6 is a functional block diagram of a sequential element (SEL) of the hybrid chaos number generator;
[0028] FIG. 7 shows one example of an implementation of an SEL;
[0029] FIG. 8 shows another example of an implementation of an SEL;
[0030] FIG. 9 is an experimental histogram of number values produced by a specific example of the hybrid chaos generator of FIG. 5;
[0031] FIG. 10 is a block diagram of a hybrid chaos generator incorporating a randomisation subsystem, which forms a random number generator according to the invention;
[0032] FIG. 11 is an experimental histogram of number values produced by a specific example of the random number generator of FIG. 10;
[0033] FIG. 12 is a scatterplot of consecutive and overlapping pairs of numbers produced by the specific example of the random number generator; and
[0034] FIG. 13 is a block diagram of a modification which can be applied to the random number generator of FIG. 10.
[0035] FIG. 2 is a functional block diagram of a generic hybrid chaos generator 2 according to the invention, comprising a random bit generator (RBG) 4, supplying a primary binary sequence
to a finite serial-in-parallel-out shift register (FSR) 6 with suitable feedback; both RBG 4 and FSR 6 are driven by a train of suitable clock pulses (CLK) supplied on line 8. Because the feedback
operates on a finite number of bits, the hybrid chaos generator can only approximate in some sense its analogue prototype. However, because the operations performed on bits by the feedback circuit
include bit reversal, the hybrid chaos generator is capable of producing at its parallel output 10 substantially uniformly distributed numbers.
[0036] One convenient and inexpensive method to generate random binary sequences is based on level crossings of a random signal generated by a physical noise source. FIG. 3 shows an example
of a generator of random binary sequences. The generator comprises a physical noise source (PNS) 12, a zero-crossing detector (ZCD) 14, which can be a comparator or a hard limiter, and a D-type
flip-flop (DFF) 16 triggered by a train of suitable clock pulses (CLK) on line 8.
[0037] FIG. 4 shows a typical realization of a noise signal s(t), a random binary waveform b(t) defined by zero crossings of that signal and a sequence of random bits obtained by using the
DFF 16 to sample the binary waveform b(t) at the time instants determined by the clock pulses (CLK).
[0038] FIG. 5 shows a specific example of a hybrid chaos generator 50 according to the present invention. This uses a similar technique for generation of a primary binary sequence. The
generator 50 also comprises a wideband physical noise source (PNS) 12 followed by a zero crossing detector (ZCD) 14, and a suitable clock generator 15. The binary waveform from the ZCD 14 is sampled
by the first of a plurality of sequential elements (SEL's) 52 forming a shift register with feedback. The number M of sequential elements employed can be equal to, or preferably larger, than the
number N of bits used to represent output N-bit numbers, i.e., preferably M=N+X.
[0039] A functional block diagram of a sequential element (SEL) is shown in FIG. 6. Each SEL is a flip-flop with suitable auxiliary circuits and has a binary data input BDI, a binary data
output BDO, a binary control input BCI and a clock input CLI.
[0040] The operation of the hybrid chaos generator 50 is summarised as follows.
[0041] When the most significant bit (MSB) of an output number assumes value zero, the bit pattern stored in the register is shifted to the right, thus implementing the “multiply-by-two”
operation required to realize the first branch of the tent map, defined by the upper line of equation (1). The random bit, 0 or 1, shifted into the leftmost SEL is supplied by sampling the primary
binary sequence obtained from the ZCD 14; this ‘appended’ random bit represents an error resulting from the feedback mechanism operating only on finite number of bits.
[0042] When the most significant bit (MSB) of an output number assumes value one, all the bits stored in the register are inverted (reversed) and such modified bit pattern is shifted to the
right, thus implementing the “complement and multiply-by-two” operation required to realize the second branch of the tent map, defined by the lower line of equation (1). The random ‘error’ bit, 0 or
1, shifted into the leftmost SEL is now equal to an inverted bit supplied by sampling the output of the ZCD 14.
[0043] Irrespective of the distribution of bits occurring in the primary binary sequence supplied by the ZCD 14, the errors introduced when implementing the two branches of the tent map
will cancel out in the long run because each output binary number will result with equal probability from either branch of the tent map. This property of ‘equal probability’ for the two branches of
the tent map follows from the chaos mechanism exploited in the present invention, including ‘stretching’ (i.e., bit shifting, equivalent to multiplying by two) and ‘folding’ (i.e., bit reversal).
[0044] FIG. 7 is a specific example of an implementation of an SEL 52. The SEL 52 comprises a D-type flip-flop (DFF) 54 and an Exclusive-OR gate (XOR) 56 driving the input of the flip-flop.
When the control bit at BCI is zero, the input bit of XOR 56 at BDI is transferred to the flip-flop at the time instant determined by the clock pulse at CLI. When the control bit at BCI is one, the
input bit of XOR 56 at BDI is inverted and then transferred to the flip-flop at the time instant determined by the clock pulse at CLI.
[0045] FIG. 8 is another example of an implementation of an SEL 52 according to the invention. The SEL 52 comprises a D-type flip-flop (DFF) 58 with complementary outputs (BDO and
{overscore (BDO)}) and a demultiplexer, or a data selector, (DMX) 60. When the control bit at BCI is zero, the bit from the non-inverted input BDI of DMX is transferred to the flip-flop at the time
instant determined by the clock pulse at CLI. When the control bit at BCI is one, the bit from the inverted input {overscore (BDI)} of DMX is transferred to the flip-flop at the time instant
determined by the clock pulse at CLI.
[0046] Many other implementations of SEL's can be developed by those skilled in the art.
[0047] The hybrid chaos generator of FIG. 5 comprising eleven SEL's 52 has been simulated. Ten consecutive outputs, including the most significant bit (MSB), have been used to represent
1024 binary numbers from the interval (0, 1, . . . , 1022, 1023). It was assumed that the bits supplied by the output of ZCD 14 were not equiprobable; the fraction of ones was equal to 0.75. A set of
10,240,000 numbers produced by the hybrid chaos generator was recorded and analyzed. The expected number of each value from the interval (0, 1, . . . , 1022, 1023) was equal to 10,000. An
experimental histogram of observed values is presented in FIG. 9 along with the 95% confidence interval shown in broken line. As seen, the distribution of generated numbers is uniform and the
statistical error is within the calculated confidence interval.
[0048] Although the hybrid chaos generator is capable of producing uncorrelated and uniformly distributed numbers, those numbers are not strictly speaking random. This follows from the fact
that since the number generation procedure is based on a deterministic non-linear equation (2), the next value can be predicted from the current value either precisely or with a relatively small
error. Such high predictability is not acceptable in most practical applications.
[0049] Numbers with greater unpredictability can be obtained when each number is produced by a different segment of the primary binary sequence from the ZCD 14. However, because such
‘decimation’ would limit the speed of operation of the hybrid chaos generator, such a technique is not preferred.
[0050] According to a preferred embodiment of the invention the uncorrelated and uniformly distributed numbers produced by the hybrid chaos generator are made completely unpredictable by
XOR'ing them bit by bit with truly random, and not necessarily uniformly distributed, bits supplied by an auxiliary binary sequence (ABS). The resulting truly random numbers are uniformly distributed
and they also exhibit maximum unpredictability.
[0051] FIG. 10 shows a random number generator 100 incorporating the hybrid chaos generator 50 of FIG. 5 and a randomisation subsystem 102. The randomisation subsystem comprises an
auxiliary physical noise source 104, an auxiliary zero-crossing detector 106 supplying an auxiliary, binary sequence, a serial-in-parallel-out shift register (SIPO) 108 and a plurality of
exclusive-OR gates 110. Each bit of an uncorrelated and uniformly distributed number, produced by the hybrid chaos generator, is XOR'ed with a respective one of the random bits stored in the SIPO 108
and supplied by the auxiliary binary sequence from ZCD 106. The bits obtained at the outputs of the exclusive-OR gates 110 form a uniformly distributed random number with maximum unpredictability.
[0052] The random number generator 100 comprising eleven SEL's has been simulated. Ten consecutive outputs, including the most significant bit (MSB), have been used to represent 1024 binary
numbers from the interval (0, 1, . . . , 1022, 1023). It was assumed that bits supplied at the output of ZCD 14 were not equiprobable; the fraction of ones was equal to 0.7. It was also assumed that
bits supplied by the auxiliary binary sequence at the output of ZCD 106 were not equiprobable and the fraction of ones was equal to 0.6. A set of 10,240,000 numbers produced by the hybrid chaos
generator incorporating the randomisation subsystem was recorded and analyzed. The expected number of each value from the interval (0, 1, . . . , 1022, 1023) was equal to 10,000. An experimental
histogram of observed values is presented in FIG. 11 along with the 95% confidence interval shown in broken line. As seen, the distribution of generated numbers is uniform and the statistical error
is within the calculated confidence interval.
[0053] In order to demonstrate maximum unpredictability of random numbers produced by the hybrid chaos generator incorporating the randomisation subsystem, consecutive and overlapping pairs
of numbers {(vk+1, vk); k=0, 1, . . . , 6000} were plotted as points filling the 1023×1023 square. As seen from the scatterplot, shown in FIG. 12, the points do not form any
distinctive pattern to facilitate predictability of generated numbers.
[0054] FIG. 13 is a block diagram of a modification which can be applied to the random number generator 100 of FIG. 10. The modification uses a single physical noise source to produce both
the preliminary binary sequence (PBS) and the auxiliary binary sequence (ABS). The system comprises a physical noise source 130, a zero-crossing detector 132, a biphase clock generator (BCG) 134 and
two one-bit buffers, 136 and 138. Random bits supplied by the ZCD 132 are transferred respectively to buffers 136 and 138 at time instants determined by alternating clock pulses supplied by the BCG
[0055] Various modifications to the described embodiments are possible.
[0056] For example, although it is preferred as in the above embodiments for the primary binary source to be truly random, a deterministic source, for example a pseudorandom bit source
which may use recirculating or feedback shift registers, could be employed. Preferably, however, the continuous source produces a non-repeating sequence, or at least a sequence having a long
repetition period.
[0057] Instead of using a separate source to randomise the uniformly-distributed sequence, the sequence outputs can themselves be manipulated in order to increase unpredictability. For
example, each bit could be exclusive-OR'ed with another bit, and the bits which are combined in this way can be varied from number to number, possibly in a random fashion.
[0058] If desired, two (or more) generators could be used to provide respective uniformly-distributed sequences, and the outputs could be combined (e.g. by interspersing or
exclusive-OR'ing) to reduce predictability.
[0059] It is preferred, as in the above embodiments, for the shift register formed by SEL's 52 to have more than N stages, for an N-bit output, and for at least the least significant stage
not to form part of the N-bit output. The reason for this is that it reduces the error caused by performing the tent map algorithm on a finite number of bits, and hence increases the uniformity of
the distribution of numbers. However, it is not essential that this technique be used.
[0060] In the above embodiments, the most significant bit is used to control the selective inversion of the bits in the SEL's during the shifting operation. However, instead, it would be
possible to use a plurality of bits, for example the most significant bit and the next-most significant bit, for collectively determining whether inversion should occur. In such a situation, however,
it may be desirable for the shift operation to be a two-bit shift, with two new bits from the continuous sequence being fed to the shift register to replace the two bits shifted out. (This is
equivalent to the operation of the embodiments described above, but taking only alternate outputs of the SEL's 52.)
[0061] Bits other than the most significant bit could be used to control the selective inversion operation. However, the resulting operation would then differ from that represented by the
tent map of FIG. 1, though it would still be representable by a piece-wise linear map, as obtained by “stretching and folding”. In some cases, this would give rise to a sequence which is uniformly
distributed over a subset of the possible outputs, the subset being determined by the initial value in the SEL's 52. Nevertheless, this may be useful, and indeed the effective memory of such a system
may have advantages in some situations. Alternatively, a supplementary mechanism may be used to alter the value in the SEL's 52 to shift between subsets.
[0062] In the above embodiments, the non-repeating uniformly-distributed sequence is obtained by taking, for each number in the sequence, the bits stored in N adjacent SEL's 52. However,
the N bits need not be acquired from adjacent SEL's. Nor is it necessary, as in the arrangement of FIG. 10, for each bit of the number to be XOR'ed with the corresponding bit of the random number
from the auxiliary bit source, the XOR'ing could take place between non-corresponding bits, and indeed the pairs of bits which are combined in this way could be changed from number to number.
[0063] If desired, the uniformly-distributed sequence could be obtained from a single output of the shift register formed by SEL's 52, either by forming each N bit number from N successive
outputs of this stage, or by using N shift registers, and forming the N bit output by taking one bit from each shift register.
[0064] Although the invention has been described in context of shifting bits into and out of a shift register, it would be possible instead to use digits of higher order, for example by
using parallel arrangements of shift registers. The digits at each stage may be stored in any convenient form; for example, decimal digits could be stored in BCD form.
[0065] The system can be arranged so that the complement formed of the value currently stored in the register is either the base complement or the base-minus-one complement.
[0066] A random binary waveform generated in accordance with the present invention is particularly suited for constructing a probing or interrogating signal for use in a shift determination
system according to UK patent application no. 9828693.3, the contents of which are incorporated herein by reference.
1. A method of generating a non-repeating sequence of numbers having a substantially uniform distribution, the method comprising using successive digits from a continuous sequence thereof to replace
digits shifted out of a shift register, and using the contents of at least one stage of the shift register to control whether the contents of the shift register are replaced by their complement, so
that the shift register has changing contents which can be used to derive said non-repeating sequence.
2. A method as claimed in claim 1, wherein each number of the non-repeating sequence is derived from N stages of the shift register, wherein N is an integer greater than 1.
3. A method as claimed in claim 2, wherein the shift register has N+X stages, wherein X is an integer of one or more.
4. A method as claimed in claim 3, wherein the N stages of the shift register from which each number is derived excludes at least the least significant stage of the shift register.
5. A method as claimed in any preceding claim, wherein the contents of the most significant stage of the shift register is used to control whether the contents of the shift register are replaced by
their complement.
6. A method as claimed in any preceding claim, wherein the operations of shifting the contents of the shift register and selectively replacing the contents by their complement are such as to cause
alternate stretching and folding of the contents.
7. A method as claimed in any preceding claim, wherein the operations of shifting the contents of the shift register and selectively replacing the contents by their complement are such that the shift
register contents are altered in accordance with a tent map.
8. A method as claimed in any preceding claim, wherein the continuous sequence of digits is a non-repeating sequence.
9. A method as claimed in any preceding claim, wherein the continuous sequence of digits is derived from a non-deterministic source.
10. A method of generating a non-repeating sequence of numbers having a substantially uniform distribution within a predetermined interval, the method comprising providing a sequence of digits and
repeatedly doubling the value of a group of said digits and replacing the result by its complement in dependence on said value, the group being successively replenished by digits from said sequence.
11. A method of generating a non-repeating sequence of numbers having a substantially uniform distribution, the method being substantially as herein described with reference to FIGS. 2 to 6,
optionally in combination with FIG. 7 or FIG. 8, of the accompanying drawings.
12. A method of generating a sequence of random numbers having a substantially uniform distribution, the method comprising generating a non-repeating sequence in accordance with a method as claimed
in any preceding claim, and combining numbers of that sequence with substantially random numbers in order to randomise the non-repeating sequence.
13. A method as claimed in claim 12, wherein the random numbers are produced by a non-deterministic source.
14. A method as claimed in claim 13, wherein the non-deterministic source is a thermal noise source.
15. A method as claimed in any one of claims 12 to 14, including a common source for producing both the continuous sequence of digits and the random numbers.
16. A method as claimed in any one of claims 12 to 15, wherein the combination of numbers from the non-repeating sequence with the random numbers is performed by exclusive-or'ing each bit of the
random number with a respective bit of a number from the non-repeating sequence.
17. A method of generating random numbers, the method being substantially as herein described with reference to FIG. 10, optionally in combination with FIG. 13, of the accompanying drawings.
18. Apparatus for generating a non-repeating sequence of numbers having a substantially uniform distribution, the apparatus being arranged to operate in accordance with a method as claimed in any one
of claims 1 to 11.
19. Apparatus for generating random numbers, the apparatus being arranged to operate in accordance with a method as claimed in any one of claims 12 to 17.
20. A method of digitally generating a non-repeating sequence of uniformly-distributed numbers, the method including the step of:
repeatedly changing a rational number according to a predetermined algorithm which, if applied to an irrational seed number, would produce an endless non-repeating sequence of
uniformly-distributed irrational numbers;
each rational number change also involving adding to the number one or more randomly-generated bits.
21. A method of digitally generating a non-repeating sequence of uniformly-distributed numbers, the method comprising:
(a) producing a sequence of randomly-generated bits:
(b) applying an algorithm to a predetermined number of the bits, the algorithm involving shifting the bits;
(c) adding one or more bits from the randomly-generated sequence to replace the shifted bits; and
(d) repeating steps (b) and (c).
Patent History
Publication number
: 20020035586
: Dec 12, 2000
Publication Date
: Mar 21, 2002
Patent Grant number
6751639 Inventor
W. J. Szajnowski
Application Number
: 09734181 | {"url":"https://patents.justia.com/patent/20020035586","timestamp":"2024-11-10T02:25:24Z","content_type":"text/html","content_length":"86239","record_id":"<urn:uuid:e20dd477-bbe3-4dad-af52-91ca99d8f828>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00415.warc.gz"} |
Question ID - 103055 | SaraNextGen Top Answer
Statement 1: A lighter gas diffuses more rapidly than a heavier gas
Statement 2: At a given temperature, the rate of diffusion of a gas is inversely proportional to the square root of its density
a) Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
b) Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
c) Statement 1 is True, Statement 2 is False
d) Statement 1 is False, Statement 2 is True | {"url":"https://www.saranextgen.com/homeworkhelp/doubts.php?id=103055","timestamp":"2024-11-08T20:28:31Z","content_type":"text/html","content_length":"17404","record_id":"<urn:uuid:79965a6f-5659-415f-9aa4-f3a2f391941a>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00395.warc.gz"} |
Mathematics of infinity: Calculus and infinity
Calculus and Infinity
It is argued that both Newton and Leibnitz invented Calculus. However, despite its practical uses there were
some analysts/mathematicians/philosophers, who were critical of the foundation of calculus, for example the
use of Infinitesimals and the limiting processes involved in obtaining the final results.
Analyse the criticism given by either George Berkeley or Karl Marx and the solution offered to these criticisms by either Cauchy et al with the limits or Robinson with the use of hyperreal numbers.
You may wish to give the notation used by Newton and/or Leibnitz and the modern notation used to study calculus. | {"url":"https://therealwriters.com/mathematics-of-infinity-calculus-and-infinity-5/","timestamp":"2024-11-07T00:28:01Z","content_type":"text/html","content_length":"26986","record_id":"<urn:uuid:1321be6c-90a0-473b-bfe0-a4384346d3f5>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00771.warc.gz"} |
[Solved] Let n be any natural number such that 5n-1 < 3n+1. T
Let n be any natural number such that 5^n-1 < 3^n+1. Then, the least integer value of m that satisfies 3^n+1 < 2^n+m for each such n, is
Answer (Detailed Solution Below) 5
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It is given that the inequality \(5^{n-1} < 3^{n+1}\) holds when n is a natural number, specifically for n = 1, 2, 3, 4, and 5.
Now, we need to find the least integer value of m that satisfies the inequality \(3^{n+1} < 2^{n+m}\)
For n=1, the least integer value of m is 2.
For n=2, the least integer value of m is 3.
For n=3, the least integer value of m is 4.
For n=4, the least integer value of m is 4.
For n=5, the least integer value of m is 5.
Hence, the least integer value of m such that for all the values of n, the equation holds is 5.
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More Number System Questions
More Quantitative Aptitude Questions | {"url":"https://testbook.com/question-answer/let-n-be-any-natural-number-such-that-5n-1l--6687cc1c68c04b9a79ee80a2","timestamp":"2024-11-08T20:57:58Z","content_type":"text/html","content_length":"214810","record_id":"<urn:uuid:e4e4dca5-fae4-42fa-9016-043f4ed171ce>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00099.warc.gz"} |
Top 2-way interaction strength tables for
The top 2-way interaction strength tables identify the pairs of variables that have the strongest interactions. The interaction tables display the % of total squared error and/or the % of squared
error for the strongest 2-way interactions. Use the % of total squared error to describe the strength of the interaction relative to the variation in the data. Use the % of squared error for the
specific pair of variables to describe the strength of the interaction relative to the strength of the main effects of the variables.
Interactions are not possible when a tree has only 2 terminal nodes. Thus, the maximum terminal nodes per tree must be 3 or larger. You can set this on the
Minitab does not display the interaction tables if all interactions have a % of total squared error or a % of squared error less than 10%.
In this example, the six strongest 2-way interactions are the same for both tables; however, the ordering varies slightly. In the first table, the interaction between Major Vessels and Thal is the
strongest 2-way interaction. The percent of total squared error is 6.05581 which means that 6.05581% of the total squared error is explained by the main effects of Major Vessels and Thal and their
2-way interaction effect.
For the same 2-way interaction between Major Vessels and Thal, the percent squared error for the predictor pair with main and interaction effects is 11.08252%.
To calculate, 11.08252% = Component 3 / ( Component 1 + Component 2 + Component 3) * 100%
• Component 1 = the squared error explained by the first main effect, Major Vessels
• Component 2 = the squared error explained by the second main effect, Thal
• Component 3 = the squared error explained by the interaction between Major Vessels and Thal and their main effects
Top 2-Way Interaction Strength
6.05581 Major Vessels Thal
6.04284 Chest Pain Type Major Vessels
4.94873 Thal Old Peak
4.42358 Major Vessels Cholesterol
3.95660 Thal Age
1.44046 Age Max Heart Rate
11.47879 Major Vessels Cholesterol
11.39675 Thal Old Peak
11.38103 Thal Age
11.08252 Major Vessels Thal
10.94302 Chest Pain Type Major Vessels
10.23224 Age Max Heart Rate | {"url":"https://support.minitab.com/en-us/minitab/help-and-how-to/statistical-modeling/predictive-analytics/how-to/treenet-classification/interpret-the-results/top-2-way-interaction-strength-tables/","timestamp":"2024-11-02T19:09:30Z","content_type":"text/html","content_length":"18363","record_id":"<urn:uuid:ecf9ba0f-4559-4f6f-9753-11bb87b5bb28>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00338.warc.gz"} |
Re: st: cumulative average moving through time
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
Re: st: cumulative average moving through time
From David Kantor <[email protected]>
To [email protected]
Subject Re: st: cumulative average moving through time
Date Wed, 06 Oct 2004 16:33:51 -0400
At 03:59 PM 10/6/2004 -0400, Dan Egan wrote, among other things:
by sort pid (ob):gen cave = sum(calc)/ob
1) Where/When did -sum()- become an acceptable argument to
-generate-!?!? I have only ever seen it in the context of -egen-.
Looking at the help for -generate-, there are no arguments that are
explicitly stated to be useable. It is only at the very bottom of the
examples that one sees an function -uniform- and then -sum- used with
gen. Are the others? I know that using many egen arguments with -gen-
will return errors (e.g. count).
(I suppose you meant "bysort pid...".)
sum() has always been a basic function. It is not a matter of a "argument" to -generate-. It's a function, and any function can appear in an expression (subject to type compatibility). Thus it can
appear anywhere an expression is accepted, such as in -generate-, among others.
See -help mathfun- for details.
The egen sum() program makes use of it. See (in the appropriate ado directory) _gsum.ado to see how it works.
Finally, the "functions" that -egen- accepts are just those that have programs written for them. (They all have "_g" as a prefix to their names. And you can write your own egen functions if you
desire.) Some of these happen to have the same name as ordinary functions such as sum(). But there is no explicit connection between the two.
In case you wanted to know, egen is a program that calls another program. If you type
egen myvar = somefunction blah blah blah
then egen will call _gsomefunction with blah blah blah as arguments.
I hope this is useful.
David Kantor
Institute for Policy Studies
Johns Hopkins University
[email protected]
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2004-10/msg00120.html","timestamp":"2024-11-10T20:47:00Z","content_type":"text/html","content_length":"9574","record_id":"<urn:uuid:b091cb46-ca2b-40b4-aa9e-e2af8fc874e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00079.warc.gz"} |
Directed homology
Catarina Faustino
CMAT, Univ. Minho
Directed algebraic topology is a relatively recent area that emerged at the interface between algebraic topology and the field of concurrency theory in computer science. An important line of research
in directed algebraic topology is directed homology. In this talk, I will discuss two concepts of directed homology for precubical sets that have been introduced in the literature. Unfortunately,
neither of these concepts behaves satisfactorily with respect to the tensor product of precubical sets, which models the parallel composition of independent concurrent systems and is therefore an
important categorical construction for precubical sets. In my PhD thesis, I will try to define a notion of directed homology that is compatible with the tensor product. | {"url":"https://cmat.uminho.pt/events/directed-homology","timestamp":"2024-11-09T12:34:10Z","content_type":"text/html","content_length":"12947","record_id":"<urn:uuid:80c6f533-cd5a-4d78-ad68-1309c7fd09c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00551.warc.gz"} |
DailyPuzzle | New Puzzles Everyday
The Mislabeled Boxes Riddle
There are three boxes. One contains 2 black marbles, one contains 2 white marbles, and one contains 1 black marble and 1 white marble. The boxes are labeled BB, WW, and BW. However, each box is
labeled incorrectly. You may take one marble at a time out of any box, but you are not allowed to look inside. What is the quickest way to determine the contents of each box?
You only need to take out 1 marble.
The key to solving this riddle is knowing that every box is mislabeled. The first step is to take a marble from the box labeled BW. Let's say you draw a black marble, you then know that the other
marble in the box must be black too, because otherwise the label would be correct. Now that you have identified the box with 2 black marbles, you can automatically determine the contents of the box
labeled WW. You know it does not contain 2 white marbles, because the label must be wrong. You know it does not contain 2 black marbles, because you've already identified that box. Therefore, it must
contain 1 black marble and 1 white marble. Then, of course, the last box must contain 2 white marbles. | {"url":"https://www.dailypuzzle.com/brain-teaser/147/The+Mislabeled+Boxes+Riddle","timestamp":"2024-11-02T21:27:39Z","content_type":"text/html","content_length":"11263","record_id":"<urn:uuid:76739508-8472-4231-a6d8-76c2446c2c34>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00416.warc.gz"} |
The Surrogate Matrix Methodology - TUM
The Surrogate Matrix Methodology
We revisit the classical lowest-order Bubnov-Galerkin finite element method and analyze a modification of it which is strongly amenable to stencil-based matrix-free computation. Roughly speaking, it
requires performing quadrature for only a small fraction of the trial and test basis function interactions and then approximating the rest through, for example, interpolation. The main idea was first
introduced in the context of first-order finite elements in [Bauer et al., 2017]. Thereafter, applications to peta-scale geodynamical simulations were presented in [Bauer et al., 2019] and a
theoretical analysis was given in [Drzisga et al., 2018]. In the massively parallel applications, it is natural to work with so-called macro-meshes as well as a piecewise polynomial space for
resolving the surrogate matrices. Recently, we presented a simple methodology to avoid over-assembling matrices in Isogeometric Analysis in [Drzisga et al., 2019]. In contrast, the surrogate matrices
in this work are computed using a B-spline interpolation space.
Problem Setting and Motivation
Let \(V\) be a suitable space and \(V_h\subsetneq V\)be a finite-dimensional subspace.
Consider a bilinear form \(a\), its surrogate bilinear form \(\tilde{a}\), and a linear functional \(F\).
Many physical problems may be written in the form
\text{Find } u_h \in V_h \text{ satisfying }
\text{for all } v_h\in V_h
\text{Find } \tilde{u}_h \in V_h \text{ satisfying }
\text{for all } v_h\in V_h
These discrete variational problems induce matrix equations of the form
\(Au = f\) and \(\tilde{A}\tilde{u} = f\).
The surrogate matrix \(\tilde{A}\)should satisfy following properties:
* \(\tilde{A}\) should be as close as possible to the standard matrix \(A\)
* \(\tilde{A}\) should be computationally cheaper to obtain than \(A\)
* \(\tilde{A}\) should preserve properties of the standard matrix \(A\), e.g., symmetry
Stencil functions
Assume that the bilinear form is of the form
a(u,v) = \int_\Omega G(x,u(x),v(x)) \mathrm{d} x
\qquad \text{for all } u, v\in V,
e.g., Poisson problem for \(G(x,u,v) = {\nabla u^\top K(x) \nabla v}\) and \(K = \frac{D\varphi^{-1} D\varphi^{-\top}}{|\det{\left(D\varphi^{-1}\right)}|}\(.
Let \(\phi\in V\) be a test function and \(\phi_\delta(x) = \phi(x-\delta)\). We define the stencil function \(\Phi^\delta(x)\) with a shift \(\delta\) as
\Phi^\delta(x) := \int_{\Omega_{\delta}} \!\!G(x+y,\phi(y),\phi_{\delta}(y)) \,\mathrm{d} y
Observe that the stencil functions \(\Phi^\delta\) are smooth functions, as can be seen in the picture below.
Therefore, replace stencil functions by surrogates which allow fast evaluation: \(\tilde{\Phi}^\delta = \Pi^\delta\, \Phi^\delta\)
If \(x_j - x_i = \delta\), observe that
\(A_{ij} = a(\phi_j, \phi_i) = \Phi^\delta(x_i)\)
For semi-structured meshes, the set \(\mathcal{D}(x_i) = \{x_j-x_i \,:\, \, a(\phi_{j},\phi_{i}) \neq 0\}\) is small, thus only a small number of stencil functions is required to describe the whole
With \(\tilde{\Phi}^\delta = \Pi^\delta\, \Phi^\delta\), we thus define the surrogate matrix as
\tilde{\Phi}_\delta(x_i) & \text{if } x_i,x_j\in\tilde{\Omega},
\\ A_{ij} &\text{otherwise.}
This definition may be easily extended to preserve symmetries and kernels of \(A\).
\(p\)-Laplacian diffusion example
We considered the time-dependent non-linear \(p\)-Laplacian diffusion problem, given in strong form as
\frac{\partial u}{\partial t} - \mathrm{div}\left(|\nabla u|^{p-2} \cdot u\right) &= f &&\quad \text{in } \Omega \times (0,T_{\mathrm{end}}]\,,
\\ u &= 0 &&\quad \text{on } \partial \Omega \times (0,T_{\mathrm{end}}]\,, \\
\\ u &= u_0 &&\quad \text{in } \Omega \times \{0\}\,.
Stokes' lid driven cavity flow in deformed domain
We consider a lid-driven cavity benchmark on a deformed domain where the fluid satisfies following equations:
-\Delta u + \nabla p &= f &&\quad \text{in } \Omega,
\\ \mathrm{div}(u) &= 0 &&\quad \text{in } \Omega,
\\ u &= g &&\quad \text{on } \Gamma_{\text{D}}
In this scenario the fluid is driven on the top edge by constant velocity \(g = (1,0)^\top\) and we assume no-slip boundary conditions \(g = 0\) on the remaining parts of the boundary.
The degrees of freedom corresponding to the nodal basis functions in the top left and top right corner are set to zero.
Furthermore, the volume forces are neglected, i.e., \(f = 0\).
Below, we show the velocity streamlines which were computed using a standard IGA approach on a mesh with \(320 \times 320\) control points.
Furthermore, the effect of different surrogate approaches on the velocity streamlines may be observed.
In the case where the interpolation order is given by \(q = 3\), the streamlines show the same behavior as in the standard approach even for a skip paramater \(M = 100\).
For other values of \(q\) and \(M\), the streamline behavior is different, but the streamlines are getting closer to the reference solution the larger \(q\) and the smaller \(M\) becomes.
We note that actually using an interpolation order of \(q=1,2,3\) in computation is still probably not recommended for standard practice.
For instance, assembly using \(q=5\) took roughly the same time as either of these lower-order choices and, in this case, the surrogate solution should be expected to be even more accurate.
(Keine Dokumente in dieser Ansicht) | {"url":"https://www.math.cit.tum.de/math/forschung/gruppen/numerical-analysis/research/the-surrogate-matrix-methodology/","timestamp":"2024-11-09T05:51:13Z","content_type":"text/html","content_length":"31612","record_id":"<urn:uuid:9415bc1f-dbc5-447a-84b9-9d6ac5e59958>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00603.warc.gz"} |
Checksum: Error Detection Code - Coding Aunty
Checksum: Error Detection Code
Checksum is an error detection method used by the TCP/IP protocol suite to detect errors in a block of data. Like LRC, checksum works on the entire data block. However, instead of using parity (like
in the case of LRC), checksum uses addition, making it more reliable.
A checksum is an error detection code, calculated by the sender and transmitted along with the original data. This calculated checksum is compared with the transmitted checksum by receiver. It helps
the receiver identify errors or corrupt datablocks.
NOTE: The checksum code is added as redundant bits to the data to be transmitted.
Calculation of Checksum:
• The original message is broken into ‘k’ number of blocks with ‘n’ bits each. Typically ‘n’ is 16bits.
• All the ‘k’ number of data blocks are added.
• If there is any carry, it is added to the sum.
• 1’s compliment of the sum is then calculated. This is called the checksum.
NOTE: This checksum is also of ‘n’ bits.
NOTE: This addition is binary addition. Here are some rules of binary addition:
Let’s take an example:
Suppose data to be transmitted is: 11001010 00101100 10111001
• Step1: Divide the original data into k blocks of n bits each.
In the above example, say, we decide to divide the data into 3 blocks of 8 bits each:
Thus, k = 3 and n = 8.
Divividing data into datablocks.
• Step2: Sum all the ‘k’ data blocks:
In this case k = 3. We sum all three data blocks together
110101111 is the result of adding all 3 data blocks
• Step 3: Add carry, if any.
The final checksum is of ‘n’ bits. In this case, n = 8. Whereas, our result is of 9 bits. Thus , there is an extra bit in the result of the addition. This extra bit is the carry we got duing the
As there is a carry, we add the carry to the sum.
We add the carry to the sum
This results in a sequence of ‘n = 8’ bits. The sequence we got is 10110000
Step 4: 1’s compliment of the result of addition and carry.
Calculating the 1’s compliment of the final sequence after addition gives us the checksum code.
NOTE: To get 1’s complement, we invert the bit. So:
In this case:
One’s compliment to calculate checksum code.
Final checksum is : 01001111
Transmission of Data with Checksum
The final checksum is appended to the left of the data. This data is then transmitted to the receiver such that, the checksum is the last block to be sent.
In this example, the checksum code, 01001111, is appended such that it is on the left of the original data blocks:
Checksum appended to the original data block.
This data is then sent in the order such that the checksum is the last data block received.
Receiver’s Side
On receiving the transmitted data, the receiver:
• Collects all the datablocks, including the checksum
• Adds all the datablocks together
• If there is any carry, it is added to the result.
• If the final result is all 1 bits, the data is accepted. If however, this is not the case, the data was corrupted during network transmission.
Let’s assume in our example, that data is transmitted over the network with no errors. Then the receiver would do the following calculation:
Data block 1 1 1 0 0 1 0 1 0
Data block 2 0 0 1 0 1 1 0 0
Data block 3 1 0 1 1 1 0 0 1
Checksum 0 1 0 0 1 1 1 1
Adding the data blocks and the checksum together gives 111111110 as the result.
This leaves a carry. This carry is added back to the answer:
Result 1 1 1 1 1 1 1 0
Carry 1
Final Result 1 1 1 1 1 1 1 1
Final result after addition of received data.
As the final result consists of all 1-bits, the receiver knows that no error occured during network transmission. It thus, accepts the data.
Advantages of Checksum:
• Efficiency: Checksums provide a quick and efficient method of error detection, requiring minimal computational resources.
• Simplicity: Checksum calculations are relatively straightforward, making them widely applicable and compatible with different systems and protocols.
• Wide Adoption: Checksums are extensively used in various domains, including networking, file transfers, and storage systems, ensuring data integrity across diverse environments.
Limitations of Checksum:
• Limited Error Detection: Checksums may fail to detect certain types of errors, especially when they result in the same checksum value as the correct data.
• Inability to Correct Errors: Checksums can only detect errors but lack the capability to correct them. Additional mechanisms are required for error correction.
• Can checksum work on messages of any length?
Yes. Checksum can work on messages of any length. It would break the message into ‘k’ blocks of ‘n’ length each and then continue with it’s normal protocol.
• What value of ‘n’ is usually used?
16 bits. ‘n’ is the length of bits in each segment of data. So we can say, data is divided into ‘k’ number of datablocks of 16 bits each.
• What are some various checksum algorithms used for error detection?
Internet Checksum, CRC (Cyclic Redundancy Check), Adler-32 Checksum, Fletcher’s Checksum, MD5 and SHA-1 are some checksum algorithms
• Is checksum more reliable than LRC?
Yes. As checksum uses addition over parity used in LRC, checksum is better than LRC.
• Is checksum more reliable than VRC?
Yes. As checksum uses addition over parity used in VRC, checksum is better than VRC. Checksum error detection detects both odd and even number of bits error.
• Is checksum more reliable than CRC?
No. The error detection capabilities of a CRC make it better than checksum.
• Can a checksum correct errors?
No. Checksums can only detect errors. It can’t correct them.
• Is checksum and crc the same thing?
No. CRC is a type of checksum algorithm.
Yes, MD5 (Message Digest Algorithm 5) is commonly used as a checksum in addition to its cryptographic applications.
Yes. While SHA-1 (Secure Hash Algorithm 1) is primarily designed as a cryptographic hash function, it can be used as a form of checksum for error detection purposes.
1 thought on “Checksum: Error Detection Code”
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non cum
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CalR Updates and Known Bugs
CalR Version 2 Updates
• Plotting can now be done accorrding to the highest frequency of data collected.
• Timeplots can now be vizualized with a smoothed rolling average.
• Quality control plot has been added to plot mass change vs total energy balance.
• Added the option to manually add mass change for quality control plots.
• Reworked various functions to perform more efficiently such as timeplots and remove outliers.
• Export hourly data to GraphPad Prism files
• Added power calculations to help vizualize needed sample sizes.
• Energy Balance has been added to statistical analysis.
• Timeplots ui tools that will help improve vizualization.
• Changelog tab was added which details frequent updates.
• C13 data from isotope analyzers from a "Sable" system can be read into CalR for plotting.
• Enviornmental columns from "Sable" systems can be read into CalR for plotting.
• Added two different methods of combining runs: continuous and split.
• Bar plots have been replaced by box plots
• Improved speed and performance by plotting with Javascript.
• Regression plots display kcal/hr instead of kcal/period.
CalR Version 1.3 Updates
New Features:
Cumulative Energy Expenditure
Cumulative Energy Balance
Small differences in food intake or energy expenditure over time can accumulate into changes in body weight. However due to small effect sizes and the variability between animals the changes in food
intake or energy expenditure often do not reach statistical significance. By showing cumulative changes in energy balance over time, we can extract more meaning from each experiment.
In CalR version 1.2 we introduced the energy balance visualization. Energy balance uses the food intake in kcal (the caloric content of the food multiplied by mass of food consumed) and subtracts the
energy expenditure over that time period. Now in Legacy CalR.3 we introduce the cumulative energy balance. We also now provide cumulative energy expenditure as a useful comparator to cumulative food
Please see below for examples:
Known & persisting issues:
• System: CLAMS
□ If recording start on exactly 00 seconds then it is possible that an error in CalR reading the time stamps may occur.
• System: TSE
□ Some files appear to report cumulative food intake instead of instantaneous food intake. CalR does not handle this situation well.
□ TSE files that contain semicolons in their name will be disconnected from the server.
□ TSE generated files may contain a "Start Time" row that will cause the file to disconnect from the server.
Please report any new issues to us at wbernard@bidmc.harvard.edu Bug Fixes:
• Fixed multiple occasions where light cycles were displayed incorrectly.
• Fixed spacing for analysis page where analysis titles would overlap analysis output.
• Fixed bug where bar plot y-axis labels would overlap with plot ticks.
• Fixed bug where bar plots would load twice before being displayed.
• Fixed Energy Balance bar plots.
• Fixed fence post error in time series selection
□ Many users select hours 0-24 assuming that they have selected one day (24 hours) when actually all 25 hours were selected and displayed as hour 0 is included. Previously only the “Bar Plots”
excluded the final selected hour (analysis of 24 hours), but other tabs (regression plots, analysis plots) included all 25 hours in the analysis. We have extended the removal of the last hour
selected to all functions in CalR to appropriately reflect what the user believes they selected. Please note that to replicate analysis performed under Legacy CalR.2, the user must extend the
time selected by an extra hour.
New features:
Increased speed of analysis following code optimization.
Time series plots:
• Added cumulative energy expenditure and cumulative energy balance.
• Added check for xytot or locomotor activity to distinguish if it is cumulative or not to allow files from older systems to be analyzed correctly.
• Added ability for wheel counts to be read in from TSE files.
• The hourly differences plot and the de-trended plot button have been deprecated.
• Added mass plotting based off of continuous mass recordings in TSE and Sable files.
Average Plots:
• “Average plots” was changed to “Bar Plots” to better represent what is may be displayed as not all plots before were technically averaged bar plots (e.g. cumulative values, not averages).
• Added variable selection and plot button to allow users to perform selection on the bar plots tab independently of the variable selected on the time series plots.
• Added more “info buttons” to briefly explain how plots are generated and what is being shown.
• Changed plotting from interactive to static plotting due to computational speed complaints.
• Removed selection of cumulative bar plots and instead overall cumulative plots are displayed in addition to average plots for variables that are derived from or already used to calculate
cumulative variables. (food, drink, EE, EB, Wheel)
• Added Titles to all bar plots as well as more informative axis labels.
Regression Plots:
• Removed Energy Balance from regression plot selection.
□ Please see analysis page changes for more in-depth explanation.
• Changed the unit scale of some regression plots from per hour to per phase period to better align with new bar plotting.
• Added more informative plot labels
Analysis Page:
• Removed cumulative GLM analysis as they were incorrect.
□ Previous versions of CalR displayed p-values not derived from the max values of the cumulative variable.
• Removed Energy Balance from GLM analysis.
□ We’ve decided that we need further study to understand how to analyze statistical significance of energy balance. Our initial approach followed this reasoning: for values that depend on mass
(e.g. energy expenditure) we use GLM. For mass-independent values (e.g. RER) we use ANOVA. Since energy balance is calculated as the difference of two mass-dependent variables (food intake
minus energy expenditure) energy balance therefore should also be a candidate to be analyzed via GLM. Yet so far our analysis of experiments has not shown a relationship between energy
balance and mass suggesting that it might be more appropriate for analysis by ANOVA. Until this is resolved, CalR does not report statistical significance values for energy balance. We look
forward to hearing input from the user community.
Anticipated features in the next versions of CalR
• Plotting with greater frequency than one point per hour
• Improved speed and performance
• Export summary data to GraphPad Prism files
• Time series spectral analysis & metabolic flexibility
• Estimating statistical power
• Making CalR source code available to the community on GitHub
• Making available additional R code for plotting data and analyzing CalR files
• Pilot version of a database for indirect calorimetry experiments using CalR files
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What is actually Space Complexity ? – Study Algorithms
The term Space Complexity is misused for Auxiliary Space at many places. Following are the correct definitions of Auxiliary Space and Space Complexity.
Auxiliary Space is the extra space or temporary space used by an algorithm.
Space Complexity of an algorithm is total space taken by the algorithm with respect to the input size. Space complexity includes both Auxiliary space and space used by input.
We can also say that the way in which the amount of storage space required by an algorithm varies with the size of the problem it is solving. Space complexity is normally expressed as an order of
magnitude, e.g. O(N^2) means that if the size of the problem (N) doubles then four times as much working storage will be needed.
For example, if we want to compare standard sorting algorithms on the basis of space, then Auxiliary Space would be a better criteria than Space Complexity. Merge Sort uses O(n) auxiliary space,
Insertion sort and Heap Sort use O(1) auxiliary space. Space complexity of all these sorting algorithms is O(n) though.
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How to count excel cells that contain specific part of text ?
To count the number of cells that contain specific part of text, you can use the function COUNTIF and you have to include
into criteria. For example to count the number of cells, that include "London", you have to use the following formula:
This formula will find 4 matching cells. It counts the cells, that contain "London" at the beginning or at the end or in the middle. | {"url":"https://www.answertabs.com/how-to-count-excel-cells-that-contain-specific-part-of-text/","timestamp":"2024-11-02T18:08:13Z","content_type":"text/html","content_length":"12853","record_id":"<urn:uuid:a3a0c660-b9ef-4487-98cb-db54135967b8>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00854.warc.gz"} |
Stretching it a bit – Nonlinear Fibre Optics
The following is a prize-winning entry in the competition “Excellence in Communication” organised by Aston University in 2012. You can find all the prize winning entries here.
Imagine that all materials are made of springs, and networks of springs. For example, if you give a blob of jelly a nudge, it wobbles around a bit and then it relaxes. Remarkably, light nudges all
materials too, but at the atomic scale. We don’t perceive it as motion of the object, but we perceive it as the colour of the object. The question now is what happens if light, instead of a gentle
nudge, gives the material a vigorous shake-up?
Well, interesting things happen.
Imagine you are playing with a paddle pong. If you hit the pong softly, all is hunky-dory. But if you hit the pong a bit too hard, you wouldn’t be able to exactly predict how the pong would behave –
the rubber band starts to respond nonlinearly to the applied forces.
Nonlinear behaviour of a simple pendulum – The black pendulum is the small angle approximation, and the lighter gray pendulum (initially hidden behind) is the exact solution. For a large initial
angle, the difference between the small angle approximation (black) and the exact solution (light gray) becomes apparent almost immediately – from https://www.acs.psu.edu/drussell/Demos/Pendulum/
Essentially the same happens with atoms and light. When light interacts with atoms, it sets its electrons into oscillation. In the study of nonlinear optics, we hit the atoms a bit harder than usual
and see how the atomic springs interact with light. Our paddle though, is a cool laser.
But even if we have a laser, to hit the atoms hard enough we need a lot of energy focused in one place. We could use a magnifying glass (or a system of lenses) in principle to obtain a tight focus,
but the rays would diverge beyond it, and the energy density will fall.
Optical fibres help us to walk around this hurdle. These fibres confine light within themselves by the principle of total internal reflection, but with two added advantages. One – the light can
travel inside the fibre for kilometres without losing much of its energy. Two – and more importantly – it is confined to dimensions of the order of 7 to 8 microns – ten times smaller than the
diameter of the human hair.
Thus we have a medium in which we can confine a lot of light energy in a tiny space, and then make it travel for kilometres. This increases the interaction of light with the medium. So we ‘pump in’
light from a high power laser through one end of a very long optical fibre, and study what happens to it after it travels a substantial distance within it.
This essentially is the study of nonlinear fibre optics. It started with the question ‘what if…?’, yet it has resulted in many real world applications. For example, it is possible to amplify a weak
signal of one colour in a fibre, by making it interact with a stronger light signal of a different colour – a technique that is used in fibre communications. By pumping in a bit more energy than
usual, one can produce light of different colours (called higher harmonic generation), or even a super-continuum of colours (as shown in the image at the top of this blog), spanning tens of
Research into nonlinear fibre optics has also spawned fast, pulsed output fibre lasers, which are routinely used in surgery and industrial applications. At Aston University, we study the nonlinear
phenomena in optical fibres in depth, with state of the art equipment. Quite literally, by pushing the limits we hope to tap into Nature’s hidden secrets, and inch closer to understanding why things
are the way they are.
Cover image – https://www.laserfocusworld.com/articles/print/volume-52/issue-06/features/microstructured-fiber-photonic-crystal-fibers-advance-supercontinuum-generation.html
Nonlinear Optics Fibre Optics – Stretching it a bit by Srikanth Sugavanam is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
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1. Brilliant!
2. […] நேறியல் இல்லாத (Nonlinear) ஃபைபர் ஆப்ட்டிக்ஸ் என்றால் என்ன? by Srikanth Sugavanam, N. Sugavanam is licensed under a Creative Commons Attribution 4.0 International License.Based on a work at https://
www.srikanthsugavanam.com/sci-blog/stretching-it-a-bit-nonlinear-fibre-optics/. […]
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Optimal Strategy for a Game
Optimal Strategy for a Game
Problem statement
You and your friend Ninjax are playing a game of coins. Ninjax place the 'N' number of coins in a straight line.
The rule of the game is as follows:
1. Each coin has a value associated with it.
2. It’s a two-player game played against an opponent with alternating turns.
3. At each turn, the player picks either the first or the last coin from the line and permanently removes it.
4. The value associated with the coin picked by the player adds up to the total amount the player wins.
Ninjax is a good friend of yours and asks you to start first.
Your task is to find the maximum amount you can definitely win at the end of this game.
'N' is always even number.
Ninjax is as smart as you, so he will play so as to maximize the amount he wins.
Example 1:
Let the values associated with four coins be: [9, 5, 21, 7]
Let’s say that initially, you pick 9 and Ninjax picks 7.
Then, you pick 21 and Ninjax picks 5.
So, you win a total amount of (9+21), i.e. 30.
In case you would have picked up 7 initially and Ninjax would have picked 21 (as he plays optimally).
Then, you would pick 9 and Ninjax would choose 5. So, you win a total amount of (7+9), i.e. 16, which is not the maximum you can obtain.
Thus, the maximum amount you can win is 30.
Example 2:
Let the values associated with four coins be: [20, 50, 5, 10]
Let’s say that initially, you pick 10 and Ninjax picks 20.
Then, you pick 50 and Ninjax picks 5.
So, you win a total amount of (10+50), i.e. 60.
In case you would have picked up 20 initially and Ninjax would have picked 50 (as he plays optimally).
Then, you would pick 10 and Ninjax would choose 5. So, you win a total amount of (20+10), i.e. 30, which is not the maximum you can obtain.
Thus, the maximum amount you can win is 60.
Input format:
The very first line of input contains an integer T denoting the number of test cases.
The first line of every test case contains an integer ‘N’ denoting the number of coins present in the line initially.
The second line of every test case contains ‘N’ space-separated integers denoting the values associated with the coins placed by Ninjax.
Output format:
For each test case, print the required answer in a separate line.
You do not need to print anything, it has already been taken care of. Just implement the given function.
1 <= 'T' <= 10
2 <= 'N' <= 10 ^ 3
0 <= 'VALUE' <= 10 ^ 5
Where 'T' is the number of test cases, 'N' is the number of coins and 'VALUE' is the amount on each coin.
Time Limit: 1 sec
Sample Input 1:
Sample Output 1:
Explanation For Sample Input 1:
For the first test case, you can pick the maximum value between 7 and 8, which is 8. Thus, Ninjax has to pick up 7.
So, you win a total amount of 8.
For the second test case, first, you pick 1, Ninjax picks 5. Then, you pick 30 and Ninjax picks 4, which is the only coin left. So, you win a total amount of (1 + 30) 31.
Sample Input 2:
Sample Output 2:
Reset Code
Full Screen Mode
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Reaction Rate - Chemistry Steps
General Chemistry
Average Rate of a Reaction
Chemical kinetics studies the rates of chemical reactions which is a measure of how fast they occur.
Similarly, the rate of a chemical reaction is measured as a change in the amounts of reactants or products (usually in molarity units) divided by the time the change occurred.
For example, let’s say in a reaction between hydrogen and chlorine gases, the concentration of H[2] depressed from 2 mol/l to 1.4 mol/l in 180 seconds.
H[2](g) + Cl[2](g) → 2HCl(g)
The rate of the reaction can be written as:
\[{\rm{rate}}\;{\rm{ = }}\;{\rm{ – }}\;\frac{{{\rm{\Delta }}\left[ {{{\rm{H}}_{\rm{2}}}} \right]}}{{{\rm{\Delta t}}}}\;{\rm{ = }}\; – \frac{{\left( {{\rm{1}}{\rm{.4}}\;{\rm{ – }}\;{\rm{2}}} \right)\,
{\rm{mol/L}}}}{{{\rm{50}}\;{\rm{s}}}}\, = \,0.012\,M/s\]
Note that we add a negative sign because the concentration of hydrogen decreases and without adding it, the rate would also be negative. This applies for the rate of disappearance of any reactant and
therefore, a negative sign is always added.
Now, let’s look at how the concentration and the rate of formation of the product HCl can be represented. It is important to pay attention to the stoichiometric ratio of the reactants and components
in the chemical equation.
For example, in this reaction, the molar ratio of H[2 ]and HCl is 1:2 which means there is always going to be two times more HCl forming than H[2] reacting.
If 0.6 mol/L H[2] reacted in 50 s, there is going to be 1.2 mol/l HCl formed and if we divide this number by 50 s, the rate would be 0.024 M/s. We got two different numbers and the question is which
one is the correct rate of the reaction?
So here is the thing, when we divide the change of concentration of any reactant or product, we are getting its depletion or formation rate which can be different if their coefficients are different.
Therefore, to keep the reaction rate applicable and identical to all the reactants, and components, we divide their rates of depletion and formation by their coefficients in the chemical reaction.
The general formula for the reaction rate, based on the coefficients, can be written as:
aA + bB → cC + dD
So, for our reaction, the reaction rate would be:
H[2](g) + Cl[2](g) → 2HCl(g)
\[rate\; = \;{\rm{ – }}\;\frac{{{\rm{\Delta }}\left[ {{{\rm{H}}_{\rm{2}}}} \right]}}{{{\rm{\Delta t}}}}\;{\rm{ = – }}\;\frac{{{\rm{\Delta }}\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{{\rm{\
Delta t}}}}\;{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\frac{{{\rm{\Delta }}\left[ {{\rm{HCl}}} \right]}}{{{\rm{\Delta t}}}}\;\]
And if we plug the numbers, the rate of the reaction is now identical regardless of which component ewe use to calculate it:
\[{\rm{rate}}\;{\rm{ = }}\;{\rm{ – }}\;\frac{{{\rm{\Delta }}\left[ {{{\rm{H}}_{\rm{2}}}} \right]}}{{{\rm{\Delta t}}}}\;{\rm{ = }}\; – \frac{{\left( {{\rm{1}}{\rm{.4}}\;{\rm{ – }}\;{\rm{2}}} \right)\,
{\rm{mol/L}}}}{{{\rm{50}}\;{\rm{s}}}}\, = \,0.012\,M/s\]
\[rate\; = \;\frac{{\rm{1}}}{{\rm{2}}}\frac{{{\rm{\Delta }}\left[ {{\rm{HCl}}} \right]}}{{{\rm{\Delta t}}}}\; = \;\frac{{\rm{1}}}{{\rm{2}}}\frac{{{\rm{1}}{\rm{.2}}\;{\rm{mol/L}}}}{{{\rm{50}}\,{\rm
{s}}}}\, = \;0.012\,M/s\]
Instantaneous Rate
The rate of the reaction is not constant, and it may be changing with the concentration of the reactants. Just like when we divide the distance we drove by the time we spent on it, we get the average
speed which was not necessarily constant all the time. Therefore, what we discussed so far is the average rate of the reaction and the equation using the coefficients is to calculate the average rate
of the reaction.
To describe the rate of the reaction at specific moment, we use what is called the instantaneous rate of a reaction. The instantaneous rate of a reaction is the rate at a particular instant during
the reaction. It can be determined from the slope of the curve at a particular point in time.
For example, let’s say we are studying the rate of the following hypothetical reaction:
A + B → C + D
To determine the instantaneous rate at, for example, 80 s, we draw the tangent for that pointing going connecting any distinguishable time points and calculate the slope of the tangent:
In this case, we draw the tangent from 60 to 120 s, so to determine the instantaneous rate, we locate the points showing the concentration change of reactant “A” and it looks to be 38 mol/L and 22
The slope is then calculated by the ratio Δy/ Δx:
\[{\rm{slope}}\;{\rm{ = }}\;\frac{{{\rm{\Delta y}}}}{{{\rm{\Delta x}}}}\;{\rm{ = }}\;\frac{{{\rm{\Delta [A]}}}}{{{\rm{\Delta T}}}}\;{\rm{ = }}\; – \frac{{{\rm{(22 – 38)}}\,M}}{{\left( {{\rm{120}}\,{\
rm{ – }}\,{\rm{60}}} \right)\,s}}{\rm{ = }}\;{\rm{0}}{\rm{.27}}\,M/s\]
*Just noticed that I forget to change the concretions to smaller numbers as molarity couldn’t be 80 M. This does not change anything we discussed conceptually and the only difference is that smaller
concentrations would give smaller rates.
The instantaneous rate at the beginning of the reaction (t = 0) is called the initial rate of the reaction.
The tangent line falls from [A]= 80 M to 50 M in the time change from 0 s to ~37 s. Therefore, the initial rate would be:
\[{\rm{slope}}\;{\rm{ = }}\;\frac{{{\rm{\Delta y}}}}{{{\rm{\Delta x}}}}\;{\rm{ = }}\;\frac{{{\rm{\Delta [A]}}}}{{{\rm{\Delta T}}}}\;{\rm{ = }}\; – \frac{{{\rm{(50 – 80)}}\,M}}{{\left( {{\rm{37}}\,{\
rm{ – }}\,{\rm{0}}} \right)\,s}}{\rm{ = }}\;{\rm{0}}{\rm{.81}}\,M/s\]
Comparing the initial rate with the rate at 80s, we can see that it goes down as the concentration decreases. This is a general trend for most reactions; the rate is faster in the beginning when
there is more reactants present and it slows down as the concentrations decrease. For some reactions though, the rate does not change with concentration.
Now, how exactly the rate depends on the concentration is described by rate laws which we will discuss in the next article.
Here is a 77-question, Multiple-Choice Quiz on Chemical Kinetics:
Check Also
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CSCI567 Homework #4
1 Boosting (20 Points)
In this problem, you will develop an alternative way of forward stagewise boosting. The overall
goal is to derive an algorithm for choosing the best weak learner ht at each step such that it best
approximates the gradient of the loss function with respect to the current prediction of labels. In
particular, consider a binary classification task of predicting labels yi ∈ {+1, −1} for instances
xi ∈ R
, for i = 1, . . . , n. We also have access to a set of weak learners denoted by H = {hj , j =
1, . . . , M}. In this framework, we first choose a loss function L(yi
, yˆi) in terms of current labels and
the true labels, e.g. least squares loss L(yi
, yˆi) = (yi −yˆi)
. Then we consider the gradient gi of the
cost function L(yi
, yˆi) with respect to the current predictions ˆyi on each instance, i.e. gi =
We take the following steps for boosting:
(a) Gradient Calculation (4 points) In this step, we calculate the gradients gi =
(b) Weak Learner Selection (8 points) We then choose the next learner to be the one that can
best predict these gradients, i.e. we choose
∗ = arg minh∈H
(−gi − γh(xi))2
We can show that the optimal value of the step size γ can be computed in the closed form in
this step, thus the selection rule for h
can be derived independent of γ.
(c) Step Size Selection (8 points) We then select the step size α
that minimizes the loss:
∗ = arg minα∈R
, yˆi + αh∗
For the squared loss function, α
should be computed analytically in terms of yi
, ˆyi
, and h
Finally, we perform the following updating step:
yˆi ← yˆi + α
In this question, you have to derive all the steps for squared loss function L(yi
, yˆi) = (yi − yˆi)
2 Neural Networks (20 Points)
(a) (8 points) Show that a neural network with a single logistic output and with linear activation
functions in the hidden layers (possibly with multiple hidden layers) is equivalent to the
logistic regression.
(b) (12 points) Consider the neural network in figure 1 with one hidden layer. Each hidden
layer is defined as zk = tanh(P3
i=1 wkixi) for k = 1, . . . , 4 and the outputs are defined as
yj =
k=1 vjkzk for j = 1, 2. Suppose we choose the squared loss function for every pair,
i.e. L(y, yb) = 1
(y1 − yb1)
2 + (y2 − yb2)
, where yj and ybj represent the true outputs and our
estimations, respectively. Write down the backpropagation updates for estimation of wki and
Figure 1: A neural network with one hidden layer
Deep Learning (60 Points)
In this programming problem, you will be introduced to deep learning via hands on experimentation. We will explore the effects of different activation functions, training techniques, architectures
and parameters in neural networks by training networks with different architectures and hyperparameters for a classification task.
For this homework, we highly recommend using the Google Cloud to run your code since training
neural networks can take several tens of hours on personal laptops. You will need all the multi-core
speedup you can get, to speed things up. We will only work with Python this time (no MATLAB),
since all the deep learning libraries we need, are freely available only for Python.
There is an accompanying code file along with this homework titled hw_utils.py. It contains four
functions which are all the functions you will need for the homework. You will not have to write any
deep learning code by yourself for this homework, instead you will just call these helper functions
with different parameter settings. Go over the file hw_utils.py and understand what each of the
helper functions do.
(a) Libraries: Launch a virtual machine on the Google Cloud (please use a 64-bit machine with
Ubuntu 16.04 LTS and the maximum number of CPU cores you can get). Begin by updating
the package list and then installing libraries
• Update package list: sudo apt-get update
• Python Package Manager (pip): sudo apt-get install python-pip
• Numpy and Scipy: Standard numerical computation libraries in Python. Install with:
sudo apt-get install python-numpy python-scipy
• Theano: Analytical math engine. Install with:
sudo apt-get install python-dev python-nose g++ libopenblas-dev git
sudo pip install Theano
• Keras: A popular Deep Learning library. Install with:
sudo pip install keras
• Screen: For saving your session so that you can run code on the virtual machine even
when you are logged out of it. Install with:
sudo apt-get install screen
Next, configure Keras to use Theano as its backend (by default, it uses TensorFlow). Open
the Keras config file and change the backend field from tensorflow to theano. The Keras
config file keras.json can be edited on the terminal with nano:
nano ~/.keras/keras.json
(b) Useful information for homework: We will only use fully connected layers for this homework in all networks. We will refer to network architecture in the format: [n1, n2, · · · , nL]
which defines a network having L layers, with n1 being the input size, nL being the output
size, and the others being hidden layer sizes, e.g. the network in figure 1 has architecture:
[3, 4, 2].
Checkout the various activation functions for neural networks namely linear (f(x) = x),
sigmoid, ReLu and softmax. In this homework we will always use the softmax activation for
the output layer, since we are dealing with a classification task and output of softmax layer
can be treated as probabilities for multi-class classification.
Have a look at the last part of the homework (hints and tips section) before you start the
homework to get some good tips for debugging and running your code fast.
A brief description of the functions in the helper file hw_utils.py is as follows:
• genmodel(): Returns a neural network model with the requested shape, activation
function and L2-regularizer. You won’t need to call this method at all.
• loaddata(): Loads the dataset for this homework, shuffles it, generates labels, bifurcates
the data and returns the training and test sets.
• normalize(): Normalizes the training and test set features.
• testmodels(): It takes the following parameters: your training and test data, a list of
model architectures, activation function (hidden layers and last layer), list of regularization coefficients, number of epochs for stochastic gradient descent (SGD), batch size for
SGD, learning rate for SGD, list of step size decays for SGD, list of SGD momentum
parameters, boolean variable to turn nesterov momentum on/off, boolean variable to
turn early stopping on/off and another boolean variable to turn the verbose flag on/off.
The method generates a model of appropriate size and trains it on your training data.
It prints out the test set accuracy on the console. In case of list of parameters, it trains
CSCI567 Fall 2016 Homework #4 Due 11/2
networks for all possible combinations of those parameters and also reports the best configuration found (i.e. the configuration which gave the maximum test accuracy). This
is the method that you will have to call a lot in your code.
Lastly, try running the experiments multiple times if needed, since neural networks are
often subject to local minima and you might get suboptimal results in some cases.
(c) Dataset and preprocessing: We will use the MiniBooNE particle identification dataset
from the UCI Machine Learning Repository. It has 130064 instances with 50 features each
and each instance has to be classified as either ”signal” or ”background”.
Download the dataset and call loaddata() in your code to load and process it. The function
loads the data, assigns labels to each instance, shuffles the dataset and randomly divides
it into training (80%) and test (20%) sets. It also makes your training and test set labels
categorical i.e. instead of a scalar ”0” or ”1”, each label becomes a two-dimensional tuple;
the new label is (1,0) if the original label is ”0” and it is (0,1) if the original label is ”1”. The
dimension of every feature is din = 50 and the dimension of output labels is dout = 2. Next,
normalize the features of both the sets by calling normalize() in your code.
(d) Linear activations: (5 Points) First we will explore networks with linear activations. Train
models of the following architectures: [din, dout], [din, 50, dout], [din, 50, 50, dout], [din, 50,
50, 50, dout] each having linear activations for all hidden layers and softmax activation for
the last layer. Use 0.0 regularization parameter, set the number of epochs to 30, batch size
to 1000, learning rate to 0.001, decay to 0.0, momentum to 0.0, Nesterov flag to False, and
Early Stopping to False. Report the test set accuracies and comment on the pattern of test
set accuracies obtained. Next, keeping the other parameters same, train on the following
architectures: [din, 50, dout], [din, 500, dout], [din, 500, 300, dout], [din, 800, 500, 300, dout],
[din, 800, 800, 500, 300, dout]. Report the observations and explain the pattern of test set
accuracies obtained. Also report the time taken to train these new set of architectures.
(e) Sigmoid activation: (5 Points) Next let us try sigmoid activations. We will only explore
the bigger architectures though. Train models of the following architectures: [din, 50, dout],
[din, 500, dout], [din, 500, 300, dout], [din, 800, 500, 300, dout], [din, 800, 800, 500, 300, dout];
all hidden layers with sigmoids and output layer with softmax. Keep all other parameters
the same as with linear activations. Report your test set accuracies and comment on the
trend of accuracies obtained with changing model architectures. Also explain why this trend
is different from that of linear activations. Report and compare the time taken to train these
architectures with those for linear architectures.
(f) ReLu activation: (5 Points) Repeat the above part with ReLu activations for the hidden
layers (output layer = softmax). Keep all other parameters and architectures the same, except
change the learning rate to 5 × 10−4
. Report your observations and explain the trend again.
Also explain why this trend is different from that of linear activations. Report and compare
the time taken to train these architectures with those for linear and sigmoid architectures.
(g) L2-Regularization: (5 Points) Next we will try to apply regularization to our network. For
this part we will use a deep network with four layers: [din, 800, 500, 300, dout]; all hidden
activations ReLu and output activation softmax. Keeping all other parameters same as for
CSCI567 Fall 2016 Homework #4 Due 11/2
the previous part, train this network for the following set of L2-regularization parameters:
, 5 × 10−7
, 10−6
, 5 × 10−6
, 10−5
]. Report your accuracies on the test set and explain the
trend of observations. Report the best value of the regularization hyperparameter.
(h) Early Stopping and L2-regularization: (5 Points) To prevent overfitting, we will next
apply early stopping techniques. For early stopping, we reserve a portion of our data as a
validation set and if the error starts increasing on it, we stop our training earlier than the
provided number of iterations. We will use 10% of our training data as a validation set
and stop if the error on the validation set goes up consecutively six times. Train the same
architecture as the last part, with the same set of L2-regularization coefficients, but this
time set the Early Stopping flag in the call to testmodels() as True. Again report your
accuracies on the test set and explain the trend of observations. Report the best value of the
regularization hyperparameter this time. Is it the same as with only L2-regularization? Did
early stopping help?
(i) SGD with weight decay: (5 Points) During gradient descent, it is often a good idea to
start with a big value of the learning rate (α) and then reduce it as the number of iterations
progress i.e.
αt =
1 + βt
In this part we will experiment with the decay factor β. Use the network [din, 800, 500,
300, dout]; all hidden activations ReLu and output activation softmax. Use a regularization
coefficient = 5 × 10−7, number of epochs = 100, batch size = 1000, learning rate = 10−5
and a list of decays: [10−5
, 5 × 10−5
, 10−4
, 3 × 10−4
, 7 × 10−4
, 10−3
]. Use no momentum and
no early stopping. Report your test set accuracies for the decay parameters and choose the
best one based on your observations.
(j) Momentum: (5 Points) Read about momentum for Stochastic Gradient Descent. We will
use a variant of basic momentum techniques called the Nesterov momentum. Train the
same architecture as in the previous part (with ReLu hidden activations and softmax final
activation) with the following parameters: regularization coefficient = 0.0, number of epochs
= 50, batch size = 1000, learning rate = 10−5
, decay = best value found in last part, Nesterov
= True, Early Stopping = False and a list of momentum coefficients = [0.99, 0.98, 0.95, 0.9,
0.85]. Find the best value for the momentum coefficients, which gives the maximum test set
(k) Combining the above: (10 Points) Now train the above architecture: [din, 800, 500, 300,
dout] (hidden activations: ReLu and output activation softmax) again, but this time we will
use the optimal values of the parameters found in the previous parts. Concretely, use number
of epochs = 100, batch size = 1000, learning rate = 10−5
, Nesterov = True and Early Stopping
= True. For regularization coefficient, decay and momentum coefficient use the best values
that you found in the last few parts. Report your test set accuracy again. Is it better or
worse than the accuracies you observed in the last few parts?
(l) Grid search with cross-validation: (15 Points) This time we will do a full fledged search
for the best architecture and parameter combinations. Train networks with architectures
[din, 50, dout], [din, 500, dout], [din, 500, 300, dout], [din, 800, 500, 300, dout], [din, 800, 800,
CSCI567 Fall 2016 Homework #4 Due 11/2
500, 300, dout]; hidden activations ReLu and final activation softmax. For each network use
the following parameter values: number of epochs = 100, batch size = 1000, learning rate
= 10−5
, Nesterov = True, Early Stopping = True, Momentum coefficient = 0.99 (this is
mostly independent of other values, so we can directly use it without including it in the
hyperparameter search). For the other parameters search the full lists: for regularization
coefficients = [10−7
, 5 × 10−7
, 10−6
, 5 × 10−6
, 10−5
], and for decays = [10−5
, 5 × 10−5
, 10−4
Report the best parameter values, architecture and the best test set accuracy obtained.
Hints and tips:
• You can use FTP clients like FileZilla for transferring code and data to and fro from the
virtual machine on the Google Cloud.
• Always use a screen session on the virtual machine to run your code. That way you can
logout of the VM without having to terminate your code. A new screen session can be started
on the console by: screen -S . An existing attached screen can be detached
by pressing ctrl+a followed by d. You can attach to a previously launched screen session by
typing: screen -r . Checkout the basic screen tutorial for more commands.
• Don’t use the full dataset initially. Use a small sample of it to write and debug your code on
your personal machine. Then transfer the code and the dataset to the virtual machine, and
run the code with the full dataset on the cloud.
• While running your code, monitor your CPU usage using the top command on another
instance of terminal. Make sure that if you asked for 24 CPUs, your usage for the python
process is showing up to be around 2400% and not 100%. If top consistently shows 100%,
then you need to setup your numpy, scipy and theano to use multiple cores. Theano (and
Keras) by default make use of all cores, but numpy and scipy will not. Since raw numpy and
scipy computations will only form a very small part of your program, you can ignore their
parallelization for the most part, but if you so require you can google how to use multiple
cores with numpy and set it up.
• Setting the verbose flag for fit() and evaluate() methods in Keras as 1 gives you detailed
information while training. You can tweak this by passing verbose=1 in the testmodels()
method in the hw_utils.py file.
• Remember to save your results to your local machine and turn off your machine after you
have finished the homework to avoid getting charged unnecessarily.
Submission Instructions: You need to submit a soft copy and a hard copy of your solutions.
• All solutions must be typed into a pdf report (named CSCI567 hw4 fall16.pdf). If you
choose handwriting instead of typing, you will get 40% points deducted.
• For code, the only acceptable language is Python2.7. You MUST include a main script
called CSCI567 hw4 fall16.py in the root directory of your code folder. After running
this main script, your program should be able to generate all the results needed for the
programming assignment, either as plots or console outputs. You can have multiple files
(i.e your sub-functions), however, once we execute the main file in your code folder, your
program should execute correctly.
• The soft copy should be a single zip file named [lastname] [firstname] hw4 fall16.zip.
It should contain your pdf report (named CSCI567 hw4 fall16.pdf) having answers to
all the problems, and the folder containing all your code. It must be submitted via
Blackboard by 11:59pm of the deadline date.
• The hard copy should be a printout of the report CSCI567 hw4 fall16.pdf and must
be submitted to locker #19 at PHE building 1st floor by 5:00pm of the deadline date.
Collaboration You may collaborate. However, collaboration has to be limited to discussion only
and you need to write your own solutions and submit separately. You also need to list the
names of people with whom you have discussed. | {"url":"https://codingprolab.com/answer/csci567-homework-4/","timestamp":"2024-11-12T02:13:54Z","content_type":"text/html","content_length":"139764","record_id":"<urn:uuid:4f5aa3f0-b19c-4efe-9320-671330bb0514>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00880.warc.gz"} |
Square Roots of Perfect Squares
Lesson Video: Square Roots of Perfect Squares Mathematics • First Year of Preparatory School
In this video, we will learn how to find square roots of perfect square integers, fractions, and decimals.
Video Transcript
Let’s take a look at how we would find the square root of a perfect square. You might ask the question, what is a perfect square? Before we answer that, lets back up and answer, what is the square of
any number? The square of a number is the product of a number and itself. A perfect square is a product of an integer and itself. Let’s look at some examples of a perfect square. Here are a few
Twenty-five is the product of five times five. Twenty-five equals five squared. Four is also a perfect square. Four is the product of two times two, also known as two squared. Can you think of an
integer multiplied by itself that equals forty-nine? It’s seven, seven times seven equals forty-nine. Forty-nine equals seven squared. And now for one hundred, do you have any ideas? Ten times ten
equals one hundred, or ten squared.
If you remember the title of the video though, square roots of perfect squares, we’re not just talking about perfect squares. We wanna talk about how to find the square roots of perfect squares.
Let’s start by defining square roots. The factors multiplied to form squares are called square roots. Let me read that one more time. The factors multiplied to form squares are called square roots.
Let’s go back to our example of twenty-five. Twenty-five equals five squared, or five times five. The factors multiplied to form the square of twenty-five is a five. So we say, the square root of
twenty-five is five.
We use this symbol to denote finding the square root of something. This symbol is called a radical sign. These are the symbols we would use if we wanted to say the square root of twenty-five is five.
First, we have the radical sign, put the perfect square inside the radical, and our solution, five, is the square root of twenty-five. Here are two examples.
The first one says, find the square root of eighty-one. And the second one says, find the square root of two hundred and twenty-five.
Let’s start here. We know we’re looking for some integer multiplied by itself. I know ten multiplied by itself equals one hundred and that eighty-one is smaller than that. Then I recognize nine times
nine equals eighty-one; the square root of eighty-one has to be nine. So we have the final answer. The square root of eighty-one equals nine. Let’s try our next example.
You’re thinking the square root of two hundred and twenty-five is some number multiplied by itself that equals two hundred and twenty-five. And twelve times twelve equals one hundred and forty-four,
thirteen times thirteen equals one hundred and sixty-nine. You probably memorized those values at some point. But now you’re just wondering what can I do, do I have to keep guessing and checking my
answer? There are some strategies that can help you solve this mentally.
Notice that in twelve times twelve equals one hundred and forty-four, you see that two times two equals four and that’s the last digit in the number. You can also notice in thirteen times thirteen
equals one hundred and sixty-nine, the same pattern is there. So we’re gonna be looking for something that multiplies together and has a five in that position. If you look at one times one, two times
two, three times three, four times four, all the way to five times five, the only thing that ends with an integer of five is five times five.
We’re looking for a number that’s larger than thirteen and ends in five. So it would be smart to check fifteen as your next value. In fact, fifteen times fifteen does equal two hundred and
twenty-five, which makes the square root of two hundred and twenty-five, fifteen. Now, you’re ready to use mental math strategies to recognize and find the square roots of perfect squares. | {"url":"https://www.nagwa.com/en/videos/947135930431/","timestamp":"2024-11-06T09:06:10Z","content_type":"text/html","content_length":"255091","record_id":"<urn:uuid:9be21531-4820-4245-8e2d-bb1d9f12bd8e>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00298.warc.gz"} |
MATLAB Toolbox | Guide to Different Toolbox in Matlab with Examples
Updated July 3, 2023
Introduction to MATLAB Toolbox
In this article, we will study toolboxes in MATLAB. The toolboxes in MATLAB comprise a wide range of functions that are integrated into MATLAB’s computing environment.
Here are a few toolboxes in MATLAB:
1. ‘Curve Fitting’
2. ‘Regression learner’
3. ‘Image processing’
These toolboxes can be accessed using the ‘APPS’ icon in the MATLAB ribbon. Let us now understand the use of a couple of toolboxes in MATLAB:
Curve Fitting Toolbox
• The curve fitting toolbox enables users to fit surfaces and curves to input data by employing techniques such as interpolation, regression, and smoothing.
• This Toolbox provides us with functions and an application to fit curves to our data.
• This toolbox is very helpful in data analytics as it helps in performing EDA (exploratory data analysis), data processing, and removing outliers
Let us now understand using the Curve fitting toolbox using an example.
In this example, we will use 3 metrics ‘x’, ‘y’, ‘z’ and will fit a curve in them using the Curve fitting toolbox. We will follow the following steps:
1. Create the 3 matrices using rand function
2. Set the ‘X Data’, ‘Y Data’, ‘Z Data’ in Curve fitting tool to our inputs, ‘x,’ ‘y’, ‘z’, respectively
Code: (to be executed in Command Window)
x = rand (5)
y = rand (5)
z = rand (5)
[Creating the 3 input matrices]
Once we execute the above code in ‘Command Window’, we will get the 3 variables created in our ‘WORKSPACE’.
Steps to Use Curve Fitting toolbox
Step 1: Click on APPS icon
Step 2: Select ‘Curve Fitting Tool’
Step 3: A pop-up window will open like below:
Step 4: Now set the ‘X Data’, ‘Y Data’, and ‘Z Data’ in this pop-up window to our inputs, ‘x,’ ‘y’, and ‘z,’ respectively. We can immediately see that Curve Fitting Toolbox will create a curve. The
equation for this curve can be seen in the Result section. We can use a custom equation using the dropdown on the top of the curve.
Next, let us learn how Regression Learner Toolbox works in MATLAB
Regression Learner Toolbox
• The Regression Learner toolbox is used to perform regression
• It is used to train a model automatically
• It can also be used to compare different options amongst linear regression, support vector machines, regression trees & visualize the results
Let us now understand the use of the Regression Learner toolbox using an example.
In this example, we will use an inbuilt dataset provided by MATLAB, ‘carbig’. We will upload this dataset to the ‘Regression Learner Toolbox’ and explore possible options. We will follow the
following steps:
• Load the inbuilt dataset ‘carbig’
• Create a table using this dataset to load it into ‘Regression Learner Toolbox’
Code: (to be executed in Command Window)
load carbig
[Loading the ‘carbig’ dataset into the Workspace]
newTable = table (Cylinders, Acceleration, Displacement,...
Model_Year, Horsepower, Weight, Origin);
[Creating the table using the dataset to make it compatible with Regression Learner Toolbox]
Once we execute the above code in ‘Command Window’, we will get the ‘newTable’ created in our ‘WORKSPACE.’
Steps to Use Regression Learner Toolbox
Step 1: Click on APPS icon
Step 2: Select ‘Regression Learner Toolbox’
Step 3: A pop-up window will open like below:
Step 4: Click on New Session in the left, which will open a new window prompt
Step 5: From the ‘Data Set Variable’ dropdown, select the ‘newTable’ table created by us
Step 6: This will load all the predictor variables under the section ‘Predictors’
Step 7: Now we can select the predictor variables as per our requirement
Step 8: Click on ‘Start Session’ to start analyzing the data
We can immediately see a response plot created by Regression Learner Toolbox. As per our requirement, we can train this data and get a response plot, residual plot, and min MSE plot using the
available options.
Let us now understand the use of the Image processing toolbox using an example.
Image Processing Toolbox
Below we will learn about the image processing toolbox:
In this example, we will use one of the inbuilt images provided by MATLAB, ‘moon.tiff. We will upload this dataset to ‘Image Processing Toolbox’ and explore possible options.
moonImage = imread ('moon.tif')
[Loading the ‘moon.tif’ image into the Workspace]
imtool (moonImage)
[Using ‘imtool’ function to start the Image processing toolbox. Alternatively, we can also select it from the APP section]
Once we execute the above code in ‘Command Window,’ we will get the ‘moonImage’ in our ‘WORKSPACE’.
Options Provided by Image Processing Toolbox
• Pixel Information
• Distance between the 2 pixels
• Details about the image
• Adjust contrast
• Crop Image
• Zoom tool
• Scroll bars
As we can see in the output, we have obtained an image of the moon that can be processed using the icons in the ribbon.
Conclusion – MATLAB Toolbox
The toolboxes in MATLAB are a collection of numerous functions MATLAB provides various toolboxes to perform functionalities like data analytics, image processing, curve fitting, etc.
Recommended Articles
This is a guide to MATLAB Toolbox. Here we discuss the three different toolboxes in MatLab with examples and outputs. You can also go through our other related articles to learn more – | {"url":"https://www.educba.com/matlab-toolbox/","timestamp":"2024-11-06T05:52:03Z","content_type":"text/html","content_length":"316301","record_id":"<urn:uuid:004c26da-6003-40ef-bc22-a75fa235badf>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00663.warc.gz"} |
Bracing a Two Storey Building
Set the Wind & Earthquake Zone…
1. Navigate to the Plan view that you wish to brace. From the Cadimage menu choose Bracing > Wind Earthquake Zone.
2. In Wind & Earthquake Zones pop up window set the Wind Zone, Earthquake Zone & Soil Class.
Note that there are two ways to calculate the Wind Zone, use the arrows in the centre of the dialog to choose one method or the other.
Brace the lower storey (Lower of two):
3. Navigate to the Plan view that you wish to brace. Select the walls of the building you want to brace, and from the Cadimage menu choose Bracing > Calculate Bracing.
A Bracing Settings dialog will come up.
4. Expand the Calculation Stages settings, and create a new Calculation Set - Enter in the ID and a description.
If you need to add a second set of braces, lines and table anywhere in your project you’ll need to use a different calculation set for each one. (see advanced section)
5. In the Calculation Stages settings, and choose the Storey Location. (Lower of two)
• Choose to Place Brace Lines and to Place Braces.
• Choose a Brace Type Set from the list.
6. Expand the Building Dimensions settings and enter the dimensions and roof pitch of the building.
Set the angle that is parallel with the length of the building. This angle is measured from the horizontal in a counter-clockwise direction. For the example building, the angle is zero.
7. Expand the Cladding Weights settings and choose cladding weights for Roof, Upper Cladding and Subfloor. Also, select the Floor Load and Construction Settings.
8. Expand the Brace Settings settings and set:
• Maximum Brace Length.
• Minimum Brace Length.
• Truncate Braces To - this setting ensures that brace lengths will be set to sensible values like 1.3, 1.8 rather than 1.304, 1.796.
Also set the stud height for braces, and choose whether or not braces can be reduced in height for placement on shorter walls.
8. Finally, choose the layer on which to place all the bracing symbols, and hit OK
Brace lines will be placed on the building, braces placed on those lines, so as to satisfy the requirements of NZS3604.
9. You must then indicate the location of the Brace Table, Click to place the brace table. This will be the top left-hand corner.
If there was a problem, and the required bracing could not be achieved, the Brace Table will report this, as will messages that come up throughout the process. See advanced section on how to add/edit
your braces
Once the table has been placed move to the next storey to continue bracing your building.
Brace the Upper storey (Upper of Two):
10. Navigate to the Plan view that you wish to brace. Select the walls of the building you want to brace, and from the Cadimage menu choose Bracing > Calculate Bracing.
A Bracing Settings dialog will come up.
11. Expand the Calculation Stages settings, and create a new Calculation Set - Enter in the ID and a description.
If you need to add a second set of braces, lines and table anywhere in your project you’ll need to use a different calculation set for each one. (see advanced section)
12. In the Calculation Stages settings, and choose the Storey Location. Upper of Two
• Choose to Place Brace Lines and to Place Braces.
• Choose a Brace Type Set from the list.
13. Expand the Building Dimensions settings and enter the dimensions and roof pitch of the building.
Set the angle that is parallel with the length of the building. This angle is measured from the horizontal in a counter-clockwise direction. For the example building, the angle is zero.
14. Expand the Cladding Weights settings and choose cladding weights for Roof, Upper Cladding and Subfloor. Also, select the Floor Load and Construction Settings.
15. Expand the Brace Settings settings and set:
• Maximum Brace Length.
• Minimum Brace Length.
• Truncate Braces To - this setting ensures that brace lengths will be set to sensible values like 1.3, 1.8 rather than 1.304, 1.796.
Also, set the stud height for braces, and choose whether or not braces can be reduced in height for placement on shorter walls.
16. Finally, choose the layer on which to place all the bracing symbols, and hit OK
Brace lines will be placed on the building, braces placed on those lines, so as to satisfy the requirements of NZS3604.
17. You must then indicate the location of the Brace Table, Click to place the brace table. This will be the top left-hand corner.
If there was a problem, and the required bracing could not be achieved, the Brace Table will report this, as will messages that come up throughout the process. See advanced section on how to add/edit
your braces.
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Article is closed for comments. | {"url":"https://cadimage.zendesk.com/hc/en-us/articles/360000547376-Bracing-a-Two-Storey-Building","timestamp":"2024-11-15T01:05:45Z","content_type":"text/html","content_length":"32912","record_id":"<urn:uuid:19076384-53c7-4275-90d9-2ac9de4d28ca>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00637.warc.gz"} |
Experiments: Measure the impact of your A/B testing - Mixpanel Docs
Experiments: Measure the impact of your A/B testing
The Experiments report analyzes how A/B test variants impact your metrics. Experiments does this by calculating the difference between variant groups and the effects of the variants on selected
Experiments requires an A/B test, its variant, and a dashboard that contains the metrics you are measuring. An experiment query calculates the variants’ effects on the dashboard metrics by
calculating the delta and the lift between the two variants.
To access Experiments, click on Applications in the top right navigation, then select Experiments.
Quick Start
Step 1: Prepare a Board
To use Experiments you must have a board that contains the various reports you wish to analyze your experiment.
Step 2: Select an Experiment
Custom Experiment - This option allows you to define the control and variant groups of the experiment. These groups can be defined by cohort, user profile property, or event property filters.
Tracked Experiments - This option is available if you have experiments in your implementation. Mixpanel automatically detects any experiments that began in the last 30 days, and the report detects
and displays them in the dropdown.
Step 3: Choose Control and Variant Group
Select the group of users that represents your control group and your variant group.
In a Custom Experiment, the control and variant groups can be a cohort of any other users filtered by events and properties.
It is important to ensure that the groups are mutually exclusive. For example, in onboarding flow testing, users exposed to the original, not new, onboarding flow should be the control. Introducing
users that could qualify under the control and variant group may abstract the report results.
Step 4: Select a Date Range
Select the date range of the experiment. In most cases, you should choose the date your experiment began as the start date.
All events tracked by users within the date range will be included in the Experiment report, even if those events took place before the experiment started.
Supported Metrics
Experiments will run calculations on the following supported metrics:
• Insights - line charts with “Total” count, including charts with breakdowns.
• Insights - line charts with “Unique” count, including charts with breakdowns.
• Insights - line chart with “Sum of property values”, including charts with breakdowns.
• Funnels - funnels with “Unique” count, including charts with breakdowns and any number of steps.
Calculation Details
The following section describes the equations used in the Experiments report.
Control and Variant Group Rate
The group rate is calculated for both control and variant groups. It is calculated differently depending on the selected metric type.
If calculating using totals in Insights, then the group rate is calculated as:
$Group\,Rate= {{ (\# \,of\,events) \over (\# \,of\,users)} \over (time)}$
If calculating using uniques in Insights, then the group rate is calculated as:
$Group\,Rate= { (\# \,of\,users\,who\,performed\,metric\,event) \over (\# of\,users\,in\,group)}$
This value is a percentage because the maximum possible value is 1. We therefore display the percentage of users in the control group who performed the metric event.
If calculating using funnels, then the rate is the overall conversion rate of the funnel for users in the group.
Lift and Lift Trend
Lift is the percentage difference between the control group and variant group rates. Lift is calculated as (variant rate - control rate) / control rate.
$Lift= { (variant \,group\,rate - control \,group\,rate) \over (control \,group\,rate)}$
You can also switch between lift and the delta, which is the absolute difference in rates, variant rate minus control rate.
Confidence is the probability that the lift or delta between your control and variant groups is significant.
For conversions we calculate a standard confidence score for binomial outcomes, and for event counts we calculate a standard confidence score for Poisson outcomes.
The trend line in the column displays how confidence has changed over the selected date range.
Adding Breakdowns
You can choose to segment all the metrics right from the Experiment report by selecting “Breakdown” -> “Select a property”, and then selecting what you want to breakdown the metrics by.
Please note: Even if a metric is already segmented, this breakdown will override the initial breakdown and show a segmented view on all the metrics by the selected property/cohort. Clicking into a
report from the Experiments report will carry forward the segmentation selected into the report.
Interpreting Results
The Experiments report locates significant differences between the Control and Variant groups. Metric rows in the table are highlighted when any difference is calculated with 95% or greater
• Positive differences, where the variant rate is higher than the control rate, are highlighted in green.
• Negative differences, where the variant rate is lower than the control rate, are highlighted in red.
• Statistically insignificant results remain gray.
Confidence Score
Confidence scores come from the hypothesis testing framework in the field of statistics. In hypothesis testing, you first choose a null hypothesis. In Mixpanel, the null hypothesis is that two
groups of users behave the same on average for a given metric. The groups of users might be variant and control groups in an A/B test, or they might just be two different cohorts of users. The
alternative hypothesis is that the two groups of users behave differently for the metric.
When Mixpanel compares a metric for two cohorts of users, we calculate the probability that we would observe a metric difference equal to or greater than the difference between the two cohorts. That
probability is called a p-value. Generally speaking, the smaller the p-value, the more likely it is that the null hypothesis is false, and the alternative hypothesis is true.
The confidence score is 1-p-value, expressed as a percentage. So the higher the confidence score, the more likely it is that the alternative hypothesis is true (meaning that the two cohorts really do
behave differently for the metric in question). We follow the traditional threshold of 95% for the confidence score, so we highlight results above 95% confidence in green for positive differences and
in red for negative differences.
Confidence Score Calculation
For event counts, we assume under the null hypothesis that each user cohort has a total event count that follows a Poisson distribution, where the parameter θ = cohort size * λ, and where λ is the
same for both cohorts. For conversion rates, we assume under the null hypothesis that each user is a Bernoulli trial with the same parameter p. For both event counts and conversion rates, Mixpanel
calculates the z-score, and the confidence score in the standard way. See this article for more information about the formulas Mixpanel uses for z-score calculations, and Poisson and binomial
Interpreting a Confidence Score
Generally speaking, higher confidence results mean that it is more likely that two cohorts of users differ significantly on your chosen metric. You can use the confidence score as a metric to quickly
interpret large numbers of results. The higher the number of metrics you are analyzing, the higher percentage of those results may be false positives.
If you are using our color-coded thresholds of 95%, there is a 5% chance that any individual result is a false positive. So if you are looking at 20 metrics at once, it is more likely that a larger
number of those metrics could be false positives. If you want more precision in decision-making, we recommend that you calculate your sample size before running an A/B test, and then only use the
results you see in the Experimentation Report once you achieve that sample size. Higher confidence results are less likely to be false positives.
Add Experiments to an Implementation
Mixpanel will automatically populate the Experiment, Control, and Variant dropdowns within the report if sent in the proper format.
Mixpanel scans for experiments that began in the date range you’ve selected for the report. If any are found, then they will appear under the “Tracked Experiments” sub-header. To do this you must
send data in the following format:
Event Name: “$experiment_started”
Event Properties:
• “Experiment name” - the name of the experiment to which the user has been exposed
• “Variant name” - the name of the variant into which the user was bucketed, for that experiment
An example track call would appear like this:
mixpanel.track('$experiment_started', {'Experiment name': 'Test', 'Variant name': 'v1'}) | {"url":"https://docs.mixpanel.com/docs/reports/apps/experiments","timestamp":"2024-11-11T00:53:50Z","content_type":"text/html","content_length":"217659","record_id":"<urn:uuid:0c59af3f-a248-4368-b0be-fac88d09c7d3>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00572.warc.gz"} |
How do you divide (2x^3+7x^2-5x-4) / (2x+1) using polynomial long division? | HIX Tutor
How do you divide #(2x^3+7x^2-5x-4) / (2x+1)# using polynomial long division?
Answer 1
Numerator #->color(white)("d")2x^3+7x^2-5x-4# #color(magenta)(x^2)(2x+1)-> color(white)("d")ul(2x^3+x^2larr" Subtract"# #color(white)("dddddddddddddd") 0+6x^2-5x-4# #color(magenta)(3x)(2x+1)-> color
(white)("dddddd") ul(6x^2+3xlarr" Subtract")# #color(white)("ddddddddddddddddddd")0-8x-4# #color(magenta)(-4)(2x+1) ->color(white)("ddddddd.d")ul(-8x-4 larr" Subtract")# #"Remainder"-> color(white)
#color(white)("dddddddddddddd")color(magenta)( x^2+3x-4 )#
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Answer 2
Quotient is ${x}^{2} + 3 x - 4$ and remainder is $0$
Hence quotient is #x^2+3x-4# and remainder is #0#
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Answer 3
To divide (2x^3+7x^2-5x-4) by (2x+1) using polynomial long division, follow these steps:
1. Arrange the dividend (2x^3+7x^2-5x-4) and the divisor (2x+1) in descending order of exponents.
2. Divide the first term of the dividend (2x^3) by the first term of the divisor (2x). The result is x^2.
3. Multiply the divisor (2x+1) by the quotient obtained in step 2 (x^2). The result is 2x^3+x^2.
4. Subtract the product obtained in step 3 (2x^3+x^2) from the dividend (2x^3+7x^2-5x-4). The result is 6x^2-5x-4.
5. Bring down the next term from the dividend (-5x).
6. Divide the first term of the new dividend (6x^2) by the first term of the divisor (2x). The result is 3x.
7. Multiply the divisor (2x+1) by the quotient obtained in step 6 (3x). The result is 6x^2+3x.
8. Subtract the product obtained in step 7 (6x^2+3x) from the new dividend (6x^2-5x-4). The result is -8x-4.
9. Bring down the next term from the dividend (-4).
10. Divide the first term of the new dividend (-8x) by the first term of the divisor (2x). The result is -4.
11. Multiply the divisor (2x+1) by the quotient obtained in step 10 (-4). The result is -8x-4.
12. Subtract the product obtained in step 11 (-8x-4) from the new dividend (-8x-4). The result is 0.
The quotient is x^2 + 3x - 4, and the remainder is 0.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Data Analysis & Statistics Workshop - Lab 5.3 High-dimensional space: interpolation; how PCA works
Lab 5.3 High-dimensional space: interpolation; how PCA works
In the last lab (5.2), we used principle component analysis to visualize a high-dimensional space. In this case, that high-dimensional space was defined by melt curves that indicated the fluorescent
activity of probes binding to sequences of single-stranded DNA. Each melt curve profile identifies a particular sequence of DNA, and can be used to detect, among other things, which strain of
tuberculosis has infected a patient, or which sequence of a gene is found in a particular cancer.
In this lab, we tackle a real-world problem in high-dimensional data analysis, which is when data are collected slightly differently and need to be aligned by interpolation.
In the second part of the lab, we will examine principle component analysis in more detail in order to reveal the mechanism by which it finds the direction of maximum variance in a sample.
Aligning data that are sampled at different intervals
Suppose you were working in the Wangh lab, and measured the melt curve in the file mycurve.txt (after normalizing and taking the 2nd derivative):
mycurve = load('mycurve.txt','-ascii');
Now the new postdoc in the lab measures the same curve and says he would like to compare the difference between his melt curve and the melt curve that you measured. The new postdoc's curve is in
othercurve.txt. First, let's plot the melt curve that the postdoc recorded:
othercurve = load('othercurve.txt','-ascii');
hold on;
plot(othercurve(1,:),othercurve(2,:),'b-'); % let's choose blue
Do the plots look similar? They do. So far, so good. The new postdoc seems to already know what he is doing in the lab. Now let's calculate the difference between the curves. Does this work?
curvediff = othercurve(2,:) - mycurve(2,:);
Why doesn't it work? Let's look at the number of data points:
disp(['The size of my curve is ' mat2str(size(mycurve)) '.']);
disp(['The size of my curve is ' mat2str(size(othercurve)) '.']);
What?? The postdoc's curve has several more data points. Let's look in more detail. The temperature points that you both ran are the following:
Q1: Are these temperature points the same? Is the same basic range covered?
Rats!! The postdoc collected perfectly reasonable data but the temperature points don't line up. Either you all didn't agree on the data points that should have been measured before hand, or,
perhaps, the machine doesn't allow you to specify the actual temperature points that are recorded. Either way, it's out of your control now. And, why would you want to limit yourself to particular
temperature values? It seems a little lame.
So how can we compare these curves?
Fortunately, there is a family of methods that will allow us to make a reasonable "resampling" of the data called interpolation. It allows us to interpolate the values of our data between the actual
points that were measured. As long as we have sampled our data reasonably densely, and we assume that the data does not change too much between adjacent points, we can do this without impacting the
quality of our data analysis.
You may be unaware but you are already familiar with a form of interpolation. When Matlab plots lines between data points, it is doing a form of interpolation. Let's re-plot mycurve:
figure(101); % leave this figure open
box off;
Examine the plot closely. The actual data points are plotted with the plots, but Matlab also shows the points connected by lines. That is, it is displaying the values between the points as varying
linearly between the previous data point and the next data point. There is no data in your experiment that suggests this is a proper and good thing to do, the lines are fiction, totally made up.
Sometimes they are helpful, as they help us to visualize a set of discrete points as a curve, but other times they hurt, as they encourage us to imagine what might be going on between the data points
when we might not have the proper information.
Q2: Can you see the distinction between the data points and the lines that connect them in a standard Matlab line plot?
Okay, let's make our own interpolation of our data. Let's increase the density of our temperature axis points by creating a data point every for every 0.5 degrees C instead of roughly every degree.
To do this, we will use the Matlab function interp1 (see help interp1).
myT = mycurve(1,1):0.5:mycurve(1,end) % create a new vector of temperature points
myF = interp1(mycurve(1,:),mycurve(2,:),myT,'linear');
mynewcurve = [myT; myF];
hold on;
Q3: Do you see the extra data points in the new graph?
There are multiple algoritms for interpolation (see help interp1). You can use "nearest neighbor", which performs no interpolation but selects the value of the closest data point. This method has the
advantage that it makes no assumptions about the data between the raw data points. Another common algorithm is the cubic spline, which are piecewise polynomial fits that are relatively smooth near
the actual data points (see Wikipedia article).
Let's try cubic splines:
myT2 = mycurve(1,1):0.5:mycurve(1,end) % create a new vector of temperature points
myF2 = interp1(mycurve(1,:),mycurve(2,:),myT2,'spline');
mynewcurve2 = [myT2; myF2];
Cubic splines add a bit of curvature between points and for some applications they make sense. In some research settings I have seen splines add very odd twists, so I am usually scared to use
anything except linear interpolation.
So how can we get our temperature data on a common scale to solve our problem?
Let's pick a common temperature range.
T_common = 30:75;
myC = interp1(mycurve(1,:),mycurve(2,:),T_common,'linear');
otherC = interp1(othercurve(1,:),othercurve(2,:),T_common,'linear');
And now we can compute our difference:
diffC = myC - otherC;
ylabel('Different in fluorescence');
How principle component analysis works
In the previous lab (5.2) we saw that one could use principle component analysis to identify the directions of maximum variation in a data set. In this way, we could reduce the 60-dimensional space
of the melt curves as a function of temperature to 3-4 dimensions, and we could visualize the similarities and differences among samples from like and unlike strains of tuberculosis.
We presented the method by calling the Matlab function pcacov, and I told you that the purpose of this function is to find the directions of maximum variation. in a sort of magical fashion. But how
does it work exactly? Principle component analysis is a beautiful and relatively straightforward application of linear algebra, and we will go through it here.
We will use a very simple example to explain how principle component analysis works. We will restrict ourselves to working with 2-dimensional data. Let's artificially generate 3 clusters of data:
x = [[randn(10,1)/2-5; randn(10,1)/2; randn(10,1)/2+5] randn(30,1)*3]
y = x * dasw.math.rot2d(-30*pi/180);
xlabel('1st dimension');
ylabel('2nd dimension');
The data is divided into 3 clusters, and stretches out along an oblique line that is the direction of maximum variation (artificially made to be about 30 degrees off of the x-axis here). Suppose we
didn't know how the data were constructed, but we wanted to find this direction of maximum variation. What could we do?
One useful quantity to examine is the covariance of the data:
C = cov(y);
Recall from Lab 2.1 that the covariance calculates how much knowing the value of 1 variable tells you about the other variable. The diagonals of the covariance matrix C are just the variances of the
first and second dimension, respectively, while the off-diagonals are the covariance of dimension 1 and dimension 2.
The covariance matrix C can also be viewed as a linear transformation, simply because any matrix can be viewed as a linear transformation (as we learned in Lab 5.1). Recall that a linear
transformation A can be used to scale/stretch, rotate, or reflect vectors around the origin.
Linear transformations in action
I have written a function that allows one to visualize how linear transformation operates on vectors. Download the function linear_transform_explorer.m below and put it in your +dasw/+plot package.
Let's see how the scaling transformation [2 0; 0 1], which scales the first dimension by a factor of 2 and does not scale the second dimension, transforms the vectors around the unit circle. For now,
ignore the last 2 plots that are called the eigenvectors.
myscale = [ 2 0 ; 0 1]
dasw.plot.linear_transform_explorer(myscale,20,1,1); % 20 steps,pause 1 sec between plots
You can see that this linear transform stretches the unit circle in the first dimension.
We can look at another transformation, a reflection:
myrefl = dasw.math.refl2d(30 * pi/180); % 30 degrees, converted to radians
dasw.plot.linear_transform_explorer(myrefl,20,1,1); % 20 steps, pause 1 sec between plots
You can see how the transformation reflects the vectors across the 30 degree line.
These transformations both have "special vectors" that, instead of being rotated or reflected by the transformation, are merely scaled. These vectors are called eigenvectors (German: same vectors).
They occur at these "special vectors" in the transformation.
For example, in the reflection of 30 degrees, the vector [-0.866 -0.5] happens to be on the line of reflection, and is merely scaled (but not reflected) by the transform.
mypoint = [ -0.866 -0.5];
mynewpoint = mypoint * myrefl
These eigenvectors represent key features of the transformation being applied. There can be 2 eigenvectors in a 2-dimensional transform. The other eigenvector for this rotation is the vector that is
perpendicular to the line of reflection:
myotherpoint = [ 0.5 -0.8660];
mynewotherpoint = myotherpoint * myrefl % returns the same vector, scaled by -1
We can project the data onto the eigenvectors to "simplify" the linear transformation:
eigenvectors = [ mypoint' myotherpoint' ];
mysimplifiedtransform = inv(eigenvectors) * myrefl * eigenvectors
The resulting matrix still contains a reflection (a fundamental operation) but it is now a simpler reflection, along the X axis.
For a given matrix, how do you find eigenvectors?
This is something that computers do for high-dimensional spaces, but is easy to do by hand for small matrixes. The trick is to find vectors such that the transform only scales them. That is, you
solve for:
A * x = lambda * x
where A is a matrix, x is a vector, and lambda is a scalar (just a number).
Equivalently, you solve for
(A-lambda) * x = 0
If you have a small matrix A = [ a b ; c d ], then this is just solving the following equation for values of x1 and x2 (and the corresponding values of lambda):
(a-lambda)*x1 + (b-lambda)*x2 = 0
(c-lambda)x1 + (d-lambda)*x2 = 0
subject to the constraint that the eigenvector has length 1 (that is, x1^2 + x2^2 == 1).
Matlab does this very well for arbitrary dimensions with the function eig:
[myeigenvecs,myeigenvalues] = eig(myrefl)
The eigenvectors of the covariance matrix identify the fundamental components of the variation
We can view the covariance matrix as a linear transformation. It is a symmetric matrix so it is also guaranteed to have real eigenvalues (due to a theorem in math).
Let's take a look at our covariance matrix C as a linear transformation:
dasw.plot.linear_transform_explorer(C,20,1,1); % 20 steps, pause 1 sec between plots
Notice that it is a stretching operation, and notice that there is a line that is merely scaled instead of rotated or reflected. This line is exactly the line of maximum variation of the data.
We can use these eigenvectors to project the data onto a simpler space:
[V,D] = eig(C);
I'm also going to re-arrange the eigenvectors so that the vector with the largest eigenvalue (diagonal values of D) is first, so that we get the biggest fundamental component first. The function diag
returns the diagonal elements of D:
mylist = diag(D)
[sortedlist, sortedindexes] = sort(mylist,1,'descend');
V = V(:,sortedindexes); % rearrange V
D = diag(sortedlist); % rearrange D
Now let's project our data onto the principle component space:
y_prime = y * V;
xlabel('Principle component 1');
ylabel('Principle component 2');
Now the direction of maximum variation is nicely along the X axis, and the direction with the 2nd most variation is nicely along the Y axis. If we had to drop a dimension, we could choose the Y axis
knowing that most of the variation in the data is along the X axis. The corresponding eigenvalues of the covariance matrix (in D) show how much relative variation is accounted for by each component:
D / sum(D(:))
Q4: How much of the total variation is in the first principle component? The second?
Matlab functions and operators
• dasw.plot.linear_transform_explorer.m
Copyright 2012-2021 Stephen D. Van Hooser, all rights reserved | {"url":"https://dataclass.vhlab.org/labs/lab-5-3-high-dimensional-space-interpolation-how-pca-works","timestamp":"2024-11-02T21:51:36Z","content_type":"text/html","content_length":"193787","record_id":"<urn:uuid:3cc4bba6-eb82-4e04-b18a-e9088c599f5c>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00236.warc.gz"} |
MILLENNIUM - Cog Quiz
Level 1
GetTypeFlags() - This gets the flag of a thing with getting the type of the thing automatically, which means if this is used for a player, like GetTypeFlags(player); it will get its actor flags
making it same as GetActorFlags(player);
It's convenient that you do not need to specify what flag to get.
If used for projectile, like GetTypeFlags(projectile);, it will be same as using GetWeaponFlags(projectile);, not sure why LEC didn't use these anywhere.
SetTypeFlags() - This is same except it sets the flag instead of getting the flag. SetTypeFlags(player) is equal to SetActorFlags(player);
Level 2
• Gets the number of sectors in a level. - GetSecotrCount();
• Gets the number of surfaces in a level. - GetSurfaceCount();
• Gets the number of things in a level. - GetThingCount();
• Gets the number of certain templates used in a level. - GetThingTemplateCount();
• Sets a color map to a sector. - SetSectorColormap(sector, CMPInt);
• Sets a mat to a surface. - SetSurfaceMat(surface, MatRef);
Level 3 | {"url":"https://millennium.massassi.net/cogquiz/cogquiz.phtml?Form=Answer&Week=7","timestamp":"2024-11-10T04:40:06Z","content_type":"application/xhtml+xml","content_length":"2777","record_id":"<urn:uuid:b7fb0454-f6af-4f5f-a6e5-68c3be831166>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00036.warc.gz"} |
Unveiling Vertical Shifts in Trigonometric Functions
[December 8, 2023 by JoyAnswer.org, Category : Mathematics]
How to find the vertical shift of a trig function? Explore techniques to identify and calculate the vertical shift in trigonometric functions. Understand how this shift impacts the function's
How to find the vertical shift of a trig function?
The vertical shift of a trigonometric function refers to the upward or downward displacement of the graph along the y-axis. This shift is determined by a constant added to the function, and it
affects the position of the entire graph without altering its shape.
The general form of a trigonometric function with a vertical shift is:
$f(x) = A \cdot \sin(Bx + C) + D$
$f(x) = A \cdot \cos(Bx + C) + D$
• $A$ is the amplitude (the vertical distance from the midline to the maximum or minimum),
• $B$ is the frequency (controls the period of the function),
• $C$ is the phase shift (horizontal shift),
• $D$ is the vertical shift.
To find the vertical shift ($D$), you need to look at the constant term that is added to the function. If the original function is $f(x) = \sin(x)$ or $f(x) = \cos(x)$, then the vertical shift is the
constant term added to it.
For example:$f(x) = \sin(x) + 2$
In this case, the vertical shift ($D$) is 2 units upward. If the constant term were -2, the shift would be 2 units downward.
Similarly, for a cosine function:$g(x) = \cos(x - \pi) - 3$
Here, the vertical shift ($D$) is -3 units downward.
To summarize, identify the constant term in the trigonometric function, and the value of that constant will tell you the vertical shift and its direction (upward or downward).
Determining the Vertical Shift in Trigonometric Functions
The vertical shift of a trigonometric function is the amount it moves up or down from its standard position. There are several methods to determine this shift:
1. Analyzing the Equation:
The standard equation of a sine function is y = sin(x), and the standard equation of a cosine function is y = cos(x). Any constant added or subtracted to the function will cause a vertical shift.
• Positive constant: Adding a positive constant shifts the function up by the value of the constant.
• Negative constant: Subtracting a constant shifts the function down by the absolute value of the constant.
For example, the equation y = sin(x) + 2 represents a sine function that has been shifted two units upwards.
2. Graphing:
We can also determine the vertical shift by graphing the function and comparing it to the standard graph.
• If the graph is above the standard graph, it has been shifted upwards. The vertical distance between the two graphs represents the amount of shift.
• If the graph is below the standard graph, it has been shifted downwards. The vertical distance between the two graphs represents the amount of shift.
3. Amplitude and Midline:
The amplitude and midline of a trigonometric function can also provide information about the vertical shift.
• Amplitude: The amplitude of a function is the absolute value of the coefficient multiplying the trigonometric term. It determines the vertical distance between the peaks and troughs of the graph.
• Midline: The midline is the horizontal line that passes exactly in the middle of the function's range.
For example, the function y = 2 sin(x) + 1 has an amplitude of 2 and a midline of y = 1. This tells us that the function has been stretched vertically by a factor of 2 and shifted up by 1 unit.
How Coefficients Affect the Vertical Shift:
The vertical shift of a trigonometric function is primarily affected by the constant term added or subtracted to the function. This constant directly determines the amount and direction of the shift.
• Coefficient of the trigonometric term: This coefficient affects the amplitude of the function, but not the vertical shift. Changing the coefficient will stretch or compress the graph vertically
but will not move it up or down.
• Phase shift: A phase shift moves the graph horizontally to the left or right. It does not affect the vertical position of the graph.
In summary, determining the vertical shift of a trigonometric function involves analyzing the equation, considering the amplitude and midline, and examining the constant term added or subtracted. By
understanding the impact of different coefficients, we can interpret and analyze various trigonometric functions with greater ease. | {"url":"https://joyanswer.org/unveiling-vertical-shifts-in-trigonometric-functions","timestamp":"2024-11-11T18:17:16Z","content_type":"text/html","content_length":"43588","record_id":"<urn:uuid:24eab7f3-9271-42c7-86fa-45f4b011de6d>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00040.warc.gz"} |
Chapter 23 Predicting one continuous variable from two (or more) things | Applied Biostats – BIOL3272 UMN – Fall 2022
Chapter 23 Predicting one continuous variable from two (or more) things
This text (roughly) follows Chapter 18 of our textbook.
The reading below is required, Whitlock and Schluter (2020)
is not.
Motivating scenarios: We have numerous explanatory variables and want to develop an synthetic model.
Learning goals: By the end of this chapter you should be able to
• Write down and interpret longer and more complex linear models.
• Interpret a polynomial regression and run one in R.
• Interpret and run two factor ANOVAs in R.
• Calculate TYPE I sums of squares.
23.1 Review of Linear Models
A linear model predicts the response variable, as \(\widehat{Y_i}\) by adding up all components of the model.
\[$$\hat{Y_i} = a + b_1 y_{1,i} + b_2 y_{2,i} + \dots{} \tag{21.2}$$\]
Linear models we have seen
• One sample t-tests: \(\widehat{Y} = \mu\)
• Two sample t-tests: \(\widehat{Y_i} = \mu + A_i\) (\(A_i\) can take 1 of 2 values)
• ANOVA: \(\widehat{Y_i} = \mu + A_i\) (\(A_i\) can take one of more than two values)
• Regression \(\widehat{Y_i} = \mu + X_i\) (\(X_i\) is continuous)
So far we’ve mainly modeled a continuous response variable as a function of one explanatory variable. But linear models can include multiple predictors – for example, we can predict Dragon Weight as
a function of both a categorical (spotted: yes/no) and continuous variable in the same model.
23.1.1 Test statistics for a linear model
• The \(t\) value describes how many standard errors an estimate is from its null value.
• The \(F\) value quantifies the ratio of variation in a response variable associated with a focal explanatory variable (\(MS_{model}\)), relative to the variation that is not attributable to this
variable (\(MS_{error}\)).
23.1.2 Assumptions of a linear model
Remember that linear models assume
• Linearity: That observations are appropriately modeled by adding up all predictions in our equation.
• Homoscedasticity: The variance of residuals is independent of the predicted value, \(\hat{Y_i}\) is the same for any value of X.
• Independence: Observations are independent of each other (aside from the predictors in the model).
• Normality: That residual values are normally distributed.
• Data are collected without bias as usual.
23.2 Polynomial regression example
We note that linear models can include e.g. squared, and geometric functions too, so long as we get out predictions by adding up all the components of the model.
A classic example of a linear model is a polynomial regression, in which we predict some response variable as a function of a predictor and higher order terms of the predictor. The most common
polynomial regression includes the explanatory variable and its square value (23.1).
\[$$\begin{split} \widehat{Y_i} = a + b_1 \times y_{1,i} + b_2 \times X_{2,i}\\ \\ \text{ where } X_{2,i} = X_{1,i}^2 \end{split} \tag{23.1}$$\]
Often including a cubic, or even a quadratic term is useful – but be thoughtful before adding too many in – each additional term takes away from our degrees of freedom, complicates interpretation,
and may overfit the data. Let your biological intuition and statistical reasoning guide you.
23.2.1 Polynomial regression example
Let’s revisit our example polynomial regression example predicting the number of species from the productivity of the plot to work through these ideas. Recall that
• A simple linear regression did not fit the data well AND violated assumptions of a regression, as residuals were large and positive for intermediate predictions and large and negative for large
or small predictions.
• Including a squared term improved the model fit and had the data meet assumptions.
Let’s write a descriptive equation for each model
\[\text{N.SPECIES = CONSTANT + PRODUCTIVITY}\] \[\text{N.SPECIES = CONSTANT + PRODUCTIVITY + PRODUCTIVITY}^2\]
We present these models in Figure 23.1. See that we can add a polynomial fit to our ggplot by typing formula = y ~ poly(x, 2, raw = TRUE) into the geom_smooth function.
bmass <- tibble( Biomass = c(192.982,308.772,359.064,377.778,163.743,168.421,128.655,98.246,107.602,93.567,83.041,33.918,63.158,139.181,148.538,133.333,127.485,88.889,138.012), n.species = c(25.895,12.729,8.342,2.885,21.504,20.434,18.293,16.046,16.046,11.655,12.725,9.515,7.16,16.042,16.042,11.655,12.725,2.88,8.338))
base_plot <- ggplot(bmass, aes(x = Biomass, y = n.species))+ geom_point()+ xlab("Productivity (g/15 Days)" )
linear_plot <- base_plot + labs(title = "Linear term") +
geom_smooth(method = 'lm')
polynomial_plot <- base_plot + labs(title = "Linear and squared term") +
geom_smooth(method = 'lm',formula = y ~ poly(x, 2, raw = TRUE))
plot_grid(linear_plot, polynomial_plot, labels = c("a","b"))
23.2.1.1 Fitting polynomial regressions in R
Fitting a model with a linear term in R should look familiar to you. linear_term <- lm(n.species ~ Biomass, bmass)
There are a bunch of ways to add a polynomial term.
• lm(n.species ~ poly(Biomass, degree = 2, raw = TRUE), bmass) Is what we typed into our geom_smooth function above. If we typed degree = 3, the model would include a cubed term as well.
• lm(n.species ~ Biomass + I(Biomass^2), bmass) Is a more explicit way to do this. When doing math to variables in our linear model we need to wrap them in I() or R gets confused and does weird
• lm(n.species ~ Biomass + Biomass2, bmass %>% mutate(Biomass2 = Biomass^2)) Or we can mutate to add a squared transform the data before making our model. NOTE: I did not pipe the mutate into lm().
That’s because lm() does take things from the standard %>% pipe. If you want to pipe into lm(), you will need the magrittr package and then you can use a special pipe, %$%, so… bmass %>% mutate
(Biomass2 = Biomass^2) %$% lm(n.species ~ Biomass + Biomass2) will work.
23.2.1.2 Interpretting the output of a polynomial regression – model coefficents
So, let’s look at this polynomial regression
## Call:
## lm(formula = n.species ~ Biomass + Biomass2, data = bmass %>%
## mutate(Biomass2 = Biomass^2))
## Residuals:
## Min 1Q Median 3Q Max
## -8.939 -1.460 -0.345 2.591 7.353
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.288706 4.442073 -0.52 0.61343
## Biomass 0.202073 0.051157 3.95 0.00115 **
## Biomass2 -0.000488 0.000116 -4.20 0.00068 ***
## ---
## Signif. codes:
## 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## Residual standard error: 4.38 on 16 degrees of freedom
## Multiple R-squared: 0.532, Adjusted R-squared: 0.473
## F-statistic: 9.08 on 2 and 16 DF, p-value: 0.00232
The output of this model should look familiar. Our rows are
• (Intercept) – the number of species we would have if we followed our curve to 0 productivity. That this value is -2.2.9 highlights the idea that we should not make predictions outside of the
range of our data. Of course, we wouldn’t predict a negative number of species ever…
• Biomass – This describes how the number of species changes with a linear increase in productivity. It’s critical to see that this DOES not mean that the number of species always increase with
productivity. That’s because of the next term,
• Biomass2 – This describes how the number of species changes with productivity squared. The negative sign means that the number of species decreases with the square of productivity. Polynomial
regressions are often used in these cases where intermediate values are largest or smallest, so it’s normal to see contrasting signs for the linear and squared terms.
Writing down this equation, we predict species number as
\[\widehat{n.species}_i = -2.29 + 0.202 \times Biomass_i -0.000488 \times Biomass_i^2\] So, for example if we had a plot with a productivity of 250 g/15 Days, we would predict it had \[$$\begin
{split} \widehat{n.specie}{s_{|Biomass=250}} &= -2.29 +0.202 \times 250 -0.000488 \times 250^2\\ &= 17.71 \end{split} \tag{23.1}$$\]
A value which makes sense, as it seems to be where out curve intersects with 250 in Figure ??B.
Our columns, Estimate, Std. Error, t value, and Pr(>|t|) should also feel familiar, all interpretations are the same as usual. The standard error describes the uncertainty in the estimate, the t
describes how many standard errors away from zero the estimate is, and the p-value describes the probability that a value this many standard errors away from zero would arise if the null where true.
One thing though
The p-values in this output do not describe the statistical significance of the predictors!! DO NOT INTERPRET THESE P-VALUES AS SUCH
One way to think about this is to just look at our simple linear model which shows basically no association between biomass and species number (and the association it shows it slightly negative).
broom::tidy(linear_term) %>% mutate_at(.vars = c("estimate"), round, digits = 4 ) %>%mutate_at(.vars = c("std.error"), round, digits = 3 ) %>% mutate_at(.vars = c("statistic"), round, digits = 4 )%>% mutate_at(.vars = c("p.value"), round, digits = 5 ) %>%kable()
term estimate std.error statistic p.value
(Intercept) 14.4251 2.777 5.1942 0.0001
Biomass -0.0078 0.015 -0.5102 0.6165
Note The summary.lm() output still usefully provides our estimates and uncertinaty in them – so don’t ignore it!
23.2.1.3 An ANOVA approach
So how do we get significance of each term? We look at the ANOVA output!
## Analysis of Variance Table
## Response: n.species
## Df Sum Sq Mean Sq F value Pr(>F)
## Biomass 1 10 10 0.52 0.48329
## Biomass2 1 339 339 17.64 0.00068 ***
## Residuals 16 307 19
## ---
## Signif. codes:
## 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We now conclude that the association between n.species and the linear term of Biomass would be quite expected from the null. How do we square these ideas? I think of the significance of the linear
term as how weird it would be to see a non-zero linear estimate in the absence of a squared term. However, this is not fully correct, as this P-value differs from the one above with just the linear
term. To make sense of this, let’s dig into how we calculate the sums of squares for these larger models.
“Sequential” Type I Sums of squares
We’ll see in this and the next section that there’s a real issue in which variable we attribute our sums of squares to in larger linear models.
In many cases (see below) Sequential “Type I” sums of squares make the most sense. Here we
• Calculate \(SS_{error}\) and \(SS_{total}\) as we always do! (Figure 23.2A,D)
• Calculate the \(SS_{thing1}\) (in this case Biomass), as if it where the only thing in the model, \(\widehat{Y_{i|bmass}}\). (Figure 23.2B).
• Calculate the \(SS_{thing2}\) (in this case \(Biomass^2\)), as the deviation of predicted values from a model with both things in it, \(\widehat{Y_{i|bmass,bmass^2}}\), minus predictions from a
model with just thing1 in it, \(\widehat{Y_{i|bmass}}\) (Figure 23.2C).
We can calculate these sums of squares in R as follows, and then compute mean squares and p-values. Before I do this, I make a tibble with prediction from both the simple linear model with just a
linear term, and the fuller linear model with the linear and squared term.
combine_models <- full_join(augment(linear_term) %>%
dplyr::select(n.species, Biomass, .fitted_lin = .fitted, .resid_lin = .resid),
augment(poly_term) %>%
dplyr::select(n.species, Biomass, .fitted_full = .fitted, .resid_full= .resid),
by = c("n.species", "Biomass"))
combine_models %>%
summarise(ss_tot = sum( (n.species - mean(n.species))^2 ),
ss_bmass = sum( (.fitted_lin - mean(n.species))^2 ),
ss_bmass2 = sum( (.fitted_full - .fitted_lin )^2 ),
ss_error = sum( (n.species - .fitted_full)^2 ),
df_bmass = 1, df_bmass2 = 1, df_error = n() - 3,
ms_bmass = ss_bmass / df_bmass ,
ms_bmass2 = ss_bmass2 / df_bmass2 ,
ms_error = ss_error / df_error,
F_bmass = ms_bmass / ms_error, F_bmass2 = ms_bmass2/ ms_error,
p_bmass = pf(q = F_bmass, df1 = df_bmass, df2 = df_error, lower.tail = FALSE),
p_bmass2 = pf(q = F_bmass2, df1 = df_bmass2, df2 = df_error, lower.tail = FALSE)) %>% mutate_all(round,digits = 4) %>%DT::datatable( options = list( scrollX='400px'))
You can scroll through the output above to see that our calculations match what anova() tells us!!!
23.3 Type I Sums of Squares (and others)
Calculating Sums of Squares sequentially, as we did in Figure 23.2, is the default way R does things.
Sequential Type I sums of squares calculate the sums of squares for the first thing in your model first, then the second thing, then the third thing etc… This means that while our
• Sums of square, Mean squares, F values, and p-values might change, depending on the order in which variables are entered into our model.
• Parameter estimates and uncertainty in them will not change with order.
In general sequential sums of squares make the most sense when
• We are not interested in the significance of the earlier terms in our model, which we want to take account of, but don’t really care about their statistical significance.
• Designs are “balanced” (Figure 23.3), as in these cases, we get the same SS, F and P values regardless of the order that we put terms into the model.
In the next class, we will look into other ways to calculate the sums of squares.
23.4 Two categorical variables without an interaction
We saw that paired t-tests increase our power because they control for extraneous variation impacting each pair. We often want to use a similar study design for a study with more than two explanatory
For example, in a randomized “Controlled Blocked Design” each “block” gets all treatments, and by including treatment in our model we can explain variability associated with block unrelated to our
main question. In such models we don’t care about the statistical significance of the block, we just want to use block to explain as much variation as possible before considering treatment.
In the study below, researchers wanted to know if the presence of a fish predator impacted diversity pf the marine zooplankton in the area. To find out they introduced the zooplankton with no, some,
or a lot of fish, in mesh bags in a stream. Each stream got three such bags – one with no, one with some, and the other with many fish. This was replicated at five streams, so each stream is a
23.4.1 Estimation and Uncertainty
The raw data are presented below, with means for treatments in the final column.
treatment Block: 1 Block: 2 Block: 3 Block: 4 Block: 5 mean_diversity
control 4.1 3.2 3.0 2.3 2.5 3.02
low 2.2 2.4 1.5 1.3 2.6 2.00
high 1.3 2.0 1.0 1.0 1.6 1.38
We can conceive of this as a linear model, in which we predict diversity as a function of block and treatment
\[DIVERSITY = BLOCK + TREATMENT\] We enter the model into R as follows
fish_dat <- read_csv("https://whitlockschluter3e.zoology.ubc.ca/Data/chapter18/chap18e2ZooplanktonDepredation.csv") %>%
mutate(treatment = fct_relevel(treatment, c("control", "low", "high")))
wrong_fish_lm <- lm(diversity ~ block + treatment, data = fish_dat)
broom::tidy(wrong_fish_lm) %>% mutate_at(.vars = c("estimate"), round, digits = 3 ) %>%mutate_at(.vars = c("std.error"), round, digits = 3 ) %>% mutate_at(.vars = c("statistic"), round, digits = 3 )%>% mutate_at(.vars = c("p.value"), round, digits = 7 ) %>%kable()
term estimate std.error statistic p.value
(Intercept) 3.50 0.384 9.109 0.0000
block -0.16 0.099 -1.613 0.1351
treatmentlow -1.02 0.344 -2.968 0.0128
treatmenthigh -1.64 0.344 -4.772 0.0006
NOTE Uhhohh. Something went wrong here. Why is there only one value for block, when there are five? It’s because R thought bock was a number and ran a regression. Let’s clarify this for R by mutating
black to be a factor.
fish_dat <- mutate(fish_dat, block = factor(block))
fish_lm <- lm(diversity ~ block + treatment, data = fish_dat)
broom::tidy(fish_lm) %>% mutate_at(.vars = c("estimate"), round, digits = 3 ) %>%mutate_at(.vars = c("std.error"), round, digits = 3 ) %>% mutate_at(.vars = c("statistic"), round, digits = 3 )%>% mutate_at(.vars = c("p.value"), round, digits = 6 ) %>%kable()
term estimate std.error statistic p.value
(Intercept) 3.42 0.313 10.938 0.0000
block2 0.00 0.374 0.000 1.0000
block3 -0.70 0.374 -1.873 0.0979
block4 -1.00 0.374 -2.676 0.0281
block5 -0.30 0.374 -0.803 0.4453
treatmentlow -1.02 0.289 -3.524 0.0078
treatmenthigh -1.64 0.289 -5.665 0.0005
We take these estimates and uncertainty about them seriously, but fully ignore the t and p-values, as above. From here we predict diversity in
• Control predation in block one is just the intercept: \(3.42 - 0 - 0 = 3.42\).
• Control predation in block three is the intercept minus the mean difference between block three and block one: \(3.42 - 0.70 - 0 = 2.72\).
• High predation in block one is the intercept minus the mean difference between high predation and the control, \(3.42 - 0 - 1.64 = 1.79\).
• Low predation in block four is the intercept, minus the mean difference between block four and block one, minus the mean difference between low predation and the control treatment: 3.42 - 1.00 -
1.02 = 1.40.
23.4.2 Hypothesis testing with two categorical predictors
In this model, the null and alternative hypotheses are
• \(H_0:\) There is no association between predator treatment and zooplankton diversity – i.e. Under the null, we predict zooplankton diversity in this experiment as the intercept plus the
deviation associated with stream (Figure 23.5a).
• \(H_A:\) There is an association between predator treatment and zooplankton diversity – i.e. Under the alternative, we predict zooplankton diversity in this experiment as the intercept plus and
the deviation associated with stream plus the effect of one or more treatments (Figure 23.5b).
Again we test these hypotheses in an ANOVA framework
## Analysis of Variance Table
## Response: diversity
## Df Sum Sq Mean Sq F value Pr(>F)
## block 4 2.34 0.58 2.79 0.1010
## treatment 2 6.86 3.43 16.37 0.0015 **
## Residuals 8 1.68 0.21
## ---
## Signif. codes:
## 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We conclude that predation treatment impacts zooplankton diversity, with diversity decrease as there are more predators. Because this is an experimental manipulation, we can conclude that predation
decreased diversity.
23.4.2.1 Calcualating sums of squares
We again use the sequential method to calculate sums of squares because we first want to account for block. The code below shows you how anova() got its answer. But in our case we ignore the F-value
and significance of block, as it’s in the model to soak up shared variation, not to be tested.
block_model <- lm(diversity ~ block, fish_dat)
full_model <- lm(diversity ~ block + treatment, fish_dat)
combine_models <- full_join(augment(block_model ) %>%
dplyr::select(diversity, block, .fitted_block = .fitted, .resid_block = .resid),
augment(full_model) %>%
dplyr::select(diversity, block, treatment, .fitted_full = .fitted, .resid_full= .resid),
by = c("diversity", "block"))
combine_models %>%
summarise(ss_tot = sum( (diversity - mean(diversity))^2 ),
ss_block = sum( (.fitted_block - mean(diversity))^2 ),
ss_treat = sum( (.fitted_full - .fitted_block)^2 ),
ss_error = sum( (diversity - .fitted_full)^2 ),
df_block = n_distinct(block) - 1, df_treat = n_distinct(treatment) -1,
df_error = n() - 1,
ms_block = ss_block / df_block,
ms_treat = ss_treat / df_treat ,
ms_error = ss_error / df_error,
F_treat = ms_treat / ms_error,
p_bmass2 = pf(q = F_treat, df1 = df_treat, df2 = df_error, lower.tail = FALSE)) %>% mutate_all(round,digits = 4) %>%DT::datatable( options = list( scrollX='400px'))
23.4.3 Post-hoc tests for bigger linear models
So we rejected the null hypothesis and conclude that predator abundance impacts zooplankton diversity. Which treatments differ? Again we, conduct a post-hoc test.
Instead of using the aov() function and piping the output to TukeyHSD(), here I’ll show you how to conduct a posthoc test with the glht() function in the multcomp package.
In the code below, I say I want to look at all pairwise comparisons between treatments, using the Tukey-Kramer method from Chapter 20.
## Simultaneous Tests for General Linear Hypotheses
## Multiple Comparisons of Means: Tukey Contrasts
## Fit: lm(formula = diversity ~ block + treatment, data = fish_dat)
## Linear Hypotheses:
## Estimate Std. Error t value
## low - control == 0 -1.020 0.289 -3.52
## high - control == 0 -1.640 0.289 -5.67
## high - low == 0 -0.620 0.289 -2.14
## Pr(>|t|)
## low - control == 0 0.0192 *
## high - control == 0 0.0012 **
## high - low == 0 0.1425
## ---
## Signif. codes:
## 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## (Adjusted p values reported -- single-step method)
We conclude that diversity in both low and high predation differ significantly from both the control, no predation treatment, treatment. But we fail to reject the hypothesis that low and high
predation treatments differ.
Whitlock, Michael C, and Dolph Schluter. 2020. The Analysis of Biological Data. Third Edition. | {"url":"https://bookdown.org/ybrandvain/Applied_Biostats_Fall_2022/morenova.html","timestamp":"2024-11-06T08:30:57Z","content_type":"text/html","content_length":"179804","record_id":"<urn:uuid:6f195162-5e0d-438f-9801-35cd4f177d1f>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00656.warc.gz"} |
How To Measure Time Series Similarity in Python
In this tutorial, we’ll explore some practical techniques to measure the similarity between time series data in Python using the most popular distance measures.
To make sure that the results are not affected by noise or irrelevant factors, we’ll apply techniques such as scaling, detrending, and smoothing.
Once the data is preprocessed, we can use simple distance measures like Pearson correlation and Euclidean distance to measure the similarity of two aligned time series.
However, in real-world scenarios, time series may not be aligned, or they may have different lengths.
In such cases, we’ll explore a more advanced technique called Dynamic Time Warping (DTW).
Get ready to dive deep into the world of time series similarity measures and learn some exciting techniques to boost your Python skills.
Let’s get started!
Preprocessing Time Series Data For Similarity Measures In Python
Before we can measure the similarity between two time series, it’s a good idea to preprocess the data to make sure that the results will not be affected by noise or factors that have nothing to do
with the similarity.
The three most common preprocessing techniques are scaling, detrending, and smoothing.
Let’s load data from yearly rice production of Brazil, China, and India to understand how these techniques work.
import pandas as pd
import numpy as np
from matplotlib import pyplot as plt
import os
data = pd.read_csv(os.path.join(path, 'rice production across different countries from 1961 to 2021.csv'))
selected = ['Brazil', 'India', 'China']
data = data.loc[data['Area'].isin(selected)]
Scaling is the process of subtracting the mean of the time series from each value and then dividing the result by the standard deviation of the series.
This is the same transformation that is applied to the data when using the z-score normalization.
You can find it named time series normalization, scaling, or standardization.
scaled = dict()
for area in data['Area'].unique():
series = data[data['Area']==area]['Value'].values
scaled[area] = (series - series.mean()) / series.std()
scaled = pd.DataFrame(scaled)
Brazil China India
1961 -1.86774 -2.50948 -1.33578
1962 -1.79355 -2.28856 -1.42553
1963 -1.71083 -2.03712 -1.28679
1964 -1.43779 -1.81721 -1.20203
1965 -0.880431 -1.70387 -1.52196
We scaled, but we still have a trend.
Sometimes you want to keep the trend if it’s important for your analysis. In any case, I will show you how to remove it.
Detrending is the process of removing any trend from the time series.
If we want to measure the similarity of the underlying patterns, not the trend, we can perform detrending using scipy.signal.detrend.
from scipy.signal import detrend
detrended = dict()
for area in selected:
detrended[area] = detrend(scaled[area])
detrended = pd.DataFrame(detrended, index=sorted(data['Year'].unique()))
Brazil India China
1961 -0.403679 0.335164 -0.936089
1962 -0.378289 0.189715 -0.767612
1963 -0.344379 0.272754 -0.568622
1964 -0.12014 0.301818 -0.401155
1965 0.38842 -0.0738108 -0.340266
This will fit a least-squares model to the data and then subtract it.
Here, scaled is the scaled data from the previous step, and detrended is the detrended data.
You can see the data between Brazil and India seems easier to compare, while China is still a bit different.
Another step we can take is to smooth the data to remove noise that can affect the similarity measures.
Smoothing is the process of removing noise from the time series.
It can be as simple as taking a moving average of the data.
smoothed = detrended.rolling(5).mean().dropna()
Smoothing with a moving average comes from the assumption that the average of the data is a better representation of the underlying pattern than the individual data points.
This code will calculate a rolling average of the data using a window size of 5. You can try different window sizes to see how it affects the results.
I dropped the first 4 rows because they contain NaN values.
Brazil India China
1965 -0.171613 0.205128 -0.602749
1966 -0.18346 0.111086 -0.458108
1967 -0.12075 0.0876422 -0.36545
1968 -0.0871858 0.0522224 -0.318492
1969 -0.131532 0.00475607 -0.314201
You can see the data is less “jagged” now.
A fancier way to smooth the data is to use a Savitzky-Golay filter.
from scipy.signal import savgol_filter
savgol = dict()
for area in selected:
savgol[area] = savgol_filter(detrended[area], window_length=5, polyorder=1)
savgol = pd.DataFrame(savgol, index=sorted(data['Year'].unique()))
To use this filter, we need to specify the window length and the polynomial order.
The window length is the number of data points used to calculate the filter.
The polynomial order is the order of the polynomial used by the filter to fit the data.
Which Preprocessing Method Is Best?
I like to take the “practical machine learning” approach to this problem.
You don’t need to use all the preprocessing methods I showed you, or even use them in the same order.
Try different preprocessing methods and see which one gives you the best results according to your metrics or experience.
Results that move too far from your expectations are usually wrong or demand a lot more investigation before you can trust them.
Another method you can use is differencing.
Simple Distances For Time Series Similarity
After preprocessing the data, we can use simple distance measures to measure the similarity between two time series.
Pearson Correlation
Pearson correlation is a measure of the linear correlation between two time series.
It was used as an official metric by two financial forecasting competitions on Kaggle: the G-Research Crypto Forecasting and the Ubiquant Market Prediction.
We can calculate the Pearson correlation coefficient directly in pandas with df.corr.
Brazil India China
Brazil 1 -0.554765 0.205345
India -0.554765 1 -0.00622666
China 0.205345 -0.00622666 1
This function gives us the correlation between all pairs of time series in an organized table, which is very convenient.
Pearson correlation varies between -1 and 1. -1 means the time series are perfectly negatively correlated (one goes up when the other goes down), 1 means the time series are perfectly positively
correlated (one goes up when the other goes up), and 0 means the time series are not correlated at all.
Anything in between means they are correlated (the direction depends on the signal), but not perfectly.
For example, the correlation between Brazil and India is -0.55, which means that rice production in Brazil tends to go down when rice production in India goes up and vice versa.
But be careful! This is not a measure of causation. It only means that this is what happened in the historical data.
Euclidean Distance
The Euclidean distance is a widely used technique to measure the similarity between time series.
It is a simple and intuitive measure that calculates the distance between two time series as the straight-line distance between their corresponding points.
We can easily calculate it with the euclidean_distances function from scikit-learn.
from sklearn.metrics.pairwise import euclidean_distances
euc_dist = euclidean_distances(smoothed.T)
pd.DataFrame(euc_dist, index=selected, columns=selected)
Brazil India China
Brazil 0 2.50487 2.8451
India 2.50487 0 2.66397
China 2.8451 2.66397 0
Notice that we pass the transpose of the dataframe to the function. This is because the function expects the time series to be in the rows, not the columns.
And remember we are calculating a distance, not a similarity. The smaller the distance, the more similar the time series are.
One of the benefits of using the Euclidean distance is its simplicity and ease of implementation. It is a straightforward measure that requires minimal computation, making it a suitable choice for
large datasets.
However, it has some drawbacks. The Euclidean distance is sensitive to outliers and it is not scale-invariant. This means that the distance between two time series will be different depending on the
scale of the data.
In our example, the data is already scaled, so this is not a problem.
Dynamic Time Warping
Dynamic Time Warping (DTW) is a more advanced technique for measuring the similarity of time series.
DTW can handle time series that are not aligned and also series with different lengths, which happens a lot in real-world data.
In this dataset, some countries have been producing rice for a long time, and others have only started recently, although our selected countries have full data.
DTW works by warping the time axis of one time series to match the time axis of the other time series.
This technique was very successful in the Driver Telematics Analysis competition, where the goal was to identify drivers by looking at their driving behavior.
The main drawback of DTW is that it is computationally expensive. It can take a long time to calculate the distance between two time series, which can be a problem if you have many comparisons to
Anyway, the DTAIDistance library implemented a fast version of DTW in C that we can use in Python.
from dtaidistance import dtw
dtw_dist = dtw.distance_matrix_fast(smoothed.T.values)
pd.DataFrame(dtw_dist, index=selected, columns=selected)
Brazil India China
Brazil 0 1.74072 1.44652
India 1.74072 0 2.17045
China 1.44652 2.17045 0
Notice that we need to transpose the dataframe, because the function expects the time series to be in the rows instead of the columns, and we need to convert it to a numpy array with the .values
Here, the smaller the distance, the more similar the time series are.
If you compare the results of the Euclidean distance and the DTW, you will notice that they are different.
Keep this in mind when you are choosing a distance measure. Each can give you different results.
We can plot the warping path between two time series with the plot_warping function.
from dtaidistance import dtw_visualisation as dtwvis
fig, ax = plt.subplots(2,1,figsize=(1280/96, 720/96))
path = dtw.warping_path(smoothed['Brazil'].values, smoothed['India'].values)
dtwvis.plot_warping(smoothed['Brazil'].values, smoothed['India'].values, path,
fig=fig, axs=ax)
ax[0].set_title('DTW Warping Path Between Brazil and India')
This shows us which points in one time series are best aligned with which points in the other time series.
It can be useful to identify shifted patterns in the time series. | {"url":"https://forecastegy.com/posts/how-to-measure-time-series-similarity-in-python/","timestamp":"2024-11-07T13:04:43Z","content_type":"text/html","content_length":"62762","record_id":"<urn:uuid:7e189291-9559-4062-8c7c-33715c3bae92>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00359.warc.gz"} |
xx+y=y24….Iyx+y=x6… II
Solve for x and y
... | Filo
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Question Text Solve for and
Updated On Aug 24, 2023
Topic Logarithms and exponential functions
Subject Mathematics
Class Grade 12
Answer Type Video solution: 1
Upvotes 80
Avg. Video Duration 11 min | {"url":"https://askfilo.com/user-question-answers-mathematics/solve-for-and-35343731333230","timestamp":"2024-11-11T11:26:04Z","content_type":"text/html","content_length":"194341","record_id":"<urn:uuid:8add3628-bc31-4c9f-b5b0-99a6860e0c5a>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00740.warc.gz"} |
KeelinCoefficients( values, percentiles, I, K, lb, ub, nTerms, flags )
This fits a Keelin distribution, also known as a MetaLog distribution, to data, or to a set of (x,p) value - percentile level pairs, and returns a vector of coefficients indexed by «K». The vector of
coefficients can be a much shorter description of the distribution than the data itself. This vector of coefficients can then be passed to the functions Keelin, DensKeelin, CumKeelin and CumKeelinInv
, reducing the computation time required by those functions.
The Keelin distribution is a versatile continuous distribution that can assume the shape of almost any standard unbounded, semi-bounded on bounded continuous distribution. If you have univariate
continuous data and don't know what distribution to use to model that data, with no reason to believe from first principles that the data needs to be of a particular distribution class, then the
Keelin distribution is likely to be a good choice. There is no need to figure out whether your data best matches a LogNormal, Gamma, Beta or some other distribution type -- if it does happen to match
one of those closely, the Keelin will usually find the same shape; however, it is capable of virtually the entire space of Skewness/Kurtosis combinations, and can even sometimes discover meaningful
multimodal distinctions.
The Keelin distribution is introduced in the paper:
• «values»: This can be either: (1) A representative sample of data points, with «percentiles» omitted, or (2) a collection of fractile estimates (corresponding to the quantile levels in
• «percentiles»: (Optional): The percentile levels (also called quantile or fractile levels) for the values in «values», also indexed by «I». Each number must be between 0 and 1. For example, when
a value in «percentiles» is 0.05, the corresponding value in «values» is the 5th percentile.
• «I»: (Optional): The index of «values» and «percentile». This can be omitted when either «values» or «percentile» is itself an index.
• «lb», «ub»: (Optional) Upper and lower bound. Set one or both of these to a single number if you know in advance that your quantity is bounded. When neither is specified, the distribution is
unbounded (i.e., with tails going to -INF and INF). When one is set the distribution is semi-bounded, and when both are set it is fully bounded.
• «nTerms»: (Optional) The number of basis terms used for the fit. This should be 2 or greater. See Keelin#Number of terms below.
• «flags»: (Optional) A bit-field, where any of the following flags can be added together.
1 = «values» contains coefficients (as obtained from the KeelinCoefficients function). When not set, «values» contains sample values.
2 = Return the basis (see #Returning the basis).
4 = Return the derivative of the basis.
8 = Do not issue a warning when infeasible. (See Infeasibility). No validation of feasibility is performed when coefficients are passed in (i.e., when «flags»=1 is set).
16 = Return the coefficients even when infeasible. When this bit is not set, Null is returned if infeasible. When this is set, a mid-value or sample is returned anyway.
32 = Disable tail constraints. The use of tail constraints is an improvement to the algorithm published in the original Keelin (2016) paper that reduces the frequency of infeasible fits to
data. You can disable these to reproduce the original Keelin algorithm. See Infeasibility for more details.
To use
To use this function, you should create two indexes to pass to «I» and «K». Your «I» index indexes your data points. The «K» index will be used for the result, and typically its length determines the
number of basis terms used (see Number of terms). The result is indexed by «K». If you omit «K», the function will create a local index named .K.
In some cases you may want to create a "panel" of distributions, where you fit the same data, but vary the number of basis terms. Since you will likely want these is a single array, you want them to
share the same «K» index, even though the number of terms varies. In this case, you should make your «K» index long enough for the largest basis, and then pass the «nTerms» parameter explicitly
(usually you will pass it a vector, varying «nTerm» across yet another index). For example, assuming you have a vector estimate of estimates for a series of percentile levels given in percentile with
both estimate and percentile indexed by I:
In this case, the result is null-padded where NumTerms is less that IndexLength(K).
Your distribution data will be in one of two forms:
• A representative sample of points for your quantity, «values». In this case, omit the «percentiles» parameter.
• A set of ( «values», «percentiles » ) pairs. This is also equivalent to specifying points on the Cumulative Probability curve.
The first case is equivalent to the second case, when the «percentiles»s are evenly spaced.
The result
The result of the function is a coefficient vector indexed by «K». This vector can then be passed directly to any of the Keelin-distribution functions, namely:
In all these cases, the vector returned from KeelinCoefficients is passed as the «xi» parameter, and you «K» index must be passed as the index parameter «I» of these functions. Also, you must pass
the 1 bit to the «flags» parameter. All of these functions name data parameter «values», but they all also support an alias name for that parameter of «ai», so that you have an option of passing the
coefficients using the named parameter convention using ai: to emphasize that these are coefficients, like this:
We don't recommend spending much time trying to interpret the coefficients. The first coefficient will always be the median of the distribution, but from there the others are less obvious. The second
coefficient tends to track to Variance, the third is tends to track Skewness and the fourth tends to track Kurtosis. They are not, however, these actual moments. It is possible to compute all moments
of the distribution directly from these parameters, see the Keelin (2016) reference, cited above.
When your quantity is unbounded, its distribution will have tails in both direction. In this case, you should omit the «ub» and «ub» parameters. If you know your quantity is bounded from below, then
specify «lb», and if you know that your quantity is bounded from above, specify «ub». The distribution supports all combinations of unbounded, bounded and semi-bounded distributions in this way.
When you compute the coefficients with a particular combination of «lb» and «ub», you must specify the same «lb» and «ub» parameters when passing these coefficients to Keelin, DensKeelin, CumKeelin,
or CumKeelinInv.
Returning the basis
Unless you are doing research on the Keelin distribution itself, you probably won't have a reason to access the basis. But, if you have a need, you can use this function to return the "basis" for the
distribution. This is a 2-D matrix indexed by «I» and «K» and is a function of «percentiles», but does not depend of «values». For example:
For an unbounded Keelin MetaLog, the values can be obtained from the basis and coefficients using
For the semi-bounded case with a lower bound, Ln(x-lb) is equal to Sum( coeffs * sampBasis, K ), hence you would use
For the upper-bounded case, Ln(ub-x) is equal to Sum( coeffs * sampBasis, K ), hence you would use
And for the fully-bounded case, Ln( (x-lb) / (ub-x)) is equal to Sum( coeffs * sampBasis, K ), hence use
See Also | {"url":"https://docs.analytica.com/index.php/KeelinCoefficients","timestamp":"2024-11-11T16:36:48Z","content_type":"text/html","content_length":"31245","record_id":"<urn:uuid:d8598487-ea2b-4deb-ad44-1aad0acd6a5f>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00202.warc.gz"} |
Using Experimental Probability to Determine the Expected Number of Outcomes of an Event
Question Video: Using Experimental Probability to Determine the Expected Number of Outcomes of an Event Mathematics • Second Year of Preparatory School
A factory produces two types of shirts: A and B. The table shows how many shirts of each type were sold in 5 samples of 100 shirts from 5 different shopping malls. If the factory sells 3,000 shirts,
how many of them do you expect to be of type A?
Video Transcript
A factory produces two types of shirts: A and B. The table shows how many shirts of each type were sold in five samples of 100 shirts from five different shopping malls. If the factory sells 3,000
shirts, how many of them do you expect to be of type A?
In order to answer this question, we need to recall the formula for expected value. This is equal to the probability of an event occurring multiplied by the number of trials or experiments. We will
begin by calculating the experimental probability of selecting a shirt of type A from the table. As there were five samples, one in each shopping mall, of 100 shirts, we know that there were 500
shirts in total. Adding the values in the second row of our table, we see that 227 of these shirts were of type A.
Whilst it is not required in this question, it is worth noting that 273 of the shirts sold were of type B and that 227 plus 273 equals 500. Since probability is equal to the number of successful
outcomes over the total number of possible outcomes, where each outcome is equally likely to be selected, we know that the probability that a shirt from this sample was of type A is equal to 227 over
500. Since the factory sells 3,000 shirts, the expected value is equal to 227 over 500 multiplied by 3,000. Both 500 and 3,000 are divisible by 500. So our calculation simplifies to 227 multiplied by
six, which is equal to 1,362.
We can conclude that from the information in the sample that if the factory sold 3,000 shirts, we would expect 1,362 of them to be of type A. | {"url":"https://www.nagwa.com/en/videos/737101049041/","timestamp":"2024-11-02T12:46:08Z","content_type":"text/html","content_length":"250193","record_id":"<urn:uuid:db8a39c0-0139-4a6f-9b48-be1ca16db8ef>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00242.warc.gz"} |
Magic Square Jigsaws
Drag the jigsaw pieces onto the frame so that the numbers form a magic square.
The sum of the numbers in each row, column and diagonal should be the same
Easier Recommended Suggested Don't miss
3 by 3 Magic Square Perfect Magic Square Magic Square Puzzle Octagram Star
If the four by four magic square is too big for Arrange the sixteen numbers on the 4 Find all of the possible ways of making the magic total Arrange the sixteen numbers on the octagram so that
you why not try this smaller, easier puzzle. by 4 grid to make a magic square. from the numbers in this four by four magic square. the numbers in each line add up to the same total.
The short web address is: The short web address is: The short web address is: The short web address is:
Transum.org/go/?to=magic Transum.org/go/?to=perfectmagic Transum.org/go/?to=magicsquarepuzzle Transum.org/go/?to=octagramstar
Description of Levels
Level 1 - Three piece puzzle
Level 2 - Four piece puzzle
Level 3 - Five piece puzzle
Level 4 - Six piece puzzle
Level 5 - Seven piece puzzle
Level 6 - Eight piece puzzle
Level 7 - Nine piece puzzle
Level 8 - Ten piece puzzle
Perfect Magic Square - If you can do the jigsaw magic squares, this is your next challenge.
More Puzzles including lesson Starters, visual aids, investigations and self-marking exercises.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a
teacher, tutor or parent. | {"url":"https://www.transum.org/Maths/Activity/Jigsaw/Magic_Square.asp?Level=4","timestamp":"2024-11-03T06:46:02Z","content_type":"text/html","content_length":"44169","record_id":"<urn:uuid:1194c9d4-ec62-4f5d-b2f5-9b4265cd9fcb>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00281.warc.gz"} |
What If Approximating the Future Can Be Simple?
Get curious about innovative methodologies with Adolfo Espíritu as he applies his curiosities about conceiving the future in the third part of a series on approximating the future for the Corporation
of Tomorrow.
A model can be thought as the skeleton of a problem: it has the minimum amount of information that is needed to describe a problem and its dynamics (by dynamics I mean how it evolves and behaves with
respect to some other parameters); skeletons--and models--give a representation of some aspect of the real thing. To make this possible, we simplify reality under assumptions, which will determine
the quality of our predictions. Three basic steps can be followed to create a model (Blanchard, et. al., 1998):
1. Clearly state the assumptions under which you are going to state the model; they should describe the relationships among quantities that will be studied. This has to do with the observation part,
since here we are trying to explain how the problem works; it has to do with understanding the problem and giving the most important aspects of the problem. Physical laws, definitions and empirical
observations come into play. For example, if we throw a ball on Earth with certain angle and speed, we know that it is going to go up and then down (that is what we see). However, to make things
easier, we are going to ignore friction between the ball and air. We want to predict where the ball land and its maximum height will.
2. Create a full list of the variables and parameters that are going to be used in the model; these are the key players of your predictions.Here, the main idea is to quantify and measure the
important aspects of your problem. Here, we distinguish three types of data:
• Independent variables: Their value is not influenced by other quantities.
• Dependent variables: They depend on other variables; a rule is needed to compute the dependent variable’s value given the independent variable’s value.
• Parameters: They are quantities that remain constant in the model.
For example, returning to our example of the ball, the distance travelled (dependent variable) will depend on the elapsed time (independent variable), and we can set as fixed values its initial
velocity and position (parameters). This is the least amount of information that we need to derive the model (I claim least because based on observations, we know that the ball is moving, and if we
change its velocity and initial position, we are going to get different results. For example, the mass of the object does not affect the result, so it is discarded).
3. Use your assumptions to relate the stated quantities through equations. Finally, we describe the relationship between variables and parameters through equations. Some good advice here is to keep
the algebra as simple as possible. In our example, I will omit the model (for further reference, check projectile motion), but what is important in these steps is to get the idea of how to get a
representation of our problem.
Finally, once the model is made, it is time for making predictions. The next step is to compare your predictions to your experimental data to determine how good your model is--which can be judged by
how accurate your calculations are to the real thing.
We make models because we want to make educated guesses and deliver the solution with fewer iterations; however, it is important to remember that models are not perfect, but approximate. Curiosity is
the main fuel of this process, since it is involved in the first step (understanding the problem), stating the input and output data (determine the variables and parameters), and, finally,
establishing how the data relates (equations).
What I want you to realize is that this method can help us solve problems at enterprises, since a problem can be modeled (maybe you do not use equations, but this process helps you to understand the
essence of the problem and its dynamics), so by identifying these contradictions, it’s easier to focus on the solution because you already know what’s not going to work based on the behavior of the
problem. This is one of the essences of the Blue Ocean Strategy, which is an innovation method, so next time you see a model remember that even though it is not perfect, it is a very good starting
point for developing your solution.
Want to see a new perspective of innovation through computation lenses? Stay tuned for the next series.
We'll be exploring questions like these live with curious people from all over the world to connect and collaborate in building the Corporation of Tomorrow (along with the School of Tomorrow), today!
Adolfo Arana Espíritu Santo is a student at the Monterrey Institute of Technology & Higher Learning and loves getting curious about, Math, Physics, and Quantum Computers.
Want to stay curious? What if you feed your curiosity and the Corporation of Tomorrow!
Thank you for your curiosity and stay curious with more Curiosity-Based Learning content and activities from What If Curiosity!
P. Blanchard, et. al. (1998). Differential Equations. USA: Brooks/Cole Publishing Company. | {"url":"https://www.whatifcuriosity.com/post/what-if-approximating-the-future-can-be-simple","timestamp":"2024-11-12T13:12:55Z","content_type":"text/html","content_length":"1050514","record_id":"<urn:uuid:d42c3481-4f73-4429-9403-5d581bb30c90>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00224.warc.gz"} |
ilometers to Nanometers
Kilometers to Nanometers Converter
Enter Kilometers
β Switch toNanometers to Kilometers Converter
How to use this Kilometers to Nanometers Converter π €
Follow these steps to convert given length from the units of Kilometers to the units of Nanometers.
1. Enter the input Kilometers value in the text field.
2. The calculator converts the given Kilometers into Nanometers in realtime β using the conversion formula, and displays under the Nanometers label. You do not need to click any button. If the
input changes, Nanometers value is re-calculated, just like that.
3. You may copy the resulting Nanometers value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Kilometers to Nanometers?
The formula to convert given length from Kilometers to Nanometers is:
Length[(Nanometers)] = Length[(Kilometers)] × 1000000000000
Substitute the given value of length in kilometers, i.e., Length[(Kilometers)] in the above formula and simplify the right-hand side value. The resulting value is the length in nanometers, i.e.,
Calculation will be done after you enter a valid input.
Consider that a high-end electric car has a maximum range of 400 kilometers on a single charge.
Convert this range from kilometers to Nanometers.
The length in kilometers is:
Length[(Kilometers)] = 400
The formula to convert length from kilometers to nanometers is:
Length[(Nanometers)] = Length[(Kilometers)] × 1000000000000
Substitute given weight Length[(Kilometers)] = 400 in the above formula.
Length[(Nanometers)] = 400 × 1000000000000
Length[(Nanometers)] = 400000000000000
Final Answer:
Therefore, 400 km is equal to 400000000000000 nm.
The length is 400000000000000 nm, in nanometers.
Consider that a private helicopter has a flight range of 150 kilometers.
Convert this range from kilometers to Nanometers.
The length in kilometers is:
Length[(Kilometers)] = 150
The formula to convert length from kilometers to nanometers is:
Length[(Nanometers)] = Length[(Kilometers)] × 1000000000000
Substitute given weight Length[(Kilometers)] = 150 in the above formula.
Length[(Nanometers)] = 150 × 1000000000000
Length[(Nanometers)] = 150000000000000
Final Answer:
Therefore, 150 km is equal to 150000000000000 nm.
The length is 150000000000000 nm, in nanometers.
Kilometers to Nanometers Conversion Table
The following table gives some of the most used conversions from Kilometers to Nanometers.
Kilometers (km) Nanometers (nm)
0 km 0 nm
1 km 1000000000000 nm
2 km 2000000000000 nm
3 km 3000000000000 nm
4 km 4000000000000 nm
5 km 5000000000000 nm
6 km 6000000000000 nm
7 km 7000000000000 nm
8 km 8000000000000 nm
9 km 9000000000000 nm
10 km 10000000000000 nm
20 km 20000000000000 nm
50 km 50000000000000 nm
100 km 100000000000000 nm
1000 km 1000000000000000 nm
10000 km 10000000000000000 nm
100000 km 100000000000000000 nm
A kilometer (km) is a unit of length in the International System of Units (SI), equal to 0.6214 miles. One kilometer is one thousand meters.
The prefix "kilo-" means one thousand. A kilometer is defined by 1000 times the distance light travels in 1/299,792,458 seconds. This definition may change, but a kilometer will always be one
thousand meters.
Kilometers are used to measure distances on land in most countries. However, the United States and the United Kingdom still often use miles. The UK has adopted the metric system, but miles are still
used on road signs.
A nanometer (nm) is a unit of length in the International System of Units (SI). One nanometer is equivalent to 0.000000001 meters or approximately 0.00000003937 inches.
The nanometer is defined as one-billionth of a meter, making it an extremely precise measurement for very small distances.
Nanometers are used worldwide to measure length and distance in various fields, including science, engineering, and technology. They are especially important in fields that require precise
measurements at the atomic and molecular scale, such as nanotechnology, semiconductor fabrication, and materials science.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Kilometers to Nanometers in Length?
The formula to convert Kilometers to Nanometers in Length is:
Kilometers * 1000000000000
2. Is this tool free or paid?
This Length conversion tool, which converts Kilometers to Nanometers, is completely free to use.
3. How do I convert Length from Kilometers to Nanometers?
To convert Length from Kilometers to Nanometers, you can use the following formula:
Kilometers * 1000000000000
For example, if you have a value in Kilometers, you substitute that value in place of Kilometers in the above formula, and solve the mathematical expression to get the equivalent value in Nanometers. | {"url":"https://convertonline.org/unit/?convert=kilometers-nanometers","timestamp":"2024-11-06T22:00:05Z","content_type":"text/html","content_length":"91044","record_id":"<urn:uuid:2c2b3fc4-5f84-44e1-b0d4-1739127ad129>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00891.warc.gz"} |