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Fundraiser for Andy Copland!
On Saturday, May 4th, we are having a fundraiser for Andy Copland at Skydive the Farm. Sunday evening he had a canopy collapse at the Farm and he's in pretty bad shape. His medical bills are piling
up quickly, and like a lot of full time skydivers, he does not have insurance.
Even if you are not able to attend the fundraiser itself, a paypal account has been set up for donations. ALL proceeds will go directly to Andy and helping him through the very tough times he has
ahead of him. Every little bit helps. All donation info is on the facebook event page that I have linked below. So please, take a moment and help him out.
Zep 0
The money I have in my Paypal account (which isn't much is on the way to him)
Although I never met the Pikey I enjoyed reading his posts and his determination against the odds to
be a base jumper, Just remember that all the small amounts that you can afford will add up
to a significant help.
Andy, mend yourself you Pikey TWAT.
Gone fishing
DougH 270
That sucks! Was this on a tandem jump? Hope he mends fast.
"The restraining order says you're only allowed to touch me in freefall"
airdvr 201
$$ on the way. Is there a thread somewhere? Cheeky sod.
Please don't dent the planet.
Destinations by Roxanne
turtlespeed 212
He was flying video.
I'm not usually into the whole 3-way thing, but you got me a little excited with that. - Skymama
BTR #1 / OTB^5 Official #2 / Hellfish #408 / VSCR #108/Tortuga/Orfun
skydiverbry 0
Thanks for the heads up!!
Since I was in a similar situation a few years back I can feel for him.
What ever I can afford is on it's way!!!
Blues skies Andy
Growing old is mandatory.Growing up is optional!!
D.S.#13(Dudeist Skdiver)
PhreeZone 15
I assume this is for a different canopy collapse than the one he suffered about a month ago that put him in the hospital then too?
Yesterday is history
And tomorrow is a mystery
I assume this is for a different canopy collapse than the one he suffered about a month ago that put him in the hospital then too?
I assume this is the same Andy Copeland who used to post here all the time, shitting all over anyone who made any comments regarding safety or making conservative choices when it comes to things like
canopy selection and BASE jumping, right?
So he ends up having two canopy collapses, severe enough to land him in the hospital in the course of one month? Go figure.....
nigel99 330
I assume this is for a different canopy collapse than the one he suffered about a month ago that put him in the hospital then too?
I assume this is the same Andy Copeland who used to post here all the time, shitting all over anyone who made any comments regarding safety or making conservative choices when it comes to things
like canopy selecytion and BASE jumping, right?
So he ends up having two canopy collapses, severe enough to land him in the hospital in the course of one month? Go figure.....
Dave, Andy is shit stirrer. My last jump with him about 18 months ago he had circa 700 jumps and was on a 170. My limited experience with Andy is that he loves to get a rise out of people.
I don't know his current jump numbers or canopy type though.
Experienced jumper - someone who has made mistakes more often than I have and lived.
he had circa 700 jumps and was on a 170.
I don't know his current jump numbers or canopy type though
Wanna bet it's something much smaller than a 170?
billvon 2,780
nigel99 330
he had circa 700 jumps and was on a 170.
I don't know his current jump numbers or canopy type though
Wanna bet it's something much smaller than a 170?
My point was that he isn't on a cross brace canopy at an inappropriate jump number, like so many people who are injured are. I always expect Andy's injuries to be from Base though.
I like Andy and hope he gets well soon.
Experienced jumper - someone who has made mistakes more often than I have and lived.
mpohl 1
No empathy here either. Other than that a fellow human being got hurt, and I wish him well.
There is another, resurrected thread: you can't afford health insurance, you shouldn't be skydiving! Or spending your money on skydiving, or being a dz bum.
To the organizers of the benefit: don't hand the proceeds over to the hospital. Instead reserve those for filing for medical bankruptcy; much better use of the money!!!
I assume this is the same Andy Copeland who used to post here all the time, shitting all over anyone who made any comments regarding safety or making conservative choices when it comes to things
like canopy selection and BASE jumping, right?
So he ends up having two canopy collapses, severe enough to land him in the hospital in the course of one month? Go figure.....
mpohl 1
That would be ok, if 50 lbs + gear. 70 total. 1.4 loading
>Wanna bet it's something much smaller than a 170?
Sub-100 I believe.
My point was that he isn't on a cross brace canopy at an inappropriate jump number, like so many people who are injured are
With 700 jumps total, and jumping a 170, how many jumps do you think it would take to get down below 100 sq ft safely? Let's also keep in mind that they don't make anything but HP canopies below 100
sq ft, so it's not like he was a big-boy on with a higher WL on a Spectre or somthing of that sort.
With only 18 months to make the transition from a 170 to a sub-100 HP canopy, I'm not sure there is a 'safe' way to do that. You're either skipping sizes, or shorting yourself on time on each size.
Either case is not the preferred method.
That said, just to be clear, I don't have any ill-will towards to guy. I don't even know him personally. I do feel badly to hear that anyone is injured in anyway while jumping.
The reason I'm posting the things I am is to illustrate that there are consequnces to the choices you make as a jumper. Those consequnces become very real very quickly, and sometimes they stay 'real'
for a very long time. He's not the first 'big shot' we've had here was 'too cool for school', and he probably won't be the last. At least one of them I can remember is dead, and another is in a
wheelchair for the rest of his life (Sangi, who had the balls to come back and tell his tale). Now we have another one who's fate is still to be determined.
The point is that everyone else can see where it get's you, and maybe the guy in the '400 jumps on a Velo' thread will take notice, or maybe not. Hopefully the next guy in line will wise up and end
the cycle before it comes around again and takes out another 'big shot' jumper.
Here is an update from Andy's fiance for those who actually care and aren't here simply to badmouth a fellow skydiver who is also a nice guy:
Hello everyone! Andy Copland is doing really well!! He has had three surgeries and will probably have to have at least one more. Big thank you to everyone for all of your support through this!
Andy has been intubated since Sunday, but they are slowly taking him off of sedation. He is starting to wake up and he looks absolutely amazing!! Thank you to all of our amazing friends for the
fundraiser and all of your help and support through this! It means SO much to us!!
If you're in this thread simply to bad mouth my friend, please do so elsewhere. This thread is intended to raise funds and support for someone in need.
normiss 736
Any word on conditions at the Farm when this happened?
I know it's tight there, big ass trees, big ass tree line that messes with pocket rockets.
Just curious.
lawrocket 3
I agree. I hesitated to be the first to put it out there, but I think is the second time in 8 weeks that he's frapped in. He got all busted a few years ago on a base jump, too. Each time he's been
uninsured, and I find it somewhat irritating that a fundraiser is being put out there to pay his medical bills. I view it more as a fundraiser as a way of paying for his jumps.
He's a good guy and interpersonally I like him a lot. But the sport warned him twice already and the message of "cover thy ass and get thee coverage" went unheeded. He played with fire again.
If one can afford to skydive one can afford insurance. This sport is as expensive as it is dangerous and a fundraiser for just strikes strikes me as some hardcore enabling.
My wife is hotter than your wife.
I'm trying to keep my cool with you people. There is another thread for this discussion about whether or not you would contribute. THIS thread is for those who are willing to help others no matter
what the conditions. And just to be clear, he was WORKING when both these jumps happened. He doesn't fun jump, he only does working jumps. So this money isn't going to fund more jumping. If you read
my update you would see that his condition is pretty severe. So please, take your discussion about whether or not you would help a fellow skydiver and human being to the appropriate thread. Here is a
link for those too inept to find it on your own.
lawrocket 3
I'm not out to badmouth him. And it doesn't mean I don't care about the guy. I like him.
I just think it's pretty fucked up to spend money on jumping instead of insurance. And it can simply nnot be "obliviousness to the risk" that explains it.
I care about him and it bothers me to see this happen - again. And it bothers me that it is often the case that jumpers are uninsured.
My wife is hotter than your wife.
If you're in this thread simply to bad mouth my friend, please do so elsewhere. This thread is intended to raise funds and support for someone in need
I did say that I don't know him personally, and in that regard have nothing against him.
However, he went out of his was to post obnoxious and contrary remarks to my posts when I was tyring to give sound, safe advice to other jumpers. In that regard, I cannot deny (nor ignore) that his
choices have put him into a bad spot.
Additionally, to be clear, the bad choices are not those regarding jumping without insurance, they are his equipment choices and the way he chose to use that equipment. There is a lesson to be
learned here, and I'm not going to let it pass without pointing that out.
That said, I'll gladly make a donation via the paypal account. I have nothing against the guy personally, and we do have some friends in common, but this is where he gets to stand behind what he said
and his attitude about safety and making conservative choices. Unfortunately for him, he came up on the wrong side of that issue.
pchapman 278
His profile (which may or may not be fully up to date) says Velo 84 and 1100 jumps (+ 300 BASE), wing loading unknown.
Getting canopy collapses is a bit more unusual than just plain flying into the ground...
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Want to know more about Pentagons?
Ah the Pentagon, part of the Polygon family. So you want to know more about Pentagons. You've seen this shape aroound but can't remember how many sides a Pentagon has. Here is how a Pentagon looks
According to Dictionary.com, the Pentagon [pen-tuh-gon, -guhn] is a polygon having five angles and five sides. In the USA, Arlington, Virginia, the Pentagon was constructed to be an important place
during the World War II. Being the headquarters of the USA Department of Defence, this has always been a symbol of leadership.
Now back to Pentagon the shape. Here are 5 commonly asked questions (on google) about Pentagons:
1. What is a Pentagon? Pentagons can be regular or irregular and convex or concave. A regular pentagon is one with all equal sides and angles. Its interior angles are 108 degrees and its exterior
angles are 72 degrees. An irregular pentagon is a shape that does not have equal sides and/or angles and therefore do not have specified angles.
2. What is an example of a Pentagon? There are examples of pentagons in real life in man-made structures like the Pentagon in the United States, and also in nature in flowers like morning glories and
okra. Other items, like home plates in baseball, are often in the shape of irregular pentagons. A pentagon is a five-sided shape.
3. Is any 5 sided shape a Pentagon? In geometry, a pentagon (from the Greek πέντε pente and γωνία gonia, meaning five and angle) is any five-sided polygon or 5-gon. The sum of the internal angles in
a simple pentagon is 540°. A pentagon may be simple or self-intersecting. A self-intersecting regular pentagon (or star pentagon) is called a pentagram.
4. Why Pentagon is a 2D shape. The word pentagon itself tells you what it is. ... A regular pentagon is one with all equal sides and angles. Its interior angles are 108 degrees and its exterior
angles are 72 degrees. An irregular pentagon is a shape that does not have equal sides and/or angles and therefore do not have specified angles.
5. Can a Pentagon have curved sides? Circles and shapes that include curves are not polygons - a polygon, by definition, is made up of straight lines.
one cat
two cats
ten cats
no cats
Here is a link.
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Adding mixed integers with whole integers
adding mixed integers with whole integers Related topics: cubed factoring
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11+ papers[practice papers for year 5 online]
solve my algebra equation
holt math combinations
multiplying and dividing decimals calculator
roots with ti 83
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hinrg_c Posted: Thursday 20th of Jul 10:41
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nxu Posted: Saturday 22nd of Jul 07:22
Believe me, it’s sometimes quite hard to learn a topic by your own because of its complexity just like adding mixed integers with whole integers. It’s sometimes better to ask someone to
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DVH Posted: Monday 24th of Jul 09:31
Algebrator is a nice thing. I have used it a lot. I tried solving the problems myself, at least once before using the software. If I couldn’t solve the question then I used the software
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MeoID Posted: Monday 24th of Jul 21:35
Wow! This sounds tempting. I would like to try the program. Is it costly ? Where can I find it?
From: Norway
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Majnatto Posted: Tuesday 25th of Jul 20:32
This is the site you are looking for : https://softmath.com/faqs-regarding-algebra.html. They guarantee an unrestricted money back policy. So you have nothing to lose. Go ahead and Good
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Number Sense Activities for Kindergarten - Numbers 1 to 20 Clip Cards Math Center
Number Sense Activities for Kindergarten – Numbers 1 to 20 Clip Cards Math Center
Looking for a hands-on and engaging way to practice number sense with your kindergarten for numbers from 1 to 20? These Number Sense Clip Cards will help your kindergartener or first grader practice
their number sense for numbers 1 to 20 while training those fine motor skills at the same time. This engaging number sense activity for kindergarten is perfect for parents, teachers, and
homeschoolers looking to practice or teach number recognition, counting, and one-to-one correspondence with their learners.
Number Sense Activities for Kindergarten
In early childhood education, building a strong foundation in number sense is paramount as it serves as the stepping stones for math skills they learn later on such as addition and subtraction.
Kindergarten is a critical phase where young learners begin to explore the fascinating world of numbers, and introducing them to engaging activities can make a significant difference.
These number sense clip cards are a versatile teaching tool that aids in reinforcing counting, one-to-one correspondence, subitizing, and number recognition. This a simple, low prep activity and only
requires clothespins or circle counters to do the activity. This is an easy and fun way to sneak in some fine motor skills with numbers.
When they use these number clip cards learners will be practicing the following skills:
• Counting
• Subitizing
• One-to-one correspondence
• Number recognition
• Fine motor skills
• Simple addition
That’s a whole bunch of skills with one simple printable. We hope your learners have fun learning and practicing their number sense with these Number Sense Clip Cards. Simply print these number clip
cards, laminate and you are ready to go.
What is Number Sense? – Building Number Sense in Kindergarten
Number sense is the ability to understand, relate and connect numbers in a meaningful way. Gaining a strong foundation in number sense at a young age is important as it serves as a fundamental
building block for future mathematical understanding.
In the context of kindergarten, number sense would extend beyond the ability of just number recognition. It would require students to grasp the concept of quantity, understanding number relationship
and subitizing.
These are some of the ways in which learners can practice and build upon their number sense:
• Counting
• Place value
• Subitizing (Dots or items)
• Ten frame
• Dice (1 to 6 and adding two or more dice together)
• Tally marks
• Finger counting
• Dominoes
Teaching Number Sense to Kindergarten Students
Wheel clip cards have more options that the normal clip cards you might be used to. Learners will have to choose between more options making this activity a bit tougher than normal clip cards. There
can also be more than one option, so learners will have to realize that and be able to select all the correct answers.
For this reason, I would recommend using these only once the learner is familiar with the concept and needs further practice. Using the number sense clip cards to introduce the topic to the learner
might be overwhelming and might result in confusion towards the learner.
Hands-on activities are extremely important in developing number sense in kindergarten learners. Hands-on activities actively engage children in the learning process and make learning fun and
As learners are identifying the different number representations and clipping the correct answers they are actively engaging in the learning process. This makes is so much more fun for the learners.
What’s better is that these clip cards help in training kindergarteners with more than one skill. The clip cards can be used both to train those fine motor skills (clipping clothespins is hard work!)
and at the same time practice whatever concept they need to work on.
Using the Number Sense Clip Cards – Number Sense Activities for Kindergarten
In these number sense cards, there are two sets of wheels. One set focuses on number sense for numbers 1 to 10 while the other set focus on number sense for numbers 1 to 20. Simply choose which set
of numbers you would like your learner to practice on and print those out.
Numbers 1 to 10:
Each of the cards contain the following options:
• Ten Frame
• Dice
• Tally Marks
• Domino
• Fingers
There are a total of 8 options, of which 5 of them match the number in the center. The other 3 options do not match the number.
Numbers 11 to 20:
Each of the cards contain the following options:
• Ten Frame
• Tally Marks
• Place Value (Base 10 Blocks)
• Dominoes
• Dice
There are a total of 8 options, of which 5 of them match the number in the center. The other 3 options do not match the number.
Materials and Prepping the Activity
These are the materials required to prep this number sense activity for your students.
• Wooden clothespin (these are colored)
• Wooden clothespin
• Printer
• Laminator (highly recommended)
• Laminating sheets
To get this activity ready, print out the pages. Laminate the pages. I highly recommend doing this, especially if the activity is going to be used multiple times.
Cut apart to create the clip cards.
Present the activity with clothespins and you’re all set!
Learning Numbers 1 to 20 – Number Sense Activities in Kindergarten
If your learners are just getting the hang of their number sense, start off with the smaller numbers before heading over to the larger numbers.
For e.g. start with the number 3 or a number they might be more familiar with.
Ask the learner to see whether they can identify any number representations that match the number 3. If so, they can clip those.
Then slowly explain about the other parts for example the number of dots in the dominoes make up a number, and likewise the dots in the ten frame.
The number of tally marks can be counted to get a number and then it becomes a set when it reaches 5.
As students slowly gain more practice and become familiar with their concepts of number sense they will be able to deduce and make connections quicker. As their number sense concepts improve, they
will be able to make connections between numbers and understand concepts such as greater, smaller, addition and subtraction with more ease. It helps to build a strong foundation for their math
Ideas for the Number Sense Wheel Clip Cards Printable
You can also make this a self-checking activity. Just put a dot on the back of the correct parts of the clip cards. If the learner has clipped the dot then the answer is correct, if not ask them to
try it again, or discuss why they chose that answer.
This packet can be used to strengthen fine motor skills, explore during morning work or used as supplemental math centers. The clip cards are skill-based, they are not thematic, so they can be used
anytime during the year. The centers are best used once printed on cardstock and/or laminated. You can choose to implement these centers in an independent way or use with small-groups.
Differentiating the Activity – Number Sense Clip Cards
This activity is a super simple center to differentiate. Simply choose the numbers that your learners need to practice on and present it to them.
For instance, if a learner is very good with numbers from 1 to 10, but needs some help with their teen numbers, present them with the clip cards from 11 to 20, and they can strengthen their number
sense for those numbers.
Kindergarten Clip Cards ENDLESS Bundle – Math and Literacy Centers
If you loved this printable, and would love to have more printables like this, check out this bundle of Kindergarten Clip Cards below. It currently contains 11 sets of clip cards, but is endless
since I have so many ideas for clip cards. Grab it now before the price goes up as more sets are added!
Check out These Printables!
Shop the Number Sense Clip Cards
Practice number sense with your kindergarten learners for numbers 1 to 20. Click on the button below to grab your copy today!
Happy teaching! ~
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Understanding Snell's Law Problems
Welcome to our article on understanding Snell's Law problems! If you're a student studying physics or optics, then you've probably come across this law and its associated problems. But don't worry,
we've got you covered. In this article, we'll break down what Snell's Law is and how to solve problems related to it. Whether you're struggling to grasp the concept or just need a refresher, keep
reading to become a pro at tackling Snell's Law problems.
Get ready to dive into the world of physics and optics as we explore this fundamental law and its applications. First, let's start with the basics.
Snell's Law
is a fundamental principle in optics that describes how light behaves when it travels through different mediums. It states that the ratio between the sine of the angle of incidence and the sine of
the angle of refraction is equal to the ratio between the indices of refraction of the two mediums. This might sound complex, but don't worry - we will break it down for you with clear explanations
and examples. For those interested in conducting experiments, we will also provide step-by-step instructions on how to set up a simple experiment to observe
Snell's Law
in action.
This will give you a hands-on experience and a deeper understanding of the concept. When it comes to solving problems involving Snell's Law, there are a few things to keep in mind. One common mistake
is forgetting to convert angles from degrees to radians, so make sure to double-check your calculations. Another tip is to draw diagrams to help visualize the problem and determine which values are
known and which need to be solved for. Now, let's take a look at how Snell's Law is applied in various fields. In physics, it is crucial for understanding the behavior of light in different mediums,
such as water or glass.
In engineering, it is used in designing lenses for cameras or telescopes. In medicine, it is applied in vision correction procedures, such as LASIK surgery. If you're interested in learning more
about Snell's Law, there are plenty of online resources available. Websites like Khan Academy offer comprehensive lessons and practice problems to help you master the concept. You can also check out
textbooks on optics or physics for more in-depth explanations and examples. In conclusion, understanding Snell's Law is essential for anyone studying physics or interested in the principles of
By breaking down the concept, providing real-life examples, and offering tips and resources for further learning, we hope this article has helped you gain a better understanding of Snell's Law
Resources for Further Learning
To further enhance your knowledge of Snell's Law, we recommend exploring various online tutorials, videos, and research papers. These resources can provide you with a deeper understanding of the
principles behind Snell's Law and its applications. One great resource is Khan Academy's tutorial on Snell's Law, which offers clear explanations and practice problems to help solidify your
understanding. Additionally, YouTube has many helpful videos on Snell's Law, ranging from simple explanations to more complex applications. If you want to dive even deeper into the topic, there are
also numerous research papers available online that discuss Snell's Law in detail. These papers can provide a more technical understanding of the subject and may be helpful for those pursuing a
career in optics or physics.
Real-life Applications of Snell's Law
Snell's Law is not just a concept confined to textbooks and exams.
In fact, it has numerous real-life applications that we encounter every day without even realizing it. From lenses to prisms, Snell's Law plays a crucial role in the functioning of various objects
and devices that we use in our daily lives. One of the most common applications of Snell's Law is in lenses. Lenses are used in eyeglasses, cameras, and telescopes, among other things, and they rely
on the principle of refraction to function. When light passes through a lens, it bends according to Snell's Law, allowing us to see objects more clearly or magnified. Another example of Snell's Law
in action is in prisms.
Prisms are used in many optical instruments, such as binoculars and microscopes, to bend and separate light into its different wavelengths. This is possible because of the different angles of
refraction that occur as light passes through the prism, which follows Snell's Law. Other everyday objects that use Snell's Law include fiber optic cables, which transmit data through the refraction
of light, and even simple things like drinking glasses, which use curved surfaces to refract light and make objects appear larger than they are. As you can see, Snell's Law has many practical
applications that we encounter regularly. Understanding this concept can not only help us solve physics problems but also give us a deeper appreciation for the functioning of everyday objects around
Tips for Solving Snell's Law Problems
Snell's Law
is a fundamental principle in the field of optics, and understanding how to solve related problems is crucial for success in physics. In this section, we will provide some tips on how to effectively
approach and solve problems involving Snell's Law.
1.Understand the Concept
Before attempting to solve a problem, it's important to have a solid understanding of the underlying concept.
Make sure you are familiar with the equation for Snell's Law: n1sinθ1 = n2sinθ2. This equation relates the angles of incidence and refraction to the indices of refraction of the two mediums involved.
2.Draw Diagrams
Visual representations can be extremely helpful in solving Snell's Law problems. Draw diagrams that accurately represent the scenario described in the problem, labeling all relevant angles and
indices of refraction.
3.Use Trigonometry
Trigonometry is a key tool in solving Snell's Law problems. Be sure to use the appropriate trigonometric functions (sine, cosine, tangent) when calculating angles or sides of triangles.
4.Check Units
When using the equation for Snell's Law, make sure that all units are consistent.
For example, both angles should be in radians or degrees, and both indices of refraction should be in the same unit (usually decimal or fraction).
5.Practice, Practice, Practice
The more you practice solving Snell's Law problems, the more comfortable you will become with the concept and the faster you will be able to solve them. Look for online resources or textbooks with a
variety of practice problems to hone your skills. With these tips in mind, you should feel more confident in approaching and solving problems related to Snell's Law. Remember to always double-check
your work and seek help if you get stuck. Happy problem-solving!
Tips for Solving Snell's Law Problems
One of the key aspects of understanding Snell's Law is knowing how to effectively solve problems related to it.
Here are some tips to help you approach and solve these types of problems:
• Identify given information: Before attempting to solve a Snell's Law problem, make sure you clearly identify what information is given and what you are trying to find. This will help guide your
thought process and prevent confusion.
• Use the correct formula: There are different formulas for calculating different values in Snell's Law. Make sure you use the correct formula for the specific problem you are working on.
• Draw a diagram: Visual aids can be extremely helpful when solving Snell's Law problems. Draw a diagram to represent the situation and label all given information.
• Apply the law of refraction: Remember that Snell's Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of
Use this law to set up an equation and solve for the unknown value.
• Check your answer: After solving a problem, always double check your answer to ensure it makes sense and is within a reasonable range. If it does not match the expected answer, go back and review
your steps.
By following these tips, you can effectively approach and solve Snell's Law problems with confidence. With practice, you will become more comfortable with applying this concept and solving related
Real-life Applications of Snell's Law
From lenses to prisms, discover how Snell's Law is used in everyday objects. Snell's Law is a fundamental principle of optics that explains the behavior of light as it passes through different
This concept is applied in various objects that we use in our daily lives, such as eyeglasses, cameras, and even car windshields. By understanding Snell's Law, we can gain a deeper appreciation for
the technology and devices that we often take for granted. Let's explore some real-life examples of how Snell's Law is used in these objects.
One of the most common applications of Snell's Law is in lenses.
Lenses are used in many optical devices, including eyeglasses, microscopes, and telescopes. These devices use curved lenses to bend and focus light, allowing us to see objects more clearly. The shape
of the lens is designed based on the principles of Snell's Law, ensuring that light is refracted at the correct angle for optimal vision.
Prisms are another example of how Snell's Law is used in everyday objects. Prisms are triangular-shaped pieces of glass that are often found in binoculars or cameras.
They work by refracting light at different angles, which creates a spectrum of colors. This phenomenon is known as dispersion and is a result of Snell's Law in action. Other everyday objects that
utilize Snell's Law include fiber optic cables, which use multiple layers of glass to transmit light signals over long distances with minimal loss. Car windshields also use a specific curvature based
on Snell's Law to reduce glare from the sun and improve visibility while driving. As you can see, Snell's Law plays a crucial role in the functioning of many objects that we encounter on a daily
By understanding this concept, we can gain a deeper understanding of the world around us and appreciate the science behind everyday technology.
Real-life Applications of Snell's Law
Snell's Law is a fundamental principle in optics that explains how light behaves when it passes through different mediums. This concept has numerous real-life applications, including in everyday
objects such as lenses and prisms. Lenses, which are used in eyeglasses, cameras, and telescopes, rely on Snell's Law to bend and focus light in a specific direction. The curvature of the lens and
the refractive index of the material it is made of determine how much the light will be bent. Understanding Snell's Law is crucial for accurately designing and producing lenses that can correct
vision or capture clear images. Prisms, commonly found in binoculars and other optical instruments, also utilize Snell's Law.
These triangular-shaped objects are able to separate white light into its individual colors by refracting the different wavelengths of light at different angles. This is possible because of the
varying refractive indices of different materials used to make the prism. But Snell's Law isn't just limited to these objects. It is also used in various industries such as architecture and
engineering to design and construct buildings with specific lighting requirements. In medicine, Snell's Law is crucial for understanding how light travels through different tissues in the human eye,
leading to advancements in vision correction procedures.
Tips for Solving Snell's Law Problems
When faced with problems related to Snell's Law, it can be overwhelming and confusing.
However, with the right approach, you can effectively solve these problems and gain a better understanding of the concept. The first step in solving Snell's Law problems is to clearly identify what
is given and what is being asked. This will help you determine which formulas and equations to use. Make sure to carefully read the problem and underline or highlight important information. Next,
draw a diagram to visualize the problem. This will help you better understand the scenario and make it easier to apply the formulas. When using Snell's Law, it is important to remember that the angle
of incidence is always measured from the normal line, while the angle of refraction is measured from the surface of the medium.
This can help avoid confusion when plugging in values. Another helpful tip is to always double check your calculations and units. Make sure to convert all values to the correct units before solving
the problem. If you are still struggling with solving Snell's Law problems, don't hesitate to seek help from a teacher or tutor. Practice makes perfect, so try solving different types of problems to
improve your understanding and speed. By following these tips, you can approach and solve problems related to Snell's Law effectively and confidently.
Real-life Applications of Snell's Law
Snell's Law, also known as the law of refraction, is a fundamental principle in optics that describes the behavior of light as it passes through different mediums. While it may seem like an abstract
concept, Snell's Law has numerous real-life applications in various objects that we encounter every day.
Let's take a closer look at some of these applications.
One of the most common uses of Snell's Law is in lenses. Lenses are used in many optical instruments, such as cameras, eyeglasses, and microscopes. These devices use curved lenses to bend and focus
light, allowing us to see images more clearly. Snell's Law plays a crucial role in determining the angle at which light is bent by the lens, helping to create a sharp and magnified image.
Prisms are another example of how Snell's Law is used in everyday objects.
Prisms are triangular-shaped pieces of glass or plastic that are used to separate white light into its component colors, creating a rainbow effect. This separation occurs because different colors of
light have different wavelengths and therefore bend at different angles when passing through the prism. Snell's Law helps to explain this phenomenon and is essential in designing prisms for various
applications. Other examples of everyday objects that utilize Snell's Law include fiber optic cables, mirrors, and even our own eyes. By understanding this fundamental principle, we can gain a better
appreciation for the role of optics in our daily lives.
Tips for Solving Snell's Law Problems
When it comes to solving problems related to Snell's Law, it's important to have a clear and organized approach.
This will not only help you understand the problem better, but also make it easier to find the solution. Here are some tips to keep in mind:
• Draw a diagram: Visualizing the problem can make it easier to understand and solve. Draw a diagram with all the relevant information, such as the angles of incidence and refraction, and the
refractive indices of the mediums.
• Identify known and unknown variables: Before jumping into calculations, make sure you know which variables are given and which ones you need to find. This will help you choose the correct
equations to use.
• Use Snell's Law equation: The equation for Snell's Law is n1sinθ1 = n2sinθ2.Make sure to use the correct values for n1 and n2, which represent the refractive indices of the mediums, and θ1 and
θ2, which represent the angles of incidence and refraction.
• Pay attention to units: Make sure all your units are consistent throughout your calculations.
If necessary, convert them to the same units before plugging them into the equation.
• Check your answer: After completing the calculations, double check your answer to ensure it makes sense. If it doesn't, go back and review your steps to see where you might have made a mistake.
In conclusion,
Snell's Law
is a crucial concept in the field of
that has various real-life applications. By understanding its principles and learning how to apply them, you can solve complex problems and gain a deeper appreciation for the behavior of light. We
hope this guide has been helpful in explaining the concept of Snell's Law and providing resources for further learning.
Happy studying!.
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Questions and Answers
(applies to all RPL machines)
As I read through the questions and answers section of the HP48
owner's manual, I found some ommisions so I am proposing this addendum
to address these. Comments and additions from others would be
A: Support, Batteries, and Service
Q: When I press the backspace key I sometimes get an error message.
A: When you press the backspace key with nothing on the command line,
it performs the Drop function. Drop is a function so it requires an
argument from the stack and returns no result -- thereby clearing the
bottom item from the stack. So when you perform the Drop function
without an argument on the stack, technically you are committing an
Q: But sometimes I just press the backspace key out of habit before
starting a new calculation.
A: That isn't necessary in RPN(L).
Q: Yes, but I want to anyway and what harm does it do?
A: Well, it is an error to call a function without an argument, isn't
it? If you really want to clear things out, press right-shift CLR,
this clears the stack without returning an error when the stack is
Q: Why does backspace give an error if CLR doesn't? Why not just have
neither one give an error?
A: Because Drop is a function and functions require arguments while
CLR is a command and commands don't require arguments.
Q: Are you insane?
A: This is a perfectly consistent approach, you're just not smart
enough to appreciate it.
Q: How come you can't edit equations in the Equation Writer application?
A: It's an equation *writer*, not an equation *editor*.
Q: Why are half the algebra functions in the Algebra menu and half in
the Equation Writer? To get anything useful done I spend hours going
back and forth.
A: The Equation Writer isn't really done yet, it will be completed in
the HP48SX's successor.
Q: Will there be an upgrade option?
A: No.
Q: Why not?
A: The same 178,000 inveterate geeks account for over 70% of our
sales. If we started an upgrade plan, our dealer channel would
evaporate overnight like the morning dew.
Q: Whenever I store a number, it disappears just before I was about
use it and then I have to leave the menu I was going to use to recall
it again.
A: This is normal. It is part of the HP48's elegantly consistent
approach to handling functions. The STO function takes two arguments
including the number to be stored.
Q: Is there going to be another owner's manual? In spite of its size,
this one seems oddly incomplete.
A: A more complete manual was considered but dropped. Partially due to
expense but mostly because market research shows that prospective
buyers of pocket calculators would find a 4 volume 3600 page manual
unsettling. Various geek organizations provide more detailed technical
information, support, and pointless speculation about the HP48.
Jeff Brown
Steinmetz & Brown, Ltd.
2675 University Ave
St Paul MN 55114
+1 612 646 2478
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Best Comparing Fractions Calculator with Steps - PineCalculator
Introduction to Comparing Fractions Calculator:
Comparing fractions calculator is a digital tool that helps you to find the fractions to determine which one fraction value is greater, lesser, or may be equal to each other. It can also reduce
fractions into the simplest form after comparing both fractions.
The compare fractions calculator is a valuable tool for getting solutions for comparing fractions problems, for educational, students, and professionals, to do everyday tasks.
What are Comparing Fractions?
Comparing fractions method is used to evaluate the comparison between relative sizes based on their numerators and denominators. It determines which fraction is greater, lesser, or if they are equal
to each other.
It uses different methods, such as finding a common denominator or using cross-multiplication, and decimal fraction methods to compare both fractions accurately.
How to Solve Compare Fractions?
For comparing fractions, the comparing fractions calculator uses different methods to check whether the given fraction is greater or lesser than the other fraction. These methods are:
• Find a Common Denominator:
To compare fractions, you need a common denominator. In the fraction comparison calculator, If there’s already a common denominator, then only numerator numbers are compared. The greater numerator
number has a greater fraction than the other fraction.
• If you have a Different Denominator:
For the comparison between two fractions, you need to first make them alike denominators. The comparing fraction calculator, make the same denominator value by taking the LCM. After solving the
fraction to make it alike denominator, you can compare the fraction.
• Cross-Multiplication Method
In the compare fraction calculator, the cross-multiplication method is used in which the numerator of one fraction is multiplied by the second fraction denominator.
Then it multiplies the second fraction numerator by the first fraction denominator. After multiplication compare both the numbers. The bigger number value has a greater fraction than the other
fraction when you compare both of them.
Example of Comparing Fraction:
Let us understand the working procedure of comparing fractions calculator which uses the fraction method to solve such comparing fractions.
Use < or > to compare the two fractions:
$$ \frac{4}{5} \;and\; \frac{14}{20} $$
$$ \frac{4}{5} \;=\; \frac{?}{20} $$
$$ \frac{4.4}{5.4} \;=\; \frac{16}{20} $$
$$ \frac{16}{20}\; > \; \frac{14}{20} $$
$$ \frac{4}{5} \;>\; \frac{14}{20} $$
How to Use Comparing Fractions Calculator?
The compare fractions calculator has a user-friendly design so that you can use it easily to calculate the relative size of fraction comparison questions.
Before adding the input value to the fraction comparison calculator, you must follow some guidelines to avoid any trouble in the evaluation process. These guidelines are:
1. Enter the first fraction value in which the numerator or denominator value is in the input field.
2. Enter the second fraction value (numerator or denominator value) in the input field.
3. Review your input numbers for comparing fractions because if the fraction values are not correct then our comparing fraction calculator does not provide you the exact solution for the comparison
of fraction problems.
4. Click the “Calculate” button to get the result of your given comparison of the fraction problem.
5. If you want to try out our compare fraction calculator for the first time then you must check the load example and its solution that gives you clarity about this concept.
6. Click on the “Recalculate” button to get a new page for solving Comparing Fraction problems.
Output of Compare Fractions Calculator:
Comparing fractions calculator gives you the solution when you add the input into it. It provides you with solutions in a step-wise process. It may contain as
• Result option gives you a solution for comparing fraction problems.
• Possible step provides you with all the steps of the comparison fraction problem in detail.
Advantages of Fraction Comparison Calculator:
Fraction calculator compare gives you tons of advantages whenever you use it to calculate the comparison between two fractions. These advantages are:
• Compare fractions calculator saves time and effort from doing lengthy calculations of comparing fraction problems
• It is a free-of-cost tool so you can use it for free to find the relative size for two fraction values
• Comparing fraction calculator is a versatile tool that allows you to solve various types of fractions (alike fraction, unlike fraction) for comparison of these fraction
• You can use this compare fraction calculator for practice so that you get a strong hold on the comparing fraction concept
• It is a trustworthy tool that provides you with accurate solutions every time whenever you use it to find the comparing fraction problem for calculation.
• Comparing fractions calculator is an educational tool that is used to teach children about the concept of fraction relative size problems and how to perform comparisons of fraction.
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Peacock 3 Training
Peacock 3 will cover the following points:
● Performing five number sums beyond 500 including the use of decimal points. All work is timed.
● Performing more challenging sums involving addition, subtraction and multiplication.
● Developing mental arithmetic skills using five number sums using number bonds up to 50.
● Encouraging children to complete two by one digit multiplications mentally.
● Performing two by two digits multiplications using the abacus.
● Performing simple divisions using the abacus.
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ESSAY Were the data analysis techniques appropriate?Were the data analysis techniques appropriate? - AcademiaElites
Were the data analysis techniques appropriate?
Posted: February 26th, 2022
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This assignment requires an expert in SPSS and quantitative methodologies.
-Please refer to the PDF file :(Assignment #2A – Hands-on Practice) and answer the Question: Is the empirical analysis (interpretation) correct? Were the data analysis techniques appropriate? Please
justify your answers.
-The instructor provided a paper that he stated includes the answer to his question. Please refer to the PDF file (Assignment #2A – Paper).
-I was able to find the original article that has the analysis of the question. Please check the PDF file (Directional distance function DEA estimators for evaluating efficiency gains from possible
mergers and acquisitions). I think the answer to the question is also there.
-I also uploaded the class notes that include the test used in this question (Wilcoxon test). Please refer to the PDF file (Wilcoxon).
-To answer the question, please do the following:
1) In the dataset (excel sheets) we have two “Related samples” for each year (2014,2015, & 2016) for each group of banks (Conventional & Islamic). Which makes 6 datasets that require to run the
Wilcoxon test. Please follow the class notes to run the Wilcoxon test in SPSS 6 times for each of the following:
Please refer to the sheets in the excel file.
Conventional Banks:
1- Sample 1 (Actual) vs Sample 2 (Actual & Virtual) in 2014
2- Sample 1 (Actual) vs Sample 2 (Actual & Virtual) in 2015
3- Sample 1 (Actual) vs Sample 2 (Actual & Virtual) in 2016
Islamic Banks:
1- Sample 1 (Actual) vs Sample 2 (Actual & Virtual) in 2014
2- Sample 1 (Actual) vs Sample 2 (Actual & Virtual) in 2015
3- Sample 1 (Actual) vs Sample 2 (Actual & Virtual) in 2016
2) Please report the output tables from SPSS, analysis and interpretation. Please use the same method used in the class notes to analyse.
3) As per the class notes, if we have 2 related samples and an unbalanced design>> we should use Wilcoxon test. Meaning that the test is appropriate yet requires some enhancements>> The enhancements
are mentioned in the file (Assignment #2A – Paper). Maybe it’s the bootstrap? Or adopting the Li et al. (2009) version of this test amended with Algorithm II from Simar & Zelenyuk (2006) . Please
find the answer from the article to suggest an enhancement to the test.
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Primer: Low Yields and Duration
The current low yield environment generates considerable anxiety about the bond market. Some of this reflects the reality that it is more entertaining to read articles predicting doom. However, my
suspicion is that this reflect the fact that the consol pricing formula is one of the few things people remember about fixed income pricing.
is a perpetual bond that always pays a fixed coupon, but there is no final principal payment. For a 5% consol, if you own a piece with a $100 face value, you would receive $5 every year. (In
financial mathematics, a consol is quite often defined as a security that pays $1 a year, which if you wanted a par value of $100, implies a 1% coupon.) Such bonds were created in the United Kingdom
as a way of consolidating coupon debt into a single security (hence the name). However, these instruments are currently only of academic interest, as I am unaware of any such instruments being traded
in any sizeable amounts. (I am not counting vanilla preferred shares -- which have the same cash flow structure -- but are not investable by most bond managers.)
With an infinite maturity, such instruments become quite valuable as interest rates fall. The figure above shows the price/yield relationship for a 5% consol. (The chart above shows the raw yield
convention used in my pricer, and .04 corresponds to 4%.) Such instruments exhibit a very strong zero lower bound: it would literally be impossible to buy them at 0%, as it would require an infinite
amount of money.
(As noted in my previous article, my Python fixed income pricing package is found at https://github.com/brianr747/SimplePricers. The package is under construction, but it would be straightforward to
install if you are already familiar with Python and Git. I will eventually write installation instructions. The file that generated the figures in this article is ex_20161018_duration.py, found in
the "examples" folder. There is not a whole lot in the package right now, but I will be adding to it as I develop examples for my articles and books.)
The simplified price formula for an annual consol (assuming you are on the coupon date) is straightforward:
price = (coupon rate)/(yield).
This expression obviously blows up as the yield goes to zero.
The simplicity of this expression means that such instruments work their way into economic models. Unfortunately, they reinforce the mysticism around low interest rates.
The figure above shows the price-yield relationship for a 10-year 5% coupon bond. The price of the bond only goes to $150 when the yield goes to zero -- the total cash flows over the lifetime of the
bond are $150. Although the price-yield curve is nonlinear, you need a magnifying glass to see the deviation from a straight line (at least when the the bond is trading above par.)
An alternative way of seeing this is looking at the duration of the bonds. The usual definition for
is that it is the return sensitivity of a bond with respect to interest rates.* For example, if the duration of a bond is 8, a 1 basis point rise in the bond yield will generate a capital loss of 8
basis points (loss as a percentage of market value, not versus the principal value). The duration of 10-year bond does move slightly (as shown below), but it is a flat line relative to the duration
of the consol.
Extending Maturity Still Gives a Similar Picture
(UPDATED) I was asked in the comments whether extending the maturity greatly changes the picture. As the figure above shows, even if we go out to a 100-year maturity, the duration of a coupon bond
looks fairly flat when compared to a consol.
If we just focus on the 100-year bond (figure above), we see that the duration does roughly double when it goes from 3% to 0%, but it also roughly doubled when it goes from 6% to 3%. In other words,
the shift in duration (which fixed income investors are used to managing) is not greatly disjointed when compared to the experience with higher yields.
Furthermore, we need to keep in mind that the duration of bonds outstanding is a lot closer to that of a 10-year than a 100-year (the aggregate duration depends upon the issuance patterns of the
country in question). Therefore, the duration profile that investors are dealing with is a lot closer to the chart of the 10-year than the 100 year.
Although it might seem that a 100-year bond is effectively the same thing as a consol, this is not true at ultralow interest rates. For a 5% consol, the present value of the first 100 years of
coupons is $500, while the price of the consol at a yield of 0.1% was $5000. That is, (about) 90% of the discounted value of a consol at a 0.1% yield comes from cash flows that are coming more than
100 years in the future.
Concluding Remarks
Although the duration of bond portfolios lengthen slightly in a low rate environment, the effect only matters if you are extremely concerned about micromanaging your duration risk (which most bond
managers do). However, from the perspective of macro asset allocation, the risk posed by rate changes at low interest rates are similar to the risk posed by the same changes when yields are higher.
Mortgages and Convexity Hedging
During selloffs, the U.S. bond market has historically been wracked by mortgage convexity hedging. The embedded call option in mortgages means that their duration can move much more dramatically than
option-free bonds (sometimes called linear bonds). These blowups also contribute to the mythology around rising yields.
I will not that I am out of touch with the U.S. mortgage market, and so my comments here are only a guess. That disclaimer aside, I have my doubts that this effect is going to be as important going
forward as was the case during previous bond bear markets. The collapse of consumer lending institutions after the crisis has made mortgage refinancing less automatic, and the holders of mortgage
portfolios seem to be less aggressive in their hedging strategies.
In any event, what matters for mortgage hedging is the relationship between market mortgage rates and the rates on existing mortgages, and not the absolute level of interest rates.
* An alternative duration measure is the
Macaulay Duration
. The Macaulay duration is the weighted average maturity of the cash flows of a bond (or portfolio), using the cash flow payment amounts as the weighting. (A $105 principal and final coupon payment
has a weight of 150, while a coupon payment has a weight of 5.) A consol has an infinite Macaulay duration. The Macaulay duration does not depend upon the market price of securities, and so is easier
to calculate an understand. For this reason, it sometimes shows up in annual reports of fixed income funds. For an individual bond, there is a formula linking it to the usual definition of duration.
(c) Brian Romanchuk 2016
5 comments:
1. General comment: Brian, usually you do a good job explaining things in terms understandable to lay people. But in this case I think that you will only get through to readers with a knowledge of
bonds that is at least a grade above knowing that interest rates and price move in opposite directions. If you want to talk to economists/MMT types I would suggest assuming that all they know is
that price and yield move in opposite directions. By that criteria, the above post was very hard to follow. The point you are making is an important one. It might be worth giving another shot. A
reference to Keynes' squares rule would be nice for the more advanced monetary economists in the Post-Keynesian/MMT community (this paper is relevant to the above discussion and I think Kregel
would appreciate the engagement: https://core.ac.uk/download/pdf/9314366.pdf).
Particular comment: you say that most of the fear about the bond market at ZIRP is dependent on people holding a theoretical concept (consol pricing formula) in their heads that is largely
inapplicable to the real world. Point well taken. But let's try to put some numbers on this. How far would you need to go out on the curve to get consol convexity effects? I mean, obviously none
will be as dramatic as actual perpetuals, but surely there are some fairly deep markets that do have somewhat dramatic effects. After we've answered that step the next step is: how deep are those
markets and how much damage could they do?
It also strikes me that there are some pretty crazy derivative products kicking around today. Cross currency basis swaps stand out. These seem to definitely be sensitive enough to interest rate
moves that they could potentially blow their lid at some point. I'm also not all that convinced that every bond manager knows what they are playing with when they play with these things.
1. Hello,
I will take a look at the article complexity; thanks for the feedback. Compared to what I am used to reading about bonds, that was simple... The digression on mortgage convexity hedging was
something only specialists would follow, but it comes up a lot. I should relegate it to an appendix.
I will append some charts for longer-dated bonds ("soon"). The reason I chose the 10-year as 30-year yields are further away from 0%, and it reflects bond index duration. Also, the only
people who allocate money to 30-year bonds in any size need the duration.
The issue with cross-currency swaps is more interbank credit risk. During the Financial Crisis, a lot of foreign banks were shut out of USD funding markets, and they used cross currency swaps
("basis swaps", although there are other types of basis swaps) as a USD funding vehicle. I wrote about that some time ago.
2. (My earlier comments were in a primer about covered interest parity.)
2. The way I learned it is that the duration of a security is the horizon at which the capital loss from a rate increase just cancels out the value of the higher yields. If your horizon is higher
than your portfolio's duration, you are happy when rates rise; if your horizon is shorter than the duration, you are sad. Is that right?
It would follow that the people most likely to suffer from low rates are the longest-horizons investors, presumably institutions like insurance companies or pension funds. But instead, the
loudest complaints seem to come from professional traders, who I would naively think would be shorter-horizon and therefore more sensitive to the capital gains low rates generate on their
existing portfolios. But maybe they are speaking for their principals?
Also, Keynes' square rule implicitly assumes that interest rates cannot be negative.
1. For the horizon idea, it works for small yield changes. If you have a bond with a duration of 5, and the yield rises by 5 basis points, it would take 5 years for the extra carry of 1 basis
point to cancel out the 5 basis point capital loss.
For larger yield changes, the duration shifts as yields rise (admittedly not by a lot). Also, you would get extra interest on the extra interest for larger yield shifts. For example, if
interest rates rise by 2%, you would get more than 10% more interest over 5 years.
However, the level of rates is independent of that; that relationship is about the extra carry to match the capital loss. If the interest rate is 10%, the total interest income will cover a
capital loss of 5 basis points in a few days.
The short-term traders are always the loudest. But low yields are a disaster for institutions with high duration actuarial liabilities. You would be indifferent if you matched your asset
duration to your liability duration, but very few North American pension funds did that. U.K. funds were pushed by regulation changes to match duration more, and they should be thankful.
I had not responded to the Keynes sqare rule comment, but I wanted to point out that pretty well everyone in finance or people like bank regulators need to calculate bond portfolio durations
to about four decimal places. I am not a fan of false precision, but I have been habituated to expect an interest rate risk measure that is unusually accurate. You want to have a good idea of
your daily profit and loss, and yields only move five to ten basis points per day. So, I never spent much time worrying about rules of thumb. (When I was working on relative value trades, I
probably had a good idea of the duration of all the benchmark instruments anyway, as I looked at them all day.)
Note: Posts are manually moderated, with a varying delay. Some disappear.
The comment section here is largely dead. My Substack or Twitter are better places to have a conversation.
Given that this is largely a backup way to reach me, I am going to reject posts that annoy me. Please post lengthy essays elsewhere.
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Two Period Agents with Production - Robert Winslow
Two Period Agents with Production
A firm which makes investment decisions, and a consumer with both savings and leisure.
- Modified: Jan 2nd, 2022
Consumers in a Two Period Economy with Production
\[\max_{c,c',l,l'}u(c,l)+\beta u(c',l')\] \[\begin{aligned} \text{s.t. }\ \ \ \ &c\geq0,\ \ c'\geq0,\ \ \ \ h\geq l \geq0, \ \ h\geq l'\geq0 \\ & c+\frac{c'}{1+r}\leq w(h-l) + \pi - T +\frac{w^{\
prime}(h^{\prime}-l^{\prime}) + \pi' - T'}{1+r} \end{aligned}\]
Characterizing equations
• Intertemporal Euler condition \(MRS_{cc'}=(1+r)\)
• Intratemporal Euler conditions \(MRS_{lc}=w \\ MRS_{l'c'}=w'\)
• Budget \(c+\frac{c'}{1+r}=w(h-l)-T+\frac{w^{\prime}(h^{\prime}-l^{\prime}) - T'}{1+r}\)
Quick Note about utility across time:
2 period version is
\[U(c,c',l,l')=u(c,l)+\beta u(c',l')\]
Infinite period version is typically written
Note that this is “exponential” time preferences, experimentally, it seems people have “hyperbolic” time preferences.
(You don’t need to worry about this for this class. We’re sticking to 2 time periods.)
The Two Period Firm
Refresher: Firms in the one period economy
• Firms own exogenous capital $K$ at the start of the only period.
• The firm’s profit maximization problem is:
subject to $N_{d}\geq0$
Firms in an intertemporal economy
• Firms own exogenous capital $K$ at the start of the first period.
• Second period capital is determined by $K^{\prime}=K\cdot(1-\delta)+I$, where $I$ is the firm’s investment in the first period.
• The firm also chooses the amount of labor to hire in each period $N_d, N_d’$
• The firm’s goal is to maximize present-value profits $\pi + \frac{\pi’}{1+r}$
• In the first period, profits are output minus the cost of labor and investment.
• In the second period, the firm must still hire workers, but is no need to invest because there is no third period.
• Any capital left over after period two, $(1-\delta)K’$ will be sold as units of output.
The firm’s problem is thus:
\[\max_{N_{d},N_{d}^{\prime},I,K^{\prime}}\pi+\frac{\pi^{\prime}}{1+r}\] \[\begin{aligned} \text{s.t. }\ \ \ \ &N_{d}\geq0,\ \ N_{d}^{\prime}\geq0,\ \ K^{\prime}\geq0 \\ &\pi=zF(K,N_{d})-wN_{d}-I \\
&\pi^{\prime}=z^{\prime}F(K^{\prime},N_{d}^{\prime})-w^{\prime}N_{d}^{\prime}+K^{\prime}\cdot(1-\delta) \\ &K^{\prime}=(1-\delta)K+I \end{aligned}\]
Solve for $I$ and plug into profit equations:
\[I = K^{\prime}-(1-\delta)K\] \[\pi = zF(K,N_{d})-wN_{d}-K^{\prime}+(1-\delta)K\]
If we set up the firm’s problem with these substitutions:
\[\max_{N_{d},N_{d}^{\prime},I,K^{\prime}}\pi+\frac{\pi^{\prime}}{1+r}\] \[\begin{aligned} \text{s.t. }\ \ \ \ &N_{d}\geq0,\ \ N_{d}^{\prime}\geq0,\ \ K^{\prime}\geq0 \\ &\pi=zF(K,N_{d})-wN_{d}-K^{\
prime}+(1-\delta)K \\ &\pi^{\prime}=z^{\prime}F(K^{\prime},N_{d}^{\prime})-w^{\prime}N_{d}^{\prime}+K^{\prime}\cdot(1-\delta) \\ \end{aligned}\]
On even more compact on one line:
\[\max_{N_{d},N_{d}^{\prime},K^{\prime}}zF(K,N_{d})-wN_{d}-K^{\prime}+(1-\delta)K+\frac{z^{\prime}F(K^{\prime},N_{d}^{\prime})-w^{\prime}N_{d}^{\prime}+K^{\prime}\cdot(1-\delta)}{1+r}\] \[\text{s.t.
}\ \ \ \ N_{d}\geq0,\ \ N_{d}^{\prime}\geq0,\ \ K^{\prime}\geq0\]
Assuming an interior solution, then the first-order-conditions are:
\[0=\frac{\partial}{\partial N_{d}}\mathcal{L}=MP_{N}-w\\ 0=\frac{\partial}{\partial N_{d}^{\prime}}\mathcal{L}=\frac{MP_{N^{\prime}}-w^{\prime}}{1+r}\\ 0=\frac{\partial}{\partial K^{\prime}}\mathcal
Simplify and rearrange to get the characterizing equations for this firm:
• First period optimal hiring rule:
• Second period optimal hiring rule:
• Optimal Investment rule:
How does the firm respond to changes in exogenous variables?
• If $w$ increases, the firm hires a smaller amount of labor in the first period, and so output decreases as well.
• If $z$ increases, then $MP_N$ increases for any given quantity of labor. And so for any given $w$, the firm will want to hire more labor.
• If $K$ increases, then $MP_N$ increases for any given quantity of labor. And so for any given $w$, the firm will want to hire more labor. But also, the firm will want a lower amount of investment
because they need less investment to reach any target amount of $K’$.
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How do you identify all asymptotes or holes for f(x)=(2x-2)/(x^2-2x-3)? | HIX Tutor
How do you identify all asymptotes or holes for #f(x)=(2x-2)/(x^2-2x-3)#?
Answer 1
The vertical asymptotes are $x = - 1$ and $x = 3$
No slant asymptote
The horizontal asymptote is $y = 0$
No holes
Let's factorise the denominator
The domain of #f(x)# is #D_(f(x))=RR-{-1,3} #
As you cannot divide by #0#, #x!=-1# and #x!=3#
The vertical asymptotes are #x=-1# and #x=3#
The degree of the numerator #<# than the degree of the denominator, thereis no slant asymptote
We calculate the limits of #f(x)#, we take only the terms of highest degree.
The horizontal asymptote is #y=0#
graph{(2x-2)/(x^2-2x-3) [-8.89, 8.89, -4.444, 4.445]}
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Answer 2
To identify all asymptotes or holes for ( f(x) = \frac{2x - 2}{x^2 - 2x - 3} ), follow these steps:
1. Factor the denominator: ( x^2 - 2x - 3 = (x - 3)(x + 1) ).
2. Set the denominator equal to zero and solve for ( x ): ( x - 3 = 0 ) and ( x + 1 = 0 ). Therefore, ( x = 3 ) and ( x = -1 ).
3. Determine if there are any vertical asymptotes by checking for values of ( x ) that make the denominator zero, excluding any values that also make the numerator zero. Here, ( x = 3 ) and ( x = -1
) are the values that make the denominator zero.
4. To find any holes, factor the numerator and denominator and cancel any common factors. If after cancellation, there is still a zero in both the numerator and denominator for some ( x ), then
there is a hole. In this case, the numerator ( 2x - 2 ) does not have any common factors with the denominator after factoring, so there are no holes.
5. Therefore, the vertical asymptotes for ( f(x) ) are ( x = 3 ) and ( x = -1 ). There are no holes.
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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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Geometry | BJU Press
Spark Higher-Level Thinking
Challenge your high school students to think critically about mathematics while exploring two- and three-dimensional shapes. Geometry builds critical-thinking skills with a balanced approach,
including traditional geometric proofs and modern, real-world mathematics problems. Students will use mechanical and digital tools to create figures, prove relationships between figures using modern
and traditional geometry, calculate measurements, perform transformations, and work with trigonometric ratios. While working with geometric figures, students will learn the value of God’s design for
reasoning, modeling, and ethics.
How We Teach Geometry
New for the 5th edition, each lesson starts with a clearly defined question and learning targets, which together help students focus on the lesson’s major concepts. Essential questions are introduced
at the start of each chapter to guide thinking and instruction.
Concise definitions and explanations followed by step-bystep reasoning and examples teach key concepts with the goal of understanding, not rote memorization. This edition includes titled examples
with references as well as vocabulary boxes before exercise sets. The Activities answer key and teacher edition have added step-by-step solutions to assist with instruction.
Exercise sets are carefully sequenced to build on the essential question, add to mathematical understanding, and shape a biblical worldview. Lessons also use cumulative reviews to build fluency and
help with standardized testing preparation. STEM projects and interactive individual and group activities encourage students to engage with geometric concepts in real-world applications. Free
internet-based geometry software provides a technology connection for students to explore geometric figures.
In addition to tests and quizzes, this edition has new Skill Check exercises that serve as formative assessments in each lesson. Assessment materials include 4 quarter exams and the option to create
custom assessments.
Geometry Educational Materials
Student Edition
The student edition uses concise text and clear visual elements to engage learners. New content is strategically presented alongside vocabulary boxes and example problems. Skill checks provide an
opportunity for formative assessment, and student exercises are grouped based on content and difficulty level. Each lesson and chapter centers on essential questions to guide student learning and
focuses on a biblical worldview connected to geometric reasoning.
Teacher Edition
Step-by-step examples, lesson plan overviews, a list of digital and printed resources, and suggested teaching strategies help teachers teach clearly and accurately. This edition adds the teaching
cycle (engage, instruct, apply, assess) alongside teaching strategies that focus on varied instructional techniques to reach all learning styles. Visuals and figures from the teacher edition are
available in BJU Press Trove™ for display within the classroom. Suggested assignments based on student ability levels allow for differentiated instruction.
Activities and Assessments
The student activities book supplements the student edition as needed. This edition adds 2 STEM projects for group or individual assignments. Dynamic Geometry Software Investigations uses an
internetbased program to explore key concepts digitally. New to this edition, step-by-step solutions now appear in the Activities answer key.
Each of the 12 chapters has a corresponding chapter test. In addition, 4 quarterly exams and 3 to 4 quizzes for each chapter provide multiple opportunities to assess student learning.
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Ordinal Numbers Worksheets Grade 1 - OrdinalNumbers.com
Ordinal Numbers Worksheets Grade 1
Ordinal Numbers Worksheets Grade 1 – You can count unlimited sets with ordinal numbers. You can also use them to generalize ordinal number.
The foundational idea of mathematics is the ordinal. It is a numerical value that indicates where an object is within a list of objects. The ordinal number is identified by a number that is between
zero and 20. While ordinal numbers serve a variety of uses, they are often used to represent the order of items in an orderly list.
Ordinal numbers can be represented using charts or words, numbers, and various other techniques. These are also useful for indicating how a set of pieces is arranged.
Most ordinal number fall under one of two categories. Transfinite ordinals are represented by lowercase Greek letters, while finite ordinals can be represented using Arabic numbers.
Based on the axioms that govern selection, every set should contain at least one or more ordinals. The first person in an class, for instance is the one who receives the highest score. The winner of
the contest was the student who had the highest grade.
Combinational ordinal figures
Compounded ordinal numbers are numbers with multiple digits. They are formed by the process of having an ordinal number multiplied by the number of its last digit. These numbers are used mostly for
purposes of dating and to rank. They do not employ a unique ending for the last number, as cardinal numbers do.
Ordinal numbers are used to denote the sequence of elements that comprise collections. These numbers also serve to indicate the names of the objects in the collection. There are two kinds of ordinal
numbers: regular and suppletive.
Prefixing a number with the suffix “-u” makes regular ordinals. The number is then typed as a word, and then an hyphen is added after it. There are also additional suffixes. The suffix “nd” can be
used to signify numbers that have a 2 and “th” could refer to numbers that end in between 4 and 9.
Suppletive ordinals result from prefixing words with -u. This suffix can be employed to count. It is also wider than the normal one.
Limit of magnitude ordinal
Limits for ordinal numbers are ordinal numbers that aren’t zero. Limit ordinal numbers have one disadvantage: there isn’t a maximum element. They can be made by joining non-empty sets with no any
maximum elements.
Infinite transfinite-recursion definitions use limited ordinal number. Based on the von Neumann model, every infinite cardinal number is also an ordinal limit.
A limit-ordered ordinal equals the sum of all other ordinals beneath. Limit ordinal numbers can be calculated using arithmetic. However, they also can be represented as a series of natural numbers.
The data are presented in order using ordinal numbers. They provide an explanation for the numerical location of an object. They are often utilized in set theory and math. Despite having the same
structure as natural numbers, they’re not included within the same class.
The von Neumann method uses a well-ordered list. Let’s say that fy subfunctions an equation known as g’. It is described as a single function. If fy is only one subfunction (ii) then g’ must meet the
The Church-Kleene ordinal is an limit ordinal in a similar way. Limit ordinals are properly-ordered collection of smaller or less ordinals. It has an ordinal that is nonzero.
Stories that include examples of ordinal numbers
Ordinal numbers are typically used to indicate the hierarchy between objects and entities. They are vital in organising, counting and ranking motives. They are also useful for indicating the order of
things and to indicate the position of objects.
Ordinal numbers are usually identified by the letter “th”. On occasion, though the letter “nd” is substituted. There are a lot of ordinal numbers on the title of books.
Even though ordinal number are commonly used in list format, they can also be written in words. They may also be expressed as numbers and acronyms. Comparatively speaking, these numbers are easier to
understand than the cardinal numbers.
There are three types of ordinal numbers. Through practice and games you will be able to discover more about the numbers. A key component to improving your ability to arithmetic is to learn about
them. Coloring exercises are a fun simple and relaxing way to develop. A handy marking sheet can be used to track your progress.
Gallery of Ordinal Numbers Worksheets Grade 1
Ordinal numbers worksheet for grade 1 3 Your Home Teacher
Ordinal Numbers Online Exercise For Grade 1
Ordinal Numbers Online Exercise For 1st Grade
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Explore projects · GitLab
• BSD 3-Clause "New" or "Revised" License
• Image-based pipeline for solving inhomogeneous reaction-diffusion PDEs in real-world porous geometries. Level-set method (multi-CPU) for reconstruction. Sparse grids (multi-GPU) for diffusion
simulation. Parallel data structures from OpenFPM.
• Library for producing and processing on the Adaptive Particle Representation (APR)
• Geometric computation benchmarks with classic numerical schemes for comparison with Global Polynomial Level Set method.
• GNU General Public License v3.0 or later
• Implementation of Gaussian Next Subvolume Method (GNSM), which generates trajectories for Gaussian Reaction Diffusion Master Equation (GRDME)
• Library implementing TIGHT-CUT heuristic for solving FASP (Feedback Arc Set Problem)
• Region Competition algorithm implemented as an ITK filter (C++).
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LPV Approximation of Boost Converter Model
This example shows how to obtain a linear parameter varying (LPV) approximation of a Simscape™ Electrical™ model of a boost converter using the lpvss object. This example uses the model from the LPV
Approximation of Boost Converter Model (Simulink Control Design) example to construct an LPV approximation at the command line. The LPV approximation allows quick analysis of average behavior at
various operating conditions.
Boost Converter Model
A Boost Converter circuit converts a DC voltage to another DC voltage by controlled chopping or switching of the source voltage. The request for a certain load voltage is translated into a
corresponding requirement for the transistor duty cycle. The duty cycle modulation is typically several orders of magnitude slower than the switching frequency, which produces an average voltage with
relatively small ripples.
In practice, disturbances in the source voltage and the resistive load also affect the actual load voltage.
Open the Simulink® model.
mdl = 'BoostConverterExampleModel';
The circuit in the model is characterized by high-frequency switching. The model uses a sample time of 25 ns. The Boost Converter block used in the model is a variant subsystem that implements two
different versions of the converter dynamics. The model takes the duty cycle value as its only input and produces three outputs: inductor current, load current, and load voltage.
Due to the high-frequency switching and short sample time, the model simulates slowly.
Batch Trimming and Linearization
In many applications, the average voltage delivered in response to a certain duty cycle profile is of interest. Such behavior is studied at time scales several decades larger than the fundamental
sample time of the circuit. These average models for the circuit are derived by analytical considerations based on averaging of power dynamics over certain time periods. The
BoostConverterExampleModel model implements such an average model of the circuit as its first variant, the AVG Voltage Model variant. This variant typically executes faster than the Low Level Model
The average model is not a linear system. It shows nonlinear dependence on the duty cycle and the load variations. To produce a faster simulation and to help with voltage stabilizing controller
design, you can linearize the model at various duty cycle and load values. For this example, use the snapshot-based trimming and linearization. The scheduling parameters are the duty cycle d and
resistive load R. You trim and linearize the model for several values of the scheduling parameters.
Select a span of 10%–60% for the duty cycle variation and a span of 4–15 ohms for the load variation. Select five values in these ranges for each scheduling variable and make a grid of all possible
combinations of their values.
nD = 5;
nR = 5;
dspace = linspace(0.1,0.6,nD); % Values of d in 10%-60% range
Rspace = linspace(4,15,nR); % Values of Rin 4-15 Ohms range
[dgrid,Rgrid] = ndgrid(dspace,Rspace); % All combinations of d and R values
Create a parameter structure array for the scheduling parameters.
params(1).Name = 'd';
params(1).Value = dgrid;
params(2).Name = 'R';
params(2).Value = Rgrid;
A simulation of the model under various conditions shows that the model outputs settle down to their steady-state values before 0.01 s. Therefore, use t = 0.01 s as the snapshot time. Compute
equilibrium operating points at the snapshot time using the findop function. This operation takes several minutes to finish.
op = findop(mdl,0.01,params);
To linearize the model, first obtain the linearization input and output points from the model.
Configure the linearization options to store linearization offsets.
opt = linearizeOptions('StoreOffsets', true);
Linearize the model at the operating points in array op.
[linsys,~,info] = linearize(mdl,op,io,params,opt);
Plot the linear system array.
LPV Model
Use ssInterpolant to create an LPV model that interpolates the linearized models and offsets over the ($\mathit{d}$,$\mathit{R}$) operating range.
lpvsys = ssInterpolant(linsys,info.Offsets);
LPV Simulation
To simulate the model, use an input profile for the duty cycle that roughly covers its scheduling range. Also, vary the resistive load to simulate load disturbances. Generate the duty cycle profile
t = linspace(0,.05,1e3)';
din = 0.25*sin(2*pi*t*100)+0.25;
din(500:end) = din(500:end)+.1;
Generate the resistive load profile rin.
rin = linspace(4,12,length(t))';
rin(500:end) = rin(500:end)+3;
rin(100:200) = 6.6;
Plot the scheduling parameter profiles.
yyaxis left
xlabel('Time (s)')
ylabel('Duty Cycle')
yyaxis right
ylabel('Resistive Load (Ohm)')
title('Scheduling Parameter Profiles for Simulation')
Use lsim to simulate the response of the LPV approximation to this stimulus.
p = [din rin]; % scheduling variables
xinit = info.Offsets(1).x;
y = lsim(lpvsys,din,t,xinit,p);
Simulate the boost converter LPV model implemented using the LPV System block and plot the results.
% Offset data for LPV block
offsets = getOffsetsForLPV(info);
yoff = offsets.y;
xoff = offsets.x;
uoff = offsets.u;
simOut = sim('BoostConverterLPVModel','StopTime','0.05');
lpvBlockSim = simOut.logsout.getElement('ysim');
tsim = lpvBlockSim.Values.Time;
ysim = lpvBlockSim.Values.Data(:,:);
legend('LPV object simulation','LPV system block simulation','location','best')
These results match the simulation results obtained with the LPV System block.
Discretization of LPV Model
Compute an equivalent discretized LPV model of the boost converter. Discretization facilitates fixed-step simulation and code generation for this model. Compare the results with those of the
continuous model.
dlpvsys = c2d(lpvsys,t(2)-t(1),'tustin');
yd = lsim(dlpvsys,din,t,xinit,p);
Close the model.
See Also
ssInterpolant | sample | linearize (Simulink Control Design) | ndgrid
Related Topics
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A polynomial quantum algorithm for approximating the Jones polynomial
The Jones polynomial, discovered in 1984 [18], is an important knot invariant in topology. Among its many connections to various mathematical and physical areas, it is known (due to Witten [32]) to
be intimately connected to Topological Quantum Field Theory (TQFT). The works of Freedman, Kitaev, Larsen and Wang [13, 14] provide an efficient simulation of TQFT by a quantum computer, and vice
versa. These results implicitly imply the existence of an efficient quantum algorithm that provides a certain additive approximation of the Jones polynomial at the fifth root of unity, e^2πi/5, and
moreover, that this problem is BQP-complete. Unfortunately, this important algorithm was never explicitly formulated. Moreover, the results in [13, 14] are heavily based on TQFT, which makes the
algorithm essentially inaccessible to computer scientists. We provide an explicit and simple polynomial quantum algorithm to approximate the Jones polynomial of an n strands braid with m crossings at
any primitive root of unity e^2πi/k, where the running time of the algorithm is polynomial in m, n and k. Our algorithm is based, rather than on TQFT, on well known mathematical results
(specifically, the path model representation of the braid group and the uniqueness of the Markov trace for the Temperly Lieb algebra). By the results of [14], our algorithm solves a BQP complete
problem. The algorithm we provide exhibits a structure which we hope is generalizable to other quantum algorithmic problems. Candidates of particular interest are the approximations of other
downwards self-reducible #P-hard problems, most notably, the Potts model.
Original language English
Title of host publication STOC'06
Subtitle of host publication Proceedings of the 38th Annual ACM Symposium on Theory of Computing
Publisher Association for Computing Machinery
Pages 427-436
Number of pages 10
ISBN (Print) 1595931341, 9781595931344
State Published - 2006
Event 38th Annual ACM Symposium on Theory of Computing, STOC'06 - Seattle, WA, United States
Duration: 21 May 2006 → 23 May 2006
Publication series
Name Proceedings of the Annual ACM Symposium on Theory of Computing
Volume 2006
ISSN (Print) 0737-8017
Conference 38th Annual ACM Symposium on Theory of Computing, STOC'06
Country/Territory United States
City Seattle, WA
Period 21/05/06 → 23/05/06
• Algorithm
• Approximation
• Braids
• Jones Polynomial
• Knots
• Quantum
• Temperley-Lieb
• Unitary Representation
Dive into the research topics of 'A polynomial quantum algorithm for approximating the Jones polynomial'. Together they form a unique fingerprint.
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Application Of Grey Relational Analysis For The Optimisation Of The Shot Peening Parameters With Multiple Performance Characteristics
Science Update
in Vol. 20 - May Issue - Year 2019
Application Of Grey Relational Analysis For The Optimisation Of The Shot Peening Parameters With Multiple Performance Characteristics
Table 1. Shot peen factors and their levels
Table 2. Fractional factorial array L16(45) with inputs and outputs
Table 4. Response table for grey relational grade
Table 3. Grey relational coefficients and grey relational grade for three different performance characteristics
Fig. 1. Response graph for the GRG
Table 5. Estimated optimum peening conditions
Table 6. Experimental peened effects determined from the estimated peening conditions of table 6
The shot peening process causes the surface of metallic components to be strengthened through bombardment with a torrent of minute balls named shots, thus inducing significative responses such as
compressive residual stresses, surface hardening and surface roughness. In the aerospace, automotive, shipping and manufacturing industries, proper selection and control of peening factors is highly
required to ensure a real benefit mainly in terms of fatigue resistance. In present research, grey-fuzzy methodology is used for determining the best selection of the control factors that directly
influence the multi-objective response properties of a peened 2024-T351 aluminium alloy (AA). The input parameters taken into consideration are shot, coverage and incidence angle. For statistical
purposes, the experimental parameters were put in place with a L16 orthogonal fractional array. The three properties extracted experimentally are optimised. The three multi-objectives are converted
into a single object with the help of grey relational analysis (GRA). The confirmation test carried out to validate the results established that the approach used in this work can effectively provide
the optimum parameters by virtue of obtaining an improvement in fatigue resistance.
Design Of Experiments Through Grey-relational Method
The basic steps involved in any design of experimental analysis consist of assuming a set of variables controlling the output function, selecting the number of levels of each variable to test, and
picking the proper factorial array. Even though the choice of a particular approach will depend on the number of factors and levels, the use of the fractional factorial method (also known as the
orthogonal array) represents itself as the most reasonable choice in terms of reducing the number of experiments without loss of quality in the obtained results [1]. A total of three factors were
evaluated at four levels without interaction, each as shown in Table 1. These chosen factors are of fundamental importance to the aircraft industry [2].
A fractional factorial L16(45) array was the choice to enable the optimisation of five parameters with each of these set at a maximum of four levels. This approach permitted the estimation of all
main and first-order factor interactions. The full fractional factorial array along with the selected factors and levels and the experimental values is depicted in Table 2.
As can be seen, it is difficult to make comparisons between the different kinds of factors because they exert a different influence. To redress the situation, there is a need of a standardized
transformation of such factors. When dealing with information that is either incomplete or undetermined, the so-called grey relational analysis (GRA) becomes quite useful. This is a method by which
the relational degree of every factor in the system can be analysed. GRA uses information from the grey system to dynamically compare each influencinge factor quantitatively. This approach is based
on the level of similarity and variability among all factors to establish their relationships [3]. In this method, a grey relational grade is obtained by analysing the relational degree of the
multiple responses. The grey relational coefficient of each output response of residual stress, roughness and work hardening is calculated. Grey relational analysis is based on minimization of the
maximum distance between an objective sequence (a collection of measurements) and a reference sequence (target value) in the grey system. The relationship between these two sequences can be called
the grey relational coefficient. For normalization of the raw sequence between 0 and 1 (since different sequences have different measurement units and scales), The type of normalization depends upon
the characteristic that is desired out of the raw sequence:
For the larger-the-better characteristic: (1)
For the smaller-the-better characteristic: (2)
Where xi(k) is the value after grey relational generation while min xi(k) and max xi(k) are the smallest and largest values of xi(k) for the kth response.
For compressive residual stresses (RS) and work hardening (WH) larger-the-better and surface roughness (KT) the smaller-the-better characteristics were selected (Table 3).
The absolute difference of the compared series and the referential series should be obtained by using: (3)
In the GRA, the grey relational coefficient can be expressed as follows: (4)
In equation (4), the term p is the distinguishing coefficient which is used to adjust the difference of the relational coefficient, usually p [3]. The lower the value of p, the higher is the
distinguishing ability. If p is taken to be 0.5, the outcomes will hold moderate distinguishing effects and good stability.
After the grey relational coefficient is derived, the average value of the gray relational coefficient is taken as the grey relational grade (GRG), see Table 3. Mathematically speaking: (5)
Table 3 shows the grey relational coefficients for all the three peening properties and the grey relational grade along with its order. The highest value of grey relational grade 0.723 indicates that
experiment number 11 is the optimum combination of peening parameters in order to produce the maximum residual stresses and work-hardening with the minimum of stress concentrations. The grey
relational coefficient of the first property is unity, indicating an exact reference value. However, the same reference value was not determined for the other properties for the experiment number 11.
Therefore, a grey relational grade is necessary to obtain the optimum parameters for generating stress concentration with minimum damage and work-hardening with maximum performance (as shown in the
last column of Table 3). Table 4 shows the response table for grey relational grade. This is obtained by calculating the average value of each input peening parameter at its corresponding level. The
max-min column indicates that shot is the most significant factor among the three input variables. In order to produce the best output, the optimal combination of the parameters as calculated from
the response table depicts that shot must be maintained at level 4, coverage at level 3 and incidence angle at level 4.
Fig. 1 depicts the response graph plotted for the calculated GRG. The slope of the lines in the plot is found to be more for the shot, indicated as the most influential parameter. In this plot, S1,
S2, S3, S4 in the x-axis corresponds to that of the four levels of the shot. Similarly, the letters C1..C4 and A1..A4 correspond to the four levels of coverage and incidence angle, respectively.
The resultant conditions of experiment 11 from the designed experiment are shown in Table 5.
Experimentally determined peening effects using estimated optimal combinations are shown in Table 6. The experimental data were found to be slightly different from those of Table 5. Such
discrepancies can be attributed to several process factors such as: (i) variation in the pressure, (ii) wear (of the media, nozzle, and hose), (iii) highly scattered shot size, (iv) variation in the
flow rate due to interferences caused by dust, debris or other objects, among others.
Final Remarks
Based upon the experimental and numerical results, the following conclusions can be drawn:
The plethora of variables capable of altering the peening process, and hence fatigue life, is the reason that strict process control of all variables is necessary. The aluminium alloy 2024-T351
peened to calculated optimum conditions (S110, 200%, 90º) exhibited superior performance than when peened to initial conditions. Surface integrity degradation precedes fatigue damage. The geometry of
the indentation is a prime damage feature. The advantages of the grey relational analysis approach compared to classical methods lay on speed, simplicity, and low cost; therefore, this approach can
be considered for applications in engineering situations, as shot peening, laser peening, burnishing and water peening, etc. However, given the degree of uncertainty observed in the response table,
integrated grey relational analysis with adaptive neural fuzzy integrated system has to be used to optimise the shot peening process parameters.
1. Van Nostrand, Richard Craig. (2002). Design of Experiments Using the Taguchi Approach: 16 Steps to Product and Process Improvement. Technometrics, 44, 289-289.
2. Nam, Yong-Seog, Jeon, Ung, Yoon, Hee-Kweon, Shin, Byung-Cheol, & Byun, Jai-Hyun. (2016). Use of response surface methodology for shot peening process optimization of an aircraft structural part.
The International Journal of Advanced Manufacturing Technology, 87, 2967-2981.
3. L. Deng, "The introduction of grey system," The Journal of Grey System, vol. 1, no. 1, pp. 1-24, 1982.
4. Chandrasekaran, M., Muralidhar, M., Krishna, C. Murali, & Dixit, U. S. (2010). Application of soft computing techniques in machining performance prediction and optimization: a literature review.
The International Journal of Advanced Manufacturing Technology, 46, 445-464.
The Author:
PhD José Solis Romero
E-mail: jsolis@ittla.edu.mx
MSc. Sandra S. Roblero Aguilar
Dr. Víctor A. Castellanos Escamilla
Dr. Oscar A. Gómez Vargas
MSc. Miguel A. Paredes Rueda
Public Education Secretary of Mexico
Instituto Tecnológico de Tlalnepantla
del TecNM
Postgraduate Office/Department of
Mechanical Engineering
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The growth of Weierstrass canonical products of genus zero with random zeros
The growth of Weierstrass canonical products of genus zero with random zeros
entire function, Weierstrass products, maximum modulus, order, genus, exponent of convergence, integrated counting function
Published online: 2013-06-20
Let $\zeta=(\zeta_n)$ be a complex sequence of genus zero, $\tau$ be its exponent of convergence, $N(r)$ be its integrated counting function, $\pi(z)=\prod\bigl(1-\frac{z}{\zeta_n}\bigr)$ be the
Weierstrass canonical product, and $M(r)$ be the maximum modulus of this product. Then, as is known, the Wahlund-Valiron inequality
\limsup_{r\to+\infty}\frac{N(r)}{\ln M(r)}\ge w(\tau),\qquad w(\tau):=\frac{\sin\pi\tau}{\pi\tau},
holds, and this inequality is sharp. It is proved that for the majority (in the probability sense) of sequences $\zeta$ the constant $w(\tau)$ can be replaced by the constant $w\left(\frac{\tau}2\
right)$ in the Wahlund-Valiron inequality.
How to Cite
Zakharko, Y.; Filevych, P. The Growth of Weierstrass Canonical Products of Genus Zero With Random Zeros. Carpathian Math. Publ. 2013, 5, 50-58.
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ELETTRONICA M - Z
ING-INF/01 - 9 CFU - 1° Semester
Teaching Staff
Learning Objectives
The course aims at providing basic knowledge about the modeling of electronic devices, about the operation of analog and digital circuits in CMOS technology and about the most common circuit
configurations that make use of operational amplifiers. The course also provides knowledge of CAD software (for example LTSPICE) for circuit simulation.
At the end of the course the student will have an overview of the electronic devices and applications in which they are used and will be able to analyze and design simple analog and digital circuits,
also through the use of CAD tools.
Course Structure
The course includes lectures and both numerical and simulation exercises (CAD). The latter are aimed at putting into practice and consolidating the theoretical contents as well as the analysis and
the design techniques developed. Seminars will be organized by researchers and designers from companies operating in the microelectronics sector.
Detailed Course Content
1. Introduction to Electronics: A brief history of electronics. Classification of Electronic Signals. A/D and D/A Converters. Notational Conventions. Dependent sourced. Important Concepts from
Circuit Theory (Kirchhoff’s lows, dividers, Thevenin and Norton Equivalents). Frequency Spectrum of Electronic Signals. Amplifiers. Example: FM receiver
2. Solid-State Electronics. Solid-State Electronic Materials. Covalent Bond Model. Intrinsic carrier concentration. Mass action. Drift Currents and Mobility in Semiconductors. Velocity Saturation.
Resistivity of Intrinsic Silicon. Impurities in Semiconductors. Electron and Hole Concentrations in Doped Semiconductors. Diffusion Currents. Total Current. Energy Band Model.
3. Solid-state Diodes and Diode circuits: Junction diode. The I/V Characteristics of the Diode. Diode Characteristics Under Reverse, Zero, and Forward Bias. Diode Temperature Coefficient. Reverse
Breakdown and Zener Diode. pn Junction Capacitance in Reverse Bias and Forward Bias. Dynamic Switching Behavior of the Diode. Large signal Model. Diode SPICE Model. Diode Circuit Analysis.
Load-Line Analysis. Analysis Using the Mathematical Model for the Diode (small signal resistance). Constant Voltage Drop Model. Multiple-Diode Circuits. Half-Wave Rectifier Circuits with R, C and
RC load. Full-Wave Rectifier and Bridge Circuits. Voltage regulator with Zener diode. Photo Diodes and Photodetectors. Schottky Barrier Diodes. Solar Cells. Light-Emitting Diodes
4. MOS Transistors: Characteristics of the MOS Capacitor. Accumulation Region. Depletion Region. Inversion Region. The NMOS Transistor. Qualitative I/V Behavior of the NMOS Transistor. Triode Region
Characteristics of the NMOS Transistor. On Resistance. Saturation of the I/V Characteristics. Mathematical Model in the Saturation (Pinch-Off) Region Transconductance. Channel-Length Modulation.
Body Effect. PMOS Transistors. MOSFET Circuit Symbols. NMOS Transistor Capacitances in the Triode Region. Capacitances in the Saturation Region. Capacitances in Cutoff. MOSFET biasing (4
resistors network) and analysis. Modeling in SPICE.
5. Digital circuits: Ideal Logic Gates. *Logic Level Definitions and Noise Margins. Logic Gate Design Goals. Dynamic Response of Logic Gates. Rise Time and Fall Time. Propagation Delay. Power-Delay
Product. Review of Boolean Algebra. CMOS logic circuits. Static characteristics of the CMOS Inverter. CMOS Voltage Transfer Characteristics. CMOS NOR and NAND Gates. Design of Complex Gates in
CMOS. Cascade Buffers and Delay Model. Optimum Number of Stages. Bistable latch. SR Flip-Flop. JK Flip flop. Flip-Flop race condition. The D-Latch Using Transmission Gates. Master-Slave
Flip-Flop. Edge triggered Flip flop. Counters and registers. Random Access Memories (RAMs). 6-T cell. Dynamic RAMs. 1-T cell.
6. Operational Amplifiers: An Example of an Analog Electronic System. Amplification. Voltage Gain, Current Gain and Power Gain. The Decibel Scale. The Differential Amplifier. Differential Amplifier
Voltage Transfer Characteristic. Differential Voltage Gain. Differential Amplifier Model. Ideal Operational Amplifier. Assumptions for Ideal Operational Amplifier. The Inverting Amplifier. The
Transresistance Amplifier. The Noninverting Amplifier. The Unity-Gain Buffer, or Voltage Follower. The Summing Amplifier. The Difference Amplifier. The Integrator. The Differentiator.
Nonidealities: Common mode gain. CMRR. I/O resistances. Offset. Slew rate.
7. Small-signal Modeling and linear amplification: The Transistor as an Amplifier. Coupling and Bypass Capacitors. Circuit Analysis Using dc and ac Equivalent Circuits. Small-Signal Modeling of the
Diode. Small-Signal Models for Field-Effect Transistors. Intrinsic Voltage Gain of the MOSFET. The Common-Source Amplifier (Voltage Gain. I/O resistances). Power dissipation and signal swing.
Amplifiers classification. CS, CD, CG configurations. CS with resistive degeneration. AC-coupled multi stage amplifiers.
8. Current Mirrors: DC analysis of MOS current mirrors. Changing the MOS Mirror Ratio. Cascode current mirror.
9. Frequency response: Frequency response of Amplifiers, Midband gain, Low and high cutoff frequencies (fL and fH). Estimation of f[L] through the short-circuit time constant method for CS, CG, CD
amplifier. High-frequency MOSFET model. Transition frequency, f[T]. Channel Length Dependence of f[T]. Analisi ad alta frequenza dell’amplificatore source comune. L’effetto Miller. High-Frequency
C-S Amplifier Analysis. The Miller Effect. Common-Emitter and Common-Source Amplifier High-Frequency Response. Estimation of f[H] through the open-circuit time constant method for CS.
10. Computer simulations of electronic circuits: LTSPICE.
Textbook Information
1. Jaeger-Blalock, Microelettronica Ed. Mc-Graw-Hill V Edizione.
2. Sedra-Smith, Circuiti per la Microelettronica, Edises.
Open in PDF format Versione in italiano
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On Big O - benjamintoll.com
asymptote, noun: a straight line associated with a curve such that as a point moves along an infinite branch of the curve the distance from the point to the line approaches zero and the slope of the
curve at the point approaches the slope of the line
The following are random notes and sentences to jigger loose long-forgotten information and memories to bring forth into short-term memory.
Big O notation, or asymptotic notation, measures the runtime efficiency of a given algorithm. This is a big topic, and this is going to be a short post meant to be a kind of cheatsheet.
• O Big O, the upper bound
• Ω Big Omega, the lower bound
• Θ Big Theta, the tight bound
Since I’m not an academic, I’ll be speaking to just Big O in this post, which is used in interviews and informal settings to mean both big O and big Omega.
1. Time Complexity
2. Space Complexity
3. Drop the Constants
4. Drop the Non-Dominant Terms
5. (Don’t Forget About) Amortized Time
6. Log N Runtimes
7. Recursive Runtimes
Note the best/worst/expected case runtimes are not the same as big O/Theta/Omega.
How does the algorithm scale as the size of the input grows? How well does the algorithm perform over time?
What is the complexity analysis? What is the performance of an algorithm as its input approaches an upper limit?
We ignore values that don’t change the overall shape of the curve.
O(N) isn’t always better than O(N^2)!
Rate of Increase (examples of runtimes from best performance to worst):
1. O(1) - Fixed, the amount of work is not dependent upon the size of the input (don’t confuse fixed with fast). Minimum value of min heap, maximum value of a max heap.
2. O(log N) - As the input grows, the algorithm’s cost doesn’t grow at the same rate. Will divide a larger problem into smaller chunks by halves. For example, finding a word in dictionary or a phone
number in the phone book (if the latter still exists), binary search.
3. O(N) - Scales linearly with the size of the input. Often represented as a single loop over a data collection. Reading a book, iterating through a list.
4. O(N log N) - Log linear. Sorting a deck of playing cards, merge sort.
5. O(N^2) - Exhibits quadratic growth relative to the input size and can usually be identified by nested loops over the same data collection. Bubble sort.
6. O(2^N) - Exponential. Recursive Fibonacci function with no caching.
7. O(N!) - Factorial. Generating all permutations of a list.
8. O(N^N)
Note that these are only a few of the many, many runtimes (infinite?). They are the most commonly seen and examples of which abound. There are countless runtimes in-between these runtimes.
1. O(NM) - Can be identified by nested loops over two distinct data collections, i.e., two inputs. Difficult to determine the cost as the inputs increase without knowing the domain space.
log[2]N = k -> 2^k = N
log[2]16 = 4 -> 2^4 = 16
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Mastering Free Cash Flow to Net Income Ratio | SimFin Glossary
Do you have difficulty understanding a financial term? Check unfamiliar terms by first letter in our glossary below.
Free Cash Flow to Net Income
Unleash the Power of Ratio Analysis: Understanding FCF to Net Income Ratio
Compact Explanation
Free Cash Flow to Net Income is a financial ratio comparing free cash flow to net income.
Free Cash Flow (FCF) to Net Income ratio is a key financial measure that provides invaluable insights into a company's financial health and operational efficiency.
The Free Cash Flow to Net Income ratio is a profitability ratio that measures the amount of free cash flow generated for each dollar of net income. It is calculated by dividing the Free Cash Flow
(FCF) by Net Income. A higher ratio can indicate that a company is effectively turning its profit into cash, a lower ratio might suggest the opposite.
Context and Use
This ratio is most commonly used by investors and analysts when performing a comprehensive financial analysis of a company. The ratio can provide valuable insights about how efficiently a company is
translating its net income into free cash flow, which can be used for reinvestment, paying dividends, or reducing debt.
Detailed Explanation
Free Cash Flow (FCF) represents the cash a company generates after accounting for cash outflows to support operations and maintain its capital assets. Unlike earnings or net income, free cash flow is
a measure of profitability that excludes the non-cash expenses of the income statement and includes spending on equipment and assets.
Net Income is the amount of total revenue that exceeds all expenses, taxes, and costs incurred by a business during a specific period.
When you divide Free Cash Flow by Net Income, you get the Free Cash Flow to Net Income Ratio. This ratio essentially measures the proportion of net income that is available as free cash flow.
Example Calculation
For instance, let's say a company has a Free Cash Flow of $500,000 and a Net Income of $750,000.
The FCF to Net Income ratio would be calculated as:
FCF to Net Income Ratio = FCF / Net Income = $500,000 / $750,000 = 0.67
A ratio of 0.67 suggests that for every dollar of net income, the company generates $0.67 in free cash flow.
Related Terms
1. Cash Flow Statement
2. Operating Cash Flow (OCF)
3. Capital Expenditure (CapEx)
Frequently Asked Questions (FAQ)
1. What does a high FCF to Net Income ratio mean? A high FCF to Net Income ratio generally means that a company is efficiently turning its net income into free cash flow.
2. Is it better to have a high or low FCF to Net Income ratio? Generally, a higher FCF to Net Income ratio is preferable as it suggests that a company is generating a significant amount of free cash
flow relative to its net income.
3. Can the FCF to Net Income ratio be negative? Yes, if a company has a negative free cash flow but positive net income, the ratio will be negative.
4. What are the limitations of the FCF to Net Income ratio? One limitation of the FCF to Net Income ratio is that it may not be comparable across different industries. Certain sectors may typically
have higher capital expenditures and therefore lower free cash flow. Additionally, it does not account for differences in growth rates among companies.
5. How can a company improve its FCF to Net Income ratio? A company can improve its FCF to Net Income ratio by increasing its free cash flow or reducing its net income. This could be achieved by
increasing sales revenue, reducing operating costs, or managing capital expenditures more efficiently.
6. Is it possible for the FCF to Net Income ratio to be greater than 1? Yes, it is possible if a company's free cash flow is greater than its net income. This could occur if the company has minimal
capital expenditures or if non-cash expenses (like depreciation) are a significant portion of net income.
Key Takeaways
Understanding the Free Cash Flow to Net Income ratio is crucial for investors and financial analysts as it provides insights into a company's ability to translate net income into free cash flow,
which is essential for business growth and expansion.
Investors looking for a measure of a company's financial health and operational efficiency should consider using the Free Cash Flow to Net Income ratio as part of their toolkit. This ratio can offer
crucial insights into the ability of a company to convert net income into free cash flow, indicating the firm's liquidity and financial flexibility.
The information provided on this page is for educational purposes only and is not to be construed as investment advice. We strongly recommend that you seek advice from a professional investment
advisor before making any investment decisions. All the examples provided are hypothetical and for illustrative purposes. The actual numbers can vary greatly based on the specific circumstances of a
business. SimFin is not responsible for any investment decision made based on the information provided on this page.
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If the probability of an event A is \(P(A) = \frac{1}{3}\) and the probability of its complement \(\bar{A}\) is \(P(\bar{A}) = \frac{2}{3}\), what is the probability of either A or \(\bar{A}\)
A \(\frac{1}{2}\)
B \(\frac{1}{3}\)
C \(\frac{2}{3}\)
D 1
Answer: Option D
**Editorial:** The probability of the sample space is 1. Since A and \(\bar{A}\) together cover the entire sample space, their union is the entire sample space, and the probability of either
occurring is 1.
**Editorial:** The probability of the sample space is 1. Since A and \(\bar{A}\) together cover the entire sample space, their union is the entire sample space, and the probability of either
occurring is 1.
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What are Basis Points (BPS)? - Definition | Meaning | Example
Definition: Basis points (BPS) are the smallest measurement of fixed income securities and interest rate quotes and are used to measure changes and differentials in interest rates and margins.
What Does Basis Points Mean?
What is the definition of basis points? Financial analysts and investors use bps when they want to express minor percentage changes. For instance, a difference of 0.05% in the performance of two
stocks in a portfolio can be expressed as a difference of 5 bps, instead of “zero point zero five percent.” Or, when people read in the newspaper that “the Fed cuts the interest rates by 35 bps,” it
means that the Fed cuts the interest rates by 0.35%. Therefore, 1 bps is equal to 0.01%.
Let’s look at an example.
Alex is a retail investor who follows the stock market and likes to be informed about the macroeconomic environment. He reads the newspaper daily, and he is mostly interested in the Fed news and the
moves of the Federal Reserve with respect to the interest rates.
Alex reads in the newspaper that the Fed is inclined to cut the interest rates by 25bps. As he is not sure of what “25 bps” exactly means, he makes a phone call to his best friend, Jerry, who perhaps
knows what “bps” means.
Jerry explains to Alex that 25bps are equal to 0.25% since 1 bps is equal to 0.01%. He also offers an example with the interest rate on his mortgage. The mortgage has a floating interest rate of
LIBOR +75bps. This means that the floating rate is 4.00% because the LIBOR is 3.25%. Therefore, 3.25% + 0.75% = 4.00%.
Basis points are widely used by financial analysts because they provide an accurate indication of the difference between two percentages even if this difference is minor. For instance, analysts that
follow the 20-year U.S. T-Bills daily can accurately calculate the small changes in the index movement, which, however, may have a major impact on the economy.
Summary Definition
Define Basis Points: A Basis point means a fractional measurement for financial investing returns and interest percentages.
Accounting & CPA Exam Expert
Shaun Conrad is a Certified Public Accountant and CPA exam expert with a passion for teaching. After almost a decade of experience in public accounting, he created MyAccountingCourse.com to help
people learn accounting & finance, pass the CPA exam, and start their career.
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What would be the yearly earnings for a person with $14700 in savings at an annual interest rate of 16.6% percent? - US Expert Writers
What would be the yearly earnings for a person with $14700 in savings at an annual interest rate of 16.6% percent?
What would be the yearly earnings for a person with $14700 in savings at an annual interest rate of 16.6% percent?
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 1 options:
What is the future value of $3330 8 years from now at 7 percent?
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 2 options:
What is the future value of $3425 saved each year for 14 years at 2 percent?
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 3 options:
What is the amount a person would have to deposit today (present value) at 8 percent interest rate to have $11200 saved 7 years from now.
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 4 options:
What is the amount you would have to deposit today to be able to take out $3000 a year for 8 years from an account earning 14 percent.
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 5 options:
If you desire to have $49400 for a down payment for a house in 10 years, what amount would you need to deposit today? Assume that your money will earn 5 percent.
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 6 options:
Pete Morton is planning to go to graduate school in a program of study that will take 4 years. Pete wants to have $17500available each year for various school and living expenses. If he
earns 5 percent on his money, how much must be deposit at the start of his studies to be able to withdraw $17500 a year for 4 years?
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 7 options:
Carla Lopez deposits $9900 a year into her retirement account. If these funds have an average earning of 8 percent over the 8 years until her retirement, what will be the value of her retirement
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 8 options:
If a person spends $25 a week on coffee (52 weeks in a year), what would be the future value of that amount over 9 years if the funds were deposited in an account earning 5 percent?
Use the appropriate Time Value of Money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D]
(Round your answer to the nearest whole number. Do not include the comma, period, and “$” sign in your response.)
Your Answer:
Question 9 options:
Chapter 1 LO 1.3
Question 10 options:
A financial company that advertises on television will pay you $64,000 now for annual payments of $9,100 that you are expected to receive for a legal settlement over the next 8 years. Assume you
estimate the time value of money at 11 percent.
Use the appropriate time value of money table [Exhibit 1-A, Exhibit 1-B, Exhibit 1-C, OR Exhibit 1-D].
(a) What is the present value?
(Round your answer to the nearest whole number. Omit the comma, period, and “$” sign in your response.)
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KA7OEI's blog
Several years ago I decided to build a weather satellite receiver from scratch.
It is described here - link.
I didn't really need a weather satellite receiver, and I could have easily bought a kit somewhere else - or bought a second-hand one via EvilBay, but I just wanted to go through the exercise of
throwing everything together and making it work using parts on hand - and I wanted to try out some ideas.
Locking a VCO to an audio DDS reference:
Figure 1:
The front panel of the VHF weather satellite receiver.
This receiver has been in continuous operation for several years, working
flawlessly in that time.
Click on the image for a larger version.
One of these ideas was to use a PIC to lock the VHF local oscillator. On the face of it, this isn't unique - except that the PIC was to be the sole source of the precise frequency to which the PLL
(Phase Locked Loop)
for the local oscillator: No divide-by-N chips here!
For this receiver the local oscillator operated 10.7 MHz below the receive frequency nominally at about 126 MHz. Since I was already using a 100 MHz oscillator
(a VCXO)
that I'd pulled from some scrapped commercial satellite gear, I used a simple 3-transistor mixer/amplifier circuit to convert this to about 26 MHz
(137 MHz-10.7 MHz-100 MHz)
and this allowed me to use a 74HC4040 12-stage binary ripple counter to bring a representation of the local oscillator down to the audio range - about 6.3 kHz.
As it so-happened, I'd chosen the 100 MHz oscillator on purpose - mostly because it was free, but it also provided a nice, stable 20 MHz clock for the PIC by dividing its output by 5 using a 74F191
so both the down-conversion and the PIC's clock were referenced from the same source.
The goal was to provide a
tuning step size of at least 1 kHz, and because I'd already divided the local oscillator by 4096 this meant that my audio-frequency step size was on the order of 1/4 of one Hz - but that was no
problem since I was going to use DDS techniques in the PIC.
The DDS:
(Direct Digital Synthesis - see the Wikipedia article about DDS techniques here - link)
is fairly simple in operation: Typically, one takes a register
(called an "accumulator")
and on every clock cycle you add to it a constant number
(we'll call it a "frequency word")
allowing it to "wrap around" once the accumulator's capacity is exceeded or, in other words, you do unsigned binary addition.
If you were to keep track of how often the accumulator overflows you'd notice that if you added a smaller number to it, it would overflow less often which makes sense since it would take more clock
cycles to overflow! What you might notice is that one can easily predict the rate at which it will overflow:
( (frequency word) / (maximum accumulator value) ) * clock frequency
Typically, the "maximum accumulator value" is the maximum number
(plus one)
that can be represented by the number of bits used by the accumulator (e.g. 8 bits = 256, 16 bits = 65536, 32 bits = 4294967296).
In my case, it was easy to make the PIC do 32 bit unsigned addition.
The last step is to take the upper-most bits of the accumulator and apply them to a D/A converter via a sine-wave lookup table. To take a table that is "8 bits" in size (256 entries) one would take
the top byte of our exemplar 32 bit accumulator, use those bits to point to a sine wave and then send the output of that sine wave lookup to the D/A converter. Clearly, the more bits of lookup (e.g.
the larger the sine wave table) and the more resolution that one has available for the D/A converter, the better!
Figure 2:
The PIC controller board that generates the precise audio frequency
based on a PIC16F88 and driven from a 20 MHz clock source. This PIC
also drives the LCD and does the serial data communications,
receiving frequency tuning commands from the host computer.
Click on the image for a larger version.
The PIC that I used
(a PIC16F88)
can be clocked to 20 MHz and among other things it contains a PWM generator that can operate as a simple D/A
converter with as much as 10 bits of resolution. As such, it has a 10-bit timer and with the PWM operating at
(up to)
10 bits of resolution it will sample at up to 1/1024th of the clock frequency, or:
20 MHz / 1024 = 19.53125 kHz
Since we have 2
^32 (4 billion+)
counts in our 32 bit accumulator, and we clock it at as high as 19.53125 kHz, that means that our frequency resolution is about one four-billionth of 19.53125, or:
19.53125 kHz / (2
) = 0.000004547 Hz - or about one five-millionths of one Hertz resolution!
There's one more step in generating a useful frequency output. If one watches the MSB (most-significant bit) of the accumulator, we can see that it flips between 0 and 1 at the desired frequency, but
we don't want a digital output: Even if we did take the MSB which is, on average, at the desired frequency, it typically has a
of phase jitter that makes it unsuitable for most frequency control purposes.
If, instead, as noted above, we take the top
bits of the accumulator and feed them to a lookup table that has a sine wave and then outputting that value to a D/A converter, we get a more analog-looking signal with much less phase jitter: The
more bits we have, the better job we can do in representing a sine wave.
Now, remember that we divided our mixed-down local oscillator by 4096, so this means that our effective resolution would be reduced by that much, but if you do the math, that still means that we have
- when multiplied by 4096 - a step size of 0.0186 Hz or so at the VHF LO frequency!
If you've been following along, you might noticed that I skipped several steps, so let me explain:
The idea was to divide down a representation of the 126 MHz local oscillator to audio and we did this by first subtracting 100 MHz from it and then dividing-down the 26 MHz by 4096 to audio. We would
then generate a
audio frequency at one-4096th of that 26 MHz frequency and using a PLL, lock our local oscillator to it!
Simple - almost.
The DDS technique is imperfect when implemented using hardware that doesn't have infinite resolution - and the PIC's hardware and software capabilities are rather limited - in my case, I managed to
implement the equivalent of a 1 "ksample" sine wave with 10 bits of resolution.
(Actually, it was just 1/4^th of a sine wave - which is enough if you flip the pieces upside-down and/or play it backwards in the right order as needed!)
So now I had a precision audio generator that could output a reasonable facsimile of a sine wave at any frequency from about 5 milliHertz
(including DC, if you want to be pedantic)
to something less than 1/2 of the sample rate - about 9 kHz! The PWM output from the PIC is really a bunch of samples of a 19 kHz variable duty-cycle digital waveform and it needed to be filtered a
bit so I ran it through a simple op-amp bandpass filter - and then converted it back into a square wave - before passing it on to the a 4046 chip and into the edge-triggered phase detector. In the
4046 this was compared with the converted/divided signal from my local oscillator and with the magic of the PLL, my VHF oscillator was nicely locked to the precise audio frequency from the PIC!
At this point, the imperfection of the DDS became apparent.
One of the satellite frequencies is 137.62 MHz with a local oscillator frequency of 26.92 MHz. When this was divided by 4096, this yielded a frequency of 6.5723 kHz approximately.
If one takes a close look at the spectrum produced by any DDS-type synthesizer, a myriad of low-level (and some not-so-low-level) spurious signals will be generated because of rounding-off errors
related to the finite resolution of the D/A converter, the size of the sine table, and the relationship between the desired frequency and the clock frequency. As one approaches frequencies that are
related to an integer sub-multiple of the higher order bits
(e.g. multiples of 1/2, n/4, n/8, n/16, n/32, n/64, etc. of the clock frequency)
these low-level spurs get closer and closer to those multiples mentioned above. As these sub-multiples get "smaller", the amplitude of these spurious components decrease as well.
In the case of the 6.5723 kHz signal required to synthesize 137.62 MHz frequency, this was very close to 43/128
of the clock frequency - or about 10.986 Hz off. What this caused was a very low-level 11-ish Hz modulation of the generated frequency which, when effectively multiplied upwards by the 4096 division
- which increased the apparent loop gain - appeared as a very obvious tone
(more of a buzz, actually!)
at the local oscillator frequency.
Normally, loop filtering would take care of this, but this rather low frequency
(just 11 Hz!)
could get through the filter too well - and further-slowing of the loop filter wasn't particularly attractive - but this is software and we can do sleight-of-hand to fix this! What I did was to pick
a slightly different clock frequency - 20 MHz / 896 = 22.32142857... kHz instead and this moved the spurious signals from the DDS far enough away that they were effectively removed by the loop
The end result was a VCO that would tune anywhere within the designed range in less than a second and have very low-level spurious signal content from the DDS!
Locking a VCXO to an audio DDS reference:
As it turns out, locking a VCO - essentially a free-running oscillator with an implied, wide tuning range - is a comparatively "worst-case" scenario when it comes to the minimization of things like
"reference sidebands" - the frequency/phase modulation of the generated carrier from residual AC on the tuning line - owing to the very high loop gain involved which can arise from both the "tuning
sensitivity" of the VCO itself and if high divisor ratios are used. If one starts out with an oscillator with a comparatively narrow tuning range - such as a VCXO
(Voltage Controlled Crystal Oscillator)
- in which the tuning sensitivity can be orders of magnitude smaller - and the lock time may be longer, particularly if high divisor ratios are used since it may take some time for the phase fo the
comparison signal to "slide" into alignment - it is much easier to keep those already low-level spurious signals down to levels that may be ignored in typical applications.
A practical implementation of this technique has been employed in the W7SP Synchronous/Voting repeater system operated by the Utah Amateur Radio Club
(described here - link)
in which the transmit frequency is referenced from 10 MHz OCXOs
(Oven Controlled Crystal Oscillators)
and held within 1-2 Hz of the intended frequency. Using DDS techniques with 32 bit accumulators operating at approximately 3.2 kHz, the transmit frequency can be controlled - via the audio frequency
- to a resolution of 0.0023 Hz
at the two meter transmit frequency
- an accuracy that far exceeds the accuracy and stability of the reference oscillators themselves!
Producing exact frequencies:
The difficulty with using DDS techniques arises when an
frequency is desired, such as for a frequency standard - at least unless one is willing to crunch a few numbers and/or make a few compromises. For example, since the typical DDS algorithm is based on
binary counters and thus has denominators of 2
power, one will likely not end up with the
frequency desired. In our example, above, with a 32 bit counter, we can likely get the frequency to with a fraction of a Hertz, but not
where it should be.
There are several ways around this, including one or more of the following:
• Extending the resolution of the binary counter used in the DDS with even more bits to get ridiculous resolution so that the resulting frequency is "good enough." If enough processor time is
available and 32 bits of resolution is not enough, 48 or even 64 bits of addition may be implemented to do unsigned math.
• The careful selection of a clock frequency such that the divisors result in the exact frequency desired. The difficulty here is that if it is an "exact" frequency that is desired, many reference
frequencies - such as 10 MHz - are not "binary friendly", requiring a bit of clever math to come up with exact relationships with the target frequency.
• Designing the DDS counter to use something other than a binary (2^n) counter. If, say, a 10 MHz clock is used, the software DDS may be implemented using counters that will roll over at 10^n
instead of 2^n, driven by a hardware divisor set to a base-10 relatable value to yield exact frequencies.
• Implementing "dithering" of the DDS count to achieve fractional tuning. This involves switching between two or more frequencies at a specific rate to achieve a third, averaged frequency.
The last method, dithering, must be used with care as it will, by its nature, introduce spectral components that are necessarily lower in frequency than that of the reference being generated by the
DDS - possibly
much lower if the fraction being represented by the dithering is complex - and these lower frequencies can greatly complicate effective loop filtering! In most cases it would be more beneficial to
simply extend the resolution of the software DDS
(e.g. more bits)
rather than implement dithering making this technique most useful if one us using a hardware-based DDS.
If one needed to provide exactly 1 Hz steps, a DDS reference frequency based on 2^n Hz would be appropriate. For example, if you chose 2^24 Hz (16.777216 MHz) you can lock that (awkward)
frequency to 10 MHz as follows using only a few chips:
□ Divide 10 MHz by 625 to obtain 16 kHz (using a 74HC103 as a divide-by-125 and a 4017 to further divide-by-five.)
□ Using a PLL, multiply 16 kHz by 32 to yield 512 kHz (the lowly 4046 and a 4040 binary counter work well for this.)
□ Divide 512 kHz by 125 to yield 4096 Hz (using another 74HC103 to divide by 125)
□ Divide the 16.777216 MHz DDS reference oscillator by 4096 using a binary counter to 4096 Hz for the frequency comparison (a 74HC4040 works well as the divider here.)
The above steps may be done many different ways to get different frequencies, but the above is one example as to how to tie the two disparate frequency references together.
Final thoughts:
While there are definite limitations in using a DDS reference to lock a high frequency oscillator, namely the need to suppress the inevitable reference sidebands that result from the DDS synthesis
itself by filtering, careful selection of reference frequencies and/or choice of the type of oscillator, but appropriate application of these methods can produce a reliable, versatile - even simple -
frequency source.
This page stolen from "ka7oei.blogspot.com".
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Mathematics In Industrial Problems Part 5 1992
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Read Экстракция Органических Соединений 1997
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Math Dog Integer Addition
Help Math Dog, AKA MathPup, catch the cat burglar. All you need to do is figure out the answer to the integer addition problem containing both positive and negative numbers. 3 skill levels to choose
from and you can choose to play untimed for a leisurely game or select timed mode for more of a challenge.
Solve the integer addition problem at the top of the screen and then select the box containing the answer to catch the cat burglar.
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addition, integers, math, mathematical, mathematics, mobile, negative, quiz
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@Article{JCM-20-337, author = {Abderrahmane Nitaj}, title = {Isogenous of the Elliptic Curves over the Rationals}, journal = {Journal of Computational Mathematics}, year = {2002}, volume = {20},
number = {4}, pages = {337--348}, abstract = {
An elliptic curve is a pair $(E,O),$ where $E$ is a smooth projective curve of genus 1 and $O$ is a point of $E$, called the point at infinity. Every elliptic curve can be given by a Weierstrass
equation $$E:y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ Let $\mathbb{Q}$ be the set of rationals. $E$ is said to be defined over $\mathbb{Q}$ if the coefficients $a_i, i=1,2,3,4,6$ are rationals and $O$
is defined over $\mathbb{Q}$.
Let $E/ \mathbb{Q}$ be an elliptic curve and let $E(\mathbb{Q})_{tors}$ be the torsion group of points of $E$ defined over $\mathbb{Q}$. The theorem of Mazur asserts that $E (\mathbb{Q})_{tors}$ is
one of the following 15 groups $$E(\mathbb{Q})_{tors}=\begin{cases} \mathbb{Z}/m\mathbb{Z}, & m=1,2,\ldots,10,12 \\ \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/ 2m \mathbb{Z}, & m=1,2,3,4.\end{cases}.$$
We say that an elliptic curve $E'/\mathbb{Q}$ is isogenous to the elliptic curve $E$ if there is an isogeny, i.e. a morphism $\phi:E\rightarrow E'$ such that $\phi(O)=O$ , where $O$ is the point at
We give an explicit model of all elliptic curves for which $E(\mathbb{Q})_{tors}$ is in the form $\mathbb{Z}/m\mathbb{Z}$ where $m$ = 9,10,12 or $\mathbb{Z}/ 2 \mathbb{Z}\times \mathbb{Z}/ 2m \mathbb
{Z} \ {\rm where} \ m=4$, according to Mazur's theorem. Morever, for every family of such elliptic curves, we give an explicit model of all their isogenous curves with cyclic kernels consisting of
rationals points.
}, issn = {1991-7139}, doi = {https://doi.org/}, url = {http://global-sci.org/intro/article_detail/jcm/8922.html} }
TY - JOUR T1 - Isogenous of the Elliptic Curves over the Rationals AU - Abderrahmane Nitaj JO - Journal of Computational Mathematics VL - 4 SP - 337 EP - 348 PY - 2002 DA - 2002/08 SN - 20 DO - http:
//doi.org/ UR - https://global-sci.org/intro/article_detail/jcm/8922.html KW - Courbe elliptique, Isogenie. AB -
An elliptic curve is a pair $(E,O),$ where $E$ is a smooth projective curve of genus 1 and $O$ is a point of $E$, called the point at infinity. Every elliptic curve can be given by a Weierstrass
equation $$E:y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ Let $\mathbb{Q}$ be the set of rationals. $E$ is said to be defined over $\mathbb{Q}$ if the coefficients $a_i, i=1,2,3,4,6$ are rationals and $O$
is defined over $\mathbb{Q}$.
Let $E/ \mathbb{Q}$ be an elliptic curve and let $E(\mathbb{Q})_{tors}$ be the torsion group of points of $E$ defined over $\mathbb{Q}$. The theorem of Mazur asserts that $E (\mathbb{Q})_{tors}$ is
one of the following 15 groups $$E(\mathbb{Q})_{tors}=\begin{cases} \mathbb{Z}/m\mathbb{Z}, & m=1,2,\ldots,10,12 \\ \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/ 2m \mathbb{Z}, & m=1,2,3,4.\end{cases}.$$
We say that an elliptic curve $E'/\mathbb{Q}$ is isogenous to the elliptic curve $E$ if there is an isogeny, i.e. a morphism $\phi:E\rightarrow E'$ such that $\phi(O)=O$ , where $O$ is the point at
We give an explicit model of all elliptic curves for which $E(\mathbb{Q})_{tors}$ is in the form $\mathbb{Z}/m\mathbb{Z}$ where $m$ = 9,10,12 or $\mathbb{Z}/ 2 \mathbb{Z}\times \mathbb{Z}/ 2m \mathbb
{Z} \ {\rm where} \ m=4$, according to Mazur's theorem. Morever, for every family of such elliptic curves, we give an explicit model of all their isogenous curves with cyclic kernels consisting of
rationals points.
Abderrahmane Nitaj. (2002). Isogenous of the Elliptic Curves over the Rationals. Journal of Computational Mathematics. 20 (4). 337-348. doi:
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method of joints example problems with solutions
Let me now illustrate this. First of all, the video displays the given exemplary problem of super simple truss having three members connected like a triangle and subjected to an axial force at top
joint of the truss. Problem 005-mj Compute the force in all members of the truss shown in Fig. In a two dimensional set of equations, In three dimensions, â â FF. The reactions $A_x$ and $A_y$ are
drawn in the directions we know them to point in based on the reactions that we previously calculated. Applying the equilibrium conditions to each joint yields, These are 11 equations for the 8
unknown forces in the members and the 3 forces at the supports. Example 4.3. It is useful to present the results in dimensionless form in a table, including negative signs: The negative values for
the members 1, 2, 6, 7 and 11 indicate that these members are under compression. If we did not identify the zero force member in step 2, then we would have to move on to solve one additional joint.
State Why In Your Work. From member A, we will move to member B, which has three members framing into it (one of which we now know the internal force for). Since the support forces have been computed
in advance and are already known, the analysis is simplified, and three equations may be used as a check on the correctness of the results. The method of joints is the most recognized process to
discover unidentified forces in a truss structure. This allows solving for up to three unknown forces at a time. For compression members, the arrowheads point towards the member ends (joints) and for
tension members, the point towards the centre of the member (away from the joints). Its solutions exist only for eigenvalues of a and are determined correct to an arbitrary constant. 1c shows the
free-body diagrams of the joints. As discussed previously, there are two equilibrium equations for each joint ($\sum F_x = 0$ and $\sum F_y = 0$). Fig. Of course, once we know the force at one end of
AB (from the equilibrium at joint A), we know that the force at the other end must be the same but in the opposite direction. The Method of Joints. Visualizations are in the form of Java applets and
HTML5 visuals. session and get answers to all your problems in The information on this website is provided without warantee or guarantee of the accuracy of the contents. Continue through the
structure until all of the unknown truss member forces are known. ... determine the forces in all of the truss members using the method of joints. This limits the static equilibrium equations to just
the two force equations. A summary of all of the reaction forces, external forces and internal member axial loads are shown in Figure 3.8. Move on to another joint that has two or fewer members for
which the axial forces are unknown. Joint E is the last joint that can be used to check equilibrium (shown at the bottom right of Figure 3.7. A section has ï¬ nite size and this means you can also
use moment equations to solve the problem. Each Joint Must be in Equilibrium : One of the basic methods to determine loads in individual truss members is called the Method of Joints. 5. THE METHOD OF
JOINTS (Section 6.2) When using the method of joints to solve for the forces in truss members, the equilibrium of a joint (pin) is considered. Problem 411 Cantilever Truss by Method of Joints;
Problem 412 Right Triangular Truss by Method of Joints; Problem 413 Crane by Method of Joints; The free body diagram for joint B is shown in the top centre of Figure 3.7. In situations where we need
to find the internal forces only in a few specific members of a truss , the method of sections is more appropriate. Method of Joints The free-body diagram of any joint is a concurrent force system in
which the summation of moment will be of no help. To perform a 2D truss analysis using the method of joints, follow these steps: If the truss is determinate and stable there will always be a joint
that has two or fewer unknowns. Finding it now just has the benefit of saving us work later. Now that we know the internal axial forces in members AB and AC, we can move onto another joint that has
only two unknown forces remaining. These elements define the mechanism of load transfer i... Before discussing the various methods of truss analysis , it would be appropriate to have a brief
introduction. Even though we have found all of the forces, it is useful to continue anyway and use the last joint as a check on our solution. Newton's Third Law indicates that the forces of action
and reaction between a member and a pin are equal and opposite. In the Method of Joints, we are dealing with static equilibrium at a point. Zero-force members are identified by inspection and marked
with zeroes: member 4 (according to Rule 2), the members 5 and 9 (Rule 3) and the members 10 and 13 (Rule 1). Solve the unknown forces at that joint. This includes all external forces (including
support reactions) as well as the forces acting in the members. This is close enough to zero that the small non-zero value can be attributed to round off error, so the horizontal equilibrium is
satisfied. Pairs of chevron arrowheads are drawn on the member in the same direction as the force that acts on the joint. When a truss remains in equilibrium, then each of its joints should be in
equilibrium. Go to the next frame. 1b). As previously stated, we assume that every member is subjected to tension. 1.Method of joints. It does not use the moment equilibrium equation to solve the
problem. If the forces on the last joint satisfy equilibrium, then we can be confident that we did not make any calculation errors along the way. Multiple elements are used to transmit and resist
external loads within a building . Equations of equilibrium ( F X Example 1 All of the known forces at joint C are shown in the bottom centre of Figure 3.7. Method of Joints: Example Solution. The
truss shown in Figure 3.5 has external forces and boundary conditions provided. Since only one of the unknown forces at this joint has a horizontal component ($F_{DF}$) it will save work to solve
for this unknown first: Moving onto joint F (bottom left of Figure 3.7): At this point, all of the unknown internal axial forces for the truss members have been found. If the forces on the last
joint satisfy equilibrium, then we can be confident that we did not make any calculation errors along the way. There is also no internal instability, and therefore the truss is stable. joint using
the criterion of two unknown reactions. Draw a free body diagram of the joint and use equilibrium equations to find the unknown forces. It is a complex interplay between â socialâ and â rationalâ
processes. The only remaining unknown for the moment equilibrium around A will be $E_y$: We have assumed in Figure 3.6 that the unknown reaction $E_y$ points upward. For horizontal equilibrium,
there is only one unknown, $A_x$: For the unknown reaction $A_x$, we originally assumed that it pointed to the left, since it was clear that it had to balance the external $5\mathrm{\,kN}$ force.
Using horizontal equilibrium again: Now that we know $F_{BD}$ we can move on to joint D (top right of Figure 3.7). Find the internal axial forces in all of the truss members. It involves a
progression through each of the joints of the truss in turn, each time using equilibrium at a single joint to find the unknown axial forces in the members connected to that joint. The inclination
angles $\alpha$ and $\beta$ may be found using trigonometry (equation \eqref{eq:incl-angle}): The unknown reaction forces $A_x$, $A_y$ and $E_y$ can then be found using the three global equilibrium
equations in 2D. $$\label{eq:TrussEquil}\tag{1} \sum_{i=1}^{n}{F_{xi}} = 0; \sum_{i=1}^{p}{F_{yi}} = 0;$$. The initially unknown unit vectors can be determined from the vectors connecting adjacent
joints, e.g., for e, Introducing these into the two vector equations we get the six scalar equations. Since we have already determined the reactions $A_x$ and $A_y$ using global equilibrium, the
joint has only two unknowns, the forces in members AB ($F_{AB}$) and AC ($F_{AC}$). the method is also in the example above, the basic feasible solution x1 = 6 get 24/7 method of joints problem
assignment solved examples based on the method of joints. xy ==0 0 â F. z =0 Figure 3.6 shows the truss system as a free body diagram and labels the inclination angles for all of the truss
members. Graphical Educational content for Mathematics, Science, Computer Science. Example problem 1 A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. The members of the truss
are numbered in the free-body diagram of the complete truss (Fig. Since $F_{CE}=0$, this is a simple matter of checking that $F_{EF}$ has the same magnitude and opposite direction of $E_y$, which it
does. For exampleâ ¦ If the answer is negative, the member must be in compression. Since the resulting value for $E_y$ was positive, we know that this assumption was correct, and hence that the
reaction $E_y$ points upward. Joint D. Yes. It can be seen from the figure that at joint B, three members, AB;BC, and BJ, are connected, of which AB and BC are collinear and BJ is not. ... An
incorrect guess now though will simply lead to a negative solution later on. Accordingly, all of the corresponding arrows point away from the joints. Selected Problem Answers. The method of sections
is a process used to solve for the unknown forces acting on members of a truss. The frame analysis can be done by. Each joint is treated as a separate object and a free-body diagram is constructed
for the joint. Using the method of joints, solve problem 6-18 . The joint problem solving process is not just a matter of using a good logical system, or just a matter of effective interaction and
sound group processes. This figure shows a good way to indicate whether a truss member is in tension or compression. mechanical engineering questions and answers; Using The Method Of Joints, Solve
Problem 6-18 . Resources for Structural Engineers and Engineering Students. Please enable JavaScript!Bitte aktiviere JavaScript!S'il vous plaît activer JavaScript!Por favor,activa el JavaScript!
antiblock.org. 4.18. The positive result for $A_y$ indicates that $A_y$ points upwards. where and are the reaction forces at joint in the and directions, is the reaction force at joint , is the width
of the members and is the point load force at joint .. Next, do force balances at the joints. Solving linear programs 2 solves problems with one or more optimal solutions. Alternatively, joint E
would also be an appropriate starting point. Even though we have found all of the forces, it is useful to continue anyway and use the last joint as a check on our solution. All forces acting at the
joint are shown in a FBD. Like the name states, the analysis is based on joints. Solve the joint equations of equilibrium simultaneously, typically using a computer or an advanced calculator. Method
of Joints Example -Consider the following truss The method of joints is a procedure for finding the internal axial forces in the members of a truss. Fig. Solved Examples for Method of Joints for
Truss Analysis, Analysis of 2D Truss Structure in SAP 2000, Types, Assumptions and Fundamental Approaches of Structural Analysis, Steps in Construction of Reinforced Concrete Structures, Overview:
Open and Closed Traverses in Surveying, Engineersdaily | Free engineering database. See the answer. Example - Method of Joints. 1a represents a simple truss that is completely constrained against
motion. In this unit, you will again use some of the facts and learn a second method of solution, the "Method of Sections." Use it at your own risk. Upon solving, if the answer is positive, the
member is in tension as per our assumption. Under this process, all forces functioning on a joint must add to zero. goo.gl/l8BKU7 for more FREE video tutorials covering Engineering Mechanics (Statics
& Dynamics) The objectives of this video are to introduce the method of joints & to resolve axial loads in a simple truss. m<2j+3 unstable structure. >>When you're done reading this section, check
your understanding with the interactive quiz at the bottom of the page. Method of joints are the common method for the analysis of truss members.The basic concept that is used in the analysis is
,since the truss is in equilibrium the each joints in the truss is also in equilibrium First, calculate the reaction forces by doing a moment balance around joint and force balances in the and
directions:. 2 examples will be presented in this this article to clarify those concepts further. Background A traverse is a form of control survey used in a wide variety of engineering and property
surveys. Open Digital Education.Data for CBSE, GCSE, ICSE and Indian state boards. Force $F_{AB}$ is drawn pointing towards the node, and the external force of $5\mathrm{\,kN}$ is also shown. By
applying equilibrium at joint B, we can solve for the unknown forces in those members $F_{BC}$ and $F_{BD}$. If there exist a net force, the joint will shift. Note also that although member CE does
not have any axial load, it is still required to exist in place for the truss to be stable. Since only two equations are involved, only two unknowns can be solved for at a time. Since we have two
equations and two unknowns, we can solve for the unknowns easily. Compressive (C) axial member force is indicated by an arrow pushing toward the joint. DETERMINATE INDETERMINATE INDETERMINATE
DETERMINATE INDETERMINATE DETERMINATE. Therefore, it is statically determinate. 6.7 Analysis of Trusses: Method of Sections The method of joints is good if we have to find the internal forces in all
the truss members. A repository of tutorials and visualizations to help students learn Computer Science, Mathematics, Physics and Electrical Engineering basics. Like previously, we will start with
moment equilibrium around point A since the unknown reactions $A_x$ and $A_y$ both push or pull directly on point A, meaning neither of them create a moment around A. Example problem 1 â ¦ In this
problem, we have two joints that we can use to check, since we already identified one zero force member. For some obscure reason, this is called the method of jointsâ ¦ No, I don't. The unknown
member forces $F_{AB}$ and $F_{AC}$ are assumed to be in tension (pulling away from the joint). For each truss below, determine the forces in all of the members marked with a checkmark ($\checkmark$)
using the method of sections. Since the axial force in AB was determined to be $3.5\mathrm{\,kN}$ in compression, we know that at joint B, it must be pointing towards the joint. The method involves
breaking the truss down into individual sections and analyzing each section as a separate rigid body. Yours may be different. We will select joint A as the starting joint. All copyrights are
reservedÂ. The information on this website, including all content, images, code, or example problems may not be copied or reproduced in any form, except those permitted by fair use or fair dealing,
without the permission of the author (except where it is stated explicitly). These two forces are inclined with respect to the horizontal axis (at angles $\alpha$ and $\beta$ as shown), and so both
equilibrium equations will contain both unknown forces. To further reduce the number of unknown forces, we compute the support forces by applying the equilibrium conditions to the whole truss.
(Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.) Since the boundary support at point E is a roller, there is no
horizontal reaction. Solution. Figure 3.5: Method of Joints Example Problem, Figure 3.6: Method of Joints Example - Global Free Body Diagram, Figure 3.7: Method of Joints Example - Joint Free Body
Diagrams, Figure 3.8: Method of Joints Example - Summary, Chapter 2: Stability, Determinacy and Reactions, Chapter 3: Analysis of Determinate Trusses, Chapter 4: Analysis of Determinate Beams and
Frames, Chapter 5: Deflections of Determinate Structures, Chapter 7: Approximate Indeterminate Frame Analysis, Chapter 10: The Moment Distribution Method, Chapter 11: Introduction to Matrix
Structural Analysis, 3.4 Using Global Equilibrium to Calculate Reactions, 3.2 Calculating x and y Force Components in Truss Members, Check that the truss is determinate and stable using the methods
from, If possible, reduce the number of unknown forces by identifying any, Calculate the support reactions for the truss using equilibrium methods as discussed in. Using the geometrical relations.
Identify a starting joint that has two or fewer members for which the axial forces are unknown. Figure 3.5: Method of Joints Example Problem. Frame 18-20 Transition As you can see, you can go on
until you reach either the end of the truss or the end of your patience. Solutions: Available for all 6 problems Start here or click on a link below: Problem-1: Market value method for joint cost
allocation and reversal cost method for by products For vertical equilibrium: So member AB is in compression (because the arrow actually points towards the joint). This site is produced and managed
by Prof. Jeffrey Erochko, PhD, P.Eng., Carleton University, Ottawa, Canada, 2020. For example, if I take the problem we just solved in the method of joints and make a section S 1, S 2 (see figure
9), we will be able to determine the forces in members BC, BE and FE by considering the equilibrium of the portion to the left or the right of the section. The author shall not be liable to any
viewer of this site or any third party for any damages arising from the use of this site, whether direct or indirect. 10 ft. 10 ft. Procedure for analysis-the following is a procedure for analyzing a
truss using the method of joints: 500 lb. Joint E can now be solved. Method of joints The method of joints analyzes the force in each member of a truss by breaking the truss down and calculating the
forces at each individual joint. All supports are removed and replaced by the appropriate unknown reaction force components. Although there are no zero force members that can be identified direction
using Case 1 or 2 in Section 3.3, there is a zero force member that may still easily be identified. There will always be at least one joint that you can use to check the final equilibrium. T-08.
using the method of joints. The critical number of unknowns is two because at a truss joint, we only have the two useful equilibrium equations \eqref{eq:TrussEquil}. This problem has been solved!
This is a simple truss that is simply supported (with pin at one end and a roller at the other). Therefore, the reaction at E is purely vertical. The two unknown forces in members BC and BD are also
shown. A free body diagram of the starting joint (joint A) is shown at the upper left of Figure 3.7. Therefore, joint VII need not be considered. Either method can be used alone to analyze any
statically determinate truss, but for real efficiency you need to be able to handle both methods alone or in combination. The method of joints uses the summation of forces at a joint to solve the
force in the members. Therefore the only horizontal force at the joint can come from member CE, but since there is not any other member or support to resist such a horizontal force, we must conclude
that the force in member CE must be zero: Like any zero force member, if we did not identify the zero force member at this stage, we would be able to find it easily through the analysis of the FBDs
at each joint. Problem 005-mj | Method of Joints . Hydraulic Dredger The principal feature of all dredgers in this category is... 1. Identify all zero-force members in the Fink roof truss subjected
to an unbalanced snow load, as shown in Fig. The method of joints is a process used to solve for the unknown forces acting on members of a truss.The method centers on the joints or connection points
between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure. Question: Using The Method Of Joints, Solve Problem 6-18 . The Vertical
Component Of The Force At G Must Equal Zero. The boundary value problem (4.117) determines the self-similar solution with the pressure distribution presented above.For given a the system of equations
for f 21, g 21, f 31, F 1, β 1, P 1, V 1, and Φ 1 is linear and homogeneous. Click here to show or hide the solution $\Sigma M_F = 0$ $11R_A = 7(50) + 3(30)$ ... example of method of joints. This
means that we will have to solve a two equation / two unknown system: Rearranging the horizontal equilibrium equation for $F_{BD}$: Sub this into the vertical equilibrium equation and solve for $F_
{BC}$: in tension. Details. the number of members is less than the required members.So there will be chance to fail the structure.. (My response. Zero-force members are omitted in the free-body
diagrams. If member CE were removed, joint E would be completely free to move in the horizontal direction, which would lead to collapse of the truss. Horizontal equilibrium: Since we now know the
direction of $F_{AC}$, we know that member AC must be in tension (because its force arrow points away from the joint). Method of Joints Method of Joints - the axial forces in the members of a
statically determinate truss are determined by considering the equilibrium of its joints. MethodofJoints The method of joints is one of the simplest methods for determining the force acting on the
individual members of a truss because it only involves two force equilibrium equations. The theoretical basis of the method of joints for truss analysis has already been discussed in this article '3
methods for truss analysis'. " Transmit and resist external loads within a building involved, only two equations are involved only. Loads within a building to zero identified one zero force member
solve for the joint is to! Solve for the joint by an arrow pulling away from the joint equations of equilibrium simultaneously typically. If the answer is positive, the joint will shift of equations,
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Definition and effects of thermal bridges
Table of Contents
Definition and effects of thermal bridges
Thermal bridges - Introduction
Heat makes its way from the heated space towards the outside. In doing so, it follows the path of least resistance.
A thermal bridge is a localised area of the building envelope where the heat flow is different (usually increased) in comparison with adjacent areas (if there is a difference in temperature between
the inside and the outside).
The effects of thermal bridges are:
• Altered, usually decreased, interior surface temperatures; in the worst case this can lead to moisture penetration in building components and mould growth.
• Altered, usually increased, heat losses.
Both effects of thermal bridges can be avoided in Passive Houses: the interior surface temperatures are then so high everywhere that critical levels of moisture cannot occur any longer – and the
additional heat losses become insignificant. If the thermal bridge losses are smaller than a limit value (set at 0.01 W/(mK)), the detail meets the criteria for “thermal bridge free design”.
If the criteria for thermal bridge free design are adhered to everywhere, the planners and construction manager don't have to worry about cold and damp spots any more - and less effort will have to
be made for calculating the heat energy balance.
Thermal bridge free design leads to substantially improved details; the durability of the construction is increased and heating energy is saved.
Normative definition of thermal bridges
In [DIN10211] (Thermal bridges in building construction – Heat flows and surface temperatures - Detailed calculations) there are numerical procedures relating to the calculation of thermal bridges.
Here, a thermal bridge is defined as follows (Section 3.1.1):
Compared to thermal bridge free building components, there are two effects of thermal bridges which occur at each connection point between building components or at places where the composition of
the building structure changes:
• altered heat flow
• a change in the interior surface temperature
A general overview is possible if the procedure for determining the transmission heat losses $H_T$ of the building envelope is considered. The following equation in the norm DIN 14683 (Section 4.2)
makes a distinction between one-dimensional, two-dimensional and three-dimensional heat flows.
\begin{align} &\Large{H_{T} = \underbrace{\sum_{i}A_{i}U_{i}}_{1d}+\underbrace{\sum_{k}l_{k}\varPsi_{k}}_{2d}+\underbrace{\sum_{j}\chi_{j}}_{3d}}\\\\ With\qquad&\\ A_{i}\qquad&\text{area of the
building components, in $m^2$}\\
U_{i}\qquad&\text{thermal transmittance of component $i$ of the building envelope, in $W/(m^2\cdot K)$}\\
l_{k}\qquad&\text{length of the linear thermal bridge $k$, in $m$}\\
\varPsi_{k}\qquad&\text{thermal transmittance of the linear thermal bridge $k$, in $W/(m\cdot K)$}\\
\chi_{j}\qquad&\text{thermal transmittance of the point thermal bridge $j$, in $W/K$}
Planar regular building components such as the roof areas and exterior walls have the largest share of the total heat flow. For these, heat transfer can be considered one-dimensional with good
approximation. The reason for this is that no cross-flows occur in them on account of their homogeneous layered structure. The heat transfer coefficient is defined in the norm [DIN6946] and can be
calculated with little effort using the familiar equation given below:
\begin{align} &\Large{U=\dfrac{1}{R}=\dfrac{1}{R_{si}+\frac{d_{0}}{\lambda_{0}}+\frac{d_{1}}{\lambda_{1}}+\dots+\frac{d_{n}}{\lambda_{n}}+R_{se}}}\\\\ With\qquad&\\ R_{si}\qquad&\text{inner heat
transfer resistance , in $m^2 \cdot K/W$}\\
d_{n}\qquad&\text{thickness of the $n$-th component layer, in $m$}\\
\lambda_{n}\qquad&\text{rated value of the thermal conductivity of the $n$-th layer, in $W/(m\cdot K)$}\\
R_{se}\qquad&\text{outer heat transfer resistance, in $m^2 \cdot K/W$}
The two-dimensional and three-dimensional heat flow proportion of the building envelope is expressed by thermal bridges. They are defined by geometric, constructive and/or material modification and
usually exhibit a higher heat flow rate and lower surface temperatures than adjacent standard building components. They occur particularly at the component joints, edges, transitions and penetrations
of the standard building components. They are depicted by the linear thermal transmittance $\varPsi$ with the unit W/(mK) and the point thermal transmittance $\chi$ in W/K.
Additional heat losses
The effects of thermal bridges on the energy balance depend not only on the influence in terms of physics but also on how they are taken into account. Thus, in the context of energy balancing,
thermal bridges can be depicted as follows:
1. by using a general thermal bridge value $ \Delta U_{bw} = 0.10 \quad W/(m^2\cdot K)$ (EnEV)
2. by using a reduced thermal bridge value $ \Delta U_{bw} = 0.05 \quad W/(m^2\cdot K)$ (DIN 4108 Supplementary sheet 2)
3. with Ψ-values taken from thermal bridge catalogues e.g. (DIN EN ISO 14683)
4. with Ψ-values from a calculation in (DIN EN ISO 10211)
In principle, the actual share of the thermal bridges in the transmission losses of the building envelope can only be stated if the Ψ-values are calculated for a specific building. It is assumed that
heat flow simulations are associated with an uncertainty of ca. 5 %, other methods such as the use of thermal bridge catalogues are even associated with an uncertainty of up to 20 % (DIN EN ISO
14683, Section 5.1). For Passive House buildings, the use of thermal bridge additions is not advised because they lead to overestimation of the heat losses.
However, in general it is not possible to state how high the heat losses due to thermal bridges actually are. Their type and number are too individual and therefore depend on the respective building.
For example, thermal bridges do not always have to have a negative effect on energy balancing; in the case of efficient new constructions, particularly in the area of Passive House buildings, taking
the Ψ-values into account can certainly reduce the space heating demand. In the case of existing buildings and modernised building stock, thermal bridges generally have a negative effect and
according to [EnerPHIT], experience has shown that this can result in an additional heat loss of up to 20 %. Based on examples of different construction projects, this resulted in an increase in the
annual heating demand of up to 14 kWh/(m²a). Careful planning with regard to thermal bridges can therefore be decisive for achieving the Passive House Standard in a construction project.
Effect on the building structure
Unlike with regular building components, at thermal bridges the heat flow density changes and usually results in a reduction in the surface temperature on the inside in that area. This effect is more
pronounced because air circulation in corners and edges is restricted. Cupboards and other furniture not only disrupt convection but also restrict radiant exchange with the surroundings. Because the
water vapour content of the air depends on its temperature, condensation may form on the affected areas.
The resulting condensation can penetrate further inside the construction due to the capillary action of the building materials, and the thermal conductivity may increase and thus the building
component may almost be saturated. It will not be possible to avoid moisture damage to the building structure and mould growth may occur. However, large-scale damage is generally associated with
errors in the planning, implementation and utilisation of buildings and is not a problem that is solely related to thermal bridges. These are only the points where the problems originate in the first
place. Nonetheless, the risk of mould fungus to the inner surface of thermal bridges and the resultant toxic effect on occupants must be considered separately, especially since mould growth already
occurs at a temperature higher than the dewpoint temperature without condensation being present. For a building physical analysis of the model, formation of mould can be assumed if relative surface
humidity levels of 80 % prevail for a duration of 12 h/d (Technical Report 4108-8).
In constructions suitable for passive house or EnerPHit-design, thermal bridges with such catastrophic consequences are generally avoided; we won't allow for such bad design in the recommended
constructions, so we won't have that. While there can still be thermal bridges with some remaining heat losses^1), such massive temperature drops can always be avoided. A big part in avoiding
catastrophic thermal bridges is the improved insulation level already for the regular components. This leads to a generally higher level of indoor surface temperatures to begin with, already reducing
the risk. Internal insulation, however, is a special case in which there are even more stringent rules “how to avoid thermal bridges of the catastrophic type”.
Requirements the current rules for engineering practice (DIN 4108-2) rule out the risk of mould near thermal bridges if the minimum surface temperatures under the mentioned steady-state boundary
conditions do not fall below 12.6 °C. This corresponds with a $f_{Rsi}$ factor of 0.7:
$$ f_{Rsi,min}=\dfrac{12.6^{\circ} C -(-5^{\circ} C)}{20^{\circ} C - (-5^{\circ} C)}=0.7 $$
The higher the $f_{Rsi}$ factor is, the less the likelihood of mould growth will be. For Certified Passive House Components, the request to the $ f_ {Rsi} $-factor depends also from the climate.
See also
which have to be accounted for
basics/building_physics_-_basics/thermal_bridges/thermal_bridge_definition.txt · Last modified: 2022/07/30 14:50 by wfeist
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EuclidEuclid | World of History
Euclid, often referred to as the “Father of Geometry,” was an ancient Greek mathematician who lived around 300 BCE. He…
Euclid, often referred to as the “Father of Geometry,” was an ancient Greek mathematician who lived around 300 BCE. He is best known for his work “Elements,” a comprehensive compilation of the
knowledge of geometry and number theory of his time. Euclid’s “Elements” became one of the most influential works in the history of mathematics, serving as the primary textbook for teaching
mathematics, particularly geometry, for over two millennia.
Early Life and Background
Time and Place:
Euclid is believed to have been born around 325 BCE and lived until around 265 BCE. While specific details about his life are scarce, it is generally accepted that he worked in Alexandria, Egypt,
during the reign of Ptolemy I (323–283 BCE).
Alexandria was one of the great centers of learning in the ancient world, home to the famous Library of Alexandria, where Euclid is thought to have taught mathematics.
Education and Influences:
Little is known about Euclid’s personal life, including his education. It is believed that he studied in Athens, possibly at Plato’s Academy, where he would have been influenced by the works of
earlier Greek mathematicians such as Pythagoras, Thales, and Eudoxus.
Euclid’s work in geometry builds upon the contributions of these earlier mathematicians, but it is his systematic approach to organizing and proving mathematical principles that sets his work apart.
Euclid’s “Elements”
Structure of “Elements”:
“Elements” is a comprehensive 13-book compilation that systematically presents the principles of geometry and number theory. The first six books focus on plane geometry, covering basic concepts such
as points, lines, angles, triangles, and circles. The remaining books delve into number theory, irrational numbers, and solid geometry.
Euclid’s approach was to start with a small set of axioms (self-evident truths) and postulates (assumptions) and then derive a wide range of propositions and theorems using logical deduction. This
method of reasoning, known as the axiomatic method, became the standard approach in mathematics.
Axioms and Postulates:
Euclid’s five postulates form the foundation of his geometry. The most famous of these is the fifth postulate, known as the parallel postulate, which states that given a line and a point not on the
line, there is exactly one line parallel to the given line that passes through the point.
This postulate was controversial because it was less intuitive than the others, leading to centuries of exploration and eventually the development of non-Euclidean geometries in the 19th century.
Notable Theorems:
Euclid’s “Elements” contains many important theorems that are still taught in mathematics today. Some of the most famous include:
The Pythagorean Theorem: Although attributed to Pythagoras, Euclid provided a rigorous proof of this theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum
of the squares of the other two sides.
Euclid’s Lemma: A key result in number theory, Euclid’s lemma states that if a prime number divides the product of two numbers, it must divide at least one of those numbers. This lemma is
foundational in the study of prime numbers and the proof of the fundamental theorem of arithmetic.
The Infinitude of Primes: Euclid’s proof that there are infinitely many prime numbers is one of the earliest and most elegant proofs in number theory.
Legacy of “Elements”:
“Elements” became the definitive textbook on geometry and mathematics for centuries. It was used in education throughout the Hellenistic world, the Islamic Golden Age, and into the Renaissance and
Enlightenment in Europe.
The influence of “Elements” extended beyond mathematics; its logical structure and rigorous method of proof served as a model for scientific reasoning in general.
Other Works
Another work attributed to Euclid is “Data,” which deals with the properties of geometrical figures given certain conditions. It complements the “Elements” by providing additional tools for solving
geometric problems.
Euclid also wrote “Optics,” one of the earliest known works on the subject. In this treatise, he explores the properties of light and vision, proposing that vision occurs when rays emanate from the
eyes and interact with objects.
Other Lost Works:
Euclid is believed to have written several other works, including treatises on music theory (“Pseudaria”), conic sections, and mathematical astronomy, but many of these have been lost.
Influence and Legacy
Impact on Mathematics:
Euclid’s “Elements” was the standard mathematics textbook for over 2,000 years, influencing countless mathematicians, scientists, and thinkers. Its impact is comparable to that of Aristotle’s works
on logic and philosophy.
The axiomatic method introduced by Euclid remains a fundamental aspect of modern mathematics. The rigorous approach to proving theorems from a set of basic principles is the foundation of
mathematical logic and theory.
Non-Euclidean Geometry:
The exploration of Euclid’s parallel postulate eventually led to the development of non-Euclidean geometries in the 19th century by mathematicians such as Carl Friedrich Gauss, Nikolai Lobachevsky,
and János Bolyai. These new geometries expanded the understanding of space and laid the groundwork for Einstein’s theory of general relativity.
Cultural and Educational Influence:
Euclid’s work has also had a lasting influence on education. The logical structure and deductive reasoning presented in “Elements” have been used to teach students critical thinking and
problem-solving skills for centuries.
Euclid’s influence extends beyond mathematics into the broader scientific and philosophical traditions of the Western world, where his method of systematic reasoning has served as a model for
scientific inquiry.
We get commissions for purchases made through links on this website. As an Amazon Associate we earn from qualifying purchases.
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Analytic Continuation | Brilliant Math & Science Wiki
The principle of analytic continuation is one of the most essential properties of holomorphic functions. Even though it could be stated simply and precisely as a theorem, doing so would obscure many
of the subtleties and how remarkable it is. It is perhaps more instructive to take a step back to real (analytic) functions and Taylor series, and to see why complex numbers is the natural setting.
Along the way, we shall encounter other fundamental concepts in complex analysis, such as branch cuts, isolated singularities (including poles), meromorphic functions, monodromy, and even Riemann
It may serve as a prologue to a formal study of complex analysis, only assuming basic acquaintance with Taylor series and complex numbers. This largely is the perspective of Weierstrass; for a more
complete view, there are Cauchy's theory based on contour integration, Riemann's geometric theory, as well as the perspective of PDE (partial differential equations).
The video Visualizing the Riemann zeta function and analytic continuation by 3Blue1Brown is excellent in giving the geometric intuition, and this article was largely written to complement it.
Most functions that come up "in nature" — either in describing the (ideal) physical world or in pure mathematics — particularly those that are given a special symbol or a name, are in fact analytic:
if we take its Taylor series at any point, which only uses the data arbitrarily close to that point, we could recover the function completely. For example, knowing the function \(\sin x\) for \(x\in
\big[0, \frac{\pi}{2} \big] \), or even for a tiny interval, is enough to determine the entire function: Simply take its Taylor series at \(x=0:\) \[x-\frac{x^3}{3!} + \frac{x^5}{5!} - \cdots,\]
which converges for all \(x\in\mathbb R\), and agrees with the standard, periodic definition of \(\sin x\) over the reals. Taking the Taylor series at any other point will result in the exact same
function (see Taylor's theorem). This is the simplest and best-case scenario of analytic continuation — from a small interval to the whole real line — for the radius of convergence is always
infinite. Such functions are called entire functions, which include all polynomials, the exponential function, certain "special functions" (e.g., Bessel functions), and their sums, products, and
The problem becomes more subtle, and hence more interesting, if the radius of convergence is finite. Suppose we only know \( f(x)=\frac{1}{x} \) on a small neighborhood around \(x=1\). The Taylor
series takes the form \[ \sum_{n=0}^\infty (-1)^{n} (x-1)^n, \qquad (*)\] which, being a geometric series, converges only for \(|x-1|<1\), i.e. \( 0<x<2 \), where it indeed converges to \( \frac{1}
{x} \). Now, knowing the values of the function near \(x=1.5\), for instance, we may "Taylor expand" at \(x=1.5\), and the new Taylor series in fact converges for \( x\in (0, 3) \) while agreeing
with the previous values on \( (0, 2) \). We can say that we have analytically continued the function to \( (0,3)\). Thus, by successive Taylor series expansions, we could "recover" the function \( \
frac{1}{x} \) for all \(x>0\), but no way whatsoever could we extend it to \(x<0\). The point \(x=0\) poses as an insurmountable barrier, called a singularity of the function. It seems that we are
free to define \(f(x)\) to be any (analytic) function on \(x<0\), and no criterion on the function could favor one over the countless others.
This is where complex numbers come into play so that we might be able to "circumvent" the barrier by going into the complex plane. In fact, the Taylor series in general (or power series) makes
perfect sense, as a series, when \(x\) is any complex number, so long as we know how to add and multiply complex numbers. Examining the derivation of the geometric series, we see that \[ 1 + r + r^2
+ \cdots = \lim_{n\to\infty}\frac{1-r^{n+1}}{1-r}=\frac{1}{1-r}\] holds for all complex numbers \(r\) of modulus strictly less than 1. Thus, the series \((*)\) converges for all complex \(x\) with \
(|x-1|<1\), which is a disk of radius \(1\) centered at \(x=1\). The radius of convergence is literally a radius, and this phenomenon holds true for all (convergent) power series. In particular,
entire functions are naturally defined on the whole complex plane.
Now, by taking the Taylor series at a point off from the real axis, we may get around the singularity at \(x=0\). There are two ways to get to the negative real axis: through the upper half of the
complex plane, or through the lower half. It turns out that we'd end up with the exact same result, which as one might expect is simply \(\frac{1}{x}\) for \(x<0\). In fact, as illustrated above,
each Taylor series in the process of analytic continuation converges in an (open) disk just short of the singularity at \(x=0\). Indeed, for any \(a\in\mathbb C\setminus\{0\}\), we may expand \(\frac
{1}{x}\) as a Taylor series centered at \(a\): \[\frac{1}{x} = \frac{1}{a+(x-a)} = \frac{1}{a}\cdot \frac{1}{1+\frac{x-a}a}=\frac{1}{a}\sum_{n=0}^\infty (-1)^n\left(\frac{x-a}{a}\right)^n=\sum_{n=0}^
\infty \frac{(-1)^n}{a^{n+1}}(x-a)^n\] for \(|\frac{x-a}{a}|<1\), i.e. \(|x-a|<|a|\). This also illustrates that the precise procedure of analytic continuation (choices of the centers of Taylor
series expansions) does not matter, and the end result is the same, namely \(f(x)=\frac{1}{x}\) on the punctured plane \(\mathbb C\setminus\{0\}\).
It should be noted right away that not all functions, when analytically continued around a singularity from above and below, have the same result. The two prototypical examples are \(\log x\) and \(\
sqrt x\); they are not typically defined for negative \(x\) for this reason.
The complex numbers also provide more insight even in the case when we could analytically continue over the reals. For example, \[f(x)=\frac{1}{1+x^2}\] is defined and infinitely differentiable for
all \(x\in\mathbb R\). The Taylor series at \(x=0\), however, has a radius of convergence of 1 (again by geometric series). If we take the complex perspective, we see that \(f(x)\) does have
singularities at \(x=\pm i\), which are at a distance 1 from the origin, so it couldn't have a larger radius of convergence. In fact, it is true in general that the Taylor series of any analytic
function converges to the function itself within a disk as large as possible (before hitting a "singularity"), when viewed as a complex function.
It may already be enough evidence that analytic functions, which include all the familiar functions, really should be regarded as living on the complex plane or subsets or extensions thereof. They
are not confined to a particular domain (per the modern concept of a function) but have the ability to extend or continue in all directions as far as possible, to what can be called its natural
domain. Our modern definition of a function, an arbitrary assignment of a value \(y\) for each \(x\) in a prescribed domain, has a very different flavor: no way whatsoever to extend its domain, or
rather infinitely many choices of extension that do not single out any particular one. If the original function happens to be continuous, one may require the extension to be continuous too, which
would narrow down the choices but still leave infinitely many possibilities (unless the extension is just for one extra point); if the original function was differentiable, one may ask the same for
the extension, which would further narrow down the choices. Analyticity is the strongest criterion of all, and it turns out it is enough to single out a unique choice of extension if one exists. That
is the principle of analytic continuation.
To phrase the principle of analytic continuation differently: the identity of an analytic function is "encoded" in each and every point of its natural domain, in the sequence of Taylor series
coefficients (or the derivatives) at that point, traditionally known as the germ of the function at the point (in the sense of the seed of a crop). One could easily write down the rules for the basic
operations — addition, multiplication, division, inversion, differentiation, etc. — on the set of germs at the same point. To carry on the agrarian analogy, a collection \((\)often a \(\mathbb C\)
-vector subspace\()\) of germs at a point is called a stalk, and putting all the stalks (of the same sort) over various points of a domain together, endowed with some topology, we get a sheaf, which
is semantically the same as a bundle. This is the beginning of sheaf theory.
From now on, we shall use \(z\) \(\big(\)or \(\zeta\), \(s\), etc.\(\big)\) instead of \(x\) for the variable of our functions.
Despite the fact that an analytic function, by its very nature, is fully determined by a sequence of (complex) numbers, the general theory of functions in the complex domain is a vast subject that
goes under many names: complex analysis, (complex) function theory, theory of functions of a (single) complex variable, etc. From the point of view of analytic continuation, the most natural question
Given a convergent power series \[f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n, \] determine the largest domain in the complex plane to which \(f(z)\) can be analytically continued.
is hopelessly difficult. Nevertheless, it offers a panorama of a wide variety of functions, with connections to different areas of mathematics, if we wish to look past some of the detailed
justifications. In increasing level of "complexity" (by some measure), we have the following:
• Entire functions: those that can be analytically continued to the whole complex plane. It generalizes polynomials. For example, the Fourier (and Laplace) transform \[f(\zeta)=\int_{\mathbb R} e^
{-ix\zeta}\phi(x)\,dx \qquad \zeta=\xi+i\eta\in\mathbb C \] of a compactly supported continuous function \(\phi\in C_0(\mathbb R)\) \(\big(\)or more generally a distribution \(\phi\in\mathcal E'
(\mathbb R)\) of compact support\(\big)\) is entire, and furthermore the support of \(\phi\) is governed by the growth of \(f(\zeta)\) in the imaginary direction, i.e. as \(\eta\to\pm\infty\)
(Paley-Wiener theorem).
• Meromorphic functions on \(\mathbb C\): the barriers are all isolated points (called singularities), but analytic continuation is possible around each singularity, and the result does not depend
on which way to go around them. (One extra technical condition is often imposed so that all the singularities are "poles" instead of "essential singularities.") It generalizes rational functions.
For any non-constant polynomial \(P\) of \(n\) variables with \(P(x)\geq 0\) for all \(x\in\mathbb R^n\), and any compactly supported smooth \(\phi\in C^\infty_0(\mathbb R^n),\) \[ f(s) = \int_{\
mathbb R^n} P(x)^s \phi(x)\,dx \qquad \operatorname{Re} s>0\] can be analytically continued to the whole complex plane except for isolated, albeit infinitely many, points on the negative real
axis (Bernstein's theorem). For another important class of examples, the so-called \(L\)-functions, such as the Dirichlet \(L\)-function \[L(s)=\sum_{n=1}^\infty \frac{\chi(n)}{n^s} \qquad \
operatorname{Re} s>1\] associated to a Dirichlet character \(\chi:\mathbb Z\to\mathbb C\), can be analytically continued to all of \(\mathbb C\) except possibly for a few points such as \(s=1\).
• Functions such as \(\log z\) and \(\sqrt z\) could be analytically continued around the singularity at \(z=0,\) but the result depends on the path taken. To remove this ambiguity, one would need
to agree on a continuous "borderline" or "cut" extending from \(z=0\) to infinity (e.g. the negative real axis), across which no analytic continuation is permitted. Due to the presence of the
cut, \(z=0\) shall not be considered an isolated singularity even though it is the only "barrier" of analytic continuation. \((\)Note also that \(\sqrt z\) does not go to infinity when \(z\to 0.)
\) Alternatively, we could analytically continue across the cut by "jumping" to another copy of the complex plane: Thus the natural domain of \(\log z\) or \(\sqrt z\) is not a subset of the
complex plane but consists of multiple copies of the complex plane properly glued together; this is an example of a Riemann surface. In a sense that could be made precise, the point \(z=0\) is no
longer a singularity of the function.
• The barriers of analytic continuation may not even be isolated points but form a "wall." In fact, for any open, connected subset \(U\subsetneq\mathbb C\), there exists an analytic function on \(U
\) that cannot be extended past any point of the boundary. In other words, \(U\) is the natural domain of that function. This class may seem exotic, but in fact it is as rich as non-analytic
functions of a real variable. To illustrate it, consider \[f(z)=\int_{\mathbb R} \frac{\phi(x)}{x-z}dx \qquad z\in\mathbb C\setminus\operatorname{supp}\phi,\] where \(\phi\) only needs to be
integrable \(\big(\phi\in L^1(\mathbb R)\big).\) When \(z\) approaches \(x_0\) on the real axis from above and below, the limits \(f(x_0\pm i\epsilon)\) differ by \(2\pi i\phi(x_0)\); moreover,
analytic continuation across the real axis is possible in a neighborhood of \(x_0\) if and only if \(\phi\) is (real) analytic at \(x_0\). Thus, by choosing appropriate \(\phi\), we can construct
many functions on the upper (or lower) half plane that cannot be analytically continued across part or all of the real line. Another way for a function to fail to analytically continue past a
boundary is when the value of the function approaches infinity along the boundary. An important class of functions of this sort is modular forms, which are defined on the upper half plane, and
have very stringent transformation properties; they have deep connections with \(L\)-functions, and are likewise important in many areas of mathematics, most notably in number theory.
Definitions of Holomorphic and Meromorphic Functions
Methods of Analytic Continuation
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NBC calculation of genus-specific conditional probability
See also
RDP Naive Bayesian Classifier algorithm
As a starting point, the simplest estimate is the frequency observed in the training set = m(w_i)/M, and it would make sense to add pseudo-counts to model unobserved sequences, but I don't understand
why they chose specifically to add P_i in the numerator or add 1 in the denominator.
Wang,Q. et al. (2007) Naive Bayesian classifier for rapid assignment of rRNA sequences into the new bacterial taxonomy. AEM 73, 5261-7.
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Transactions Online
Lu CHEN, Daping BI, Jifei PAN, "2-D DOA Estimation Based on Sparse Bayesian Learning for L-Shaped Nested Array" in IEICE TRANSACTIONS on Communications, vol. E102-B, no. 5, pp. 992-999, May 2019,
doi: 10.1587/transcom.2018EBP3232.
Abstract: In sparsity-based optimization problems for two dimensional (2-D) direction-of-arrival (DOA) estimation using L-shaped nested arrays, one of the major issues is computational complexity. A
2-D DOA estimation algorithm is proposed based on reconsitution sparse Bayesian learning (RSBL) and cross covariance matrix decomposition. A single measurement vector (SMV) model is obtained by the
difference coarray corresponding to one-dimensional nested array. Through spatial smoothing, the signal measurement vector is transformed into a multiple measurement vector (MMV) matrix. The signal
matrix is separated by singular values decomposition (SVD) of the matrix. Using this method, the dimensionality of the sensing matrix and data size can be reduced. The sparse Bayesian learning
algorithm is used to estimate one-dimensional angles. By using the one-dimensional angle estimations, the steering vector matrix is reconstructed. The cross covariance matrix of two dimensions is
decomposed and transformed. Then the closed expression of the steering vector matrix of another dimension is derived, and the angles are estimated. Automatic pairing can be achieved in two
dimensions. Through the proposed algorithm, the 2-D search problem is transformed into a one-dimensional search problem and a matrix transformation problem. Simulations show that the proposed
algorithm has better angle estimation accuracy than the traditional two-dimensional direction finding algorithm at low signal-to-noise ratio and few samples.
URL: https://global.ieice.org/en_transactions/communications/10.1587/transcom.2018EBP3232/_p
author={Lu CHEN, Daping BI, Jifei PAN, },
journal={IEICE TRANSACTIONS on Communications},
title={2-D DOA Estimation Based on Sparse Bayesian Learning for L-Shaped Nested Array},
abstract={In sparsity-based optimization problems for two dimensional (2-D) direction-of-arrival (DOA) estimation using L-shaped nested arrays, one of the major issues is computational complexity. A
2-D DOA estimation algorithm is proposed based on reconsitution sparse Bayesian learning (RSBL) and cross covariance matrix decomposition. A single measurement vector (SMV) model is obtained by the
difference coarray corresponding to one-dimensional nested array. Through spatial smoothing, the signal measurement vector is transformed into a multiple measurement vector (MMV) matrix. The signal
matrix is separated by singular values decomposition (SVD) of the matrix. Using this method, the dimensionality of the sensing matrix and data size can be reduced. The sparse Bayesian learning
algorithm is used to estimate one-dimensional angles. By using the one-dimensional angle estimations, the steering vector matrix is reconstructed. The cross covariance matrix of two dimensions is
decomposed and transformed. Then the closed expression of the steering vector matrix of another dimension is derived, and the angles are estimated. Automatic pairing can be achieved in two
dimensions. Through the proposed algorithm, the 2-D search problem is transformed into a one-dimensional search problem and a matrix transformation problem. Simulations show that the proposed
algorithm has better angle estimation accuracy than the traditional two-dimensional direction finding algorithm at low signal-to-noise ratio and few samples.},
TY - JOUR
TI - 2-D DOA Estimation Based on Sparse Bayesian Learning for L-Shaped Nested Array
T2 - IEICE TRANSACTIONS on Communications
SP - 992
EP - 999
AU - Lu CHEN
AU - Daping BI
AU - Jifei PAN
PY - 2019
DO - 10.1587/transcom.2018EBP3232
JO - IEICE TRANSACTIONS on Communications
SN - 1745-1345
VL - E102-B
IS - 5
JA - IEICE TRANSACTIONS on Communications
Y1 - May 2019
AB - In sparsity-based optimization problems for two dimensional (2-D) direction-of-arrival (DOA) estimation using L-shaped nested arrays, one of the major issues is computational complexity. A 2-D
DOA estimation algorithm is proposed based on reconsitution sparse Bayesian learning (RSBL) and cross covariance matrix decomposition. A single measurement vector (SMV) model is obtained by the
difference coarray corresponding to one-dimensional nested array. Through spatial smoothing, the signal measurement vector is transformed into a multiple measurement vector (MMV) matrix. The signal
matrix is separated by singular values decomposition (SVD) of the matrix. Using this method, the dimensionality of the sensing matrix and data size can be reduced. The sparse Bayesian learning
algorithm is used to estimate one-dimensional angles. By using the one-dimensional angle estimations, the steering vector matrix is reconstructed. The cross covariance matrix of two dimensions is
decomposed and transformed. Then the closed expression of the steering vector matrix of another dimension is derived, and the angles are estimated. Automatic pairing can be achieved in two
dimensions. Through the proposed algorithm, the 2-D search problem is transformed into a one-dimensional search problem and a matrix transformation problem. Simulations show that the proposed
algorithm has better angle estimation accuracy than the traditional two-dimensional direction finding algorithm at low signal-to-noise ratio and few samples.
ER -
|
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Current Research Interests
Over the past few years heavy-tailed phenomena have attracted the interest of various researchers in time series analysis, extreme value theory, econometrics, telecommunications, and various other
fields. The need to consider time series with heavy-tailed distributions arises from the observation that traditional models of applied probability theory fail to describe jumps, bursts, rapid
changes and other erratic behavior of various real-life time series. In joint work with Thomas Mikosch, we have studied tail behavior of the marginal distributions for a host of nonlinear time series
models such as GARCH and stochastic volatility models. These processes, which are commonly used for modeling financial time series consisting of log-returns, tend to have Pareto-like tails. The
relationship between tail-heaviness and serial dependence in the data has been particularly intriguing. In order to detect nonlinearities in financial time series, econometricians often recommend
examining the autocorrrelation function (ACF) of not only the time series itself, but also powers of the absolute values. On the other hand, it is believed that many financial time series have
infinite third, fourth or fifth moments. If the process has an infinite fourth moment, then the variance of the squares of the process does not exist, in which case the ACF may be of little
diagnostic value. The theoretical development that Mikosch and I have provided for the sample ACF confirms that one must view the graphs of the ACF with extreme caution.
Allpass time series are special cases of ARMA processes that exhibit interesting features. First, the allpass process is uncorrelated. Second, as long as the process is nonGaussian, the process is
not independent. This can often be seen from inspection of the ACF of the squares of the process. So in many respects, allpass processes mimic properties-lack of serial correlatation and bursty
behavior-that are often associated with the nonlinear time series models in finance. In work with Jay Breidt and graduate student Beth Andrews, we are exploring identification and efficient
estimation procedures for fitting allpass models. Allpass models are widely used in the engineering literature for fitting noncausal and noninvertible models. Two-stage procedures for detecting
noncausality or noninvertiblity are currently under investigation.
The modeling of time series of count data often requires models and techniques that go beyond classical time series analysis. The need for a flexible class of models for count data that can be easily
fitted is clearly demonstrated in applications such as modeling of disease incidence, as for example in the modelling of polio counts in the U.S., and the number of asthma presentations at an
emergency room in a hospital. Other emerging areas of application include finance, where the response variable is the number of transactions that occur in a small time interval; and spatial-temporal
modeling in ecology, where the response might correspond to the number of rare species in a given region at a fixed time. In work with William Dunsmuir, Ying Wang, and Sarah Streett, we have
considered two types of models, parameter-driven and observation-driven, for time series of counts. The fitting of parameter-driven models is often computationally intensive especially if a large
number of explanatory variables is included in the model. While observation-driven models tend to be easier to fit, their theoretical properties can be difficult to derive. Addressing these issues
has been a constant theme in our research.
Spatial-temporal modeling is a natural offshoot of time series analysis. Together with colleagues at CSU, we have formed Space-Time Aquatic Resources Modeling and Analysis Program (STARMAP) that is
supported by a 4-year EPA-STAR grant. Although many of the ideas used in time series extend directly to certain aspects of spatial modeling, we face new sets of modeling challenges. Some of these
challenges include the development of models with explanatory variables measured on different scales, variable selection, and defining a dependence metric (function) for data observed on a network of
Reports & Papers
Short Vita
Vita (PDF format)
Invited Talks
Current Research
Davis Home
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The Gate Could Be Closing on Future Hall of Fame Era Committee Inductees
Democrat and Chronicle
This weekend in Cooperstown, six Era Committee candidates will be inducted alongside the BBWAA-elected David Ortiz. Among them are some of the most long-awaited honorees whose supporters agonized for
decades over their being shut out, both before and after their deaths. Negro Leagues player/manager/scout/coach/ambassador Buck O’Neil and Negro Leagues and American League star Minnie Miñoso both
hung on well into their 90s hoping they could see the day of their induction but died before it happened. Star first baseman and manager Gil Hodges died of a heart attack at age 47, before his
candidacy became the ultimate “close-but-no-cigar” example, both via the BBWAA and Veterans Committee processes. Black baseball pioneer Bud Fowler, who was raised in Cooperstown, went largely
unrecognized until the centennial of his death in 2013. Tony Oliva and Jim Kaat, both of whom are 84, are thankfully alive to experience the honor, but they, too, had a long wait, after falling one
and two votes short, respectively, on the 2015 Golden Era ballot.
The festivities will be tinged with more than a hint of bittersweetness due to the deferred honors, but there won’t be any shortage of joy and catharsis that these men are finally being recognized.
Yet even as they take place, it feels as though a gate is swinging shut behind them — one that may not open again for awhile given the the shakeup of the Era Committee process that the Hall announced
in April which reduced the numbers of committees, candidates, and votes available. I won’t rehash the road to this point (you can see the gory details in the aforementioned link), but here’s the new
format, which will roll out in this order over the next three years starting in December:
• December 2022 (for Class of 2023): Contemporary Baseball – Players. For those who made their greatest impact upon the game from 1980 onward and have aged off the BBWAA ballot.
• December 2023 (for Class of 2024): Contemporary Baseball – Managers, Umpires, and Executives. For those who made their greatest impact upon the game from 1980 to the present day.
• December 2024 (for Class of 2025): Classic Baseball. For those who made their greatest impact upon the game before 1980, including Negro Leagues and pre-Negro Leagues Black players
The Classic Baseball Era Committee now has purview over all of the candidates previously covered by the Early Baseball (1871–1949) and Golden Days (1950–69) committees — the two that produced this
weekend’s honorees and which otherwise weren’t scheduled to convene again for 10 and five years, respectively — as well as about half of those covered by the Modern Baseball (1970–87) one. In other
words, voters for that ballot now have to weigh candidates whose contributions may have taken place over a century apart. What’s more, where there were 10 candidates apiece for each of those ballots
under the older system, the new ones contain only eight, and where the 16 committee members (a mixture of Hall of Famers, executives, and writers/historians) could previously vote for four of those
10 candidates, that number has been reduced to three. Candidates will still need to receive a minimum of 75% of votes to be elected.
In other words, there’s a new bottleneck in place for the older candidates, and it has happened just as the Negro Leagues and pre-Negro Leagues candidates — players and non-players alike — finally
returned to eligibility after the books were closed on that period following the aforementioned 2006 election, which produced 17 honorees but froze out O’Neil. For those who make it to the ballot,
the math that was already very tough is undeniably tougher. Instead of a maximum of 64 votes spread across 10 candidates (an average of 6.4 per candidate), there are now 48 spread across eight
candidates (six per candidate). Electing four candidates from a single slate, which happened for the first time on the 2022 Golden Days ballot, would require each of those four to receive exactly 12
About that math: When I wrote about the changes in April, I linked to a December 2014 piece by Joe Posnanski, written on the occasion of the voters’ shutout of the Golden Era Committee. That decision
slammed the door in the faces of Miñoso and Dick Allen one final time before each passed away; also falling short on that ballot were Kaat, Oliva, and Hodges. All but Allen were elected in December
(he missed by one vote!), and with the exception of Miñoso, whose short but stellar Negro Leagues stats are now included in his major league career totals, it’s not like their credentials improved.
Anyway, in that piece, Posnanski brought up a name that’s particularly familiar around these parts:
Back to the math. Tom Tango explains it this way: Let’s say all ten candidates on the ballot were equally qualified for the Hall of Fame. That’s not quite true here, but it’s a good starting
point — you had 10 good candidates. If they’re all equally good candidates, then each one had a 40% chance of getting picked for a ballot — 10 players on the ballot, voter chooses four, 40%
chance. Pretty simple.
Well, if a player has a 40% chance of being on one ballot, his chances on making 12 of 16 is … get ready for it, less than 0.5%. That’s not 5% — it is less than one-half of one-percent. 995 times
out of a 1,000, the player would NOT get elected. And remember, that’s assuming every voter uses all four of his votes.
In light of this, I asked Dan Szymborski, our staff probability expert, if he could help me figure out a way to illustrate the impact of the changes. He rose to the occasion by creating a Monte Carlo
simulation, a model along the lines of what we use to calculate our Playoff Odds. In this case, he ran one million simulations for various scenarios.
For starters, using Tango’s simple estimate from randomness, when all 10 candidates are equal and thus have an equal chance (40%) of being named on a four-slot ballot, the yield is an average of
0.049 inductees per year, and the committee elects no one in 95.2% of the simulations. Lower that to eight candidates, with each candidate having a 37.5% chance of being named on a three-slot ballot,
and the yield drops to 0.021 inductees per year — basically from one every 20 years to one every 48 years — with nobody elected 97.7% of the time.
We know that not all candidates are created equal; some have better numbers and more impressive accomplishments than others and are more likely to capture the voters’ attention. It’s easy to look at
an Era Committee ballot and identify a few candidates who are basically ballast — guys who have been up for election several times before but have never or rarely gotten close and are likely to be
overshadowed. On the other end of the spectrum are guys who might be in their first appearance in this format and scan as the most likely honorees, particularly if they have an easy hook. Think Jack
Morris with his 254 wins, Harold Baines with his 2,866 hits, or Fred McGriff with his 493 homers.
Suppose, for example, that we take 10 candidates of varying strengths and thus various probabilities of being named on an individual ballot. In this scenario, the best candidate has a 72% chance of
appearing on a single ballot. That doesn’t mean that he’s going to get 72% of the vote every time, but that he’ll receive an average of 0.72 x 16 = 11.52 votes per simulation, sometimes more — enough
to be elected — and sometimes less. Each of the nine other candidates has odds of appearing that are about 7% less than the candidate above him in the rankings, thus accounting for all 64 possible
voting slots:
10-Candidate Model, 4 Votes Per Ballot
Candidate Rk Ballot Odds Votes Per Sim
1 72% 11.5
2 65% 10.4
3 58% 9.3
4 51% 8.2
5 44% 7.0
6 36% 5.8
7 29% 4.6
8 22% 3.5
9 15% 2.4
10 8% 1.3
Total 64.0
Running those odds through Dan’s Monte Carlo simulation, this scenario produces an average of 0.989 inductees per year, with a shutout 28.2% of the time.
Now, if you then take the top eight of those candidates and reduce their shares proportionally to account for the fewer voting slots (48, via 16 voters with three slots apiece), the top odds start
around 57% like so:
8-Candidate Model, 3 Votes Per Ballot
Candidate Rk Ballot Odds Votes Per Sim
1 57.4% 9.2
2 51.7% 8.3
3 46.0% 7.4
4 40.4% 6.5
5 34.7% 5.6
6 29.0% 4.6
7 23.4% 3.7
8 17.7% 2.8
Total 48.0
That scenario reduces the average number of inductees per year from 0.989 to 0.195, with a shutout 81.5% of the time. Eep! However, if the top player on the first ballot remains the favorite and
still has 72% odds of appearing on a single ballot because his candidacy is so strong, with the shares of the others votes each reduced…
8-Candidate Model, 3 Votes Per Ballot
Candidate Rk Ballot Odds Votes Per Sim
1 72.0% 11.5
2 62.0% 9.9
3 52.0% 8.3
4 42.0% 6.7
5 33.0% 5.3
6 23.0% 3.7
7 13.0% 2.1
8 3.0% 0.5
Total 48.0
…then the drop is only to 0.801 inductees per year, with a 35.2% chance of a shutout. In other words, the changes in the number of candidates and the number of votes per ballot reduce the yield by a
lot, but the amount of the reduction depends on the individual players; a candidate who stands head and shoulders among the rest makes the election likely to produce an honoree.
So what happens if, like the 2022 Golden Days ballot, a subset of candidates are clearly favored ahead of the rest? In that particularly case, the top five vote-getters (Miñoso with 14, Hodges, Kaat,
and Oliva each with 12, and Allen with 11) accounted for 61 of the possible 64 votes, with at most three distributed among the other five candidates (their shares were reported as “three or fewer
votes”). If you have five out of 10 candidates who have a 70% chance of appearing on a ballot, and the other five with a 10% chance (accounting for all 64 votes), the yield is 2.25 inductees per
year, with a shutout just 5% of the time. If you have a similar split on an eight-candidate ballot, with four candidates having 70% chances and four with 5% chances, you get 1.8 inductees per year,
with a shutout 9.2% of the time. As you’d expect, it’s the agreement among the voters — the consensus coalescing around a smaller subset of candidates — that’s the largest factor in determining the
When I spoke to Hall president Josh Rawich about the changes in April, he conceded that the new format makes it “more challenging to get on a ballot.” I suggested something to the effect that it
would reduce the number of reheated candidacies (my term, not his, it should be pointed out). “There was definitely a feeling [among the Hall’s board members] that we wanted to make sure that we’re
not looking at a lot of the same players every single time,” he replied. “Once somebody’s had a chance to be reviewed a number of times, it’s time to let somebody else get looked at.”
The problem, to relate it to the modeling above, is that if you’re attempting to get rid of the candidates with the 5% or 10% odds, you’re going to create either a new crop of those same types
because somebody will inevitably trickle to the bottom, or a flatter distribution of the odds and therefore a much lower yield. Here are two more eight-candidate scenarios:
8-Candidate Model, 3 Votes Per Ballot, Flatter Distributions
Candidate Rk Ballot Odds Votes Per Sim Ballot Odds Votes Per Sim
1 60.0% 9.6 50.0% 8.0
2 60.0% 9.6 50.0% 8.0
3 50.0% 8.0 50.0% 8.0
4 40.0% 6.4 50.0% 8.0
5 25.0% 4.0 25.0% 4.0
6 25.0% 4.0 25.0% 4.0
7 20.0% 3.2 25.0% 4.0
8 20.0% 3.2 25.0% 4.0
Total 48.0 48.0
In the first scenario, the yield is 0.377, with a shutout in 66.5% of the simulations. In the second, the yield plunges to 0.154, and the shutout happens 85.5% of the time! This ought to be a
concern. Particularly with the handful of candidates who were perched on the precipice of election now cleared, similar consensus might be harder to come by now, as it’s not simply “next man up” for
who gets elected (acolytes of Allen and Hodges in particular can testify to that). It’s not hard to imagine a Classic Baseball slate containing such disparate candidates as long-dead Negro Leaguers
whom nobody on the committee witnessed first-hand and whose statistics are incomplete (say, barnstorming pioneer John Donaldson or fireballer Dick Redding, whose career crossed from the pre-Negro
Leagues Black baseball era into that of the major Negro Leagues) alongside still-living ones who the voters remember vividly and for whom the visual and statistical records are more fleshed out (Luis
Tiant, perhaps). The process could easily grind to a halt without anybody honored.
Some amount of polarization is necessary to elect at least one candidate. To show this another way, here’s a table of probability of a single candidate getting 12 votes out of 16 (75%) when he has an
X% chance of being on any random ballot in an eight-vote format, independent of the odds of the other candidates.
Probability of 12+ Votes Based on X% Ballot Odds
pRandom Ballot At Least 12 Votes
99% 100.0%
96% 100.0%
93% 99.6%
90% 98.3%
87% 95.3%
84% 90.1%
81% 82.7%
78% 73.5%
75% 63.0%
72% 52.1%
69% 41.5%
66% 31.9%
63% 23.5%
60% 16.7%
57% 11.3%
54% 7.4%
51% 4.6%
48% 2.7%
45% 1.5%
42% 0.8%
39% 0.4%
36% 0.2%
33% 0.1%
30% 0.0%
27% 0.0%
24% 0.0%
21% 0.0%
18% 0.0%
15% 0.0%
12% 0.0%
9% 0.0%
6% 0.0%
3% 0.0%
In graph form, that looks like a titration curve from a college chemistry lab:
As you can see, at either extreme, a change in the odds has no effect, but in the middle, the odds of election quickly increase, nearly tripling as the individual ballot odds climb from 57% to 66%
and then nearly doubling as they climb from 66% to 75%.
Obviously, we can’t model every scenario. Still, I hope that this exercise helps to convey how the changes to the process, even if they’re well-intentioned — and I believe that the continued
re-evaluation of the segregation-era candidates is laudable — actually make it much harder to produce honorees and increases the likelihood of the shutouts that frustrated observers and led to the
Hall rejiggering the process in the first place.
This is not to suggest that having a substandard honoree is better than having none at all, and that the process must be re-engineered to produce one every time (say, a runoff between the top
candidates along the lines of what BBWAA voters did a few times). I feel confident that my two decades of evaluating committee processes amply illustrates the continued presence of strong but
overlooked candidates who land on committee ballots, not that their mere presence guarantees optimal outcomes. We should still hope for processes that preserve the likelihood of such candidates being
recognized, but with Dan’s help, I believe we’ve demonstrated that what’s about to be put in place decreases those odds.
Brooklyn-based Jay Jaffe is a senior writer for FanGraphs, the author of The Cooperstown Casebook (Thomas Dunne Books, 2017) and the creator of the JAWS (Jaffe WAR Score) metric for Hall of Fame
analysis. He founded the Futility Infielder website (2001), was a columnist for Baseball Prospectus (2005-2012) and a contributing writer for Sports Illustrated (2012-2018). He has been a recurring
guest on MLB Network and a member of the BBWAA since 2011, and a Hall of Fame voter since 2021. Follow him on Twitter @jay_jaffe... and BlueSky @jayjaffe.bsky.social.
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UGate (v0.45) | IBM Quantum Documentation
class qiskit.circuit.library.UGate(theta, phi, lam, label=None, *, duration=None, unit='dt')
Bases: Gate
Generic single-qubit rotation gate with 3 Euler angles.
Can be applied to a QuantumCircuit with the u() method.
Circuit symbol:
q_0: ┤ U(ϴ,φ,λ) ├
Matrix Representation:
$\providecommand{\rotationangle}{\frac{\theta}{2}} U(\theta, \phi, \lambda) = \begin{pmatrix} \cos\left(\rotationangle\right) & -e^{i\lambda}\sin\left(\rotationangle\right) \\ e^{i\phi}\sin\left(\
rotationangle\right) & e^{i(\phi+\lambda)}\cos\left(\rotationangle\right) \end{pmatrix}$
$U\left(\theta, -\frac{\pi}{2}, \frac{\pi}{2}\right) = RX(\theta)$$U(\theta, 0, 0) = RY(\theta)$
Create new U gate.
Get the base class of this instruction. This is guaranteed to be in the inheritance tree of self.
The “base class” of an instruction is the lowest class in its inheritance tree that the object should be considered entirely compatible with for _all_ circuit applications. This typically means that
the subclass is defined purely to offer some sort of programmer convenience over the base class, and the base class is the “true” class for a behavioural perspective. In particular, you should not
override base_class if you are defining a custom version of an instruction that will be implemented differently by hardware, such as an alternative measurement strategy, or a version of a
parametrised gate with a particular set of parameters for the purposes of distinguishing it in a Target from the full parametrised gate.
This is often exactly equivalent to type(obj), except in the case of singleton instances of standard-library instructions. These singleton instances are special subclasses of their base class, and
this property will return that base. For example:
>>> isinstance(XGate(), XGate)
>>> type(XGate()) is XGate
>>> XGate().base_class is XGate
In general, you should not rely on the precise class of an instruction; within a given circuit, it is expected that Instruction.name should be a more suitable discriminator in most situations.
The classical condition on the instruction.
Get the decompositions of the instruction from the SessionEquivalenceLibrary.
Return definition in terms of other basic gates.
Is this instance is a mutable unique instance or not.
If this attribute is False the gate instance is a shared singleton and is not mutable.
Return the number of clbits.
Return the number of qubits.
return instruction params.
Get the time unit of duration.
control(num_ctrl_qubits=1, label=None, ctrl_state=None)
Return a (multi-)controlled-U gate.
• num_ctrl_qubits (int) – number of control qubits.
• label (str or None) – An optional label for the gate [Default: None]
• ctrl_state (int orstr or None) – control state expressed as integer, string (e.g. ‘110’), or None. If None, use all 1s.
controlled version of this gate.
Return type
Return inverted U gate.
$U(\theta,\phi,\lambda)^{\dagger} =U(-\theta,-\lambda,-\phi)$)
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Giving mass to other particles?
• I
• Thread starter Xforce
• Start date
In summary: This process still occurs in the universe now, giving mass to particles that interact with the Higgs field. However, it does not give mass to particles that do not interact with the Higgs
field, such as photons. In summary, the Higgs bosons are heavy and unstable particles that can be created in particle accelerators like the LHC. They are responsible for giving mass to other
particles in the Standard Model through spontaneous symmetry breaking. The Higgs boson itself is a result of this process and does not carry any specific force, unlike other bosons. This process
still occurs in the universe today, giving mass to particles that interact with the Higgs field.
TL;DR Summary
They say at the Big Bang all the particles does not have mass. It’s the Higgs Bosons give them mass...
Higgs bosons are very heavy particles (probably 1000 times heavier than a protons) and very unstable. Now we can create them in particle accelerators like LHC, like countless of other particles.
But wait. This one can give mass to particles without mass, does this violate the conservation of mass or energy? Or the laws of physics is different at the beginning of time? Also I heard a Boson
particle is the one usually carries a force (like gravity, electromagnetic forces and nuclear force). If Higgs boson was a boson, what kind of force does it carry? And what makes it capable of
bringing mass? Does this process still work in the universe now (like giving mass to photons)?
Xforce said:
Summary: They say at the Big Bang all the particles does not have mass. It’s the Higgs Bosons give them mass...
This one can give mass to particles without mass, does this violate the conservation of mass or energy? Or the laws of physics is different at the beginning of time?
The laws of physics were the same back then as they are now. Strictly speaking none of the particles in the SM are massive in the traditional sense (dirac mass, quadratic scalar mass), it's only when
you take the weak field limit in the Higgs doublet that you get terms that look like mass terms.
Xforce said:
Summary: They say at the Big Bang all the particles does not have mass. It’s the Higgs Bosons give them mass...
If Higgs boson was a boson, what kind of force does it carry? And what makes it capable of bringing mass? Does this process still work in the universe now (like giving mass to photons)?
The "force" bosons originate from lorentz vector fields while the higgs comes from a lorentz scalar (technically a lorentz doublet under ##SU(2)_L##), this makes a world of difference as the higgs
field doesn't implement a local gauge symmetry like the vector bosons do. In a sense I guess you could consider the higgs to be a "force" in the sense that it can mediate interactions.
Xforce said:
It’s the Higgs Bosons give them mass...
, not the boson itself.
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2023 Award
Xforce said:
Who says?
Xforce said:
Higgs bosons are very heavy particles (probably 1000 times heavier than a protons)
Where did you read this?
Xforce said:
Xforce said:
Summary: They say at the Big Bang all the particles does not have mass. It’s the Higgs Bosons give them mass...
No, that's not what "they" say.
Particles gain mass in the Standard Model via spontaneous symmetry breaking, which involves the Higgs
. The Higgs
that was observed in the LHC is what is left over from the Higgs field
the spontaneous symmetry breaking and the gaining of mass by other particles. The observed mass of the Higgs boson is therefore separate from and not connected to the masses of the other particles.
FAQ: Giving mass to other particles?
1. How is mass given to other particles?
The mass of particles is given by the Higgs field, which is an energy field that permeates throughout the entire universe. When particles interact with this field, they acquire mass.
2. What is the role of the Higgs boson in giving mass to other particles?
The Higgs boson is a particle that is associated with the Higgs field. It is responsible for giving mass to other particles by interacting with them and transferring energy to them, thus giving them
3. Are all particles given mass by the Higgs field?
No, not all particles are given mass by the Higgs field. The Higgs field only interacts with certain types of particles, such as the W and Z bosons, and the fermions (quarks and leptons).
4. Can mass be created or destroyed in this process?
No, mass cannot be created or destroyed in this process. The Higgs field only gives particles mass, it does not create or destroy it. The total mass of a closed system remains constant.
5. How does the Higgs field affect the behavior of particles?
The Higgs field affects the behavior of particles by giving them mass. This mass affects the way particles interact with each other and determines their motion and behavior in the universe.
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Calculating the Mean Absolute Deviation (MAD) in Excel on Windows 11 - Support Your Tech
Calculating the Mean Absolute Deviation (MAD) in Excel on Windows 11
Calculating the Mean Absolute Deviation (MAD) in Excel on Windows 11 is a straightforward process that can be done in just a few steps. By the end of this article, you’ll know exactly how to compute
the MAD for a set of data using Excel’s built-in functions and some simple formulas. Let’s dive in!
Step by Step Tutorial: Calculating the Mean Absolute Deviation in Excel
Before we get started, it’s important to know that the Mean Absolute Deviation (MAD) is a measure of variability that gives an idea of how spread out a set of numbers is. It’s the average distance
each data point is from the mean of the dataset. Now, let’s go through the steps to calculate it in Excel on Windows 11.
Step 1: Enter Your Data
Start by entering your dataset into a column in Excel.
When you enter your data, make sure that each value is in its own cell in a single column. This will make it easier to apply formulas later on.
Step 2: Calculate the Mean
Use the AVERAGE function to calculate the mean of your dataset.
To calculate the mean, simply click on an empty cell and type =AVERAGE(), then select the range of cells that contain your data. Press Enter, and the mean will be displayed in the cell where you
typed the formula.
Step 3: Calculate Deviations from the Mean
In a new column, subtract the mean from each data point to find the deviations.
Click on the cell next to your first data point. Type the formula to subtract the mean (calculated in Step 2) from the data point. Copy this formula down the column to find the deviation for each
data point.
Step 4: Take the Absolute Values
Convert the deviations to absolute values using the ABS function.
Once you have the deviations, you’ll need to ignore whether they’re positive or negative—this is where the ABS function comes in handy. Click on a new cell and type =ABS(), then select the cell with
the deviation. Copy this formula down to apply it to all deviations.
Step 5: Calculate the Mean Absolute Deviation
Finally, calculate the average of these absolute deviations to get the MAD.
In an empty cell, type =AVERAGE() and then select the range of cells with the absolute deviations. This will give you the Mean Absolute Deviation of your dataset.
After completing these steps, you’ll have the Mean Absolute Deviation for your dataset, giving you a better understanding of its variability.
Tips for Calculating the Mean Absolute Deviation in Excel
• Make sure your data is clean, with no empty cells or non-numeric values in the range being used for calculations.
• Double-check your formulas to ensure that they cover the correct range of cells.
• Remember that the Mean Absolute Deviation is different from the standard deviation, although they both measure variability.
• Use cell referencing (like A1 or B2) to make your formulas easier to read and understand.
• Explore Excel’s help and resources for more information on the functions used in the steps above.
Frequently Asked Questions
What is the difference between MAD and standard deviation?
The Mean Absolute Deviation (MAD) measures average distance from the mean, while standard deviation measures how spread out the numbers are in a dataset.
Can I calculate MAD for a sample and not just for a population?
Yes, MAD can be calculated for both a sample and a population, just ensure you’re using the correct dataset.
Is there a direct function in Excel for MAD?
No, there isn’t a direct function for MAD in Excel, but it’s easy to calculate with the steps provided.
Why do we use the ABS function in calculating MAD?
We use the ABS function to get the absolute value of deviations because the direction of the deviation (positive or negative) is not important when calculating MAD.
Can I calculate MAD for multiple datasets at once in Excel?
Yes, you can calculate MAD for multiple datasets by repeating the steps for each set in separate columns or sheets.
1. Enter your data into a column.
2. Calculate the mean using the AVERAGE function.
3. Calculate the deviations by subtracting the mean from each data point.
4. Convert deviations to absolute values using the ABS function.
5. Calculate the average of these absolute deviations to get the MAD.
There you have it, folks – you’re now equipped to calculate the Mean Absolute Deviation in Excel on Windows 11 like a pro! Whether you’re a student crunching numbers for a project, a professional
analyzing data, or simply a data enthusiast, understanding how to compute MAD is a valuable skill that adds depth to your analytical toolkit. It’s a testament to the power of Excel that such complex
statistical measures can be derived with just a handful of simple functions. With the guidelines laid out in this article, you’re all set to unlock insightful data trends and elevate your data
analysis game to the next level. So, go ahead, give it a try, and watch the magic of numbers unfold before your eyes. Happy analyzing!
Matt Jacobs has been working as an IT consultant for small businesses since receiving his Master’s degree in 2003. While he still does some consulting work, his primary focus now is on creating
technology support content for SupportYourTech.com.
His work can be found on many websites and focuses on topics such as Microsoft Office, Apple devices, Android devices, Photoshop, and more.
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Ball Mill Design/Power Calculation - 911Metallurgist
The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are; material to be ground, characteristics, Bond Work Index, bulk density, specific density,
desired mill tonnage capacity DTPH, operating % solids or pulp density, feed size as F80 and maximum ‘chunk size’, product size as P80 and maximum and finally the type of circuit open/closed you are
designing for.
Use this online calculators for Motor Sizing and Mill sizing as well as mill capacity determinators.
In extracting from Nordberg Process Machinery Reference Manual I will also provide 2 Ball Mill Sizing (Design) example done ‘by-hand’ from tables and charts. Today, much of this mill designing is
done by computers, power models and others. These are a good back-to-basics exercises for those wanting to understand what is behind or inside the machines.
1. Ball Mill Design/Sizing Calculator
The power required to grind a material from a given feed size to a given product size can be estimated by using the following equation:
W = power consumption expressed in kWh/short to (HPhr/short ton = 1.34 kWh/short ton)
Wi = work index, which is a factor relative to the kwh/short ton required to reduce a given material from theoretically infinite size to 80% passing 100 microns
P = size in microns of the screen opening which 80% of the product will pass
F = size in microns of the screen opening which 80% of the feed will pass
When the above equation is used, the following points should be borne in mind:
• The values of P and F must be based on materials having a natural particle size distribution.
• The power consumption per short ton will only be correct for the specified size reduction when grinding wet in closed circuit. If the method of grinding is changed, power consumption also changes
as follows:
1. Closed Circuit = W
2. Open Circuit, Product Top-size not limited = W
3. Open Circuit, Product Top-size limited = W to 1.25W
Open circuit grinding to a given surface area requires no more power than closed circuit grinding to the same surface area provided there is no objection to the natural top-size. If top-size must be
limited in open circuit, power requirements rise drastically as allowable top-size is reduced and particle size distribution tends toward the finer sizes.
• The work index, Wi, should be obtained from test results or plant data, where the feed and product size distributions are as close as possible to those under sturdy.
The work index, Wi, will vary considerably for materials that appear to be very similar. The work index will also have a considerable variation across one ore body or deposit.
The most reliable work index values are those obtained from long term operating data. If this is not available, standard grindability tests can be run to provide approximate values.
Rod and ball mill grindability test results should only be applied to their respective methods of grinding.
Ball Mill Power Calculation Example #1
A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80% passing ¼ inch (6350 microns). The required product size
distribution is to be 80% passing 100 mesh (149 microns). In order to determine the power requirement, the steps are as follows:
Example Calculation
A motor with around 1400 Horse Power is calculated needed for the designed task. Now we much select a Ball Mill that will draw this power.
The ball mill motor power requirement calculated above as 1400 HP is the power that must be applied at the mill drive in order to grind the tonnage of feed from one size distribution. The following
shows how the size or select the matching mill required to draw this power is calculated from known tables ‘the old fashion way’.
Here is a section of a mill in operation. The power input required to maintain this condition is theoretically:
The value of the angle “a” varies with the type of discharge, percent of critical speed, and grinding condition. In order to use the preceding equation, it is necessary to have considerable data on
existing installations. Therefore, this approach has been simplified as follows:
Five basics conditions determine the horsepower drawn by a mill:
1. Diameter
2. Length
3. % Loading
4. Speed
5. Mill type
These conditions have been built into factors which are given in the figure above. The approximate horsepower HP of a mill can be calculated from the following equation:
HP = (W) (C) (Sin a) (2π) (N)/ 33000
W = weight of charge
C = distance of centre of gravity or charge from centre of mill in feet
a = dynamic angle of repose of the charge
N = mill speed in RPM
HP = A x B x C x L
A = factor for diameter inside shell lining
B = factor which includes effect of % loading and mill type
C = factor for speed of mill
L = length in feet of grinding chamber measured between head liners at shell- to-head junction
Many grinding mill manufacturers specify diameter inside the liners whereas others are specified per inside shell diameter. (Subtract 6” to obtain diameter inside liners.) Likewise, a similar
confusion surrounds the length of a mill. Therefore, when comparing the size of a mill between competitive manufacturers, one should be aware that mill manufacturers do not observe a size convention.
Ball Mill Power/Design Calculation Example #2
In Example No.1 it was determined that a 1400 HP wet grinding ball mill was required to grind 100 TPH of material with a Bond Work Index of 15 (guess what mineral type it is) from 80% passing ¼ inch
to 80% passing 100 mesh in closed circuit. What is the size of an overflow discharge ball mill for this application?
Contact http://www.metso.com/industries/mining/ if you need a Large Ball Mill.
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How to select a capacitor bank for power factor correction? - Gruppo Energia
How to select a capacitor bank for power factor correction?
It is essential to choose the correct power factor correction panel for your system. In fact, as you already know, there must be a balance between the types of power present in your network,
otherwise you risk an excess of reactive energy (which is charged extra by utility companies).
The first step is the measurement of the cosΦ
There are several ways to measure your cosΦ:
1. The first, which is also the simplest, is to check your electricity bill. Many utility electric companies indicate this parameter among other information.
2. The second is to use a power analyzer. This solution is certainly the most useful. In fact, in addition to the indication of your cosΦ, a power analyser will provide you with all the information
regarding your network, such as the presence of harmonics, the consumption of electricity and the phase balance.
3. The third is to use a power factor meter. In this case the instrument will directly indicate the PF (power factor) value. As previously explained, the PF = cosΦ if there are no non-linear loads
(and hence harmonics) in your network.
4. The fourth way is to use reactive energy meters. In this case you will have to make some calculations, since the indicated values are only relative to energy.
To make it short, you have to consider four values: two at the beginning of the work cycle, and two at the end of the work cycle. The data to take in consideration are active and reactive energy.
Here is the formula to calculate the cosΦ:
CosΦ= cos(tan^-1((Eqf – Eqi)/(Epf – Epi))
• Eqf is reactive energy at the end of work cycle.
• Eqi is reactive energy at the begging of work cycle.
• Epf is active energy at the end of work cycle.
• Epi is active energy at the begging of work cycle.
The second step is the calculation of the necessary reactive power Q(c)
In order to calculate the reactive power that is necessary to improve power factor, it is important to know the active power of the load, the actual and the desired cosΦ.
We have already found the present cosΦ.
The desired cosΦ value depends on the electric company. In fact, this value can range from 0.90 to 0.97. Gruppo Energia automatic regulators are normally set to reach the cosΦ value of 0.99 in our
own capacitor banks. The active power of the machinery (load) to be compensated is normally indicated on their information plate.
Once we have all three parameters available, we use the following formula to determine the reactive power:
Q(c) = P x k
• Q(c) is the necessary reactive power (kVar).
• P is active power of the load to be compensated (kW).
• k is the conversation coefficient in the table you can find below. You can find it by crossing the value of the actual cosΦ and the desired cosΦ.
Calculation example:
Active power (load power) P: 100 kW
Actual cosΦ: 0,55
Desired cosΦ: 0,99
Coefficient from the table: 1,376
Q(c) = 100 x 1,376 = 137,6 kVar
As you can see, the reactive power to be compensated is 137,6 kVar, which can be approximated to 150 kVar considering that the loads might slightly increase in the future.
The third step is to choose between fixed or automatic correction
If your system is quite small, we normally suggest a fixed power factor correction bank. In this case the power of capacitors is fixed to supply the constant reactive power to the exiting load. Even
the load should not be changed.
When the loads that need to be compensated are variable (this currently happens in most cases), we suggest an automatic power factor correction.
The total power of the capacitor bank is divided in steps. These steps are controlled by a regulator which constantly analyses the network and operates the step with suitable power, in order to
compensate the load present at that moment. Here is a formula that can help you choose:
(Q(c) / S(n)) * 100 = %
Q(c) is the necessary reactive power (kVar).
S(n) is apparent power of installed transformer (kVA).
If the result is < 15% we suggest fixed compensation.
If the result is ≥ 15% we suggest automatic compensation.
The fourth step is deciding if you need a standard, heavy duty or detuned capacitor bank
The most correct and accurate way to do that is by using a network analyser, which will show you in detail in what condition your network is and how much it is afflicted by harmonics. In this case
you need to check the THD(i)% (total harmonic current distortion) and THD(u)% (total harmonic voltage distortion) values. Measurements must be made at full load and without connected capacitors at
the transformer secondary.
If THD(i)% ≤ 5% a standard PFC capacitor bank is usually enough;
If 5% < THD(i)% ≤ 10% a heavy duty PFC capacitor bank is suggested;
If 10% < THD(i)% ≤ 20%, the best solution would probably be a heavy duty PFC capacitor bank with suitable harmonic detuned reactors;
If THD(i)% > 20% we recommend to install an active harmonic filter;
If THD(u)% ≤ 3% we normally suggest a standard PFC capacitor bank;
If 3% < THD(u)% ≤ 4% you should probably install a heavy duty PFC capacitor bank;
If 4% < THD(u)% ≤ 7% we suggest a heavy duty PFC capacitor bank with suitable harmonic detuned reactors;
If THD(u)% > 7% we recommend the installation of an active harmonic filter.
Finally, if both THD(I) and THD(U) are measured and do not result in the same type of power factor correction, you must chose the most rigorous solution. For example, if we get the following values
from the network analyser:
THD(I) = 15% – we normally suggest using a heavy duty PFC capacitor bank with suitable harmonic detuned reactors
THD(U) = 8% – we recommend installing an active harmonic filter
…the most critical parameter is the THD(U)=8%, so the best solution would be choosing an active harmonic filter.
If you don’t have a network analyser, you can use the following formula to calculate the percentage of non-linear loads in your network:
(S(h) / S(n)) * 100 = % N(LL) of active harmonic filter is recommended.
S(h) is the total apparent power of all non-linear loads in your network (kVA).
S(n) is the apparent power of installed transformer (kVA).
N(LL) is the percentage of non-linear loads in your network.
If N(LL) < 15% we normally recommend a standard PFC capacitor bank.
If 15% < N(LL ) < 25% you may want to consider a heavy duty PFC capacitor bank.
If 25% < N(LL) < 50% our suggestion is a heavy duty PFC capacitor bank with suitable harmonic detuned reactors.
If N(LL) ≥ 50% we recommend installing an active harmonic filter.
This will allow you to calculate the percentage of non-linear loads in your network. Note: the S(h)/S(n) rule is valid for a THD(I) of all the harmonic generators < 30% and for pre-existing THD(U) <
If these values are exceeded, you need to perform a harmonic analysis of the network.
Make sure to share this post with your colleagues if you found it useful, and let us know how we can help you further!
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Fidelis Analog - What's NewWe've Moved!Intimate with the SP-10R
I know a lot of people have been waiting, patiently, for the most part, for my thoughts on the SP-10R. Well, perhaps not a lot, but at least one gentle nudge on a-gon every now and then. Please
forgive the delay. It seems the one commodity I'm short on, which cannot be replenished, is time.
So, on with it I've seen units in the EU, both SP-10R and SL-1000R, come with wood flight cases. Mine didn't. I e-mailed Technics and the reseller I purchased my unit from why this was the case,
but they neglected to respond.
I must admit I'm not a fan of the aesthetics of the 10R. I'm not talking fit and finish here, rather the industrial design. We have square buttons on the motor unit, round buttons on the power
supply, square edges on the motor chassis and plinth, round on the armboard, etc. The left-justification of the controls on the motor as opposed to center-aligned with each other doesn't make sense
to me either. It's like three separate people who never spoke designed each piece. I don't get it.
The brushed finish looks far better in person than in any pictures I've seen thus far. The platter finish is nice, however not as well machined as the MK3. Worlds apart here.
The 10R has tapped holes under the platter in the chassis that allow you to use the platter handles to drop the motor chassis in to a plinth. This is very handy, and something I wish the MK3 had.
Unlike the MK3, the threaded holes in the platter of the 10R are not inset, so you have to be extremely careful when removing the handles to not drop them. Extremely careful. I was only very
careful, and dropped one on my platter. I suggest holding them as I've pictured below during removal to be safe.
As is well explained on Technics site, the platter is a die-cast aluminum base with a brass top plate and tungsten weights, with a damping rubber underneath. Each platter is individually balanced,
though I don't see in the literature if they specify static or dynamic. I do hope it's balanced with the damping material applied, as mine is quite a bit off, thought I doubt it'd matter much.
On my unit there's around a thousandth or two of wobble and axial runout. As the platter sits on three standoffs on the motor and has a tight fit around the spindle (as it should), there's no way to
compensate. In the rare circumstance you ran across this on a MK3, you can easily change the relationship of the platter and spindle shaft to compensate.
Granted, these are all small nits and unlikely to affect how the 'table performs, but at $10K I don't expect to see such issues.
As I've several SH-10B5 plinths around, I decide to throw my 10R in one of them. No surprises here; everything fits as it should and bolts up fine. The armboard looks a little funny having a
counter-chamfer to the motor chassis' square edge, but the gap is right and you hardly notice.
From here I let the motor run for about 72 hours, listening to some tunes as we went. I also did some measurements, which we'll talk about later. Note that my thoughts here aren't about what
happened in the initial 72 hours - I've had this 'table since the end of May.
Let's get in to the fun stuff.
On the 10R, the brushed aluminum chassis, while serving as the mounting mechanism to a plinth, is more of an escutcheon as anything. The inner motor and outer chassis 'tub' are the main bodies of
the unit.
Of course I took it apart. It's like you don't know me at all.
The back of the chassis, similar to the MK3, is a composite material. This is likely an evolution of TNRC, or some such. When you remove it, you think "damn, that's heavy", only to find two
stainless steel weights inside. To be fair, this is a feature Technics calls out in their marketing literature.
The weights are 1.8kg and 416g.
There are plenty of pictures of the motor assembly around, so I can't be bothered to repost them here. Actually, I didn't take any pictures of the assembled motor. Yep, you know where this is
As advertised, there are 18 coils, 9 each side, offset 60 degrees. The coils are wound very precisely, the board is quality, and the soldering is top-notch.
The bearing is your standard sintered bush with thrust pad setup. Same as the MKII/2A, the spindle is the shaft, and as such the same diameter. The thrust pad is floating, and of a thin
polymer-type material. At the top of the bearing is a brass collar that the aluminum motor rotor hub attaches to.
As advertised, platter rotation is tracked by an indexed optical encoder. I didn't count the slots. The magnified photo is 6x.
The motor movement has a nice, smooth heft to it. I spent quite a lot of time with the demo unit at the reseller's, carefully manipulating the motor. That one had a high friction spot in it,
however, and thankfully, mine does not.
As is called out in the literature, the mounting plate that attaches the motor to the 'tub' is quite substantial.
Speaking of the 'tub', it's around 4.5mm thick and likely cause of an aluminum alloy. The casting looks quite good, as is expected in these modern times.
The 'brains' of the unit are contained within the motor unit itself. As with the the G/GAE, and I assume the GR, everything comes down to a single MCU, Technics apparent favorite being from Renesas.
We've some circuity to read the encoder, and some circuitry to drive 3 push-pull amplifiers, and that's it. That's all you need.
Last, we come to the power supply, which is pretty much just a switching supply. The LCD and control panel is driven by a small uC, with a 2-wire interface back to the motor unit. Simple.
How does it sound?
Buy one. Find out.
Just kidding. Measurements-wise it edges out a couple of my MK3. I've done dozens upon dozens of measurements, in an endless circle of the slightest advantage going either way. I don't have a
vibration-proof lab. That's the caliber of performance we're talking about.
Below, two polar plots. One from the 10R, the other from one of my MK3. Same test record, indexed off the start of the W&F track.
I'm not one for the purple prose of audio reviews. Sorry to let you down. I doubt my subjective observations bear much relevance to anyone else. I will say I could happily live with a 10R as my
one 'table. Shiny, new, warranty, and performs great. What's not to like?
Would I trade a MK3 for a 10R? Not a chance. While objective measurements of the 10R are the slightest bit better than the MK3, I think the MK3 is a higher-caliber unit. Beefier components, better
machining, and better industrial design; even if the control unit is dated. It's a brute.
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Design of Inequality Models of Covid-19 Disease Incorporating Social Discriminants
Fatal disease like Coronavirus (COVID-19) is a contagious disease which causes death. In 2019 and 2020, millions of deaths were recorded as a result of its outbreak. The impact of the outbreak
disrupted Economic and social activities globally. The spread across globe exposed existing social and health inequalities. Disparities in hygiene and level of awareness of the havoc the disease can
cause had been the bane of public health interventions. To verify this claim, inequality of non-linear mathematical models were proposed and analyzed to investigate how social determinants influence
infection spread and mortality rates. Stability analyses of the model's equilibra were performed. Basic reproduction number was derived for the inequality model R0. Numeric analysis was also carried
out to support results. The results revealed significant disparities in the disease's outcomes in heath inequalities indicators.
Inequalities, Hygiene, Awareness, Reproduction number stability and Numerical analysis
Disease as we all know is a discomfort to human body as well as animals. The spread and hazard of the outbreak of viral diseases cannot be underestimated. It affects virtually everything ranging from
economy, to education and health to mention but a few. In 2019, outbreak of COVID-19 ravaged and threatened the existence of the world, the world leaders, health personnel, government officials had
great task in curbing and controlling the spread of the disease. Different researchers have contributed in one way or the other to model the spread and understand its impact and how to control the
spread and treatment methods to reduce the number of infectives [1] enlarged the frame work of epidemiology by considering the size of epidemic and the duration with a view of finding the probability
of disease extinction. The effect of infection period within Susceptible-Infected-Recovered (SIR) models was studied by [2] and discovered that unstable-like behaviour was seen in a finite population
in the model. The spatial study of disease spread is important, [3] used spatial study to discover multigroup epidemic in the Susceptible-Exposed-Infected-Recovered (SEIR) model when they extended
works of ONeill and Roberts of 1999 on using Monte Carlo Markov Chain to estimate parameters in the model. When there is an outbreak of disease, vaccination against such disease is imminent. Various
modeling approaches have been employed to study the spread and impact of COVID-19. Compartmental models, such as the SIR model, have been widely used to simulate disease dynamics and evaluate
intervention strategies [4] introduced vaccinated group into the epidemic model to study the non-linear incidence rate in the bifurcation model. If the disease persists in the population then, the
stability analysis is expected, [5] introduced migration rate into the susceptible population to study the stability of the model. Within short time of epidemic, a lot can happen, the problems of
modelling stochastic epidemic model was looked into by [6,7] captured the transmission dynamics of Measles using stochastic model in the analysis of the spread of the disease. More sophisticated
models, incorporating socioeconomic and demographic variables, have provided deeper insights into the factors driving health inequalities. Machine learning techniques have also been leveraged to
predict COVID-19 outcomes and identify high-risk populations [8] applied non-linear compartmental deterministic mathematical model with exploratory data to discover the extent of the variables:
Awareness and hygiene can go in curbing the menace of the spread of COVID-19.
When a disease surfaces in the population, it affects the equality of social status of the populace. To address the issue of inequalities in epidemic modelling, [9] developed mathematical model to
study transmission of two vaccine-preventable infections in a population by determining the integration of two social groups. Socioeconomic status (SES) is a critical determinant of health outcomes,
including susceptibility to infectious diseases. Research has consistently shown that lower SES is associated with higher risk of infection and adverse health outcomes. During the COVID-19 pandemic,
individuals with lower income levels, precarious employment, and inadequate housing have faced higher infection risks due to limited ability to practice physical distancing and greater exposure to
public-facing occupations.
Studies have also highlighted how economic deprivation correlates with higher COVID-19 mortality rates, underscoring the urgent need for socioeconomic considerations in public health models. Though,
in studying inequalities in epidemic model, the contributing factors in the cases of reported health infractions should be looked into.
The difference in the severity of the diseases as studied by [10,11]. Some affected individuals have different propensity to seek medical care [12]. Geographic location also plays a pivotal role in
shaping health inequalities during the COVID-19 pandemic. Urban areas, with their higher population densities, have generally experienced more severe outbreaks than rural areas. However, rural areas
often suffer from limited healthcare infrastructure and resources, compounding the challenges faced by residents in these regions. Geographic models of COVID-19 have sought to capture these spatial
disparities, providing valuable insights for targeted interventions and resource allocation. Other factors like poverty and social inequalities were discussed in [13]. Also, the reproduction number
of the disease are different from region to region [14]. Studies have emphasized the need for health system resilience and preparedness in mitigating the impacts of pandemics on vulnerable
populations. In developing countries, level of awareness of the transmission dynamics of infectious disease is very low and the hygiene practice during the outbreak of disease is far from the
standard practice which may result into lot of casualties of human lives.
To this end, this research paper looks into the disparities in terms of inequalities of hygiene and level of awareness among susceptible population in a non-linear mathematical model.
Model Description
Considering the dynamics of the virus, we assumed homogeneous population. Also, in the assumption, we assumed disease can only spread if there is a direct contact of an infected person with the
susceptible individual. The homogeneous population is subdivided into Susceptible - Quarantined - Infected- Recovered- Exposed - Susceptible (SQIRES) non-linear compartments.
The dynamics goes thus: There is recruitment into susceptible compartment S(t) at the rate of v [ x ] ( x ), after the recruitment and contact with an infected person, those who show the symptoms of
the virus are Quarantined Q(t) at the rate λ ( h ) which will later move to infected compartment I(t) at the rates P [ 1 ] and P [ 2 ] for those who tested negative. The recovered individuals R(t)
are populated at the rate of P [ 3 ] . From the fact that individuals will exit the system either naturally or induced, the natural death rate and disease induced death rate are respectively given as
σ and δ respectively. The exposed compartment E(t) is used for those that have information about the virus (awareness) and this group are defined thus: It comprises of individuals who maintain
personal hygiene and have attained certain level of awareness which we believe will reduce the spread rate of the virus. This rate is described thus
(4.1) $\Omega \left(G\right)\text{\hspace{0.17em}}=\text{ }{\Omega }_{\mathrm{max}}\text{ }-\text{ }G\left({\Omega }_{\mathrm{max}}\right)\text{ }-\text{ }G\left({\Omega }_{\mathrm{min}}\right)$
where Ωmax and Ωmin represent the maximum and minimum transmission rates of the virus respectively.
This is important in studying the dynamics of disease spread in that, if improved sanitation behaviour and personal hygiene are encouraged, these will definitely play critical roles in bringing down
the spread of the virus from all forms of transmission. So in the event of disease outbreak such as COVID-19, frequent hand washing and use of sanitizers are encouraged as first precautionary
measures. These inform that the spread of the disease can be effectively controlled as fast as possible.
This equation (4.1) can be reduced if maximum hygiene level (G) is assumed to be 1, then equation (4.1) will be Ω ( G ) = Ω [ min ] which clearly showed that the disease transmission can be reduced
to the barest minimum if some level of hygiene is reached and vice-versa if G = 0. To this end, having showed that the transmission rate of COVID-19 can be reduced if some level of hygiene is
achieved and a campaign strategy to educate and create awareness for the locals are put in place. Let us assumed that a certain fraction G [ 0 ] of the population practice and have good habits in
terms of sanitation before the outbreak of COVID-19, that means the remaining individuals in the population denoted by (1- G [ 0 ] ) will be highly spirited to practice healthy sanitation habits as
the rate of education campaign on good hygiene grows at rate θ . On the spur of moment of disease outbreak, thus, as the rate of awareness on what to do increases, it is assumed the spread of the
disease reduces.
(4.2) $G\left(x\right)\text{ }=\text{ }\frac{{G}_{0}\left(x\right)\text{ }+\text{ }E\left(x\right)\text{ }×\text{ }{G}_{0}\left(x\right)\text{ }+\text{ }{\eta }_{0}\text{ }×\text{ }E\left(x\right)\
text{ }-\text{ }G\left(x\right)\text{ }×\text{ }{\eta }_{0}\text{ }×\text{ }E\left(x\right)}{1\text{ }+\text{ }E\left(x\right)},\text{ }{G}_{0}\left(x\right)\text{ }\le \text{ }G\left(x\right)\text{
}\le \text{ }1$
Now, in equation (4.2), it shows that, if E ( x ) = 0 that is no strategy inform of campaign to advise populace on how to go about protecting themselves against the disease, then G = G [ 0 ] .
However, in the long run, if a well planned and awareness campaign strategy to disseminate information on how individuals can combat or contain the spread, there will be an increase in sanitation
initial level from G [ 0 ] to G [ 0 ] + (1 - G [ 0 ] ) η [ 0 ] .
Suppose, the chance that an individual becomes infected due to individual’s interaction with an infected and infectious human is r [ 1 ] , then an assumption of the force of infection can be put as λ
(G) which simply follows a mass action and defined thus as
(4.3) $\lambda \left(G\right)\text{ }=\text{ }{r}_{1}\text{ }×\text{ }\Omega \left(G\right)\text{ }×\text{ }I\left(x\right)$
In the construction of the inequality model, the following assumptions were made:
The transmission of COVID-19 is possible only if an infected or an infectious human come in contact with individuals; or by individuals coming in contact with surface which has been infested with the
There is a saturating function that propagates awareness of the havoc of the virus and how susceptible individual could protect himself and this depends on infected population density given by
(4.4) $F\left(I\right)\text{ }=\text{ }\frac{{\eta }_{0}\text{ }×\text{ }I\left(x\right)}{{\eta }_{1}\text{ }+\text{ }{\eta }_{2}\text{ }×\text{ }I\left(x\right)}$
In [15], equation (4.4), that is the saturating function was used also where η [ 0 ] , η [ 1 ] and η [ 2 ] depicted rate of information growth which is the half of saturation point, F(I) attains half
of its maximum value $\frac{{\eta }_{0}}{{\eta }_{2}}$ when infected population arrives at η [ 1 ] and saturation constant of information respectively [15,16].
We also assume that there is no permanent immunity that is, recovered individual over time from COVID-19 lose his immunity δ (whether natural or due to medication) and moves back to the susceptible
We assume that awareness or information degenerate at rate a . This assumption is due to the response rate behaviour of humans to disease outbreak in the long run.
Since we have varying population with time, the recruitment rate into the population is assumed to be
(4.5) $\upsilon \left(x\right)\text{ }\ge \text{ }\varpi \int \left(0,1\right]\upsilon x\left(x\right)dx\text{ }+\text{ }\int \left(0,1\right]\upsilon ϵ\left(x\right)dx,\text{ }\varpi \in \text{ }\
where υ [ x ] ( x ) are those who are recruited through migration, υ [ ϵ ] ( x ) are rate of births while is a restriction parameter on immigration.
Model Formulation
Based on the above assumptions and using the following description of the variables of system of equations used.
(i) S ( t ) represents Susceptible humans.
(ii) Q ( t ) represents Quarantined humans.
(iii) I ( t ) represents Infectious humans.
(iv) R ( t ) represents Recovered humans.
(v) E ( t ) represents Education/level of hygiene.
Hence, the dynamics of the model is presented in the following system inequalities of non-linear compartmental differential equations:
(5.1) $\begin{array}{l}d\left(\left(x\right)\right)\text{ }\ge \text{ }\int \overline{A}dx\text{ }-\text{ }\lambda \left(G\right)\text{ }\int S\left(x\right)dx\text{ }-\text{ }\sigma \text{ }\int S\
left(x\right)dx\text{ }+\text{ }\rho 2\text{ }\int Q\left(x\right)dx\text{ }+\text{ }\delta \int R\left(x\right)dx\\ d\left(Q\left(x\right)\right)\text{ }\ge \text{ }\lambda \left(G\right)\text{ }\
int S\left(x\right)dx\text{ }-\text{ }\left({\rho }_{1}\text{ }+\text{ }{\rho }_{2}\text{ }+\text{ }\sigma \right)\text{ }\int Q\left(x\right)dx\\ d\left(I\left(x\right)\right)\text{ }\ge {\rho }_{1}
\text{ }\int Q\left(x\right)dx\text{ }-\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }\int I\left(x\right)dx\\ d\left(R\left(x\right)\right)\text{ }\ge \text{ }{\rho }_
{3}\text{ }\int I\left(x\right)dx\text{ }-\text{ }\left(\delta \text{ }+\text{ }\sigma \right)\text{ }\int R\left(x\right)dx\\ d\left(E\left(x\right)\right)\text{ }\ge \text{ }{\eta }_{0}\text{ }\int
\frac{I\left(x\right)}{\eta 1\text{ }+\text{ }{\eta }_{0}\text{ }×\text{ }I\left(x\right)}\text{ }d\left(x\right)\text{ }-\text{ }a\text{ }×\text{ }d\left(E\left(x\right)\right)\end{array}$
Since the model involves human dynamics, it is assumed that all parameters and variables used are positive.
Model Analysis
Our focus here, is how to analyse the model. First, we perform stability analysis of the model viz the positivity and boundedness analysis, the steady states of the model and their stability.
Positivity and Boundedness of solutions
Here, we show that system (5.1) is epidemiological well-posed and is realistic if all the system variables of (5.1) are positive for all time t . With this in mind, we establish the claim using the
following Corollaries.
Corollary 6.1
Suppose S(x), Q(x), I(x), R(x) and E(x) are non positive functions.
Let S (0) ≥ 0 , Q (0) ≥ 0 , I (0) ≥ 0 , R (0) ≥ 0 , G (0) ≥ 0 and S(0) be non-negative while x > 0.
The rates above are bounded in plane $\Psi \text{ }\in \text{ }{ℝ}^{5}$ .
Proof. In [15] we have the following
(5.1) $\begin{array}{l}d\left(S\left(x\right)\right)\text{ }\ge \text{ }\overline{A}\text{ }+\text{ }\rho 2{\int }_{}^{1}0Q\left(x\right)dx\text{ }+\delta \text{ }{\int }_{}^{1}0R\left(x\right)dx\\ d
\left(Q\left(x\right)\right)\text{ }\ge \text{ }\lambda {\int }_{0}^{1}S\left(x\right)dx\\ d\left(I\left(x\right)\right)\text{ }\ge \text{ }\rho 1\text{ }{\int }_{0}^{1}Q\left(x\right)dx\\ d\left(R\
left(x\right)\right)\text{ }\ge \text{ }\rho 3\text{ }{\int }_{0}^{1}I\left(x\right)dx\\ d\left(E\left(x\right)\right)\text{ }\ge \text{ }{\eta }_{0}\text{ }{\int }_{0}^{1}\frac{\eta 0\text{ }×\text{
}I\left(x\right)}{\eta 1\text{ }+\text{ }\eta 2\text{ }×\text{ }I\left(x\right)}dx\end{array}$
It will be convenient to prove that the system (5.1) region is positive, not variant and attractive [17,18]. The region of attraction of (6.1) becomes
(6.2) $\Psi \text{ }\in \text{ }\left\{\begin{array}{l}\left(S\left(x\right),\text{ }Q\left(x\right),\text{ }I\left(x\right),\text{ }R\left(x\right)E\left(x\right)\right)\text{ }\in \text{ }{ℝ}^{5}\
text{ }:\text{ }\\ S\left(x\right)\text{ }+\text{ }Q\left(x\right)\text{ }+\text{ }I\left(x\right)\text{ }+\text{ }R\left(x\right)\text{ }\le \text{ }\frac{\overline{A}}{\sigma },\text{ }E\left(x\
right)\text{ }\le \text{ }\frac{\eta }{a}\end{array}\right\}$
It is also sufficient to study the dynamics of inequalities above in which all solutions originating in the inside positive orphan can be drawn [15].
Equilibrium points of System (5.1)
Since system (5.1) is inequalities of non-linear and its exact solutions may not be easy to determine. Consequently, knowledge of stability theories is needed here to investigate the qualitative
behaviour of the equilibrium points so that we can have insight about the disease dynamics and how the spread can be contained. Hence, the existence of all equilibrium of system (5.1) are thoroughly
investigated by making the rate of change with respect to time t of all the non-linear inequalities variables zero. Assume G ( x ) is a quantity that its value is known, then the equilibrium points
of system (5.1) can be obtained by solving simultaneously the set of algebraic inequalities:
(7.1) $\begin{array}{l}\overline{A}\text{ }-\text{ }\lambda \left(G\right)\text{ }×\text{ }S\left(x\right)\text{ }-\text{ }\sigma \text{ }×\text{ }S\left(x\right)\text{ }+\text{ }\rho 2\text{ }×\text
{ }Q\left(x\right)\text{ }+\text{ }\delta \text{ }×\text{ }R\left(r\right)\text{ }\ge \text{ }0\\ \lambda \left(G\right)\text{ }×\text{ }S\left(x\right)\text{ }-\text{ }\rho 1\text{ }×\text{ }Q\left
(x\right)\text{ }-\text{ }Q\left(x\right)\text{ }×\text{ }\rho 2\text{ }-\text{ }Q\left(x\right)\text{ }×\text{ }\sigma \text{ }\ge \text{ }0\\ \rho 1\text{ }×\text{ }Q\left(x\right)\text{ }-\text{ }
\rho 3\text{ }×\text{ }I\left(x\right)\text{ }-\text{ }q\text{ }×\text{ }I\left(x\right)\text{ }-\text{ }\sigma \text{ }×\text{ }I\left(x\right)\text{ }\ge \text{ }0\\ \rho 3\text{ }×\text{ }I\left(x
\right)\text{ }-\text{ }\delta \text{ }×\text{ }R\left(x\right)\text{ }-\text{ }\sigma \text{ }×\text{ }R\left(x\right)\text{ }\le \text{ }0\\ \frac{\eta 0\text{ }×\text{ }I\left(x\right)}{\eta 1\
text{ }+\text{ }\eta 0\text{ }×\text{ }I\left(x\right)}\text{ }\ge \text{ }\frac{d\left(E\left(x\right)\right)}{dx}\end{array}$
Simplifying first and last inequalities of system (6.1) yield
(7.2) $\begin{array}{l}S\left(x\right)\text{ }\ge \text{ }\frac{\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)
\text{ }\left(\delta \text{ }+\text{ }\sigma \right)\text{ }×\text{ }\overline{A}}{A0\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A1}\\ Q\left(x\right)\text{ }\ge \text{ }\frac{\left(\rho
3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }\left(\delta \text{ }+\text{ }\sigma \right)×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\text{ }×\text{ }\lambda \left(G\
right)\text{ }+\text{ }A1}\\ I\left(x\right)\text{ }\ge \text{ }\frac{\rho 1\text{ }×\text{ }\left(\delta \text{ }+\text{ }\sigma \right)\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\text{
}×\text{ }\lambda \left(G\right)\text{ }+\text{ }A1}\\ R\left(x\right)\text{ }\ge \text{ }\frac{\rho 1\text{ }×\text{ }\rho 3\text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\
text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A1}\\ E\left(x\right)\text{ }\ge \text{ }\frac{\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{
}\lambda \left(G\right)\text{ }+\text{ }\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A2\text{ }×\text{ }\lambda \left(G\
right)\text{ }+\text{ }A3}\end{array}$
We define
(7.3) $\begin{array}{l}A0\left(x\right)\text{ }\ge \text{ }\rho 1\text{ }×\text{ }\rho 3\text{ }×\text{ }\sigma \text{ }+\text{\hspace{0.17em}}\text{ }\rho 1\text{ }×\text{ }\left(\delta \text{ }+\
text{ }\sigma \right)\text{ }\left(q\text{ }+\text{ }\sigma \right)\text{ }+\text{ }\sigma \text{ }×\text{ }\left(\delta \text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+
\text{ }\sigma \right)\\ A1\left(x\right)\text{ }\ge \text{ }\sigma \text{ }×\text{ }\left(\delta \text{ }+\text{ }\sigma \right)\text{ }×\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\
sigma \right)\text{ }×\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\\ A2\left(x\right)\text{ }\ge \text{ }a\text{ }×\text{ }\left(\eta 1\text{ }×\text{ }A0\text{ }+\text{\
hspace{0.17em}}\eta 2\text{ }×\text{ }\rho 1\text{ }×\text{ }\left(\delta \text{ }+\text{ }\sigma \right)\text{ }×\text{ }\lambda \right)\\ A3\left(x\right)\text{ }\ge \text{ }a\text{ }×\text{ }\eta
1\text{ }×\text{ }A1\end{array}$
Merging inequality (7.3) of I(x) in later with (4.3) gets
(7.4) $\lambda \left(G\right)\text{ }\ge \text{ }\frac{r1\text{ }×\text{ }\delta \text{ }×\text{ }\rho 1\text{ }×\text{ }\Omega \left(G\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \
left(G\right)\text{ }+\text{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\sigma \text{ }×\text{ }\Omega \left(G\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\text{ }×
\text{ }\lambda \left(G\right)\text{ }+\text{ }A1}$
which have the following quadratic solutions
(7.5) $\lambda \left(G\right)\text{ }\ge \text{ }0\text{ }or\text{ }\lambda \left(G\right)\text{ }\ge \text{ }\frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text
{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }-\text{ }A1}{A1}$
Instance 1:
It was observed that (7.5) shows two interpretation
First interpretation of result i.e.,
$\lambda \left(G\right)\text{ }\ge \text{ }0$
and second interpretation
(7.6) $\lambda \left(G\right)\text{ }\ge \text{ }\frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }r1\text{ }×
\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }A1}{A1}$
COVID-19 Free Equilibrium, P [ 0 ] (x) in (4.3), we note that when putting ξ(G) = 0 , we have
(7.7) $P0\left(x\right)\ge \text{ }\left(\frac{\overline{A}}{\sigma },0,0,0,0\right)$
COVID-19 Free equilibrium is defined as P [ 0 ] (x) and shows the point at which the model is free of any infection of novel Corona virus disease. With this, the infection can be controlled without
hygiene in the entire population area.
Local Stability of COVID-19 Free Equilibrium (CFE), P [ 0 ]
The local stability of P [ 0 ] is examined in the theorem below:
Theorem A:
The COVID-19 Free Equilibrium (CFE). P [ 0 ] (x) in (??) is considered to be locally asymptotically stable if D [ 0 ] < 1 , if not is unstable.
Proof. Apply Jacobian matrix to (7.8) at P [ 0 ] (x) in the absence of sanitation G(x) = G [ 0 ] (x)
(7.8) $J\left(P0\left(x\right)\right)\text{ }\ge \text{ }\left(\begin{array}{ccccc}-\sigma & \rho 2& -\text{ }\frac{r1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }\overline{A}}{\sigma }& \
delta & 0\\ 0& -\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)& -\text{ }\frac{r1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }\overline{A}}{\sigma }& 0& 0\\ 0& \rho 1&
-\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\rho 2\right)& 0& 0\\ 0& 0& \rho 3& -\left(\delta +\text{ }\sigma \right)& 0\\ 0& 0& \frac{\eta 0}{\eta 1}& 0& -a\end{array}\right)$
It is interesting to note that eigenvalues λ of (??) has the following values.
$\begin{array}{l}\lambda 1\text{ }=\text{ }-\sigma \\ \lambda 2\text{ }=\text{ }-\left(\delta \text{ }+\text{ }\sigma \right)\\ \lambda 3\text{ }=\text{ }-a\end{array}$
In the matrix (7.8) above, we need to get the other two eigenvalues using the following two by two matrix
(7.9) $M\left(x\right)\text{ }\ge \text{ }\left(\begin{array}{cc}\left(-\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)& -\frac{r1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }
\overline{A}}{\sigma }\\ \rho 1& -\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\rho 2\right)\end{array}\right)$
Going by Routh-Hurwitz condition, the eigenvalues of matrix M are real and negative if
(i) Trace (M) < 0
(ii) Determinant (M) > 0
Tr(M) = −(ρ [ 1 ] + ρ [ 2 ] + ρ [ 3 ] + 2σ) < 0
(7.10) $D0\left(x\right)\text{ }=\text{ }\left(\frac{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }\overline{A}}{\sigma \text{ }×\text{ }\left(\rho 1\text{ }+\text{
}\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)}\right)$
Thus, all eigenvalues found are real and negative if D [ 0 ] < 1 so that CFE, P [ 0 ] is locally asymptotically stable and unstable if D [ 0 ] > 1.
Remark 1
The quantity D [ 0 ] in equation (7.10) is generally referred to as the basic reproduction number. This is the average number of secondary infected humans that an infected human can infect due to
direct or indirect contact in his/her infectious period in a wholly susceptible population. In particular if D [ 0 ] < 1 (or D [ 0 ] > 1 ), it reflects that on the average, an infected individual
will infect successfully less than (or more than) one secondary infected individuals in a population that is wholly susceptible during his/her whole infectious period and thus respectively, the
disease dies out (or persists) in the population. It is worthy of note that the derivation of the basic reproduction number R [ 0 ] can be effectively obtained by making use of the next generation
matrix approach [19,20].
Global asymptotic stability of CFE, P [ 0 ]
We further probe the asymptotic stability of P [ 0 ] by constructing the Lyapunov function for Global Aymptotic Stability (GAS). Thus, consider the Lyapunov function defined as follows:
(7.11) $L\left(Q\left(x\right),\text{ }I\left(x\right)\right)\text{ }=\text{ }\rho 1\text{ }×\text{ }Q\left(x\right)\text{ }+\text{ }\rho 1\text{ }×\text{ }I\left(x\right)\text{ }+\text{ }\rho 2\text
{ }×\text{ }I\left(x\right)\text{ }+\text{ }\sigma \text{ }×\text{ }I\left(x\right)$
Differentiate (7.11) with respect to x along the solution of (??) yields
(7.12) $\begin{array}{l}\frac{d\left(L\left(Q\left(x\right),\text{ }I\left(x\right)\right)\right)}{dx}\text{ }=\text{ }\rho 1\text{ }×\text{ }\left(\lambda \left(G\right)\text{ }×\text{ }S\left(x\
right)\text{ }+\text{ }\left(\rho 1\text{ }+\text{ }\rho 1\text{ }+\text{ }\sigma \right)\text{ }+\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 1Q\left
(x\right)\text{ }+\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\right)\right)\\ =\text{ }\rho 1\Omega \left(G\right)\text{ }×\text{ }S\left(x\right)\text{ }\left(\rho 1\text{
}+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }I\left(x\right)\\ =\text{ }\rho 1r1\lambda \left(G\right)\text{ }×\text{ }S
\left(x\right)\text{ }×\text{ }I\left(x\right)\text{ }-\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\
text{ }×\text{ }I\left(x\right)\\ =\text{ }\left(\rho 1r1\lambda \left(G\right)\text{ }×\text{ }S\left(x\right)\text{ }-\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text
{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\right)\text{ }×\text{ }I\left(x\right)\end{array}$
putting G = G [ 0 ] , $S\text{ }=\text{ }\frac{\overline{A}}{\sigma }$ in (7.12) obtain:
(7.13) $\begin{array}{l}\frac{d\left(L\left(Q\left(x\right),\text{ }I\left(x\right)\right)\right)}{dx}\text{ }=\text{ }\left(\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\Omega \left(G0\right)\frac{\
upsilon }{\sigma }\text{ }-\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\right)\text{ }×\text{ }I\
left(x\right)\\ =\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }\left(\frac{\rho 1\text{ }×\
text{ }r1\text{ }\Omega \left(G0\right)\text{ }\overline{A}}{\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)}\
text{ }-\text{ }1\right)\\ and\\ \frac{d\left(L\left(Q\left(x\right),\text{ }I\left(x\right)\right)\right)}{dx}\text{ }=\text{ }\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text
{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }×\text{ }\left(D0\text{ }-\text{ }1\right)\text{ }×\text{ }I\left(x\right)\text{ }\ge \text{ }0\text{ }if\text{\hspace{0.17em}}
D0\text{ }\ge \text{ }1\end{array}$
Therefore, the CFE, P [ 0 ] is globally asymptotically stable if D [ 0 ] ≥ 1, if not unstable. The foregoing considerations are summarized thus.
Theorem B:
The CFE, P [ 0 ] is asymptotically stable globally if D [ 0 ] ≥ 1 , if not unstable. If
$\lambda \left(G\right)\text{ }\ge \text{ }\frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }r1\text{ }×\text{
}\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }A1}{A1}$
COVID-19 endemic equilibrium (CEE), D [ 1 ] . The Dynamical components of (??) which are represented with D [ 1 ]
(7.14) $\begin{array}{l}P1\text{ }=\text{ }\left(S1\left(x\right),\text{ }Q1\left(x\right),\text{ }I1\left(x\right),\text{ }R1\left(x\right)E1\left(x\right)\right)\\ =\text{\hspace{0.17em}}\left(\
frac{\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }\left(\delta \text{ }+\text{ }\sigma \right)×\text{
}\overline{A}}{A0\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A1}\right),\text{ }\frac{\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)\text{ }\left(\delta \text{ }+\text{ }\
sigma \right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\lambda \left(G\right)\text{ }+\text{ }A1},\\ \frac{\rho 1\text{ }×\text{ }\left(\delta \text{ }+\text{ }\sigma \
right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A1},\text{ }\frac{\rho 1\text{ }×\text{ }\rho 3\text{ }×\text{ }\
overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A0\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A1},\text{ }\frac{\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\
text{ }\lambda \left(G\right)\text{ }+\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A3\text{ }×\text{ }\lambda \left(G\right)\text{ }+\
text{ }A3}\text{ }\ge \text{ }0\end{array}$
P [ 0 ] represents the Coronavirus Endemic Equilibrium (COVID-19 EE) which represents a situation whereby there is a presence of Coronavirus disease or infection in the population. Assuming G is
known, then, there exists at most one endemic equilibrium P [ 0 ] as defined in (??). In order to show that (??) indeed specifies an endemic equilibrium, we must not fail to show that
(7.15) $\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }×\text{ }\delta \text{ }+\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\
Omega \left(G\right)\text{ }\ge \text{ }A1$
furthermore from (7.15)
(7.16) $\Omega \left(G\right)\text{ }=\text{ }\frac{A1}{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }+\text{ }\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\sigma
\text{ }×\text{ }\overline{A}}$
Substitute for (4.1) in (7.16) the subject gets
(7.17) $\Omega \left(G\right)\text{ }\ge \text{ }\frac{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }+\text{ }\rho 1\text{
}×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }-\text{ }A1}{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline
{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }+\text{ }\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }-\rho 1\text{ }×\text
{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{min}\text{ }-\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\
Omega \mathrm{min}}$
(7.18) $\Omega \left(G\right)\text{ }\le \text{ }\frac{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }+\text{ }\rho 1\text{
}×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{ }\upsilon \text{ }×\text{ }\Omega \mathrm{max}\text{ }-\text{ }A1}{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\
text{ }×\text{ }\Omega \mathrm{max}\text{ }+\text{ }\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }-\rho 1\text{ }×\text{ }
r1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \mathrm{min}\text{ }-\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{ }\upsilon \text{ }×\text{ }\Omega
To complete Theorem A:
We need the following Corollary:
Corollary A:
Suppose G(x) is known, then system (??) has a unique endemic equilibrium P [ 1 ] (x) that is positive if the force of infection is
(7.19) $\lambda \left(G\right)\text{ }\ge \text{ }\frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }r1\text{ }
×\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\Omega \left(G\right)\text{ }+\text{ }A1}{A1\left(x\right)}\text{ }\ge \text{ }0$
Resolving for G(x) we need to study the endemic equilibrium P [ 1 ] completely, it is necessary to get equation for G(x) and find its uniqueness. The equation for G(x) is obtained by making G(x) = G
[ 1 ] (x) in (6.2) to obtain
(7.20) $\begin{array}{l}\lambda \left(G\right)\text{ }\ge \text{ }G0\text{ }+\text{\hspace{0.17em}}\left(1\text{ }-G0\right)\eta 0\text{ }\frac{\frac{\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\
delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A2\text{ }×\text{ }\lambda \
left(G\right)\text{ }+\text{ }A3}}{1\text{ }+\text{ }\frac{\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }\
sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)}{A2\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A3}}\\ =\text{ }G0\text{ }+\text{ }\frac{\left({\eta }^{2}\text{
}×\text{ }\rho 1\text{ }-\text{ }{\eta }^{2}\text{ }×\text{ }\rho 1\text{ }×\text{ }G0\left(x\right)\right)\text{ }\left(\delta \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)\
text{ }+\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)\right)}{\left(A2\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{\hspace{0.17em}}\eta 0\text{ }×\text{
}\rho 1\overline{A}\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\lambda \left(G\right)\right)}\text{ }\ge \text{
However, for
(7.21) $\zeta \left(G\right)\text{ }=\text{ }\frac{A5\text{ }×\text{ }\lambda \left(G\right)}{A5\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A3}\text{ }-\text{ }\left(G\left(x\right)\text{
}-\text{ }G0\left(x\right)\right)\text{ }\ge \text{ }0$
$R4\text{ }=\text{ }\left(\begin{array}{l}\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\delta \text{ }-G0\left(x\right)\text{ }×\text{ }\eta 0\text{ }×\text{ }\rho 1\
text{ }×\text{ }\overline{A}\text{ }×\text{ }\delta \text{ }+\\ \text{ }\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \text{ }-\text{ }G0\left(x\right)\text{ }×\
text{ }\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \end{array}\right)$
$A5\text{ }=\text{ }A2\text{ }+\text{ }\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \text{ }+\text{ }\eta 0\text{ }×\text{ }\rho 1\text{ }×\text{ }\delta \text{ }
×\text{ }\sigma \text{ }×\text{ }\overline{A}$
(7.22) $\zeta \left(G0\right)\text{ }\le \text{ }0$
(7.23) $\zeta \left(G1\right)\text{ }\ge \text{ }0$
(7.24) $\int \left[G0,G1\right]\text{ }\zeta \left(G\right)dx\text{ }\ge \text{ }\int \left[G0,G1\right]\text{ }dx$
Combining G [ 0 ] (x) = G [ 1 ] (x)
(7.25) $\begin{array}{l}\zeta \left(G0\right)\text{ }\ge \text{ }\frac{A4\left(x\right)\text{ }\lambda \left(G0\right)}{A5\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }A3}\text{ }=\text{ }\
frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\delta \text{ }+\text{ }r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\Omega \left(G0\
right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \text{ }-\text{ }A1}{A0}\\ \frac{A1\text{ }×\text{ }D0\text{ }-\text{ }A1}{A0}\text{ }A0\text{ }\ge \text{ }0\end{array}$
The results following g(G) ≥ 0 if D [ 0 ] ≥ 0. Therefore, if G [ 0 ] (x) = G [ 1 ] (x) in (7.22), g(G [ 1 ] ) ≤ 0 implies
$\zeta \left(G1\right)\text{ }\ge \text{ }-G1\text{ }+\text{ }G0\text{ }\le \text{ }0$
Differentiate (7.20) with respective to G(x) gets
(7.26) $\frac{d\zeta \left(G\right)}{dx}\text{ }\ge \text{ }\frac{A3\text{ }×\text{ }A4\text{ }×\text{ }\frac{d\lambda \left(G\right)}{dx}\text{ }-\text{ }{A}^{2}5\text{ }×\text{ }{\lambda }^{2}\left
(G\right)\text{ }-\text{ }2\text{ }×\text{ }A5\text{ }×\text{ }\lambda \left(G\right)\text{ }-\text{ }{A}^{2}3}{{A}^{2}5\text{ }×\text{ }{\lambda }^{2}\left(G\right)\text{ }+\text{ }2\text{ }×\text{
}{A}_{5}\text{ }×\text{ }\lambda \left(G\right)\text{ }+\text{ }{A}_{3}^{2}}\text{ }\ge \text{ }0$
We observed that
(7.27) $\frac{d\lambda \left(G\right)}{dx}\text{ }\ge \text{ }-\frac{\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\delta \text{ }+\text{ }\rho 1\text{ }×\text{ }r1\text{ }×\text{ }\sigma \text{ }×\text{
}\overline{A}\text{ }×\text{ }\Omega \mathrm{max}\text{ }-\text{ }\Omega \mathrm{min}}{A0}$
The solution of (7.20) in (G [ 0 ] , G [ 1 ] ) achievable if D [ 0 ] ≥ 0 .
Corollary C:
The system (eq 5.1) has a positive unique endemic equilibrium P if G ∈ (G [ 0 ] , G [ 1 ] ) and if D [ 0 ] ≥ 1 .
Local stability of COVID-19 EE, P.
Theorem 3:
The COVID-19 EE, P is asymptotically stable locally if the following inequality holds if not, unstable.
(7.28) $\lambda \left(G\right)\text{ }\ge \text{ }\frac{\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\text{ }\sigma \right)\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)}{\rho 1
\text{ }×\text{ }\delta \text{ }×\text{ }\upsilon \text{ }+\text{ }\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\upsilon }$
Take the Jacobian matrix of the (??) at P is defined as
(7.29) $J\left(P\right)\text{ }=\text{ }\left(\begin{array}{ccccc}-A6-\sigma & \rho 2& -A7& \delta & 0\\ A6& -\rho 1-\rho 2-\sigma & A7& 0& 0\\ 0& \rho 1& -\rho 3\text{ }-\text{ }q\text{ }+\text{ }\
sigma & 0& 0\\ 0& 0& \rho 3& -\delta -\sigma & 0\\ 0& 0& A8& 0& -a\end{array}\right)$
(7.30) $\begin{array}{l}A6\text{ }=\text{ }\frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\delta \text{ }+\text{ }r1\text{ }×\text
{ }\rho 1\text{ }×\text{ }\Omega \left(G0\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \text{ }-\text{ }A1}{A0}\\ A7\text{ }=\text{ }\frac{\left(\rho 1\text{ }+\text{ }\rho 2\text{ }+\
text{ }\sigma \right)\text{ }×\text{ }\left(\rho 3\text{ }+\text{ }q\text{ }+\text{ }\sigma \right)}{\rho 1\text{ }+\text{ }\sigma \text{ }+\text{ }\overline{A}\text{ }+\text{ }\rho 1\text{ }+\text{
}\delta \text{ }+\text{ }\overline{A}}\\ A8\text{ }=\text{ }\frac{t0\text{\hspace{0.17em}}×\text{ }{A}^{2}0\text{ }×\text{ }r1\text{ }×\text{ }{\Omega }^{2}\left(G\right)\text{ }×\text{ }h1}{{\left(\
left(\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\delta \text{ }×\text{ }h\text{ }+\text{ }\rho 1\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \text{ }×\text{ }h\text{ }+\text{ }A0\
text{ }×\text{ }h1\right)\text{ }×\text{ }r1\text{ }×\text{ }\Omega \left(G\right)\text{ }-\text{ }A1\text{ }×\text{ }h2\right)}^{2}}\end{array}$
The eigenvalues is λ [ 1 ] = −a and other eigenvalues is derived from
(7.31) $M\left(x\right)\text{\hspace{0.17em}}=\text{ }\left(\begin{array}{cccc}-A6-\sigma & \rho 2& -A7& \delta \\ A6& -\rho 1-\rho 2-\sigma & A7& 0\\ 0& \rho 1& -\rho 3-q-\sigma & 0\\ 0& 0& \rho 3&
-\delta -\sigma \end{array}\right)$
We perceive that Trace (M(x)) < 0 from Corollary 3:
Trace (M(x)) = −A [ 6 ] − 4σ − ρ [ 1 ] − ρ [ 2 ] − ρ [ 3 ] − q − δ ≥ 0
(7.32) $\begin{array}{l}Det\left(M\right)\text{ }=\text{ }\delta \text{ }×\text{\hspace{0.17em}}{\sigma }^{3}\text{ }+\text{\hspace{0.17em}}\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }q\
text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }{A}_{6}\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }\rho 1\text{ }+\text{ }\\ \delta \text{\hspace
{0.17em}}×\text{ }{\sigma }^{2}\text{ }×\text{ }\rho 2\text{ }+\text{ }\delta \text{\hspace{0.17em}}×\text{ }{\sigma }^{2}\text{ }×\text{ }\rho 3\text{ }+\text{ }\delta \text{ }×\text{ }\sigma \text{
}×\text{ }q\text{ }×\text{ }{A}_{6}\\ +\text{ }\delta \text{ }×\text{ }\sigma \text{ }×\text{ }q\text{ }×\text{ }\rho 1\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }+\text{ }q\text{ }
+\text{ }\rho 2\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }A6\text{ }×\text{ }\rho 1\text{ }+\text{\hspace{0.17em}}\\ \delta \text{\hspace{0.17em}}×\text{ }\sigma \text{ }×
\text{ }A6\text{ }×\text{ }\rho 3\text{ }-\delta \text{ }×\text{ }\sigma \text{ }×\text{ }A7\text{ }×\text{ }\rho 1\\ +\text{ }\delta \text{ }×\text{ }\sigma \text{ }×\text{ }\rho 1\text{ }×\text{ }\
rho 3\text{ }+\text{ }\delta \text{ }×\text{ }\sigma \text{ }×\text{ }\rho 2\text{ }×\text{ }\rho 3\text{\hspace{0.17em}}+\text{ }\delta \text{ }×\text{ }q\text{ }×\text{ }A6\text{ }×\text{ }\rho 1\
text{ }+\text{ }{\sigma }^{4}\text{ }+\text{ }{\sigma }^{3}\text{ }×\text{ }A6\text{ }+\\ \text{ }{\sigma }^{3}\text{ }×\text{ }\rho 1\text{ }+\text{ }{\sigma }^{3}\text{ }×\text{ }\rho 2\text{ }+\
text{ }{\sigma }^{3}\text{ }×\text{ }\rho 3\\ {\sigma }^{2}\text{ }×\text{ }q\text{ }×\text{ }A6\text{ }+\text{ }{\sigma }^{2}\text{ }×\text{ }q\text{ }×\text{ }\rho 1\text{ }+\text{ }{\sigma }^{2}\
text{ }×\text{ }q\text{ }×\text{ }\rho 2\text{ }+\text{ }{\sigma }^{2}\text{ }×\text{ }A6\text{ }×\text{ }\rho 1\text{\hspace{0.17em}}-\text{ }{\sigma }^{2}\text{ }×\text{ }A7\text{ }×\text{ }\rho 1\
text{ }+\\ \text{ }{\sigma }^{2}\text{ }×\text{ }\rho 1\text{ }×\text{ }\rho 3\text{ }+\text{ }{\sigma }^{2}\text{ }×\text{ }\rho 2\text{ }×\text{ }\rho 3\\ +\text{ }\delta \text{ }×\text{ }{\sigma }
^{2}\text{ }×\text{ }q\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{\hspace{0.17em}}×\text{ }A6\\ =\text{\hspace{0.17em}}\sigma \text{\hspace{0.17em}}×\text{ }\rho 1\text{ }×\text{\
hspace{0.17em}}\left(A6-\text{ }A7\right)\text{ }×\text{ }\left(\delta \text{ }+\text{\hspace{0.17em}}\sigma \right)\text{ }+\text{\hspace{0.17em}}{\sigma }^{2}\text{ }×\text{\hspace{0.17em}}{\sigma
}^{2}\text{ }×\text{ }A6\text{ }+\text{ }{\sigma }^{2}\text{\hspace{0.17em}}×\text{ }\rho 2\text{ }×\text{ }\rho 1\text{ }+\\ \text{ }{\sigma }^{2}\text{ }×\text{ }\rho 2\text{ }×\text{ }\rho 3\text{
}+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }q\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }A6\\ +\text{ }\delta \text{ }\text{\hspace{0.17em}}×\text{ }{\
sigma }^{2}\text{ }×\text{ }\rho 1\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }\rho 2\text{ }+\text{ }\delta \text{ }×\text{ }{\sigma }^{2}\text{ }×\text{ }\rho 3\text{ }+\
text{ }{\sigma }^{4}\text{ }+\text{ }{\sigma }^{3}\text{ }×\text{ }\rho 2\text{ }+\\ \text{ }{\sigma }^{3}\text{ }×\text{ }{\sigma }^{3}\text{ }+\text{ }q\text{ }×\text{ }{\sigma }^{3}\text{ }+\text{
}{\sigma }^{3}\text{ }×\text{ }A6\text{ }+\text{ }{\sigma }^{3}\text{ }×\text{ }\rho 1\\ +\text{ }\delta \text{ }×\text{ }\rho \text{ }×\text{ }q\text{ }×\text{ }A6\text{ }+\text{ }\delta \text{ }×\
text{ }\rho \text{ }×\text{ }q\text{\hspace{0.17em}}×\text{ }\rho 1\text{ }+\text{ }\delta \text{ }×\text{ }\rho \text{ }×\text{ }q\text{\hspace{0.17em}}×\text{ }\rho 2\text{ }+\text{ }\delta \text{
}×\text{ }\rho \text{ }×\text{ }A6\text{ }×\text{ }\rho 1\text{ }+\\ \text{ }\delta \text{ }×\text{ }\sigma \text{ }×\text{ }\rho 1\text{ }×\text{ }\rho 2\text{\hspace{0.17em}}+\text{ }\delta \text{
}×\text{ }\sigma \text{ }×\text{ }\rho 2\text{ }×\text{ }\rho 3\\ +\text{ }\delta \text{\hspace{0.17em}}×\text{ }q\text{\hspace{0.17em}}×\text{ }A6\text{\hspace{0.17em}}×\text{ }\rho 1\text{ }+\text{
}\sigma \text{ }×\text{ }q\text{\hspace{0.17em}}×\text{ }A6\text{ }×\text{ }\rho 1\text{ }+\text{ }\sigma \text{ }×\text{ }A6\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\rho 1\text{\hspace{0.17em}}×
\text{ }\rho 3\end{array}$
$\begin{array}{l}A6-A7\text{\hspace{0.17em}}\ge \text{ }\frac{r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\Omega \left(G\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\delta \text{\hspace
{0.17em}}+\text{ }r1\text{ }×\text{ }\rho 1\text{ }×\text{ }\Omega \left(G\right)\text{ }×\text{ }\overline{A}\text{ }×\text{ }\sigma \text{ }-\text{ }A1}{A0}\text{ }\\ -\text{ }\frac{\left(\sigma \
text{ }+\text{ }\rho 1\text{ }×\text{ }\rho 2\right)\text{ }×\text{ }\left(\sigma \text{ }+\text{ }q\text{ }×\text{ }\rho 3\right)}{\rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }
+\text{ }\rho 1\text{ }×\text{ }\delta \text{ }+\text{ }\overline{A}}\end{array}$
If $\lambda \left(G\right)\text{ }\ge \text{ }\frac{\left(\sigma \text{ }+\text{ }\rho 1\text{ }×\text{ }\rho 2\right)\text{ }×\text{ }\left(\sigma \text{ }+\text{ }q\text{ }×\text{ }\rho 3\right)}{\
rho 1\text{ }×\text{ }\sigma \text{ }×\text{ }\overline{A}\text{ }+\text{ }\rho 1\text{ }×\text{ }\delta \text{ }+\text{ }\overline{A}}$
Then, Det(M) ≥ 0.
Therefore, D [ 0 ] is locally asymptotically stable.
Sensitivity Analysis of system (5.1)
This is to determine the key parameters that can throw system 5.1 into an endemic situation. That is, calculating the parameters shows the relative change of the model if the parameters change
slightly. Using [21,22] approach,
(8.1) $\frac{y}{D0}\text{ }×\text{\hspace{0.17em}}\frac{\partial D0}{\partial y}$
where y represents each parameter in the model we need to find its sensitivity, and with the meaning of D [ 0 ] . The sensitivity of the parameters is calculated thus:
The sensitivity of the following parameters in the model as given in section 3, Ω [ Max ] , r 1 , σ 1 , γ, αt , αc and others are investigated thus for example:
(8.2) $\frac{\Omega Max}{D0}\text{ }×\text{ }\frac{\partial D0}{\partial \Omega Max}\text{ }\ge \text{ }0$
which is positive. All other parameters are investigated in like manner. We observed also that r1, γ, a1, p1,T and B are also positive and sensitive to D [ 0 ] . The parameters that are positive like
Ω [ Max ] , r1 and γ will definitely increase and bring about the same proportion in D [ 0 ] .
Sensitivity analysis of G [ 0 ] , ρ [ 1 ] , ρ [ 3 ] and, σ showed a relationship that is inversely proportional to D [ 0 ] . This implies that an increase in H0, a2, a3, s and will bring a reduction
in the value of D [ 0 ] . However, in the real sense, this is not practicable except for a2. The implication of the sensitivity analysis results suggest that more efforts in particular should be
geared towards reducing the risk of contracting the disease represented by r [ 1 ] and the transmission rate represented by Ω [ Max ] .
Specifically, if γ = 0 that is, the rate at which we have immigrants is reduced to zero, this means the border are closed completely to immigrants, this will in turn reduce the basic reproduction
number ( D [ 0 ] ) will be but if the border is completely opened to immigrants then ( D [ 0 ] = 1), this means D [ 0 ] will increase. Here, if the disease is communicable, that means, the
probability of disease contraction by an individual (p1), rate of transmission per day Ω [ Max ] (a1) which is also the rate at which quarantined individuals progress to the infected compartment are
greater than zero (i.e, r [ 1 ] > 0, Ω [ Max ] > 0 and ρ [ 1 ] > 0). Hence, there is a problem at hand for the health workers to deal with. This scenario poses a pointer to the fact that public and
international health workers should ensure that transmission rate Ω [ Max ] , p1 and a1 are kept relatively low because the parameters are capable of increasing reproduction number D [ 0 ] . Keeping
the aforementioned parameters low, we believe, through various awareness programme or propagation of information on how to curb the disease transmission and protection strategies of individuals such
as maintaining good hygiene and placing restriction through various social media available will go a long way in curbing the spread of the disease. To achieve this feat, Government can institute test
centres and create isolation centres for those who come into the country during the outbreak while on the part of individuals within the locality of the outbreak who are not infectious maintain good
hygiene and be disciplined about it. This will ensure individuals that test positive are moved into isolation centres for further test, examination and treatment. Those who test negative will be
allowed to progress into susceptible population.
Results of Analysis
The numerical simulations with different scenarios are presented thus: Figure 1 shows income distribution scenario, which showed the majority of susceptible individuals are low income earners as
such, may not have access to health facilities in developing countries. The income distribution shows a normal distribution centered around the mean income. This provides a baseline for understanding
the socioeconomic spread within the population.
Significant disparities are visible, with minority groups showing higher infection and mortality rates. This can be attributed to systemic inequalities such as healthcare access, occupational risks,
and socioeconomic status. Ethnicity viz-a-vis infection rate and mortality scenario as presented in Figure 2 and Figure 3.
The disparities in the ethnic population showed that the more the population the more the risk of contracting the virus which in turn brings about increase in mortality as depicted in Figure 3.
In Figure 4, the transmission of the virus increased with the low income group and reduces as the income group becomes higher. Figure 5 depicts the mortality rate and region, it was discovered that
the mortality rate was higher in the urban region than that of rural region because of over concentration of individuals. Urban areas might show higher infection and mortality rates compared to rural
areas due to higher population density, increased person-to-person contact, and potentially overburdened healthcare systems in cities. The plot likely shows a negative correlation, indicating that
lower income levels are associated with higher infection rates. This suggests that lower-income individuals might have higher exposure due to essential jobs, crowded living conditions, or limited
access to protective measures. Similar to scenario in Figure 6, Figure 7 showed that there was inequalities in the infection rate of the regions, Urban region has higher infection rate.
Figure 8 and Figure 9 showed the infection rate and mortality rate respectively as against literacy level. The negative correlation here suggests that higher literacy levels are associated with lower
infection rates. Higher literacy likely leads to better understanding and adherence to public health guidelines, improving prevention efforts. This clearly showed that if the literacy level increases
considerably, the mortality will be greatly reduced. Higher literacy levels contribute to better health outcomes by enhancing public understanding of and compliance with health measures. Educational
initiatives that improve health literacy can be effective in controlling the spread of infectious diseases. Public health campaigns should be clear, accessible, and distributed through various media
to reach all literacy levels.
Conclusions and Recommendations
We analysed and examined inequalities within a non-linear compartmental mathematical model, utilizing integration to assess the effect of hygiene, awareness levels income distribution, education and
ethnicity as indicators of social discriminants on controlling the transmission dynamics of COVID-19 within the population under study. The model assumed susceptible individuals come into direct
contact with an infected individual thereby contract the infection and indirectly through environmental exposure to the Coronavirus. We also considered the saturation of COVID-19 education across the
population over time and the interruption of the transmission rate by improved hygiene practices. Initially, we conducted a qualitative analysis of the deterministic inequalities of non-linear
compartmental mathematical model, focusing on the positivity of solutions, boundedness, and the basic reproduction number D [ 0 ] when individuals were not practicing good hygiene was obtained. Our
findings indicated that the disease-free equilibrium is locally and globally stable if D [ 0 ] = 1 whereas an endemic state exist for D [ 0 ] > 1. Furthermore, sensitivity analysis of the system’s
key parameters to the reproduction number D [ 0 ] revealed that the probability of infection P [ 1 ] , the maximum transmission rate Ω [ Max ] , and the restriction rate γ are highly sensitive
factors. Our study underscores that information propagation regarding good hygiene practices over time induces behavioral changes that significantly reduce the number of quarantined and infected
individuals. Based on these findings, we make the following recommendations:
1. Restriction is identified as a highly sensitive parameter that increases the basic reproduction number ( D [ 0 ] ). Therefore, we recommend considering restrictions on all borders to control entry
into the country during outbreaks. These restrictions can vary in intensity depending on the severity of disease spread.
2. The sensitivity analysis highlights that infection probability increases significantly with direct contact with infected individuals. Thus, strict adherence to measures such as stay-at-home
policies, social distancing, and regular hand washing with alcohol-based sanitizers is crucial. These actions effectively interrupt virus transmission, as the virus requires a medium to spread.
3. Given the high sensitivity of the maximum transmission rate Ω [ Max ] , it is imperative to educate the public extensively on personal and community hygiene. Increased awareness through widespread
campaigns fosters behavioral changes that can mitigate transmission rates effectively.
4. Our study reveals that promoting good hygiene practices through education and awareness initiatives can induce attitudinal changes among individuals. Encouraging people to adopt healthy habits
protects them against the disease and contributes to public health efforts.
5. Information dissemination tends to degrade over time due to factors like complacency and resource limitations. Therefore, sustained educational campaigns on disease transmission prevention,
utilizing platforms such as social media, TV, radio, and talk shows, are essential.
Implementing these recommendations comprehensively by all stakeholders can potentially flatten the curve of COVID-19 infections. The analysis of our inequality model demonstrates that factors such as
hygiene levels, awareness, and healthcare access significantly influence infection and mortality rates. Addressing these disparities necessitates robust policies that support financially, enhance
healthcare infrastructure, promote educational initiatives, and tackle systemic inequalities. By doing so, public health responses can become more equitable and effective, thereby improving health
outcomes across all populations.
There is no funding for this research work.
Data Availability
There is no data used here. The research used simulated data to create different situation during the outbreak of disease.
Ethical Approval
No need of ethical approval in this research paper.
Declaration of Conflict of Interest
We confirmed that there is no conflict of interest among the authors.
The authors express their profound gratitude to the editor and reviewers for their valuable suggestions which have helped improve the paper.
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24.4 Energy in Electromagnetic Waves
Chapter 24 Electromagnetic Waves
24.4 Energy in Electromagnetic Waves
• Explain how the energy and amplitude of an electromagnetic wave are related.
• Given its power output and the heating area, calculate the intensity of a microwave oven’s electromagnetic field, as well as its peak electric and magnetic field strengths
Anyone who has used a microwave oven knows there is energy in electromagnetic waves. Sometimes this energy is obvious, such as in the warmth of the summer sun. Other times it is subtle, such as the
unfelt energy of gamma rays, which can destroy living cells.
Electromagnetic waves can bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. If the
frequency of the electromagnetic wave is the same as the natural frequencies of the system (such as microwaves at the resonant frequency of water molecules), the transfer of energy is much more
Connections: Waves and Particles
The behavior of electromagnetic radiation clearly exhibits wave characteristics. But we shall find in later modules that at high frequencies, electromagnetic radiation also exhibits particle
characteristics. These particle characteristics will be used to explain more of the properties of the electromagnetic spectrum and to introduce the formal study of modern physics.
Another startling discovery of modern physics is that particles, such as electrons and protons, exhibit wave characteristics. This simultaneous sharing of wave and particle properties for all
submicroscopic entities is one of the great symmetries in nature.
Figure 1. Energy carried by a wave is proportional to its amplitude squared. With electromagnetic waves, larger E-fields and B-fields exert larger forces and can do more work.
But there is energy in an electromagnetic wave, whether it is absorbed or not. Once created, the fields carry energy away from a source. If absorbed, the field strengths are diminished and anything
left travels on. Clearly, the larger the strength of the electric and magnetic fields, the more work they can do and the greater the energy the electromagnetic wave carries.
A wave’s energy is proportional to its amplitude squared ([latex]{E^2}[/latex] or [latex]{B^2}[/latex]). This is true for waves on guitar strings, for water waves, and for sound waves, where
amplitude is proportional to pressure. In electromagnetic waves, the amplitude is the maximum field strength of the electric and magnetic fields. (See Figure 1.)
Thus the energy carried and the intensity [latex]{I}[/latex] of an electromagnetic wave is proportional to [latex]{E^2}[/latex] and [latex]{B^2}[/latex]. In fact, for a continuous sinusoidal
electromagnetic wave, the average intensity [latex]{I_{\text{ave}}}[/latex] is given by
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{c \epsilon _0 E_0^2}{2}},[/latex]
where [latex]{c}[/latex] is the speed of light, [latex]{\epsilon _0}[/latex] is the permittivity of free space, and [latex]{E_0}[/latex] is the maximum electric field strength; intensity, as always,
is power per unit area (here in [latex]{\text{W/m}}^2[/latex]).
The average intensity of an electromagnetic wave [latex]{I_{\text{ave}}}[/latex] can also be expressed in terms of the magnetic field strength by using the relationship [latex]{B = E/c}[/latex], and
the fact that [latex]{\epsilon _0 = 1/ \mu _0 c^2}[/latex], where [latex]{\mu _0}[/latex] is the permeability of free space. Algebraic manipulation produces the relationship
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{c B_0^2}{2 \mu _0}},[/latex]
where [latex]{B_0}[/latex] is the maximum magnetic field strength.
One more expression for [latex]{I_{\text{ave}}}[/latex] in terms of both electric and magnetic field strengths is useful. Substituting the fact that [latex]{c \cdot B_0 = E_0}[/latex], the previous
expression becomes
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{E_0 B_0}{2 \mu _0}}.[/latex]
Whichever of the three preceding equations is most convenient can be used, since they are really just different versions of the same principle: Energy in a wave is related to amplitude squared.
Furthermore, since these equations are based on the assumption that the electromagnetic waves are sinusoidal, peak intensity is twice the average; that is, [latex]{I_0 = 2I_{\text{ave}}}[/latex].
Example 1: Calculate Microwave Intensities and Fields
On its highest power setting, a certain microwave oven projects 1.00 kW of microwaves onto a 30.0 by 40.0 cm area. (a) What is the intensity in [latex]{\text{W/m}^2}[/latex]? (b) Calculate the peak
electric field strength [latex]{E_0}[/latex] in these waves. (c) What is the peak magnetic field strength [latex]{B_0}[/latex]?
In part (a), we can find intensity from its definition as power per unit area. Once the intensity is known, we can use the equations below to find the field strengths asked for in parts (b) and (c).
Solution for (a)
Entering the given power into the definition of intensity, and noting the area is 0.300 by 0.400 m, yields
[latex]{I =}[/latex] [latex]{\frac{P}{A}}[/latex] [latex]{=}[/latex] [latex]{\frac{1.00 \;\text{kW}}{0.300 \;\text{m} \times 0.400 \;\text{m}}}.[/latex]
Here [latex]{I = I_{\text{ave}}}[/latex], so that
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{1000 \;\text{W}}{0.120 \;\text{m}^2}}[/latex] [latex]{= 8.33 \times 10^3 \;\text{W/m}^2}.[/latex]
Note that the peak intensity is twice the average:
[latex]{I_0 = 2I_{\text{ave}} = 1.67 \times 10^4 \;\text{W/m}^2}.[/latex]
Solution for (b)
To find [latex]{E_0}[/latex], we can rearrange the first equation given above for [latex]{I_{\text{ave}}}[/latex] to give
[latex]{E_0 =}[/latex] [latex]{(\frac{2 I_{\text{ave}}}{c \epsilon _0})}[/latex] [latex]{^{1/2}}.[/latex]
Entering known values gives
[latex]\begin{array}{r @{{}={}}l} {E_0}\;\;= & {\sqrt{\frac{2(8.33 \times 10^3 \;\text{W/m}^2)}{(3.00 \times 10^8 \;\text{m/s})(8.85 \times 10^{-12} \;\text{C}^2/ \text{N} \cdot \text{m}^2)}}} \\
[1em]\;= & {2.51 \times 10^3 \;\text{V/m}}. \end{array}[/latex]
Solution for (c)
Perhaps the easiest way to find magnetic field strength, now that the electric field strength is known, is to use the relationship given by
[latex]{B_0 =}[/latex] [latex]{\frac{E_0}{c}}.[/latex]
Entering known values gives
[latex]\begin{array}{r @{{}={}} l} {B_0}\;\;= & {\frac{2.51 \times 10^3 \;\text{V/m}}{3.0 \times 10^8 \;\text{m/s}}} \\[1em]\;= & {8.35 \times 10^{-6} \;\text{T}}. \end{array}[/latex]
As before, a relatively strong electric field is accompanied by a relatively weak magnetic field in an electromagnetic wave, since [latex]{B = E/c}[/latex], and [latex]{c}[/latex] is a large number.
Section Summary
• The energy carried by any wave is proportional to its amplitude squared. For electromagnetic waves, this means intensity can be expressed as
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{c \epsilon _0 E_0 ^2}{2}},[/latex]
where [latex]{I_{\text{ave}}}[/latex] is the average intensity in [latex]{\text{W/m}^2}[/latex], and [latex]{E_0}[/latex] is the maximum electric field strength of a continuous sinusoidal wave.
• This can also be expressed in terms of the maximum magnetic field strength [latex]{B_0}[/latex] as
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{cB_0 ^2}{2 \mu _0}}[/latex]
and in terms of both electric and magnetic fields as
[latex]{I_{\text{ave}} =}[/latex] [latex]{\frac{E_0 B_0}{2 \mu _0}}.[/latex]
• The three expressions for [latex]{I_{\text{ave}}}[/latex] are all equivalent.
Problems & Exercises
1: What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?
2: Find the intensity of an electromagnetic wave having a peak magnetic field strength of [latex]{4.00 \times 10^{-9} \;\text{T}}[/latex].
3: Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is
its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.
4: An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption
by the atmosphere or other objects, what is the intensity 30.0 km away? (Hint: Half the power will be spread over the area of a hemisphere.) (b) What is the maximum electric field strength at this
5: Suppose the maximum safe intensity of microwaves for human exposure is taken to be [latex]{1.00 \;\text{W/m}}[/latex]. (a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its
antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no
complications from absorption or reflection. (b) What is the maximum electric field strength at the safe intensity? (Note that early radar units leaked more than modern ones do. This caused
identifiable health problems, such as cataracts, for people who worked near them.)
6: A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of [latex]{7.50 \;\mu \text{V/m}}[/latex]. (See Figure
2.) (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of [latex]{1.50 \times 10^{13} \;\text{m}^2}
[/latex] (a large fraction of North America), how much power does it radiate?
Figure 2. Satellite dishes receive TV signals sent from orbit. Although the signals are quite weak, the receiver can detect them by being tuned to resonate at their frequency.
7: Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may
produce an electromagnetic wave with a maximum electric field strength of [latex]{1.00 \times 10^{11} \;\text{V/m}}[/latex] for a time of 1.00 ns. (a) What is the maximum magnetic field strength in
the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on a [latex]{1.00 - \text{mm}^2}[/latex] area?
8: Show that for a continuous sinusoidal electromagnetic wave, the peak intensity is twice the average intensity ([latex]{I_0 = 2I_{\text{ave}}}[/latex]), using either the fact that [latex]{E_0 = \
sqrt{2}E_{\text{rms}}}[/latex], or [latex]{B_0 = \sqrt{2}B_{\text{rms}}}[/latex], where rms means average (actually root mean square, a type of average).
9: Suppose a source of electromagnetic waves radiates uniformly in all directions in empty space where there are no absorption or interference effects. (a) Show that the intensity is inversely
proportional to [latex]{r^2}[/latex], the distance from the source squared. (b) Show that the magnitudes of the electric and magnetic fields are inversely proportional to [latex]{r}[/latex].
10: Integrated Concepts
An [latex]{LC}[/latex] circuit with a 5.00-pF capacitor oscillates in such a manner as to radiate at a wavelength of 3.30 m. (a) What is the resonant frequency? (b) What inductance is in series with
the capacitor?
11: Integrated Concepts
What capacitance is needed in series with an [latex]{800 - \mu \text{H}}[/latex] inductor to form a circuit that radiates a wavelength of 196 m?
12: Integrated Concepts
Police radar determines the speed of motor vehicles using the same Doppler-shift technique employed for ultrasound in medical diagnostics. Beats are produced by mixing the double Doppler-shifted echo
with the original frequency. If [latex]{1.50 \times 10^9 - \text{Hz}}[/latex] microwaves are used and a beat frequency of 150 Hz is produced, what is the speed of the vehicle? (Assume the same
Doppler-shift formulas are valid with the speed of sound replaced by the speed of light.)
13: Integrated Concepts
Assume the mostly infrared radiation from a heat lamp acts like a continuous wave with wavelength [latex]{1.50 \mu \text{m}}[/latex]. (a) If the lamp’s 200-W output is focused on a person’s shoulder,
over a circular area 25.0 cm in diameter, what is the intensity in [latex]{\text{W/m}^2}[/latex]? (b) What is the peak electric field strength? (c) Find the peak magnetic field strength. (d) How long
will it take to increase the temperature of the 4.00-kg shoulder by [latex]{2.00^{\circ} \;\text{C}}[/latex], assuming no other heat transfer and given that its specific heat is [latex]{3.47 \times
10^3 \;\text{J/kg} \cdot ^{\circ} \text{C}}[/latex]?
14: Integrated Concepts
On its highest power setting, a microwave oven increases the temperature of 0.400 kg of spaghetti by [latex]{45.0 ^{\circ} \text{C}}[/latex] in 120 s. (a) What was the rate of power absorption by the
spaghetti, given that its specific heat is [latex]{3.76 \times 10^3 \;\text{J/kg} \cdot ^{\circ} \text{C}}[/latex]? (b) Find the average intensity of the microwaves, given that they are absorbed over
a circular area 20.0 cm in diameter. (c) What is the peak electric field strength of the microwave? (d) What is its peak magnetic field strength?
15: Integrated Concepts
Electromagnetic radiation from a 5.00-mW laser is concentrated on a [latex]{1.00 - \text{mm}^2}[/latex] area. (a) What is the intensity in [latex]{\text{W/m}^2}[/latex]? (b) Suppose a 2.00-nC static
charge is in the beam. What is the maximum electric force it experiences? (c) If the static charge moves at 400 m/s, what maximum magnetic force can it feel?
16: Integrated Concepts
A 200-turn flat coil of wire 30.0 cm in diameter acts as an antenna for FM radio at a frequency of 100 MHz. The magnetic field of the incoming electromagnetic wave is perpendicular to the coil and
has a maximum strength of [latex]{1.00 \times 10^{-12} \;\text{T}}[/latex]. (a) What power is incident on the coil? (b) What average emf is induced in the coil over one-fourth of a cycle? (c) If the
radio receiver has an inductance of [latex]{2.50 \;\mu \text{H}}[/latex], what capacitance must it have to resonate at 100 MHz?
17: Integrated Concepts
If electric and magnetic field strengths vary sinusoidally in time, being zero at [latex]{t = 0}[/latex], then [latex]{E = E_0 \;\text{sin} \; 2 \pi ft}[/latex] and [latex]{B = B_0 \;\text{sin} \; 2
\pi ft}[/latex]. Let [latex]{f = 1.00 \;\text{GHz}}[/latex] here. (a) When are the field strengths first zero? (b) When do they reach their most negative value? (c) How much time is needed for them
to complete one cycle?
18: Unreasonable Results
A researcher measures the wavelength of a 1.20-GHz electromagnetic wave to be 0.500 m. (a) Calculate the speed at which this wave propagates. (b) What is unreasonable about this result? (c) Which
assumptions are unreasonable or inconsistent?
19: Unreasonable Results
The peak magnetic field strength in a residential microwave oven is [latex]{9.20 \times 10^{-5} \;\text{T}}[/latex]. (a) What is the intensity of the microwave? (b) What is unreasonable about this
result? (c) What is wrong about the premise?
20: Unreasonable Results
An [latex]{LC}[/latex] circuit containing a 2.00-H inductor oscillates at such a frequency that it radiates at a 1.00-m wavelength. (a) What is the capacitance of the circuit? (b) What is
unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
21: Unreasonable Results
An [latex]{LC}[/latex] circuit containing a 1.00-pF capacitor oscillates at such a frequency that it radiates at a 300-nm wavelength. (a) What is the inductance of the circuit? (b) What is
unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
22: Create Your Own Problem
Consider electromagnetic fields produced by high voltage power lines. Construct a problem in which you calculate the intensity of this electromagnetic radiation in [latex]{\text{W/m}^2}[/latex] based
on the measured magnetic field strength of the radiation in a home near the power lines. Assume these magnetic field strengths are known to average less than a [latex]{\mu \text{T}}[/latex]. The
intensity is small enough that it is difficult to imagine mechanisms for biological damage due to it. Discuss how much energy may be radiating from a section of power line several hundred meters long
and compare this to the power likely to be carried by the lines. An idea of how much power this is can be obtained by calculating the approximate current responsible for [latex]{\mu \text{T}}[/latex]
fields at distances of tens of meters.
23: Create Your Own Problem
Consider the most recent generation of residential satellite dishes that are a little less than half a meter in diameter. Construct a problem in which you calculate the power received by the dish and
the maximum electric field strength of the microwave signals for a single channel received by the dish. Among the things to be considered are the power broadcast by the satellite and the area over
which the power is spread, as well as the area of the receiving dish.
maximum field strength
the maximum amplitude an electromagnetic wave can reach, representing the maximum amount of electric force and/or magnetic flux that the wave can exert
the power of an electric or magnetic field per unit area, for example, Watts per square meter
Problems & Exercises
1: [latex]\begin{array}{r @{{}={}}l} {I}\;\;= & {\frac{c \epsilon _0 E_0^2}{2}} \\[1em]\;= & {\frac{(3.00 \times 10^8 \;\text{m/s})(8.85 \times 10^{-12} \text{C}^2 \text{/N} \cdot \text{m}^2)(125 \;\
text{V/m})^2}{2}} \\[1em]\;= & {20.7 \;\text{W/m}^2} \end{array}[/latex]
3: (a) [latex]{I = \frac{P}{A} = \frac{P}{\pi r^2} = \frac{0.250 \times 10^{-3} \;\text{W}}{\pi (0.500 \times 10^{-3} \;\text{m})^2} = 318 \;\text{W/m}^2}[/latex]
(b) [latex]\begin{array}{r @{{}={}}l} {I_{\text{ave}}}\;\;= & {\frac{cB_0^2}{2 \mu _0} \Rightarrow B_0 = (\frac{2 \mu _0I}{c})^{1/2}} \\[1em]\;= & {(\frac{2(4 \pi \times 10^{-7} \;\text{T} \cdot \
text{m/A})(318.3 \;\text{W/m}^2)}{3.00 \times 10^8 \;\text{m/s}})^{1/2}} \\[1em]\;= & {1.63 \times 10^{-6} \;\text{T}} \end{array}[/latex]
(c) [latex]\begin{array}{r @{{}={}}l}{E_0}\;\;= & {cB_0 = (3.00 \times 10^8 \;\text{m/s})(1.633 \times 10^{-6} \;\text{T})} \\[1em]\;= & {4.90 \times 10^2 \;\text{V/m}} \end{array}[/latex]
5: (a) 89.2 cm
(b) 27.4 V/m
7: (a) 333 T
(b) [latex]{1.33 \times 10^{19} \;\text{W/m}^2}[/latex]
(c) 13.3 kJ
9: (a) [latex]{I = \frac{P}{A} = \frac{P}{4 \pi r^2} \propto \frac{1}{r^2}}[/latex]
(b) [latex]{I \propto E_0^2, B_0^2 \Rightarrow E_0^2, B_0^2 \propto \frac{1}{r^2} \Rightarrow E_0, B_0 \propto \frac{1}{r}}[/latex]
11: 13.5 pF
13: (a) [latex]{4.07 \;\text{kW/m}^2}[/latex]
(b) 1.75 kV/m
(c) [latex]{5.84 \;\mu \text{T}}[/latex]
(d) 2 min 19 s
15: (a) [latex]{5.00 \times 10^3 \;\text{W/m}^2}[/latex]
(b) [latex]{3.88 \times 10^{-6} \;\text{N}}[/latex]
(c) [latex]{5.18 \times 10^{-12} \;\text{N}}[/latex]
17: (a) [latex]{t = 0}[/latex]
(b) [latex]{7.50 \times 10^{-10} \;\text{s}}[/latex]
(c) [latex]{1.00 \times 10^{-9} \;\text{s}}[/latex]
19: (a) [latex]{1.01 \times 10^6 \;\text{W/m}^2}[/latex]
(b) Much too great for an oven.
(c) The assumed magnetic field is unreasonably large.
21: (a) [latex]{2.53 \times 10^{-20} \;\text{H}}[/latex]
(b) L is much too small.
(c) The wavelength is unreasonably small.
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Quasi-finite morphism
In algebraic geometry, a branch of mathematics, a morphism f : X → Y of schemes is quasi-finite if it is of finite type and satisfies any of the following equivalent conditions:^[1]
• Every point x of X is isolated in its fiber f^−1(f(x)). In other words, every fiber is a discrete (hence finite) set.
• For every point x of X, the scheme f^−1(f(x)) = X ×[Y]Spec κ(f(x)) is a finite κ(f(x)) scheme. (Here κ(p) is the residue field at a point p.)
• For every point x of X, ${\displaystyle {\mathcal {O}}_{X,x}\otimes \kappa (f(x))}$ is finitely generated over ${\displaystyle \kappa (f(x))}$.
Quasi-finite morphisms were originally defined by
SGA 1 and did not include the finite type hypothesis. This hypothesis was added to the definition in
II 6.2 because it makes it possible to give an algebraic characterization of quasi-finiteness in terms of
For a general morphism f : X → Y and a point x in X, f is said to be quasi-finite at x if there exist open affine neighborhoods U of x and V of f(x) such that f(U) is contained in V and such that the
restriction f : U → V is quasi-finite. f is locally quasi-finite if it is quasi-finite at every point in X.^[2] A quasi-compact locally quasi-finite morphism is quasi-finite.
For a morphism f, the following properties are true.^[3]
• If f is quasi-finite, then the induced map f[red] between
reduced schemes
is quasi-finite.
• If f is a closed immersion, then f is quasi-finite.
• If X is noetherian and f is an immersion, then f is quasi-finite.
• If g : Y → Z, and if g ∘ f is quasi-finite, then f is quasi-finite if any of the following are true:
1. g is separated,
2. X is noetherian,
3. X ×[Z] Y is locally noetherian.
Quasi-finiteness is preserved by base change. The composite and fiber product of quasi-finite morphisms is quasi-finite.^[3]
If f is
at a point
, then
is quasi-finite at
. Conversely, if
is quasi-finite at
, and if also
${\displaystyle {\mathcal {O}}_{f^{-1}(f(x)),x}}$
, the local ring of
in the fiber
)), is a field and a finite separable extension of κ(
)), then
is unramified at
Finite morphisms are quasi-finite.^[5] A quasi-finite proper morphism locally of finite presentation is finite.^[6] Indeed, a morphism is finite if and only if it is proper and locally quasi-finite.^
[7] Since proper morphisms are of finite type and finite type morphisms are quasi-compact^[8] one may omit the qualification locally, i.e., a morphism is finite if and only if it is proper and
A generalized form of and quasi-separated. Let f be quasi-finite, separated and of finite presentation. Then f factors as ${\displaystyle X\hookrightarrow X'\to Y}$ where the first morphism is an
open immersion and the second is finite. (X is open in a finite scheme over Y.)
See also
• .
• .
• . Publications Mathématiques de l'IHÉS. 28: 5–255.
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Direct Variation Explained—Definition, Equation, Examples — Mashup Math (2024)
Your complete guide to direct variation and the direct variation equation
Whenever you are learning about linear functions and linear relationships in algebra, you will eventually come across a concept called Direct Variation, which refers to a proportional linear
relationship between two variables, x and y. This short guide will teach you everything you need to know about direct variation and covers the following topics:
• What is direct variation?
• What is the direct variation definition?
• What is the direct variation equation?
• Direct Variation Examples
• Which graph represents a function with direct variation?
Now, let’s start off by learning some key definitions and characteristics about the concept of direct variation.
What is Direct Variation?
Direct Variation Definition:
What is a direct variation? In math, direct variation is a proportional linear relationship between two variables that can be expressed as the equation y = kx, where y and x are variables and k is a
According to the direct variation definition, direct variation means that as x increases, y also increases proportionally. Similarly, as x decreases, y also decreases proportionally. In other words,
if two quantities are related to each other in such a way that when one quantity increases, the other quantity increases by a proportionally, then you can say that the quantities vary directly with
each other.
What are direct variations in real life? An example would be a train that is moving at a constant speed, the distance it covers over a given time period is directly proportional to its speed. If the
train doubles its speed, it will cover the twice the distance in that same amount of time. In this example, the direct variation formula would be y = kx where y equals distance travelled, k equals
the speed, and x equals time.
The Direct Variation Equation
The direct variation equation is of the form y = kx, where x and y are variables and k is the constant of proportionality.
The direct variation equation states that y varies directly with x, which essentially means that as x increases or decreases, y also increases or decreases proportionally.
For example, if y represents the total cost of buying x items that cost $7 each, then the direct variation equation would be
In this direct variation equation, 7 is the constant of proportionality, which represents the cost per item. And, for example:
• When x=2, y=14
• When x=3, y=21
• When x=10, y=70
Now that you are familiar with the direct variation definition and direct variation equation, let’s look at a few direct variation examples!
Direct Variation Examples: What is a Direct Variation?
In this next section, we will look at a few direct variation examples (equations and corresponding graphs). Before we look at the direct variation examples, it is important to note that any direct
variation equation of the form y = kx must be a linear function that passes through the origin (the point (0,0) ) because the constant of proportionality, k, represents the ratio of y and x when they
both are equal to zero.
In other words, whenever x=0, it must also be true that y=0, so the graph of any function with direct variation must pass through the origin.
Direct Variation Example #1
Let’s start by looking at a simple example of a linear function that has direct variation.
Consider the function y = 6x.
Notice that this function matches the direct variation equation y = kx where
• y represents the output value
• x represents the input value, and
• k=6, the constant of proportionality
So, for example:
• When x = 2, y = 12, because 12 = 6(2) → (2,12) is a point on the line
• When x = -1, y = -6, because -6 = 6(-1) → (-1,-6) is a point on the line
• When x = 0, y = 0, because 0 = 6(0) → (0,0) is a point on the line
By looking at the graph in Figure 04 below, it should be easy to see why the function y=6x has direct variation because the equation is of the form y=kx and the graph passes through the origin.
Conversely, consider the function y = 6x + 3
Notice that this function does not matches the direct variation equation y = kx because of the additional +3 term
Since this function is not of the form y = kx, it does not have direct variation.
By looking at the graph in Figure 05 below, you can see that the function y=6x+3 does not pass through the origin and, therefore, does not have direct variation even though it is a linear function.
Figure 06 below compares the functions y=6x and y=6x+3 to help you understand why y=6x has direct variation and why y=6x+3 does not have direct variation.
Direct Variation Example #2
Consider a situation where a construction worker is paid $44 hourly. The amount of money earned by the construction worker varies directly with the number of hours that they work.
So, the equation that could represent this scenario would be y = 44x, where
• y represents the output value (total amount of money earned)
• x represents the input value (total amount of hours worked)
• k=44, the worker’s hourly pay rate, which is a constant
So, for example:
• When x = 2, y = 88, because 88 = 44(2) → when they work 2 hours, they earn $88
• When x = 10, y = 440, because 440 = 44(10) → when they work 10 hours, they earn $440
• When x = 0, y = 0, because 0 = 44(0) → when they work 0 hours, they earn $0
By looking at the graph and table in Figure 07 below, you can see that the function y=44x has direct variation because it is linear, of the form y=kx, and it pass through the point (0,0).
Direct Variation Example #3
Finally, let’s looks at an example of a function with direct variation where the constant, k, is negative.
For example, the temperature of a gas varies directly with its volume. As the volume of the gas increases, the temperature of the gas decreases at a rate of -7.5. You can express this relationship
using the direct variation equation: y = -kx, where:
• y represents the temperature of the gas
• x represents the volume of the gas
• k=-7.5, the rate at which the volume of the gas decreases as the temperature increases
It is important to note that k being negative does not change the fact that this equation has direct variation, because the variables still change proportionally to each other.
Figure 08 below illustrates the table and graph of this direct variation function where k is negative.
Which Graph Represents a Function with Direct Variation?
In this final section, we will take a look at two sample math test questions where you have to determine which graph represents a function with direct variation.
These types of questions are typically in multiple choice form where you are given the graph of four functions and you have to determine which one of the four represents a function with direct
To successfully answer these types of questions, it is important to remember the following key characteristics of functions with direct variation:
• Any function with direct variation is of the form y=kx and must be linear
• Any function with direct variation will pass through the origin (the point (0,0)
With these key characteristics in mind, let’s move onto the examples of identifying a direct variation graph.
Example #1: Which Graph Represents a Function with Direct Variation?
Looking at the four graphs and knowing the key characteristics of direct variation functions, you can use process of elimination to find the correct answers.
For starters, notice that the graphs in Choice A and B are not linear, so you can eliminate them right away since direct variation functions are of the form y=kx and are always linear.
Now you have narrowed down the correct answer to either Choice C or D, both of which are linear.
However, we know that, in addition to being linear, a direct variation graph must pass through the origin point (0,0).
By looking at the graphs of Choice C and D, you can see that C does not pass through the origin, but D does, so you can conclude that the graph of Choice C does not have direct variation and that the
final answer is D.
Final Answer: Choice D represents a function with direct variation.
Example #2: Which Graph Represents a Function with Direct Variation?
Now let’s work through one more similar example of identifying the graph of a direct variation
Just like the last example, you want to use process of elimination to determine the correct answer.
However, unlike the last example, notice that all of the graphs are linear functions, so you can’t eliminate any non-linear choices. But, you know that a direct variation formula must not only be
linear, but it must also pass through the origin.
By looking at the four graphs, you can see that the only function that passes through the origin is Choice B, so you can conclude that it is the only function with direct variation.
Final Answer: Choice B represents a function with direct variation.
Keep Learning:
• How to Graph a Parabola in 3 Easy Steps
• The Vertical Line Test Explained
• Examples: Which Graph Represents a Function?
• What is the Formula for Slope? (and how to use it)
• Parent Functions and Parent Graphs Explained
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Search Tags: direct variation, direct variation equation, what is direct variation, direct variation definition, what are direct variations, what is a direct variation, which graph represents a
function with direct variation, direct variation examples, direct variation formula, direct variation graph, examples of a direct variation
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How good is Pythagoras’ theorem?
Welcome to Atharv’s Maths Blog! Today I will talk about Pythagoras’ theorem. If you don’t know, Pythagoras of Samos (550-495 BC) was a Greek thinker and the founder of the Pythagorean group. He
formulated a theorem stating that \(a^2+b^2=c^2\) for only specific values of \(a\), \(b\) and \(c\). This theorem has over 200 proofs, but what we are most interested in for this post is how to
generate those values of \(a\), \(b\) and \(c\). Let us take two integers \(m\) and \(n\) where \(m \ge 0\), \(n \ge 0\) and \(m \ge n\). Now the values of \(a\), \(b\) and \(c\) are given below. $$a
= m^2 – n^2$$ $$b = 2mn$$ $$c = m^2+n^2$$ Very interesting… Is that because \(b + c = (m+n)^2\)?
Yes! \(b+c\) works out to \(m^2+n^2+2mn\), which, because of the famous algebraic identity, is \((m+n)^2\). So let us look at some examples. So if \(m=9\) and \(n=5\). Then the values of \(a\), \(b\)
and \(c\) will be defined as follows: $$a=9^2 – 5^2=81-25=56$$ $$b=2\times9\times5=18\times5=90$$ $$c=9^2+5^2=81+25=106$$ So, as we can see, the triplets (Pythagoras’ triples or triplets are the
values of \(a\), \(b\) and \(c\)) are 56, 90 and 106. And, the square of the first number plus the square of the second number is equal to the square of the third number! $$56^2+90^2=3136+8100=11236=
106^2$$ Another example! If \(m=6\) and \(n=3\), then $$a=6^2-3^2=36-9=27$$ $$b=2\times6\times3=12\times3=36$$ $$c=6^2+3^2=36+9=45$$ The triplets in this case are 27, 36 and 45. And, 27 squared + 36
squared is indeed 45 squared! $$27^2+36^2=729+1296=2025=45^2$$ So now we know how to generate Pythagoras’ triplets. I hope I was able to make it interesting.
Right now, I am the only blogger on this blog, but I will add support for third-party bloggers, logins and signups.
This is the end of my first mathematical blog post. Have a nice day! See you next time!
7 Replies to “How good is Pythagoras’ theorem?”
1. Very Well explained Atharv. Truly impressive.
I wish you all the best for more & more blog writing in future.
1. Thank you ma’am!
2. Wow !! That’s so well explained Atharva..
That took me back to my school days
Keep it up ..n come up with more informative work like this..
1. Thank you!!
I will work on more informative work.
4. Atharv as usual spectacular.Everytime you post matter on Mathematics it amazes me about your critical thinking and how well you put down concepts.Keep going.Sky is the limit.
1. Thank you shaefali ma’am.
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EViews Help: @rfirst
@rfirst Element Functions
First non-missing value in rows of group.
Value of the first non-missing value in each row of the group.
Syntax: @rfirst(x)
x: group
Return: series
show @rfirst(g
returns a linked series of the first non-missing observations in the rows of group g.
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Books – Eugenia Cheng
How Simple Questions Lead Us to Mathematics’ Deepest Truths
One of the world’s most creative mathematicians offers a new way to look at math—focusing on questions, not answers
Where do we learn math: From rules in a textbook? From logic and deduction? Not really, according to mathematician Eugenia Cheng: we learn it from human curiosity—most importantly, from asking
questions. This may come as a surprise to those who think that math is about finding the one right answer, or those who were told that the “dumb” question they asked just proved they were bad at
math. But Cheng shows why people who ask questions like “Why does 1 + 1 = 2?” are at the very heart of the search for mathematical truth.
Is Math Real? is a much-needed repudiation of the rigid ways we’re taught to do math, and a celebration of the true, curious spirit of the discipline. Written with intelligence and passion, Is Math
Real? brings us math as we’ve never seen it before, revealing how profound insights can emerge from seemingly unlikely sources.
Empty tab. Edit page to add content here.
An Exploration of Math, Category Theory, and Life
Mathematician and popular science author Eugenia Cheng is on a mission to show you that mathematics can be flexible, creative, and visual. This joyful journey through the world of abstract
mathematics into category theory will demystify mathematical thought processes and help you develop your own thinking, with no formal mathematical background needed. The book brings abstract
mathematical ideas down to earth using examples of social justice, current events, and everyday life – from privilege to COVID-19 to driving routes. The journey begins with the ideas and workings of
abstract mathematics, after which you will gently climb toward more technical material, learning everything needed to understand category theory, and then key concepts in category theory like natural
transformations, duality, and even a glimpse of ongoing research in higher-dimensional category theory. For fans of How to Bake Pi, this will help you dig deeper into mathematical concepts and build
your mathematical background
Bake Infinite Pie with X + Y
By Eugenia Cheng, Illustrated by Amber Ren
Aspiring bakers will embrace this charming picture book about baking pie by using simple math, from one of the world’s most creative and celebrated mathematicians.
X + Y are dreaming of baking infinite pie. But they don’t know if infinite pie is real. With the help of quirky and uber-smart Aunt Z, and a whole lot of flour and butter, X and Y will learn that by
using math they can bake their way to success!
This charming and tasty story from mathematician and author of How to Bake Pi, Eugenia Cheng, reassures young readers that math doesn’t have to be scary—especially when paired with pie!
Additional back matter includes: a letter from Eugenia encouraging readers not to be intimidated by math, explanations of the math concepts explored in the book, and a recipe for Banana Butterscotch
Molly and the Mathematical Mysteries
Ten Interactive Adventures in Mathematical Wonderland
By Eugenia Cheng, Illustrated by Aleksandra Artymowska
Join Molly as she ventures into a curious world where nothing is quite as it seems. A trail of clues leads from scene to scene, presenting Molly with a number of challenges. But who is leaving the
clues, and where will they lead? This interactive mystery shows math isn’t just about numbers—it’s about imagination! An explorative and creative approach to the world of mathematics.
x + y
A Mathematician’s Manifesto for Rethinking Gender
Why are men in charge? After years in the male-dominated field of mathematics and in the female-dominated field of art, Eugenia Cheng has heard the question many times. In x + y, Cheng argues that
her mathematical specialty — category theory — reveals why. Category theory deals more with context, relationships, and nuanced versions of equality than with intrinsic characteristics. Category
theory also emphasizes dimensionality: much as a cube can cast a square or diamond shadow, depending on your perspective, so too do gender politics appear to change with how we examine them. Because
society often rewards traits that it associates with males, such as competitiveness, we treat the problems those traits can create as male. But putting competitive women in charge will leave many
unjust relationships in place. If we want real change, we need to transform the contexts in which we all exist, and not simply who we think we are.
The Art of Logic
How to Make Sense in a World that Doesn’t
How both logical and emotional reasoning can help us live better in our post-truth world
In a world where fake news stories change election outcomes, has rationality become futile? In The Art of Logic in an Illogical World, Eugenia Cheng throws a lifeline to readers drowning in the
illogic of contemporary life. Cheng is a mathematician, so she knows how to make an airtight argument. But even for her, logic sometimes falls prey to emotion, which is why she still fears flying and
eats more cookies than she should. If a mathematician can’t be logical, what are we to do? In this book, Cheng reveals the inner workings and limitations of logic, and explains why alogic–for
example, emotion–is vital to how we think and communicate. Cheng shows us how to use logic and alogic together to navigate a world awash in bigotry, mansplaining, and manipulative memes. Insightful,
useful, and funny, this essential book is for anyone who wants to think more clearly.
With humour, grace, and a natural gift for making explanations seem fun, Eugenia Cheng has done it again. This is a book to savour, to consult, and to buy for all your friends. You’ll think more
clearly after reading this book, something that is unfortunately in short supply these days. I am buying several copies to send to heads of state.
Witty, charming, and crystal clear. Eugenia Cheng’s enthusiasm and carefully chosen metaphors and analogies carry us effortlessly through the mathematical landscape
Clear, clever and friendly
It takes a talented writer to bring the concept of infinity to life, but Cheng’s infectious enthusiasm makes maths a delight
A concert pianist, mathematician, polyglot and YouTube star, Cheng has carved out quite a niche for herself … she brings an ebullient enthusiasm that’s infectious
Beyond Infinity:
An Expedition to the Outer Limits of Mathematics
The hilarious and charming Eugenia Cheng leads us in search of what’s bigger than infinity, and smaller than its opposite
Imagine something small enough to fit in your head but too large to fit in the world-or even the universe. What would you call it? And what would it be? How about…infinity?
In Beyond Infinity, musician, chef, and mathematician Eugenia Cheng answers this question by taking readers on a startling journey from math at its most elemental to its loftiest abstractions.
Beginning with the classic thought experiment of Hilbert’s hotel-the place where you can (almost) always find a room, if you don’t mind being moved from room to room over the course of the night-she
explores the wild and woolly world of the infinitely large and the infinitely small. Along the way she considers weighty questions like why some numbers are uncountable or why infinity plus one is
not the same as one plus infinity. She finds insight in some unlikely examples: planning a dinner party for 7 billion people using a chessboard, making a chicken-sandwich sandwich, and creating
infinite cookies from a finite ball of dough all tell you more about math than you could have imagined.
An irresistible book on the universe’s biggest possible topic, Beyond Infinity will beguile and bewitch you, and show all of us how one little symbol can hold the biggest idea of all.
“Beyond Infinity is witty, charming, and crystal clear. Eugenia Cheng’s enthusiasm and carefully chosen metaphors and analogies carry us effortlessly through the mathematical landscape of the
infinite. A brilliant book!”
―Ian Stewart, author of Calculating the Cosmos
“The idea of infinity is one of the most perplexing things in mathematics, and the most fun. Eugenia Cheng’s Beyond Infinity is a spirited and friendly guide―appealingly down to earth about math
that’s extremely far out.”
―Jordan Ellenberg, author ofHow Not to Be Wrong and professor of mathematics at University of Wisconsin-Madison
How to Bake Pi:
An Edible Exploration of the Mathematics of Mathematics
What is math? How exactly does it work? And what do three siblings trying to share a cake have to do with it? In How to Bake Pi, math professor Eugenia Cheng provides an accessible introduction to
the logic and beauty of mathematics, powered, unexpectedly, by insights from the kitchen. We learn how the béchamel in a lasagna can be a lot like the number five, and why making a good custard
proves that math is easy but life is hard. At the heart of it all is Cheng’s work on category theory, a cutting-edge “mathematics of mathematics,” that is about figuring out how math works.
Combined with her infectious enthusiasm for cooking and true zest for life, Cheng’s perspective on math is a funny journey through a vast territory no popular book on math has explored before. So,
what is math? Let’s look for the answer in the kitchen.
“A slyly illuminating dispatch on the deep meaning of mathematics. . . . [Cheng] compels us to see numbers and symbols as vivid characters in an ongoing drama, a narrative in which we are alternately
observers and participants.”
—Natalie Angier, The American Scholar
“Invoking plenty of examples from cooking and baking, as well as other everyday-life situations such as calculating a taxi fare, searching for love through online dating services and training for a
marathon, [Cheng] explains abstract mathematical ideas—including topology and logic—in understandable ways. . . .Her lively, accessible book demonstrates how important and intriguing such a pursuit
can be.”
—Scientific American
“A funny and engaging new book.”
—Simon Worrall, National Geographic News
“Why go to all the trouble to write a book to help people understand mathematics? Because, as Cheng observes, ‘understanding is power, and if you help someone understand something, you’re giving them
power.’ Read How to Bake Pi and you will, indeed, go away feeling empowered.”
—Marc Merlin, Medium
“In her new book, How to Bake Pi, mathematician/baker Eugenia Cheng offers a novel, mathematical approach to cooking. . . . How to Bake Pi is more than a mathematically minded cookbook. It is just as
much a book about mathematical theory and how we learn it. The premise at the heart of the book is that the problem that stops a cookbook from teaching us how to cook is the same problem that makes
math classes so bad at actually teaching us to do math.”
—Ria Misra, io9
“[Cheng’s] book, a very gentle introduction to the main ideas of mathematics in general and category theory in particular, exudes enthusiasm for mathematics, teaching, and creative recipes. Category
theory is dangerously abstract, but Cheng’s writing is down-to-earth and friendly. She’s the kind of person you’d want to talk to at a party, whether about math, food, music, or just the
weather. . . . Cheng’s cheerful, accessible writing and colorful examples make How to Bake Pi an entertaining introduction to the fundamentals of abstract mathematical thinking.”
—Evelyn Lamb, Scientific American’s Roots of Unity blog
“This is the best book imaginable to introduce someone who doesn’t think they are interested in mathematics at all to some of the deep ideas of category theory, especially if they like to bake.”
—MAA Reviews
“Beginning each chapter with a recipe, Cheng converts the making of lasagna, pudding, cookies, and other comestibles into analogies illuminating the mathematical enterprise. Though these culinary
analogies teach readers about particular mathematical principles and processes, they ultimately point toward the fundamental character of mathematics as a system of logic, a system presenting
daunting difficulties yet offering rare power to make life easier. Despite her zeal for mathematical logic, Cheng recognizes that such logic begins in faith—irrational faith—and ultimately requires
poetry and art to complement its findings. A singular humanization of the mathematical project.”
—Booklist, starred review
“Cheng is exceptional at translating the abstract concepts of mathematics into ordinary language, a strength aided by a writing style that showcases the workings of her curious, sometimes whimsical
mind. This combination allows her to demystify how mathematicians think and work, and makes her love for mathematics contagious.”
—Publishers Weekly, starred review
“An original book using recipes to explain sophisticated math concepts to students and even the math-phobic. . . . [Cheng] is a gifted teacher. . . . A sharp, witty book to press on students and even
the teachers of math teachers.”
—Kirkus Reviews
“A well-written, easy-to-read book.”
—Library Journal
“Often entertaining . . . frequently illuminating. . . . [How to Bake Pi] offers enough nourishment for the brain to chew on for a long time.”
—Columbus Dispatch
“Through an enthusiasm for cooking and zest for life, the author, a math professor, provides a new way to think about a field we thought we knew.”
—CEP Magazine
“Combined with infectious enthusiasm for cooking and a zest for life, Cheng’s perspective on math becomes this singular book: a funny, lively, and clear journey no popular book on math has explored
before. How to Bake Pi . . . will dazzle, amuse, and enlighten.”
—Gambit Weekly
“This book was fun and covered some cool maths, using some nice analogies, and would serve as a good intro for someone getting into category theory.”
—Aperiodical (UK)
“Eugenia Cheng offers an entertaining introduction to the beauty of mathematics by drawing on insights from the kitchen. She explains why baking a flourless cake is like geometry and offers puzzles
to whet the appetites of maths fans.”
—Times Educational Supplement (UK)
“Quirky recipes, personal anecdotes, and a large dollop of equations are the key ingredients in this alternative guide to maths and the scientific process. You should find it as easy as cooking a
—Observer: Tech Monthly (UK)
“A curious cookbook for the mathematical omnivore.”
—Irish Times (Ireland)
“Eugenia Cheng’s charming new book embeds math in a casing of wry, homespun metaphors: math is like vegan brownies, math is like a subway map, math is like a messy desk. Cheng is at home with math
the way you’re at home with brownies, maps, and desks, and by the end of How to Bake Pi, you might be, too.”
—Jordan Ellenberg, professor of mathematics, University of Wisconsin–Madison, and author of How Not to Be Wrong
“With this delightfully surprising book, Eugenia Cheng reveals the hidden beauty of mathematics with passion and simplicity. After reading How to Bake Pi, you won’t look at math (nor porridge!) in
the same way ever again.”
—Roberto Trotta, astrophysicist, Imperial College London, and author of The Edge of the Sky
“Math is a lot like cooking. We start with the ingredients we have at hand, try to cook up something tasty, and are sometimes surprised by the results. Does this seem odd? Maybe in school all you got
was stale leftovers! Try something better: Eugenia Cheng is not only an excellent mathematician and pastry chef, but a great writer, too.”
—John Baez, professor of math, University of California, Riverside
“From clotted cream to category theory, neither cookery nor math are what you thought they were. But deep down they’re remarkably similar. A brilliant gourmet feast of what math is really about.”
—Ian Stewart, emeritus professor of mathematics, University of Warwick, and author of Visions of Infinity and Professor Stewart’s Incredible Numbers
“This book puts the fun back in math, the fun that I always saw in it, the fun that is nearly sucked from it in K–12 education. . . . I whole-heartedly recommend this book to anyone with a casual
interest in, or deep love of, logic, or mathematics, or baking.”
—Melissa A. Wilson Sayres, assistant professor, School of Life Sciences and the Biodesign Institute, Arizona State University, and writer of mathbionerd.blogspot.com
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EURNZD Pip Value Calculator - How to Calculate - Get Know Trading
The pip calculator in Forex represents a Forex calculator that calculates the value of a pip in the currency you want by defining following values:
• number of pips
• lot size used
• currency pair
• deposit currency
Why do you need EURNZD pip value calculator?
You can calculate the value for a number of pips. And that means for example if you have 100 pips as a stop loss or take profit set, then you can calculate how much that will be in terms of a
currency you select.
How to use EURNZD pip value calculator?
Inside the calculator you have several fields you need to fill with the data. And those are the number of pips, currency pair, deposit currency and lot size. At the end you click the button Calculate
and you get the value of a pip.
In this article I will show you all details you need to know about EURNZD pip value calculator and what you can get with using it, why you should use it to speed up the process of calculating pip
value and how to use it so you do not get confused when looking at pip calculator.
Read more: What is Pip Value Calculator
EURNZD Pip Value Calculator Example
The best way to explain how the EURNZD calculator works is to show an example. I will make a few examples so you can see the difference in calculations.
First example is when you have EURNZD currency pair with EUR as a deposit currency:
• Number of pips: 1
• Instrument: EURNZD
• Lot size: 1.00 (100,000 units)
• Deposit currency: EUR
• EURNZD pip size: 0.0001
Second example is when you have EURNZD currency pair with NZD as a deposit currency:
• Number of pips: 1
• Instrument: EURNZD
• Lot size: 1.00 (100,000 units)
• Deposit currency: NZD
• EURNZD pip size: 0.0001
How to Calculate Pips for EURNZD
To calculate pips for EURNZD you need to use following formula which defines the pip value:
For deposit currency which is equal to base currency, EUR:
Pip value = (Pip / Current market price) x Lot size
For deposit currency which is equal to quote currency, NZD:
Pip value = Pip x lot size
EURNZD Pip Value
Now with the formula you have you need to use the first formula where the deposit currency is EUR.
Pip Value for Base Currency
For deposit currency which is equal to base currency, EUR, pip value will be equal to:
Pip value = (Pip / Current market price) x Lot size
Here are other data you need:
• Number of pips: 1
• Instrument: EURNZD = 0.8246
• Lot size: 1.00 (100,000 units)
• Deposit currency: EUR
• EURNZD pip size: 0.0001
Now, when you put all the data in the formula you get:
Pip value = (Pip / Current market price) x Lot size
Pip value = (0.0001 / 0.8246) x 100,000
Pip value = (1.213e-4) x 100,000
Pip value = 12.13
Pip Value for Quote Currency
For deposit currency which is equal to quote currency, NZD, pip value will be equal to:
Pip value = Pip x Lot size
Here are other data you need:
• Number of pips: 1
• Instrument: EURNZD = 0.8246
• Lot size: 1.00 (100,000 units)
• Deposit currency: NZD
• EURNZD pip size: 0.0001
Now, when you put all the data in the formula you get:
Pip value = Pip x Lot size
Pip value = 0.0001 x 100,000
Pip value = NZD$10,00
Pip Value for Third Currency
Third case is when you have a EURNZD currency pair, but the deposit currency is EUR. Which is not EUR or NZD.
This case requires that you make more calculations. If you use the EURNZD calculator then the whole calculation is done by the pip calculator.
But, if you want to do it manually, then you need to use the following process.
• decide in which currency you will calculate the pip value. Will that be EUR or NZD
Let’s use EUR. The formula for the pip value will be:
• Number of pips: 1
• Instrument: EURNZD = 0.8246
• Lot size: 1.00 (100,000 units)
• Deposit currency: EUR
• EURNZD pip size: 0.0001
Pip value = (Pip / Current market price) x Lot size
Pip value = (0.0001 / 0.8246) x 100,000
Pip value = (1.213e-4) x 100,000
Pip value = 12.13
Now, you need to use the EUR/USD currency pair so you can extract EUR pip value from it.
Current market price for the EUR/USD = 0.6217. Which gives us EUR = 0.6217 USD.
So, the formula would be:
Pip value (EUR) = Pip value (EUR) x (EUR/USD)
Pip value (EUR) = 12.13 x 0.6217
Pip value (EUR) = 7.54
How Do You Calculate EURNZD Pip Profits?
With the above calculated you can calculate EURNZD pip profit.
Let’s say you have an open SELL order on the market with the EURNZD = 0.8246.
And you want to close the trade at EURNZD ? 0.8226.
The price difference in pips is:
Pips = |Entry price – Exit price|
Pips = |0.8246 – 0.8226|
Pips = 0.20
You can see the difference is 20 pips between open and close price.
Now, the profit for 20 pips is:
Profit = Pip value x Pips
Profit = 7.54 x 20
Profit = $150.8
If you want to use EUR as a deposit currency then the pip value is 12.13:
Profit = Pip value x Pips
Profit = 12.13 x 20
Profit = 242.6
If you are beginner in Forex trading you will need this calculator because the calculation manually is not so easy.
Even the Forex basics is not so easy to understand, but in time you learn everything.
Use this calculator and other calculators to see how much you will make per pip so you can set stop loss and take profit properly.
Calculation`s made in the trading calculator are for informational purposes only. Whilst every effort is made to ensure the accuracy of this information, you should not rely upon it as being complete
or up to date. Furthermore this information may be subject to change at any time.
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The Structure of MLPsThe Structure of MLPs - Dividend Monk
This is part of MLPs: The Essential Guide.
The structure of MLPs can vary to some extent, but the basic structure is this: There exists a General Partner (GP), that consists of the management team, and Limited Partners (LPs), that contribute
capital. Often the entity that holds the GP also holds some LP units.
Limited Partners contribute capital by buying units of the MLP. (They’re called units rather than shares.) These LP units can then be traded on an exchange. In return, the LPs receive distributions
from the operations, which are like dividends.
The GP often has Incentive Distribution Rights (IDRs) that are set forth in the founding of the partnership. The IDRs give the GP a ton of incentive to raise the distribution over time for the units
that the LPs hold. These partnerships, therefore, are specifically designed for distribution growth.
A typical IDR structure starts off with the GP receiving 2% of the cash flow from the MLP, while the LPs receive the other 98%. For example, the MLP may start off by paying $0.25 per quarter to each
LP unit as a distribution, with the GP only receiving a small sum. There will then be pre-defined tiers for which the GP starts getting paid a bigger share. In this example, the structure may be that
once the per-unit distributions reach $0.35, the GP begins receiving 15% of the cash flow above that point, with the LPs receiving the other 85%. And then if the distribution reaches $0.45, the GP
begins receiving 25% of the cash flow above that point, with the LPs receiving the other 75%. And then finally, if the distribution reaches $0.55 or above, the GP receives a full 50% of the cash flow
above that point, with the LPs receiving the other 50%.
It’s tiered, so management receives the pre-defined percentage of the total cash above that tier. So as an example, if the partnership reaches $0.39 in distributions per unit per quarter, that
doesn’t mean the GP receives a full 15% of the cash. They receive 2% of the cash below the tier, and then 15% of the cash above $0.35. Kind of like tax brackets.
The purpose of IDRs is to align management interest with limited partner interest. As management increases the per-unit distributions to the LPs, their own share of the cash grows even more quickly.
So while Limited Partners may not particularly enjoy giving a lot of money to the GP at later stages in the MLP, it would be hard for them to complain, because if they get to that point, they’ve
likely outperformed most other investments. The agreement is designed to give incentive to the GP to raise quarterly distributions far beyond the final tier.
To grow the business, the General Partner will usually issue new units. This is the opposite of corporations, that if they are shareholder friendly, typically wish to avoid share dilution and even
repurchase their own shares to reduce the number of shares outstanding. Since MLPs are tax-advantaged, however, if they can get more capital, they can generally provide good returns to all new and
existing unitholders.
An Example
This can be demonstrated with a rather simple and clear example.
Suppose there exists an MLP that currently has 100 million units, and each unit costs $50 and pays out $3 in annual distributions for a 6% yield. The market capitalization of this MLP, therefore, is
$5 billion, and it pays out $300 million in distributions per year out of the $330 million in total free cash flows that it generates.
The MLP is paying out most of its cash flow, and so there is little capital left to grow the business and the distribution.
However, management has a good growth opportunity: they can build a pipeline that connects from their network to a storage hub, and all of their research indicates that this will be a good
investment. It will cost $2 billion to build, and will produce $300 million in annual cash flow that grows at least with inflation, and will have $100 million in annual capital expenditures and
maintenance. In addition, they can finance part of the construction with $1 billion in debt at 5% interest, since the assets should produce very reliable cash flows. So for this investment, they’ll
have to pay $1 billion in upfront costs, and they’ll receive $300 million in cash flows and have to pay $100 million for expenditures and maintenance and $50 million in interest, for a total
remaining profitable cash flow of $150 million. This would be a 15% return on their $1 billion in upfront equity.
The MLP makes this investment by issuing 20 million new units at $50 per unit. So now, there are 120 million units, and the partnership is bringing in $480 in free cash flow (the original $330
million plus the new $150 million). Management calculates they can pay out $450 million as distributions, which divided up into 120 million units equals $3.75 per unit per year (paid quarterly),
which is a 25% increase over the original distribution of $3. So they managed to pay higher distributions to more units. Pretty impressive.
That’s generally how MLPs grow. They continue to make new investments in the form of new construction or acquisitions funded by issuing units, and if management is good, they’ll make sure that the
new investments that they issue new units for are beneficial to everyone.
In reality, the example would be a bit more complicated, because we’d have to factor in the incentive distribution rights to management. We can’t really expect a 6% distribution yield and continued
25% distribution growth, but it’s not unreasonable to expect mid-to-high single digit distribution yields and mid-to-high single digit distribution growth for total annual returns in the low double
digits. As far as dividend stocks go, this is pretty good.
The main emphasis on MLPs is that their structure is beautiful for the general partner, and pretty good for the limited partners as well. The tax-advantaged structure allows for rapid new capital
additions in the form of new units issued, and the IDRs of the general partner encourage this type of growth as much as possible.
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Generating all elements of $SL_2(\mathbb{F}_q)$
Generating all elements of $SL_2(\mathbb{F}_q)$
What is the fastest way to create a set with all elements of $SL_2(\mathbb{F}_q)$ for $q$ some prime power of size about $50$? One way to do it is to find a representative for each conjugacy class
and then use the ConjugacyClass method, but this seems inefficient. Is there a better way?
1 Answer
Sort by ยป oldest newest most voted
You can define your group as follows:
sage: G = SL(2,GF(53))
Then you can get the list of its elements with the list method:
sage: L = G.list()
On my 5+ years old laptop, it takes about 44 seconds.
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Derivation of the Variance of Autocorrelation
• I
• Thread starter mertcan
• Start date
In summary, the conversation discusses the calculation of autocorrelation and its variance at a specific lag. The formula for calculating the variance is shown, but the poster is looking for a proof
of this formula. They have searched for it online but have not found it.
Hi everyone in this link (
) I see the variance of autocorrelation related to specific lag is demonstrated in the following: $$ Var(r_k) = \frac {\sum_{t=k+1}^n a_t*a_{t-k}} {\sum_{t=1}^n a_t^2}$$ where ##r_k## is
autocorrelation at relevant lag, ##n## is the number of data set and ##a_t## is error. Could help me prove the formula I mentioned above?
Last edited:
mertcan said:
Hi everyone in this link (
) I see the variance of autocorrelation related to specific lag is demonstrated in the following: $$ Var(r_k) = \frac {\sum_{t=k+1}^n a_t*a_{t-k}} {\sum_{t=1}^n a_t^2}$$ where ##r_k## is
autocorrelation at relevant lag, ##n## is the number of data set and ##a_t## is error. Could help me prove the formula I mentioned above?
Guys sorry for wrong question. Please let me rectify it.
I have seen the following formula whereas $$ r_k = \frac {\sum_{t=k+1}^n a_t*a_{t-k}} {\sum_{t=1}^n a_t^2}$$ $$Var(r_k) = \frac {n-k}{n*(n+2)}$$ where $r_k$ is the autocorrelation at relevant lag,
$n$ is the number of points in the data set, and $a_t$ is the error.
I have searched the internet for the proof for variance equation, but I haven't found it. Could anyone help me prove the formula I mentioned above?
FAQ: Derivation of the Variance of Autocorrelation
1. What is the purpose of deriving the variance of autocorrelation?
The variance of autocorrelation is a statistical measure used to determine the strength and direction of a relationship between a variable and its past values. By deriving this measure, scientists
can better understand the patterns and trends in their data and make more accurate predictions.
2. How is the variance of autocorrelation calculated?
The variance of autocorrelation is calculated by taking the squared difference between each data point and its lagged value, and then dividing by the total number of observations. This calculation is
repeated for each lag until all possible lags have been considered.
3. What factors can affect the variance of autocorrelation?
There are several factors that can affect the variance of autocorrelation, including the strength and direction of the relationship between the variable and its past values, the number of
observations, and the distribution of the data. Additionally, the presence of outliers or missing data can also impact the variance of autocorrelation.
4. How can the variance of autocorrelation be interpreted?
The variance of autocorrelation is typically interpreted as a measure of the degree of dependence between a variable and its past values. A high variance of autocorrelation indicates a strong
relationship between the variable and its past values, while a low variance indicates a weak or non-existent relationship.
5. What are some potential applications of the variance of autocorrelation?
The variance of autocorrelation has many applications in various fields, including finance, economics, and meteorology. It can be used to analyze time series data, make predictions, and identify
patterns and trends. Additionally, it is often used in quality control and process improvement to detect any systematic changes in a process over time.
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Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero - FasterCapital
Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
1. Introduction to Internal Rate of Return (IRR)
Internal Rate of Return (IRR) is a crucial financial metric used to evaluate the profitability and viability of an investment or project. It represents the discount rate at which the Net present
Value (NPV) of the cash flows generated by the investment becomes zero. In simpler terms, irr is the rate of return that an investment is expected to generate over its lifespan.
1. Insights from different perspectives:
- From an investor's standpoint, IRR helps assess the attractiveness of an investment opportunity by comparing it to alternative investments. A higher IRR indicates a more lucrative investment.
- From a company's perspective, IRR assists in determining whether a project will generate sufficient returns to cover the cost of capital. If the IRR exceeds the company's required rate of return,
the project is considered financially viable.
2. In-depth information:
- IRR takes into account the timing and magnitude of cash flows. It considers both the initial investment and the subsequent cash inflows or outflows over the project's lifespan.
- The calculation of IRR involves finding the discount rate that equates the present value of cash inflows to the present value of cash outflows. This is done through iterative calculations or by
utilizing financial software.
- IRR is often used in conjunction with other financial metrics, such as the Payback Period and Net Present Value, to provide a comprehensive analysis of an investment opportunity.
3. Examples:
- Let's consider a hypothetical project with an initial investment of $100,000 and expected cash inflows of $30,000 per year for five years. By applying the IRR formula, we find that the IRR of this
project is 10%. This indicates that the project is expected to generate a 10% return on investment.
- Another example could be a real estate investment where the initial cash outflow is significant, but the subsequent rental income and potential property appreciation contribute to a higher IRR.
Remember, the IRR is a valuable tool for decision-making, but it does have limitations. It assumes that cash flows are reinvested at the calculated IRR, which may not always be realistic.
Additionally, IRR may produce multiple solutions or no solution at all in certain cases.
Introduction to Internal Rate of Return \(IRR\) - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
2. Understanding the Concept of Net Present Value (NPV)
Understanding the concept of Net Present Value (NPV) is crucial when evaluating the profitability of a project or investment. NPV is a financial metric that calculates the present value of future
cash flows by discounting them back to the present using a specified discount rate. It helps determine whether an investment is worthwhile by comparing the present value of cash inflows and outflows.
Insights from different perspectives shed light on NPV. From a financial standpoint, NPV considers the time value of money, recognizing that a dollar received in the future is worth less than a
dollar received today. This is because money can be invested and earn returns over time. By discounting future cash flows, NPV accounts for this opportunity cost.
1. Cash Flow Projection: NPV begins with estimating the future cash flows associated with a project or investment. These cash flows can include revenues, expenses, taxes, and salvage value.
2. Discount Rate Selection: The discount rate represents the required rate of return or the opportunity cost of capital. It reflects the risk and time preferences of the investor. The discount rate
should align with the project's risk profile.
3. Discounting Cash Flows: Each projected cash flow is discounted back to its present value using the chosen discount rate. The formula for discounting is: present Value = Future Cash flow / (1 +
Discount Rate)^n, where 'n' represents the time period.
4. NPV Calculation: The NPV is calculated by summing up the present values of all cash inflows and outflows. A positive NPV indicates that the project is expected to generate more value than the
initial investment, while a negative NPV suggests the opposite.
5. Decision Making: Based on the NPV calculation, investment decisions can be made. Positive NPV projects are generally considered favorable, as they are expected to generate returns higher than the
discount rate. Conversely, negative NPV projects may not be economically viable.
Let's illustrate the concept with an example: Suppose you are evaluating a project that requires an initial investment of $10,000. Over the next five years, it is expected to generate cash inflows of
$3,000, $4,000, $5,000, $6,000, and $7,000, respectively. Assuming a discount rate of 10%, we can calculate the present value of each cash flow and sum them up to determine the NPV.
Remember, this example is for illustrative purposes only and does not reflect real-world data. It demonstrates how NPV incorporates the time value of money and helps assess the profitability of an
Understanding the Concept of Net Present Value \(NPV\) - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
3. Defining IRR and its Significance in Project Evaluation
I'd be happy to provide you with a detailed section on "Defining IRR and its Significance in Project Evaluation" for your blog. Here's the section:
In project evaluation, the Internal Rate of Return (IRR) plays a crucial role in assessing the financial viability of an investment or project. It is defined as the discount rate that makes the Net
present Value (NPV) of a project or investment equal to zero. The IRR represents the rate of return at which the present value of cash inflows equals the present value of cash outflows.
1. Evaluating Investment Profitability: IRR helps determine the profitability of an investment by comparing the expected return with the cost of capital. If the IRR exceeds the required rate of
return or hurdle rate, the project is considered financially viable.
2. decision-Making tool: IRR serves as a decision-making tool for project selection. When comparing multiple investment opportunities, the project with the highest IRR is generally preferred, as it
indicates a higher potential for generating positive returns.
3. Assessing Risk and Uncertainty: IRR takes into account the timing and magnitude of cash flows, providing insights into the risk and uncertainty associated with an investment. A higher IRR implies
a greater potential for variability in returns, indicating higher risk.
4. Comparing Investment Alternatives: IRR enables the comparison of different investment alternatives with varying cash flow patterns. By calculating the IRR for each option, decision-makers can
identify the investment that offers the highest potential return.
Now, let's illustrate these concepts with an example:
Suppose you are evaluating two investment projects: Project A and Project B. Project A requires an initial investment of $100,000 and is expected to generate cash inflows of $30,000 per year for five
years. Project B, on the other hand, requires an initial investment of $150,000 and is expected to generate cash inflows of $40,000 per year for five years.
Calculating the IRR for both projects, we find that Project A has an IRR of 10%, while Project B has an IRR of 8%. Based on these results, Project A appears to be more financially attractive, as it
offers a higher rate of return compared to Project B.
The Internal Rate of Return (IRR) is a vital metric in project evaluation, providing insights into investment profitability, aiding decision-making, assessing risk, and facilitating comparisons
between investment alternatives. By understanding and utilizing IRR effectively, businesses can make informed investment decisions and maximize their returns.
Defining IRR and its Significance in Project Evaluation - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
4. Methods and Formulas
In this section, we will delve into the topic of calculating the Internal Rate of Return (IRR) for projects or investments. The IRR is a crucial financial metric that helps determine the discount
rate at which the Net present Value (NPV) of a project becomes zero. It is widely used in financial analysis to assess the profitability and viability of investment opportunities.
When it comes to calculating IRR, there are several methods and formulas that can be employed. Let's explore them in detail:
1. Trial and Error Method: One common approach is the trial and error method, where different discount rates are tested until the NPV equals zero. This iterative process involves adjusting the
discount rate until the present value of cash inflows matches the present value of cash outflows. While this method can be time-consuming, it provides an accurate estimation of the IRR.
2. Excel Functions: Excel offers built-in functions, such as the IRR function, which simplifies the calculation process. By inputting the cash flows and their respective timings, Excel can
automatically compute the IRR. This method is efficient and widely used in financial modeling and analysis.
3. Mathematical Formulas: There are mathematical formulas available to calculate IRR. One such formula is the Newton-Raphson method, which uses calculus to find the root of the equation representing
the NPV. This method provides a more precise solution but requires a deeper understanding of mathematical concepts.
4. Interpolation: Interpolation is another technique used to estimate the IRR. It involves linearly interpolating between two discount rates where the NPV changes sign. This method is useful when the
trial and error method becomes impractical due to complex cash flow patterns.
Now, let's consider an example to illustrate the concept of calculating IRR. Suppose we have an investment project with an initial cash outflow of $10,000 and expected cash inflows of $3,000 per year
for the next five years. By applying the trial and error method or using Excel's IRR function, we can determine the IRR of this project, which represents the discount rate at which the npv becomes
It's important to note that the accuracy of the IRR calculation depends on the quality of the underlying data and assumptions made. Additionally, the IRR should be interpreted in conjunction with
other financial metrics to make informed investment decisions.
Calculating the Internal Rate of Return (IRR) involves various methods and formulas, including trial and error, Excel functions, mathematical formulas, and interpolation. Each approach has its
advantages and considerations, and the choice of method depends on the complexity of the cash flow patterns and the available resources. By understanding and applying these techniques, financial
analysts can assess the profitability and feasibility of investment projects effectively.
Methods and Formulas - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
5. Positive, Negative, and Zero
In the realm of finance and investment analysis, the Internal Rate of Return (IRR) plays a crucial role in evaluating the profitability and viability of a project or investment. The IRR represents
the discount rate at which the Net Present Value (NPV) of a project becomes zero. Understanding how to interpret IRR results is essential for making informed financial decisions.
1. Positive IRR: A positive IRR indicates that the project or investment is expected to generate returns higher than the discount rate used. In other words, the project is deemed financially
attractive as it offers a rate of return that exceeds the cost of capital. For example, if the IRR is 15%, it implies that the project is expected to yield a return of 15% or higher.
2. Negative IRR: On the contrary, a negative IRR suggests that the project or investment is expected to generate returns lower than the discount rate used. This indicates that the project is not
financially viable and may result in a loss. Negative IRRs are typically associated with projects that have high initial costs or face significant challenges in generating sufficient cash flows to
cover expenses.
3. Zero IRR: A zero IRR implies that the project or investment is expected to break even, with the NPV equal to zero. This means that the project's cash inflows are exactly offset by its cash
outflows, resulting in no net gain or loss. While a zero IRR may not indicate profitability, it can still be considered a viable option if it aligns with other strategic objectives or serves as a
benchmark for comparison.
It's important to note that interpreting IRR results should not be solely relied upon when making investment decisions. Other factors such as risk assessment, market conditions, and sensitivity
analysis should also be taken into consideration. Additionally, IRR results should be compared with the required rate of return or hurdle rate to determine the project's feasibility.
Understanding the implications of positive, negative, and zero IRR results is crucial for evaluating the financial viability of a project or investment. By considering these insights from different
perspectives and utilizing tools like sensitivity analysis, investors can make more informed decisions and mitigate potential risks.
Positive, Negative, and Zero - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
6. Advantages and Limitations of IRR as an Investment Metric
The internal rate of return (IRR) is a widely used investment metric that helps evaluate the profitability of a project or investment. It is defined as the discount rate that makes the net present
value (NPV) of a project or investment equal to zero. While IRR offers several advantages, it also has certain limitations that should be considered.
Advantages of IRR:
1. Easy to understand: IRR provides a single percentage value that represents the potential return on investment. This makes it easier for investors to compare different projects or investments and
make informed decisions.
2. Incorporates the time value of money: By considering the timing and magnitude of cash flows, IRR takes into account the concept of the time value of money. It recognizes that a dollar received in
the future is worth less than a dollar received today.
3. Considers the entire cash flow stream: Unlike other metrics such as payback period, IRR considers the entire cash flow stream over the life of the investment. This provides a more comprehensive
view of the project's profitability.
4. Enables comparison with the cost of capital: IRR allows investors to compare the potential return on investment with the cost of capital. If the IRR is higher than the cost of capital, the project
is considered financially viable.
Limitations of IRR:
1. Multiple IRRs: In some cases, projects with unconventional cash flow patterns may have multiple IRRs, making it difficult to interpret the results. This can lead to confusion and misinterpretation
of the investment's profitability.
2. Ignores project size: IRR does not consider the absolute size of the investment. Two projects with the same IRR may have significantly different cash flows and overall profitability.
3. Relies on accurate cash flow estimates: The accuracy of IRR calculations heavily depends on the accuracy of cash flow estimates. Small errors in cash flow projections can lead to significant
variations in the calculated IRR.
4. Assumes reinvestment at the IRR: IRR assumes that all cash flows generated by the investment are reinvested at the calculated IRR. This may not always be realistic, as reinvestment opportunities
at the IRR may not be readily available.
It's important to note that while IRR is a useful metric, it should not be the sole factor in investment decision-making. Other factors such as risk, market conditions, and strategic alignment should
also be considered.
Advantages and Limitations of IRR as an Investment Metric - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
7. Comparing IRR with Other Financial Metrics
One of the challenges of evaluating the profitability and feasibility of a project or investment is choosing the appropriate financial metric to use. There are several metrics that can be used, such
as net present value (NPV), internal rate of return (IRR), payback period, profitability index, and modified internal rate of return (MIRR). Each of these metrics has its own advantages and
disadvantages, and they may not always agree on the ranking of different projects or investments. In this section, we will compare irr with other financial metrics and discuss their strengths and
weaknesses, as well as some scenarios where one metric may be more suitable than another.
Some of the points that we will cover are:
1. The relationship between IRR and NPV. IRR and NPV are both based on the concept of discounting future cash flows to their present value. However, while NPV uses a predetermined discount rate
(usually the cost of capital or the required rate of return), irr is the discount rate that makes the NPV of a project or investment zero. This means that IRR is the rate of return that a project or
investment generates over its lifetime. A project or investment is considered acceptable if its IRR is greater than or equal to the discount rate, and vice versa. For example, suppose a project
requires an initial investment of $10,000 and generates cash inflows of $3,000, $4,000, and $5,000 in the next three years. The NPV of this project at a 10% discount rate is $1,057.64, which means
that the project adds value to the firm. The IRR of this project is 16.28%, which means that the project earns a return of 16.28% on the initial investment.
2. The advantages and disadvantages of IRR. One of the main advantages of IRR is that it is easy to understand and communicate. It expresses the profitability of a project or investment as a single
percentage, which can be compared with other projects or investments, or with the cost of capital or the required rate of return. Another advantage of IRR is that it does not depend on the discount
rate, which may be difficult to estimate or vary over time. However, IRR also has some disadvantages, such as:
- It may not exist or be unique for some projects or investments, especially those with non-conventional cash flows (such as negative cash flows followed by positive cash flows, or multiple changes
in the sign of cash flows). For example, suppose a project requires an initial investment of $10,000 and generates cash inflows of $15,000, -$5,000, and $10,000 in the next three years. This project
has two IRRs: 11.80% and 48.43%, which makes it ambiguous to evaluate the project using IRR.
- It may not reflect the true profitability of a project or investment, especially when the reinvestment rate assumption is unrealistic. IRR assumes that the intermediate cash flows are reinvested at
the same rate as the IRR, which may not be feasible or consistent with the firm's opportunity cost of capital. For example, suppose a project requires an initial investment of $10,000 and generates
cash inflows of $20,000 in the first year and $5,000 in the second year. The IRR of this project is 50%, which implies that the $20,000 cash inflow in the first year is reinvested at 50% to generate
$10,000 in the second year. However, this may not be possible or realistic, as the firm may not have another project or investment that offers such a high return. A more realistic assumption is that
the intermediate cash flows are reinvested at the cost of capital or the required rate of return, which is what NPV assumes.
- It may not rank projects or investments correctly, especially when they have different scales, durations, or timing of cash flows. IRR only considers the percentage return of a project or
investment, but not the absolute amount of value added or the time value of money. For example, suppose a firm has two mutually exclusive projects, A and B, that require the same initial investment
of $10,000. Project A generates cash inflows of $15,000 in the first year and $5,000 in the second year, while project B generates cash inflows of $5,000 in the first year and $15,000 in the second
year. The IRR of both projects is 25%, which suggests that they are equally profitable. However, the NPV of project A at a 10% discount rate is $7,387.36, while the NPV of project B is $6,612.64,
which means that project A adds more value to the firm than project B. This is because project A has a higher present value of cash inflows, as it receives more cash earlier than project B.
Comparing IRR with Other Financial Metrics - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
8. Real-World Applications of IRR in Business Decision Making
In the realm of business decision making, the concept of Internal Rate of Return (IRR) plays a crucial role. IRR is defined as the discount rate that makes the Net present Value (NPV) of a project or
investment equal to zero. It is a widely used financial metric that helps organizations evaluate the profitability and feasibility of various investment opportunities.
1. Capital Budgeting: One of the primary applications of irr is in capital budgeting. By calculating the IRR of potential projects, businesses can determine which investments are likely to generate
the highest returns. This allows them to allocate their financial resources effectively and make informed decisions about which projects to pursue.
2. Project Evaluation: IRR is also used to evaluate the financial viability of individual projects. By comparing the IRR of different projects, organizations can prioritize investments based on their
potential for generating higher returns. This helps in optimizing resource allocation and maximizing profitability.
3. Investment Analysis: IRR is a valuable tool for analyzing investment opportunities. By calculating the IRR of potential investments, businesses can assess the attractiveness of different options.
Investments with higher IRRs are generally considered more favorable, as they offer greater potential for generating positive cash flows and returns.
4. Performance Measurement: IRR is often used as a performance measurement metric for evaluating the success of investment projects. By comparing the actual IRR achieved with the expected or target
IRR, organizations can assess the effectiveness of their investment decisions. This allows them to identify areas for improvement and make adjustments to their investment strategies.
5. Risk Assessment: IRR also helps in assessing the risk associated with investment projects. Higher IRRs indicate higher potential returns, but they may also come with increased risk. By considering
the IRR alongside other risk metrics, businesses can make more informed decisions about the level of risk they are willing to undertake.
6. Capital Rationing: In situations where financial resources are limited, irr can be used for capital rationing. By ranking projects based on their IRRs, organizations can allocate funds to projects
with the highest potential returns. This ensures that the available capital is utilized in the most efficient and profitable manner.
Overall, the applications of irr in business decision making are diverse and far-reaching. From capital budgeting to investment analysis and risk assessment, IRR provides valuable insights that help
organizations make informed financial decisions. By considering the IRR alongside other financial metrics, businesses can enhance their decision-making processes and drive sustainable growth.
Real World Applications of IRR in Business Decision Making - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
9. Harnessing the Power of IRR for Financial Success
The internal rate of return (IRR) is a powerful tool that can help investors and managers evaluate the profitability and efficiency of a project or investment. It can also be used to compare
different projects or investments and choose the best one. However, using the IRR alone is not enough to make sound financial decisions. There are some limitations and challenges that need to be
considered and addressed. In this section, we will discuss how to harness the power of IRR for financial success, by looking at some insights from different perspectives, such as accounting, finance,
economics, and ethics. We will also provide some tips and best practices on how to calculate and interpret the irr correctly and avoid common pitfalls.
Some of the insights that can help us use the IRR effectively are:
1. Understand the assumptions and implications of the IRR. The IRR is based on the assumption that the cash flows of a project or investment are reinvested at the same rate as the IRR. This may not
be realistic, especially for long-term projects or investments that have varying cash flows. The IRR also implies that the project or investment has a single, unique, and constant discount rate that
makes the net present value (NPV) zero. However, this may not be true for some projects or investments that have multiple or changing discount rates, such as those with non-conventional cash flows.
In these cases, the IRR may not exist, or may have multiple values, which can lead to confusion and inconsistency. Therefore, it is important to understand the assumptions and implications of the IRR
and check whether they are valid and applicable for the project or investment under consideration.
2. Use the IRR in conjunction with other criteria. The IRR is not the only criterion that can be used to evaluate the profitability and efficiency of a project or investment. There are other
criteria, such as the NPV, the payback period, the profitability index, the return on investment, and the benefit-cost ratio, that can provide complementary or supplementary information and insights.
For example, the NPV can show the absolute value of the project or investment, while the IRR can show the relative rate of return. The payback period can show the time required to recover the initial
investment, while the IRR can show the time-adjusted return. The profitability index can show the ratio of the present value of the benefits to the present value of the costs, while the IRR can show
the break-even point. The return on investment can show the percentage of the profit to the investment, while the IRR can show the percentage of the profit to the present value of the investment. The
benefit-cost ratio can show the ratio of the benefits to the costs, while the IRR can show the ratio of the present value of the benefits to the present value of the costs. Therefore, it is advisable
to use the IRR in conjunction with other criteria and compare the results and trade-offs.
3. Use the IRR with caution and care. The IRR is a useful and convenient tool, but it can also be misleading and deceptive if used incorrectly or inappropriately. There are some pitfalls and
challenges that need to be avoided and overcome when using the IRR. For example, the IRR can be affected by the scale and timing of the cash flows, which can make the comparison of different projects
or investments unfair and inaccurate. The IRR can also be influenced by the choice of the initial guess or the iteration method, which can make the calculation of the IRR unreliable and unstable. The
IRR can also be manipulated or distorted by the inclusion or exclusion of certain cash flows, such as sunk costs, opportunity costs, externalities, or taxes, which can make the evaluation of the
project or investment biased and unethical. Therefore, it is essential to use the IRR with caution and care and check the validity and reliability of the results and the integrity and morality of the
Harnessing the Power of IRR for Financial Success - Internal rate of return: IRR: The discount rate that makes the NPV of a project or investment zero
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Addition Rules And Multiplication Rule
Math, particularly multiplication, develops the keystone of various academic disciplines and real-world applications. Yet, for lots of learners, mastering multiplication can pose a difficulty. To
resolve this hurdle, educators and moms and dads have actually accepted a powerful tool: Addition Rules And Multiplication Rules For Probability Worksheet.
Introduction to Addition Rules And Multiplication Rules For Probability Worksheet
Addition Rules And Multiplication Rules For Probability Worksheet
Addition Rules And Multiplication Rules For Probability Worksheet -
It s worth noting that this formula is truly an extension of the Addition Rule Remember that the Addition Rule requires that the events E E and F F are mutually exclusive In that case the compound
event E and F E and F is impossible and so P E and F 0 P E and F 0 So in cases where the events in question are mutually
Math Addition Rules and Multiplication Rules for Probability Determine whether these events are mutullly exclusive 1 Roll a die t an even number and get a number less 3 2 a die get a prime number and
get an odd 3 a get a number greater than 3 4 Select a student No 5 Select a Sfident at UGA student is a a 6 Select school the the Fird the
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Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rules and Multiplication Rules for Probability Worksheet I Determine whether these events are mutually exclusive Roll a die get an even number and get a number less than 3 Roll a die get a
prime number and get an odd number Roll a die get a number greater than 3 and get a number less than 3
Example 4 4 1 4 4 1 Klaus is trying to choose where to go on vacation His two choices are A New Zealand A New Zealand and B Alaska B Alaska Klaus can only afford one vacation The probability that he
chooses A A is P A 0 6 P A 0 6 and the probability that he chooses B B is P B 0 35 P B 0 35
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Addition rule for probability basic One hundred students were surveyed about their preference between dogs and cats The following two way table displays data for the sample of students who responded
to the survey Find the probability that a randomly selected student prefers dogs Enter your answer as a fraction or decimal
We multiply the probabilities along the branches to find the overall probability of one event AND the next even occurring For example the probability of getting two tails in a row would be P T and T
1 2 1 2 1 4 When two events are independent we can say that P A and B P A P B Be careful
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Consistent practice sharpens multiplication effectiveness, improving overall math capacities.
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Supplying Useful Responses Responses aids in identifying locations of improvement, urging continued progression. Difficulties in Multiplication Technique and Solutions Motivation and Involvement
Difficulties Monotonous drills can result in uninterest; innovative methods can reignite motivation. Overcoming Worry of Mathematics Negative understandings around math can prevent development;
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Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Check more of Addition Rules And Multiplication Rules For Probability Worksheet below
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rules And Multiplication Rule For Probability Worksheet Answers Free Printable
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rules And Multiplication Rule For Probability Worksheet Answers Free Printable
Addition Rules In Probability And Statistics
Addition Rules And Multiplication Rules For Probability Worksheet Answer Key Designbymian
span class result type
Math Addition Rules and Multiplication Rules for Probability Determine whether these events are mutullly exclusive 1 Roll a die t an even number and get a number less 3 2 a die get a prime number and
get an odd 3 a get a number greater than 3 4 Select a student No 5 Select a Sfident at UGA student is a a 6 Select school the the Fird the
span class result type
1 The Addition Law As we have already noted the sample space S is the set of all possible outcomes of a given experiment Certain events A and B are subsets of S In the previous block we defined what
was meant by P A P B and their complements in the particular case in which the experiment had equally likely outcomes
Math Addition Rules and Multiplication Rules for Probability Determine whether these events are mutullly exclusive 1 Roll a die t an even number and get a number less 3 2 a die get a prime number and
get an odd 3 a get a number greater than 3 4 Select a student No 5 Select a Sfident at UGA student is a a 6 Select school the the Fird the
1 The Addition Law As we have already noted the sample space S is the set of all possible outcomes of a given experiment Certain events A and B are subsets of S In the previous block we defined what
was meant by P A P B and their complements in the particular case in which the experiment had equally likely outcomes
Addition Rules And Multiplication Rule For Probability Worksheet Answers Free Printable
Addition Rules And Multiplication Rule For Probability Worksheet Answers Free Printable
Addition Rules In Probability And Statistics
Addition Rules And Multiplication Rules For Probability Worksheet Answer Key Designbymian
Addition Rule Of Probability Part 2 YouTube
PPT Addition Rule for Probability PowerPoint Presentation Free Download ID 3946089
PPT Addition Rule for Probability PowerPoint Presentation Free Download ID 3946089
Addition and Multiplication Rules Worksheet Playing Cards Probability Free 30 day Trial
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Urging constant technique, supplying help, and developing a positive discovering atmosphere are advantageous steps.
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Environmental Economics
We don't feel pretty anymore with the gray banner and green background so we're playing around with new designs. Please bear with us as we experiment off and on (previews don't cut it).
This canned design is called "Hills green" or something like that. It is the most "environmental" of the designs but maybe over the top? Comments welcome.
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Vertical and horizontal span on plots
Vertical and horizontal span on plots
Is there a way to add a vertical or horizontal span to a plot in Sage? I need something similar to axhspan() and axvspan() functions in Matplotlib (example).
Thank you!
Actually,Sage's plot module is a wrapper of matplotlib, so you can use Matplotlib directly
Yes, I know I can use Matplotlib directly, but using Sage's *list_plot()* function is very convenient, and I would like to mark some range along the horizontal axis on the plot generated by several
consecutive *list_plot()* calls. Currently I just add a coloured polygon, but I have to adjust its edges to exactly fit the area I need. *axvspan()* is much easier to use in this sense.
2 Answers
Sort by ยป oldest newest most voted
There is not a direct analogue to axhspan or axvspan, though it would probably be fairly straightfoward to wrap those two matplotlib commands (patches welcome!). Alternatively, you could construct
the plot you want, then call the .matplotlib() method on the plot, and then add whatever you want to the resulting matplotlib figure. Then save the matplotlib figure using the normal matplotlib
edit flag offensive delete link more
Hmm... That sounds interesting promising. I didn't know about such possibility. I'll definitely play with it a bit. Thanks for the tip! :)
v_2e ( 2012-08-04 15:42:09 +0100 )edit
See http://www.sagemath.org/doc/reference/sage/plot/graphics.html#sage.plot.graphics.Graphics.matplotlib
Jason Grout ( 2012-08-05 02:09:55 +0100 )edit
I'm not sure I understand exactly what you are doing. But, you can specify the tickmarks using ticks.
You can specify the min and max of the axis ranges with xmin, xmax, ymin,ymax.
Does this help?
edit flag offensive delete link more
No, marking a range with the axis ticks only is not exactly what I want. What I want is to paint some part of a plot along the X-axis with some color. Just like in the Matplotlib example I gave a
link to.
v_2e ( 2012-08-04 15:44:10 +0100 )edit
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What is the value of x and y?
What is the value of xy?
This seemingly simple question has a surprisingly complex answer. In mathematics, the value of xy is the product of the two numbers x and y. However, in the real world, the value of xy can be much
more nuanced.
For example, the value of xy might represent the number of sales a company makes when it sells x units of product at a price of y dollars per unit. Or, it might represent the number of people who are
exposed to a message when an advertisement is shown x times on a website with y visitors per day.
In each of these cases, the value of xy is determined by a number of factors, including the price of the product, the number of people who are exposed to the message, and the effectiveness of the
In this article, we will explore the different factors that affect the value of xy, and we will provide some tips on how to maximize the value of your marketing campaigns.
| Column 1 | Column 2 | Column 3 |
| What is the value of xy? | x = 10, y = 20 | x * y = 200 |
The Meaning of xy
xy is the symbol for the cross product of two vectors. In mathematics, the cross product is a binary operation that takes two vectors as inputs and produces a third vector as output. The cross
product is defined as the vector perpendicular to the two input vectors, with a magnitude equal to the product of the magnitudes of the input vectors and the sine of the angle between them.
The cross product is a vector operator, which means that it takes two vectors as inputs and produces a third vector as output. The cross product is denoted by the symbol $\times$. The cross product
of two vectors $\vec{a}$ and $\vec{b}$ is written as $\vec{a} \times \vec{b}$.
The cross product of two vectors has the following properties:
• It is anticommutative, meaning that $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$.
• It is bilinear, meaning that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ and $(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{b} \
times \vec{c}$.
• It is associative, meaning that $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$.
• It is distributive over scalar multiplication, meaning that $k(\vec{a} \times \vec{b}) = k\vec{a} \times \vec{b} = \vec{a} \times k\vec{b}$.
The cross product is a useful operation in many areas of mathematics and physics. In particular, it is used in vector calculus, mechanics, and electromagnetism.
Why is xy important?
The cross product is an important operation in many areas of mathematics and physics. In particular, it is used in vector calculus, mechanics, and electromagnetism.
In vector calculus, the cross product is used to define the area of a parallelogram and the volume of a parallelepiped. It is also used to define the curl of a vector field.
In mechanics, the cross product is used to define the torque on a body and the angular momentum of a body. It is also used to define the moment of inertia of a body.
In electromagnetism, the cross product is used to define the magnetic field and the magnetic force on a charged particle. It is also used to define the Lorentz force law.
The cross product is a versatile operation that can be used to solve a variety of problems in mathematics and physics. It is a fundamental operation that is essential for understanding these fields.
How is xy calculated?
The cross product of two vectors $\vec{a}$ and $\vec{b}$ is calculated as follows:
$$\vec{a} \times \vec{b} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
where $\vec{i}$, $\vec{j}$, and $\vec{k}$ are the unit vectors in the x-, y-, and z-directions, respectively, and $a_1$, $a_2$, $a_3$, $b_1$, $b_2$, and $b_3$ are the components of the vectors $\vec
{a}$ and $\vec{b}$.
The cross product of two vectors is a vector that is perpendicular to both vectors $\vec{a}$ and $\vec{b}$. The magnitude of the cross product is equal to the area of the parallelogram formed by the
vectors $\vec{a}$ and $\vec{b}$.
Applications of xy
The cross product has a variety of applications in mathematics and physics. Some of these applications include:
• In vector calculus, the cross product is used to define the area of a parallelogram and the volume of a parallelepiped. It is also used to define the curl of a vector field.
• In mechanics, the cross product is used to define the torque on a body and
3. Problems with xy
• What are the limitations of xy?
XY has a number of limitations, including:
• It is not always accurate. XY is a statistical model, and as such, it is subject to the same limitations as any other statistical model. This means that it can produce inaccurate results if the
data used to train it is not representative of the population that it is being used to predict.
• It is not always reliable. XY is a black-box model, which means that it is not always possible to understand why it makes the predictions that it does. This can make it difficult to trust the
results of XY, especially in cases where the consequences of a wrong prediction are high.
• It can be biased. XY can be biased if the data used to train it is biased. This can lead to XY making predictions that are systematically unfair to certain groups of people.
XY can be misused in a number of ways, including:
• Using XY to make decisions about people without their consent. XY can be used to make decisions about people, such as whether they should be granted a loan or hired for a job, without their
consent. This can have a negative impact on people’s lives, especially if they are unfairly discriminated against.
• Using XY to create echo chambers. XY can be used to create echo chambers, where people are only exposed to information that confirms their existing beliefs. This can lead to people becoming more
polarized and less open to new ideas.
• Using XY to manipulate people. XY can be used to manipulate people, such as by showing them targeted advertising or by creating fake news stories. This can lead to people making decisions that
are not in their best interests.
There are a number of ways to improve XY, including:
• Making XY more accurate. XY can be made more accurate by using better data, by using more sophisticated models, and by making sure that the models are not biased.
• Making XY more reliable. XY can be made more reliable by making sure that the models are transparent and that people understand how they work. This can help to build trust in XY and to make it
more likely that people will use it to make decisions.
• Making XY less biased. XY can be made less biased by using more diverse data and by making sure that the models are not biased against certain groups of people. This can help to ensure that XY is
fair and that it does not discriminate against people.
4. The Future of xy
• What are the potential uses of xy?
XY has a number of potential uses, including:
• Predicting the future. XY can be used to predict future events, such as the weather, the stock market, and the outcome of elections.
• Personalizing experiences. XY can be used to personalize experiences, such as by showing people targeted advertising or by creating personalized recommendations.
• Improving decision-making. XY can be used to improve decision-making, such as by helping people to make better financial decisions or by helping doctors to make better diagnoses.
• How will xy change the world?
XY has the potential to change the world in a number of ways, including:
• Making the world more efficient. XY can be used to make the world more efficient by automating tasks that are currently done by humans. This can free up people’s time to do more important things.
• Making the world more informed. XY can be used to make the world more informed by providing people with access to information that they would not otherwise have. This can help people to make
better decisions and to understand the world around them.
• Making the world more equitable. XY can be used to make the world more equitable by helping to identify and address systemic biases. This can help to ensure that everyone has a fair chance to
• What are the challenges facing xy?
There are a number of challenges facing XY, including:
• The need for data. XY requires large amounts of data to train its models. This can be a challenge, especially for companies that are just starting out.
• The need for expertise. Building and using XY models requires expertise in statistics, machine learning, and other technical fields. This can be a challenge for companies that do not have the
necessary resources.
• The need for trust. People need to trust XY in order to use it to make decisions. This can be a challenge, especially given the potential for X
What is the value of xy?
The value of xy is x multiplied by y. For example, if x = 3 and y = 4, then xy = 3 * 4 = 12.
How do I find the value of xy?
To find the value of xy, multiply x by y. For example, if x = 3 and y = 4, then xy = 3 * 4 = 12.
What is the difference between xy and x + y?
xy is the product of x and y, while x + y is the sum of x and y. For example, if x = 3 and y = 4, then xy = 3 * 4 = 12, while x + y = 3 + 4 = 7.
What is the significance of xy?
The value of xy can be used to find the area of a rectangle, the volume of a cuboid, and the surface area of a sphere. For example, if x = 3 and y = 4, then the area of a rectangle with sides x and y
is 3 * 4 = 12 square units.
the value of xy is a complex and multifaceted topic. There is no one definitive answer, as the value of xy can vary depending on a number of factors, including the specific context in which it is
being used. However, the key takeaways from this discussion are that:
• The value of xy is not fixed, but rather is fluid and constantly changing.
• The value of xy is influenced by a number of factors, including the individual’s personal experiences, beliefs, and values.
• The value of xy is important to consider in order to make informed decisions and live a fulfilling life.
By understanding the value of xy, we can better understand ourselves and the world around us. We can make more informed decisions, live more fulfilling lives, and create a more just and equitable
Author Profile
Hatch, established in 2011 by Marcus Greenwood, has evolved significantly over the years. Marcus, a seasoned developer, brought a rich background in developing both B2B and consumer software for
a diverse range of organizations, including hedge funds and web agencies.
Originally, Hatch was designed to seamlessly merge content management with social networking. We observed that social functionalities were often an afterthought in CMS-driven websites and set out
to change that. Hatch was built to be inherently social, ensuring a fully integrated experience for users.
Now, Hatch embarks on a new chapter. While our past was rooted in bridging technical gaps and fostering open-source collaboration, our present and future are focused on unraveling mysteries and
answering a myriad of questions. We have expanded our horizons to cover an extensive array of topics and inquiries, delving into the unknown and the unexplored.
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Chapter 1, Introduction to SciPy, shows the benefits of using the combination of Python, NumPy, SciPy, and matplotlib as a programming environment for scientific purposes. You will learn how to
install, test, and explore the environments, use them for quick computations, and figure out a few good ways to search for help. A brief introduction on how to open the companion IPython Notebooks
that comes with this book is also presented.
Chapter 2, Working with the NumPy Array As a First Step to SciPy, explores in depth the creation and basic manipulation of the object array used by SciPy, as an overview of the NumPy libraries.
Chapter 3, SciPy for Linear Algebra, covers applications of SciPy to applications with large matrices, including solving systems or computation of eigenvalues and eigenvectors.
Chapter 4, SciPy for Numerical Analysis, is without a doubt one of the most interesting chapters in this book. It covers with great detail the definition and manipulation of functions (one or several
variables), the extraction of their roots, extreme values (optimization), computation of derivatives, integration, interpolation, regression, and applications to the solution of ordinary differential
Chapter 5, SciPy for Signal Processing, explores construction, acquisition, quality improvement, compression, and feature extraction of signals (in any dimension). It is covered with beautiful and
interesting examples from the field of image processing.
Chapter 6, SciPy for Data Mining, covers applications of SciPy for collection, organization, analysis, and interpretation of data, with examples taken from statistics and clustering.
Chapter 7, SciPy for Computational Geometry, explores the construction of triangulation of points, convex hulls, Voronoi diagrams, and applications, including the solving of the two dimensional
Laplace Equation via the Finite Element Method in a rectangular grid. At this point in the book, it will be possible to combine techniques from all the previous chapters to show state-of-the-art
research performed with ease with SciPy, and we will explore a few good examples from Material Science and Experimental Physics.
Chapter 8, Interaction with Other Languages, introduces one of the main strengths of SciPy—the ability to interact with other languages such as C/C++, Fortran, R, and MATLAB/Octave.
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Multiplying Negative Numbers Worksheet 2024 - NumbersWorksheets.com
Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet – The Bad Amounts Worksheet is a great way to start off instructing your kids the very idea of bad amounts. A poor variety is any number that is under zero. It
could be additional or subtracted. The minus sign indicates the negative variety. Also you can publish negative figures in parentheses. Beneath is actually a worksheet to help you get started off.
This worksheet has an array of adverse numbers from -10 to 10. Multiplying Negative Numbers Worksheet.
Negative phone numbers are quite a lot whose value is less than no
A poor quantity includes a worth below absolutely no. It can be indicated on a quantity series in two techniques: with the optimistic amount created because the very first digit, along with the
adverse amount created as the very last digit. A positive quantity is written with a plus indication ( ) well before it, but it is optionally available to create it doing this. If the number is not
written with a plus sign, it is assumed to be a positive number.
They can be displayed by way of a minus indication
In historic Greece, unfavorable phone numbers were actually not utilized. These were ignored, as his or her mathematics was according to geometrical principles. When European scholars started out
translating historical Arabic texts from Northern Africa, they came to identify negative numbers and appreciated them. These days, unfavorable amounts are depicted with a minus indication. To
understand more about the history and origins of adverse figures, look at this report. Then, try these good examples to discover how negative numbers have progressed as time passes.
They could be additional or subtracted
Positive numbers and negative numbers are easy to add and subtract because the sign of the numbers is the same, as you might already know. Negative numbers, on the other hand, have a larger absolute
value, but they are closer to than positive numbers are. These numbers have some special rules for arithmetic, but they can still be added and subtracted just like positive ones. You can even
subtract and add adverse numbers using a amount range and implement exactly the same regulations for subtraction and addition while you do for beneficial amounts.
They can be depicted from a number in parentheses
A poor variety is symbolized by way of a number encased in parentheses. The adverse indication is transformed into its binary comparable, along with the two’s complement is held in the same devote
memory. Sometimes a negative number is represented by a positive number, though the result is always negative. When this happens, the parentheses needs to be provided. If you have any questions about
the meaning of negative numbers, you should consult a book on math.
They are often divided by a positive amount
Negative numbers can be divided and multiplied like positive numbers. They can be divided by other negative phone numbers. However, they are not equal to one another. At the first try you flourish a
negative quantity by way of a positive variety, you will definitely get absolutely nothing because of this. To create the solution, you must select which sign your answer ought to have. It is
actually simpler to recall a negative variety when it is designed in mounting brackets.
Gallery of Multiplying Negative Numbers Worksheet
Multiplying Negative Numbers Worksheet Multiplying And Dividing Whole
Multiplying Negative Numbers Worksheet
Negative Number Multiplication Worksheet
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The error information "Equation infeasible due to rhs value"
Dear Sir,
I am learning to estimate the Tobit model and implement its statistics inference through GAMS. However, as I am calculate the inverse of hessian matrix, the error information “Equation infeasible due
to rhs value” come out all the time.
I had tried to solve this problem throuth trail and error for several day. But I still can not fix it. Attached are the files of my GAMS code and data respectively.
Tobit.gms (5.14 KB)
Data_for_Tobitall.xlsx (39.7 KB)
Are there anyone who can help me to point out the error parts of my code?
Thanks a lot.
Your model wasn’t really ready to be worked on. If you share something make sure what you send works (e.g. you include a CSV file, but you only had an Excel file attached). Also you “solved” the
model twice with convert (to get the Hessian) without explaining why. This looked messy/wrong. Nothing one can’t figure out, but if you expect help, I suggest that in the future you invest a little
more in making it easier for people to help.
The issue at hand was that you started your scalar variables at x1 and with card of m you let it go to x12. Trouble is that you forgot the objective variable, which comes first (you declared LOGIK
before BETA) and hence LOGIK was x1 and your first BETA was x2 and hence you fell short the last BETA variable. The model still ran over m and hence your Hessian had a zero row and column which made
the LP infeasible. This was easily fixed by declaring the xlist over x2*x%num_x% where num_ is card(m)+1. You could have noticed that since the Hessian parameter did not have any data for sigsqr.
Looking at results is important!
Next issue was that the inverse of the Hessian calculated with the LP solver was numerically very bad. Your Hesse matrix is numerically very challenging. If you look at the HINV you see that this is
full of zeros (and definitively not invertable which it should). An LP solver is not suited for this badly scaled problem. GAMS has tools that calculate an inverse (https://www.gams.com/latest/docs/
T_LINALG.html#linalg_invert) that pay more attention to bad numerics. If I use this, the model runs through to the end. Please find the modified model attached.
Tobit.gms (5.23 KB)
PS There are plenty female GAMS users that can potentially help. By addressing your potential helpers with “Dear Sir” you exclude many of them.
Dear Bussieck,
Firstly, I do appreciate your help on correcting my original codes. After your corrections, the error is solved finally. In addition, your explanations on the corrections are also help me to
understant the logic on GMAS coding. Thank you so much.
Secondly, I am sorry for the mistake of uploading the wrong data file. I should upload the .csv one, not the .exls one. Sorry for my carelessness in my lastt post.
Thank you again for your great help.
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Calculating Percentiles and Quantils
It is recommended to get familiar with rank() function before proceeding with this tutorial.
For this tutorial, we will leverage Data Set loaded with exam scores.
Data set includes two columns:
Percentile (or centile) is the value of a variable below which a certain percent of observations fall. For example, the 20th percentile is the value (or score) below which 20 percent of the
observations may be found.
You desire to create table showing percentile next to score for each student.
• Create new table with student ID drill-down and Score indicator.
• Create new indicator - Percentile.
• Add following formula into Indicators settings.
• Setup percentage to Unit and associate it with appropriate Format.
int records = aggregatePrevLevel(1){L_ID_COUNT}
int rank = rank() {M_SCORE}
double percentile = 1-(rank/records)
return percentile
1. line: Store the number of total records (students). Since, student drill-down is used, aggregation one level up is needed.
2. line: Obtain rank for each record.
3. line: Recalculate rank to percentile. For example, if rank is 5 from 100 students, the percentile will be: 1-(5/100) = 95%.
A value which divides a set of data into equal proportions. Examples are median, quartile and decile.
You desire to create KPI label showing the median of exam scores.
• Create new KPI label.
• Create new indicator - Quantile.
• Add following formula into Indicators settings.
• Create quantile variable, to be able to dynamically change observed quantile.
int records = L_ID_COUNT
double groups = 100/@quantile
int key = round(records-records/groups)
double median = 0
rank = rank(){M_SCORE}
if (rank == key){
median = M_SCORE
return median
1. Use first three lines to convert provided quantile variable and find the corresponding position within the set of scores.
2. Obtain rank for each score, aggregated to the level of student's ID.
3. If the current rank equals the position, store score to the median variable.
Next Steps
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Par Technology
PAR Technology Corporation through its wholly owned subsidiary ParTech, Inc., is a customer success-driven, global restaurant and retail technology company with over 100,000 restaurants in more than
110 countries using ...
How has the Par Technology stock price performed over time
How have Par Technology's revenue and profit performed over time
All financial data is based on trailing twelve months (TTM) periods - updated quarterly, unless otherwise specified. Data from
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TATTER: Two-sAmple TesT EstimatoR
TATTER (Two-sAmple TesT EstimatoR) performs two-sample hypothesis test. The two-sample hypothesis test is concerned with whether distributions p(x) and q(x) are different on the basis of finite
samples drawn from each of them. This ubiquitous problem appears in a legion of applications, ranging from data mining to data analysis and inference. This implementation can perform the
Kolmogorov-Smirnov test (for one-dimensional data only), Kullback-Leibler divergence, and Maximum Mean Discrepancy (MMD) test. The module performs a bootstrap algorithm to estimate the null
distribution and compute p-value.
Astrophysics Source Code Library
Pub Date:
June 2020
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This post is the seventh one of our series on the history and foundations of econometric and machine learning models. The first fours were on econometrics techniques. Part 6 is online here.
Boosting and sequential learning
As we have seen before, modelling here is based on solving an optimization problem, and solving the problem described by equation $(6)$ is all the more complex because the functional space $\mathcal
{M}$ is large. The idea of boosting, as introduced by Shapire & Freund (2012), is to learn, slowly, from the errors of the model, in an iterative way. In the first step, we estimate a model $m_1$ for
$y$, from $\mathbf{X}$, which will give an error $\varepsilon_1$. In the second step, we estimate a model $m_2$ for $\varepsilon_1$, from $X$, which will give an error $\varepsilon_2$, etc. We will
then retain as a model, after $k$ iterations m^{(k)}(\cdot)=\underbrace{m_1(\cdot)}_{\sim y}+\underbrace{m_2(\cdot)}_{\sim \epsilon_1}+\underbrace{m_3(\cdot)}_{\sim \epsilon_2}+\cdots+\underbrace{m_k
(\cdot)}_{\sim \epsilon_{k-1}}=m^{(k-1)}(\cdot)+m_k(\cdot)~~~(7)Here, the error $\varepsilon$ is seen as the difference between $y$ and the model $m(\mathbf{x})$, but it can also be seen as the
gradient associated with the quadratic loss function. Formally, $\varepsilon$ can be seen as $abla\ell$ in a more general context (here we find an interpretation that reminds us of residuals in
generalized linear models).
Equation $(7)$ can be seen as a descent of the gradient, but written in a dual way. The problem will then be rewritten as an optimization problem: m^{(k)}=m^{(k-1)}+\underset{h\in\mathcal{H}}{\text
{argmin}}\left\lbrace \sum_{i=1}^n \ell(\underbrace{y_i-m^{(k-1)}(\boldsymbol{x}_i)}_{\varepsilon_{k,i}},h(\boldsymbol{x}_i))\right\rbrace~~~(8)where the trick is to consider a relatively simple
space $\mathcal{H}$ (we will speak of “weak learner”). Classically, $\mathcal{H}$ functions are step-functions (which will be found in classification and regression trees) called “stumps”. To ensure
that learning is indeed slow, it is not uncommon to use a shrinkage parameter, and instead of setting, for example, $\varepsilon_1=y-m_1 (\mathbf{x})$, we will set $\varepsilon_1=y-\alpha\cdot m_1 (\
mathbf{x})$ with $\alpha\in[0.1]$. It should be noted that it is because a non-linear space is used for $\mathcal{H}$, and learning is slow, that this algorithm works well. In the case of the
Gaussian linear model, remember that the residuals $\varepsilon=y-\mathbf{x}^T\beta$ are orthogonal to the explanatory variables, $\mathbf{X}$, and it is then impossible to learn from our errors. The
main difficulty is to stop in time, because after too many iterations, it is no longer the m function that is approximated, but the noise. This problem is called overlearning.
This presentation has the advantage of having a heuristic reminiscent of an econometric model, by iteratively modelling the residuals by a (very) simple model. But this is often not the presentation
used in the learning literature, which places more emphasis on an optimization algorithm heuristic (and gradient approximation). The function is learned iteratively, starting from a constant value, m
^{(0)}=\underset{m\in\mathbb{R}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \ell(y_i,m)\right\rbracethen we consider the following learning procedure{\displaystyle m^{(k)}=m^{(k-1)}+{\underset{h\in {\
mathcal {H}}}{\text{argmin}}}\sum _{i=1}^{n}\ell(y_{i},m^{(k-1)}(\mathbf{x}_{i})+h(\mathbf{x}_{i}))}~~~(9)which can be written, if $\mathcal{H}$ is a set of differentiable functions, {\displaystyle m
^{(k)}=m^{(k-1)}-\gamma_{k}\sum _{i=1}^{n}\nabla _{m^{(k-1)}}\ell(y_{i},m^{(k-1)}(\mathbf{x}_{i})),} where {\displaystyle \gamma _{k}=\underset{\gamma }{\text{argmin }}\sum _{i=1}^{n}\ell\left(y_
{i},m^{(k-1)}( \mathbf{x}_{i})-\gamma \nabla _{m^{(k-1)}}\ell(y_{i},m^{(k-1)}( \mathbf{x}_{i}))\right).} To better understand the relationship with the approach described above, at step $k$,
pseudo-residuals are defined by setting r_{i,k}=-\left.\frac{\partial \ell(y_i,m(\mathbf{x}_i))}{\partial m(\mathbf{x}_i)}\right\vert_{m(\mathbf{x})=m^{(k-1)}( \mathbf{x})}\text{ where }i=1,\cdots,nA
simple model is then sought to explain these pseudo-residuals according to the explanatory variables $\mathbf{x}_i$, i.e. $r_{i,k}=h^\star(\mathbf{x}_i)$, where $h^\star\in\mathcal{H}$. In a second
step, we look for an optimal multiplier by solving\gamma_k = \underset{\gamma\in\mathbb{R}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \ell(y_i,m^{(k-1)}( \mathbf{x}_i)+\gamma h^\star(\mathbf{x}_i))\
right\rbrace then update the model by setting m_k (\cdot)=m_(k-1) (\cdot)+\gamma_k h^\star (\cdot) . More formally, we move from equation $(8)$ – which clearly shows that we are building a model on
residuals – to equation $(9)$ – which will then be translated as a gradient calculation problem – noting that $\ell(y,m+h)=\ell(y-m,h)$. Classically, class $\mathcal{H}$ of functions consists in
regression trees. It is also possible to use a form of penalty by setting $m_k (\cdot)=m_(k-1) (\cdot)+u\gamma_k h^\star (\cdot)$, with $u\in(0,1)$. But let’s go back a little further – in our next
post – on the importance of penalization before discussing the numerical aspects of optimization.
To be continued (keep in mind that references are online here)…
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Heat losses and heating cost | КИЕВ ГАЗОБЛОК
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The ability of autoclaved aerated concrete (AAC) blocks or any other substance to conduct heat is called the coefficient of thermal conductivity or specific thermal conductivity (λ). This value is
equal to the amount of thermal energy passing through a sample of aerated concrete (aerated concrete) with a thickness of 1 m and an area of 1 square meter per unit of time with a temperature
difference of one degree.
The product of the reciprocal of the thermal conductivity coefficient by the wall thickness (D) is called the thermal resistance (R):
R = D / λ
The amount of thermal energy (Q) arriving per unit of time (t) through a wall with an area (S) at a temperature difference (Tv-Tn) is determined by the formula:
W = Q / t = S × (TV-Tn) / R
Consider an example of calculating heat loss through a wall with an area of 120 square meters (3 m high and 40 m long, which corresponds to a house area of 100 square meters) made of aerated
concrete (AAC block) from "Stonelight" with a thickness of 0.4 m and a density of 400 kg / cubic meter. The thermal conductivity coefficient of this aerated concrete (gas block) is 0.1 W / (m ×
deg.). The thermal resistance of such a wall is equal to:
R = 0.4 / 0.1 = 4 m2 × deg / W
Let's say the temperature difference on the outer and inner sides of the wall in winter is 30 degrees. (-10 C .... +20 C), then the power of heat losses through such a wall is equal to:
W = 120 × 30/4 = 900 W
Thus, without taking into account other heat losses (through windows, roofs, doors), for round-the-clock support of such a temperature difference within a month, Q = 900 × 24 × 30 = 648 kWh of heater
energy (or 577 UAH when using an electric heater and cost electricity 0.89 UAH per 1 kWh)
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hackerrank algorithms solution
Given an array of integers, find the sum of its elements.
for example
if the array, arr = [1, 2, 3], 1+2+3 so return 6
function discription
_Complete the simpleArraySum function in the editor below. It must return the sum of the array elements as an integer.
simpleArraySum has the following parameter(s):
ar: an array of integers
input format
The first line contains an integer,n , denoting the size of the array.
The second line contains space-separated integers representing the array's element
// bin/python3
import math
import os
import random
import re
import sys
we define a function to return an interger
def simpleArraySum(ar):
# Write your code here
for i in range(0,ar_count):
x += ar[i]
return x
if name == 'main':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
ar_count = int(input().strip())
ar = list(map(int, input().rstrip().split()))
result = simpleArraySum(ar)
fptr.write(str(result) + '\n')
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How to build your own prediction algorithm
How to build your own prediction algorithm¶
This page describes how to build a custom prediction algorithm using Surprise.
The basics¶
Want to get your hands dirty? Cool.
Creating your own prediction algorithm is pretty simple: an algorithm is nothing but a class derived from AlgoBase that has an estimate method. This is the method that is called by the predict()
method. It takes in an inner user id, an inner item id (see this note), and returns the estimated rating \(\hat{r}_{ui}\):
from surprise import AlgoBase, Dataset
from surprise.model_selection import cross_validate
class MyOwnAlgorithm(AlgoBase):
def __init__(self):
# Always call base method before doing anything.
def estimate(self, u, i):
return 3
data = Dataset.load_builtin("ml-100k")
algo = MyOwnAlgorithm()
cross_validate(algo, data, verbose=True)
This algorithm is the dumbest we could have thought of: it just predicts a rating of 3, regardless of users and items.
If you want to store additional information about the prediction, you can also return a dictionary with given details:
def estimate(self, u, i):
details = {
'info1' : 'That was',
'info2' : 'easy stuff :)'
return 3, details
This dictionary will be stored in the prediction as the details field and can be used for later analysis.
The fit method¶
Now, let’s make a slightly cleverer algorithm that predicts the average of all the ratings of the trainset. As this is a constant value that does not depend on current user or item, we would rather
compute it once and for all. This can be done by defining the fit method:
class MyOwnAlgorithm(AlgoBase):
def __init__(self):
# Always call base method before doing anything.
def fit(self, trainset):
# Here again: call base method before doing anything.
AlgoBase.fit(self, trainset)
# Compute the average rating. We might as well use the
# trainset.global_mean attribute ;)
self.the_mean = np.mean([r for (_, _, r) in self.trainset.all_ratings()])
return self
def estimate(self, u, i):
return self.the_mean
The fit method is called e.g. by the cross_validate function at each fold of a cross-validation process, (but you can also call it yourself). Before doing anything, you should call the base class fit
() method.
Note that the fit() method returns self. This allows to use expression like algo.fit(trainset).test(testset).
The trainset attribute¶
Once the base class fit() method has returned, all the info you need about the current training set (rating values, etc…) is stored in the self.trainset attribute. This is a Trainset object that has
many attributes and methods of interest for prediction.
To illustrate its usage, let’s make an algorithm that predicts an average between the mean of all ratings, the mean rating of the user and the mean rating for the item:
def estimate(self, u, i):
sum_means = self.trainset.global_mean
div = 1
if self.trainset.knows_user(u):
sum_means += np.mean([r for (_, r) in self.trainset.ur[u]])
div += 1
if self.trainset.knows_item(i):
sum_means += np.mean([r for (_, r) in self.trainset.ir[i]])
div += 1
return sum_means / div
Note that it would have been a better idea to compute all the user means in the fit method, thus avoiding the same computations multiple times.
When the prediction is impossible¶
It’s up to your algorithm to decide if it can or cannot yield a prediction. If the prediction is impossible, then you can raise the PredictionImpossible exception. You’ll need to import it first:
from surprise import PredictionImpossible
This exception will be caught by the predict() method, and the estimation \(\hat{r}_{ui}\) will be set according to the default_prediction() method, which can be overridden. By default, it returns
the average of all ratings in the trainset.
Using similarities and baselines¶
Should your algorithm use a similarity measure or baseline estimates, you’ll need to accept bsl_options and sim_options as parameters to the __init__ method, and pass them along to the Base class.
See how to use these parameters in the Using prediction algorithms section.
Methods compute_baselines() and compute_similarities() can be called in the fit method (or anywhere else).
class MyOwnAlgorithm(AlgoBase):
def __init__(self, sim_options={}, bsl_options={}):
AlgoBase.__init__(self, sim_options=sim_options, bsl_options=bsl_options)
def fit(self, trainset):
AlgoBase.fit(self, trainset)
# Compute baselines and similarities
self.bu, self.bi = self.compute_baselines()
self.sim = self.compute_similarities()
return self
def estimate(self, u, i):
if not (self.trainset.knows_user(u) and self.trainset.knows_item(i)):
raise PredictionImpossible("User and/or item is unknown.")
# Compute similarities between u and v, where v describes all other
# users that have also rated item i.
neighbors = [(v, self.sim[u, v]) for (v, r) in self.trainset.ir[i]]
# Sort these neighbors by similarity
neighbors = sorted(neighbors, key=lambda x: x[1], reverse=True)
print("The 3 nearest neighbors of user", str(u), "are:")
for v, sim_uv in neighbors[:3]:
print(f"user {v} with sim {sim_uv:1.2f}")
# ... Aaaaand return the baseline estimate anyway ;)
bsl = self.trainset.global_mean + self.bu[u] + self.bi[i]
return bsl
Feel free to explore the prediction_algorithms package source to get an idea of what can be done.
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Ask Uncle Colin: The Area In Between
Dear Uncle Colin,
I have the graphs of $y=\sin(x)$ and $y=\cos(x)$ for $0 < x < 2\pi$. They cross in two places, and I need to find the area enclosed.
I’ve figured out that they cross at $\piby 4$ and $\frac{5}{4}\pi$, but after that I’m stuck!
- Probably A Simple Calculation, Absolutely Lost
Hi, PASCAL, and thanks for your message!
You’ll probably be unsurprised to learn there are several ways to do it.
$\int_{\piby4}^{\frac{5}{4}\pi} \sin(x) - \cos(x) \dx = \left[ -\cos(x) - \sin(x) \right]_{\piby4}^{\frac{5}{4}\pi}$
$\dots = \left[ \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) \right]$
$\dots = 2\sqrt{2}$.
$\sin(x) - \cos(x) \equiv \sqrt{2} \sin\left( x - \piby 4\right)$
$\int_{\piby4}^{\frac{5}{4}\pi} \sin(x) - \cos(x) \dx =\sqrt{2} \int_{0}^{\pi} \sin(t) \dt$ (after a sneaky variable change).
$\dots = \sqrt{2} \left[ -\cos(\pi) - \cos(0) \right]$
$\dots = 2\sqrt{2}$
I can’t help but feel there’s an even nicer way, though!
Hope that helps,
- Uncle Colin
A selection of other posts
subscribe via RSS
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Back to the table of contents
Acquaintance and first steps
This chapter is devoted to a first acquaintance with possibilities which Mathapar opens. The language MathPar, which is described below, may be considered as a kind of development of a TeX language.
TeX serves for writing mathematical texts and preparing them for publication. It may be called "passive" in comparison with the language MathPar, which permits to execute computations, and so is a
self-depended mathematical language. A problem definition and a result of computations are written in MathPar.
Just after computations you see the whole mathematical text as a pdf-image which is accustomed in scientific and technical publications.
The result may be further used in different ways.
(1) You may click the text with your mouse, and it would return to the initial form of the MathPar language. Then you may continue to edit the text or pass to a next task. There is another way to
change a form of a text — using a button , placed between buttons "$\blacktriangleright$" and "$+$".
(2) You may click the image of the mathematical text with the right mouse button, and the drop-down menu appears. The upper field Show-Math-As permits to pass to the choice of language. It is
suggested to chose Tex or MathML. You may open a field you need.
For example, a matrix of the size $2\times 2$ will be written in MathPar in the following way:
in TeX as follows:
A= $\backslash$left(begin{array}{${cc}$}$ a \& b$ $\backslash\backslash c \& d \backslash\backslash$ end{array}$\backslash$right).
In MathML it is much more complicated.
The text obtained in Tex or in MathML you may copy and past into TeX or HTML file and use for publication. You may also save it as an image and use in any document. It is useful, for example, when it
is necessary to save a plot or a solution of a problem
2.1 Input data and run the calculations
At the center of the screen there is an entry field where it is possible to enter mathematical expressions. To start a task press the button $\blacktriangleright$. When your cursor is disposed in the
field of input you can press the combination of keys Ctrl+Enter.
On the top of the screen you can see buttons ${\small \fbox {Help}}$ and ${\small \fbox{Handbook}}$. This is way to the help files. All the fields of the help pages are active, and you can run the
help examples. You can copy text from the samples and transfer them into the field for user input.
When you enter mathematical expressions, they must be separated by a semicolon (;) or text comments, which are enclosed in quotation marks. When you need to have a mathematical expression in the
comment as part of the comment, it must be skirted in the dollar signs ($\$$). For example, you can write a comment:
\noindent $"$ Two different notations $\$ \backslash exp(x)\$$ and $\$\backslash e \widehat{ }{} x\$$ are used for the exponential function.$"$
To obtain results it is necessary to use the command print() and to specify the names of those expressions which are required to be printed.
If the list of statement does not contain a print statement print() or any other operator (plot(), prints(), etc.), it will be shown the result obtained in the last statement. All commands or
operators should begin with the symbol "back slash" ($\backslash$). The button $\fbox{+}$ lets us add new entry fields. You can press the combination of keys Crtl+Del to remove this field or you can
press the button $\fbox{x}$ at the right side of this field on the screen. The button $\fbox{C}$ is designed to clean the values of all previously typed names. It is useful to have such button when
the numerical values are entered in some sections, and the calculations are done in other sections. Clearing all names allows you to obtain a symbolic expression rather than the number.
On the left of the screen you can see fields with a current environment and a current random access memory. Under the fields differant buttons for enter of functions are placed.
Working with files
Functions for working with files are available at the "Files" collapsible panel from menu at the left.
Here is what you can do with files:
1) Save the result of the last run section as PDF file with "Save PDF" button. You can specify desirable paper size (dimensions are in centimeters), by default page has size A4 (21x29.7 cm)
2) Upload text files to Mathpar server with "Upload file" button. Under the button there is a list of uploaded files. Files should contain Mathpar expressions or tables in specific format.
Table contains of header — it's the first row with arbitrary strings in it — and number rows. Columns are separated with tabulation symbol. Functions for working with tables are available at the
panel "Graphics and tables" (see also section 3.1 Plotting functions of help system).
3) Input Mathpar expressions from uploaded files with fromFile() function. E.g., to make an expression from file myfile.txt and assign this expression to variable $a$ run: a = fromFile('myfile.txt').
2.2 Mathematical functions
The following notations for elementary functions and constants are accepted.
$\backslash$i — imaginary unit,
$\backslash$e — the basis of natural logarithm,
$\backslash$pi — the ratio of length of a circle to its diameter,
$\backslash$infty — infinity sign.
Functions of one argument
$\backslash$ln — natural logarithm,
$\backslash$lg — decimal logarithm,
$\backslash$sin — sine,
$\backslash$cos — cosine,
$\backslash$tg — tangent,
$\backslash$ctg — cotangent,
$\backslash$arcsin — arcsine,
$\backslash$arccos — arccosine,
$\backslash$arctg — arctangent,
$\backslash$arcctg — arccotangent,
$\backslash$sh — sine hyperbolic,
$\backslash$ch — cosine hyperbolic,
$\backslash$th — tangent hyperbolic,
$\backslash$cth — cotangent hyperbolic,
$\backslash$arcsh — arcsine hyperbolic,
$\backslash$arcch — arccosine hyperbolic,
$\backslash$arcth — arctangent hyperbolic,
$\backslash$arccth — arccotangent hyperbolic,
$\backslash$exp — exponent,
$\backslash$sqrt — root square,
$\backslash$abs — absolute value of real numbers (module for complex numbers),
$\backslash$sign — number sign (returns $1$, $0$, $-1$ when number sign is $+$, $0$, $-$, correspondingly), $\backslash$unitStep$(x,a)$ — is a function which, for $ x> a $ takes the value $ 1 $, and
for $ x <a $ takes the value $ 0 $;
$\backslash$fact — factorial. It is defined for positive integers. It is equivalent to $n!$.
Functions of two arguments
$\widehat{ }{}$ — degree,
$\backslash$log — logarithm of function with given base,
$\backslash$rootOf(x, n) — root of degree n of x,
$\backslash$Gamma — the function Gamma,
$\backslash$Gamma2 — the function Gamma 2,
$\backslash$binomial — binomial coefficient.
2.3 Actions with functions
For the above functions and their compositions, you can calculate the value of the function at the point, substitute the expression into a function instead of arguments, calculate the limit of the
function, calculate derivative, etc. For this purpose, the following commands are defined.
To calculate the value of a function at a point you must run value(f, [var1, var2, …, varn]), where $f$ — function, and $var1, var2, …, varn $ — values of the variables of the ring.
For the substitution of expressions to the function you must execute the value(f, [func1, func2, …, funcn]), where $ f $ — a function $ func1, func2, …, funcn $ — expressions that are substituted for
the corresponding variables.
To calculate the limit of a function at a point you must run lim(f, var), where $ f $ — this function, and $ var $ — the point at which you want to find the limit.
In order to calculate the derivative of $f$ in the variable $y$ in the ring $\mathbb {Z} [x, y, z]$ you must run D(f, y). To find a mixed first-order derivative of the function $ f $ there is a
command D(f, [x, y]), to find the derivative of higher order you must use the command $\backslash {\mathbf {D}} (f, [x \widehat{ }{} k, z \widehat{ }{} m, y \widehat{ }{} n])$, where $ k, m, n $
indicate the order of the derivative of variables.
2.4 Solution of the algebraic equation
To obtain a solution of the algebraic equation use the command solve.
The command $FLOATPOS = N$ is used for setting the environment. It sets the number of decimal places after the decimal point $ (N) $, which should appear in the print of the numerical results of
approximate calculations. It is not connected with the process of calculation, but only with printing. By default, $ FLOATPOS = 2 $.
2.5 Solution of the algebraic inequalities
To obtain a solution of the algebraic inequalities use the command solve, which contains the inequalities. We can solve strict and not strict algebraic inequalities. Open interval is indicated in
parentheses ( ), closed interval is indicated inbrackets [ ], set is denoted by braces { }.
2.6 Solution of the algebraic inequalities systems
To obtain a solution of the algebraic inequalities systems use the command solve[In1, In2, ..., Ink], where $[In1, In2, ..., Ink]$ — vector, where contain inequalities. System contain strict and not
strict algebraic inequalities. Open interval is indicated in parentheses (), closed interval is indicated inbrackets, set is denoted by braces { }.
2.7 Operations on subsets of the real numbers
To specify a subset use the command set((a,b),(c,d]), where $a,b,c,d$ are numbers. Subset may consist of open intervals indicated by parentheses ( ), half-open intervals indicated by [ ) or ( ],
segments indicated by brackets [ ] and points indicated by braces { }, or like segments.
Simple subset is denoted by the same brackets, but you need to add a backslash ($ \backslash $) in front of each bracket. For example $ \backslash (3,4.5) \backslash] $, $ \backslash[7, 7 \backslash]
$ or $\backslash{8 \backslash}$. The operator $ \backslash {\mathbf {set}} $ is not required.
With subsets we can make the following operations: union, intersection, subtraction, calculation of the symmetric difference and complement set, using the commands $\backslash cup$, $\backslash cap$,
$\backslash setminus$, $\backslash triangle$ and symbol (') apostrophe.
2.8 Vectors and matrices
To define the row-vector you have to list its elements in square brackets.
To define the matrix you must take in square brackets a list of row vectors, for example, $ A = [[1, 2], [3, 4]] $.
Element of the matrix may be obtained by specifying the row and column number in the two lower indexes of the matrix, and an element of the vector may be obtained by specifying its number in the
lower index of the vector. The is an example for obtaining elements. You have to set $a=\backslash elementOf(A)$, and then obtain $a$_{$i, j$}. If $B$ is a vector, then you have to set $b=\backslash
elementOf(B)$, and then obtain element $b$_{$i$}. You can get a row of the matrix as a vector-row and column of the matrix as a column vector. The row vector obtained by specifying the number of row
in the first index and a sign of question (?) in the second index, for example, $a$_{$i, ?$}. Column vector obtained by specifying the number of column in the second index and the sign of question
(?) in the first index, for example, $a$_{$?, j$}. The names of non-commutative objects, such as matrices and vectors, must be written with the symbol <<back slash>> ($\backslash$) and a capital
To denote zero and identity matrix you can use the caps $\backslash O$ and $\backslash I$, with two indexes, indicating the number of rows and columns. With the help of the symbol $\backslash I $,
you can create any size square matrix whose elements on the main diagonal are equal to $ 1 $, and the remaining elements are zero. For example, $\backslash I$_{$2, 3$} and $\backslash O$_{$2, 2$}
denote the matrix $ \left(\begin {array} {ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ \end {array} \right) $ and $ \left(\begin {array} {cc} 0 & 0 \ 0 & 0 \ \end {array} \right) $. You can specify zero vectors,
indicating the index number of elements: $\backslash O$_{$3$} denote the vector $ [0, 0, 0] $ and $I$_{$3$} denotes the vector $ [ 1, 0, 0] $.
Column vector can be formed by transposing the row vector, for example, $ D = [7, 2, 3] ^ T $ — it is a column vector with three elements. Arithmetic operations are indicated by standard signs
``~+~'',~``~-~'', ``~*~''.
2.9 Generation of random elements
Mathpar can generate of random elements such as numbers, polynomials and matrices.
Generation of numbers
To create a random number you have to execute the command randomNumber(k), where $k$ is the number of bits.
Generation of random polynomial
To create a random polynomial with three variables you have to execute the command randomPolynom(d1, d2,…, ds, dens, bits), where $dens$ is a polynomial density, $bits$ is a number of bits in
numerical coefficients, and $d1, d2, …, ds$ denote the highest degrees of variable. If $dens=100$, you get a polynomial that has all coefficients non-zero, all $(d1 +1) (d2 +1) \cdot (ds +1)$
non-zero terms. When $ dens < 100$, only $ dens \% $ coefficients are nonzero, and the remaining $ (100-dens) \% $ will be zero.
Generation of random matrix
To create arandom numerical matrix you have to execute the command randomMatrix(m, n, dens, bits), where last two arguments are the density of matrix and the number of bits in numerical elements of
matrix, and first two arguments denote the sizes of a matrix.
To create a polynomial matrix you have to execute the command randomMatrix(m, n, dens, d1, d2, …, ds, pol_dens, pol_bits)), where first three arguments denote the size of a matrix and its density,
last two arguments are the density of polynomials and the number of bits in numerical coefficients, the numbers $d1, d2,…, ds$ set the highest degrees of polynomial variables.
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A is a matrix of 3 rows and 2 columns and B is a matrix of 2 ro... | Filo
Question asked by Filo student
is a matrix of 3 rows and 2 columns and is a matrix of 2 rows and 3 columns. If and , then
a. determinants of and are always equal
b. determinant of is zero
c. determinant of is zero
d. none of these
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Question Text is a matrix of 3 rows and 2 columns and is a matrix of 2 rows and 3 columns. If and , then
Updated On Apr 25, 2023
Topic Matrices and Determinant
Subject Mathematics
Class Class 12
Answer Type Video solution: 1
Upvotes 90
Avg. Video Duration 5 min
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Excel: How to Use XLOOKUP with Multiple Criteria | Online Tutorials Library List | Tutoraspire.com
Excel: How to Use XLOOKUP with Multiple Criteria
by Tutor Aspire
You can use the following XLOOKUP formula in Excel to look up cells that meet multiple criteria:
This particular formula will look for the cell in the range D2:D13 where the following criteria is all met:
• The value in cell range A2:A13 is equal to the value in cell F2
• The value in cell range B2:B13 is equal to the value in cell G2
• The value in cell range C2:C13 is equal to the value in cell H2
The following example shows how to use this formula in practice.
Example: XLOOKUP with Multiple Criteria in Excel
Suppose we have the following dataset that contains information about various basketball players:
Now suppose we would like to look up the points value for the player who meets all of the following criteria:
• Team = Cavs
• Position = Guard
• Starter = Yes
We can use the following formula to do so:
We can type this formula into cell I2 and then press Enter:
This XLOOKUP formula is able to look up “Cavs” in the Team column, “Guard” in the Position column, “Yes” in the Starter column, and return the points value of 30.
We can check the original dataset and confirm that this is the correct points value for the player that meets all of these criteria:
Note that we used three criteria in this particular XLOOKUP formula, but you can use similar syntax to include as many criteria as you would like.
Additional Resources
The following tutorials explain how to perform other common operations in Excel:
Excel: How to Find Duplicates Using VLOOKUP
Excel: How to Use VLOOKUP to Return All Matches
Excel: How to Use VLOOKUP to Return Multiple Columns
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An Etymological Dictionary of Astronomy and Astrophysics
Hermitian conjugate
همیوغ ِاِرمیتی
hamyuq-e Hermiti
Fr.: conjugé hermitien
Math.: The Hermitian conjugate of an m by n matrix A is the n by m matrix A^* obtained from A by taking the → transpose and then taking the complex conjugate of each entry. Also called adjoint
matrix, conjugate transpose. → Hermitian operator.
Hermitian, named in honor of the Fr. mathematician Charles Hermite (1822-1901), who made important contributions to number theory, quadratic forms, invariant theory, orthogonal polynomials, elliptic
functions, and algebra. One of his students was Henri Poincaré; → conjugate.
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Air Leaks and Gaps: Impact on Insulation Effectiveness in context of insulation efficiency
10 Sep 2024
Title: The Devastating Effects of Air Leaks and Gaps on Insulation Efficiency: A Critical Analysis
Air leaks and gaps are a ubiquitous problem in building envelopes, compromising the effectiveness of insulation systems. This article delves into the impact of these imperfections on insulation
efficiency, exploring the underlying physics and providing a comprehensive analysis of the consequences.
Insulation plays a crucial role in maintaining thermal comfort and energy efficiency in buildings. However, the presence of air leaks and gaps can significantly compromise the performance of
insulation systems, leading to heat transfer and energy losses. This article aims to investigate the effects of air leaks and gaps on insulation effectiveness, shedding light on the underlying
mechanisms and providing insights for improvement.
The Physics of Heat Transfer:
Heat transfer occurs through three primary modes: conduction, convection, and radiation. In the context of insulation, heat transfer is primarily governed by conduction and convection:
Q = k \* A \* (T1 - T2) / d
where Q is the heat flux, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures on either side of the insulation, and d is the thickness of the insulation.
The Impact of Air Leaks:
Air leaks can significantly compromise the effectiveness of insulation systems by introducing convective heat transfer:
Q_conv = h \* A \* (T1 - T2)
where Q_conv is the convective heat flux, h is the convective heat transfer coefficient, and A is the surface area.
The presence of air leaks can lead to a significant increase in heat transfer rates, compromising the performance of insulation systems.
The Impact of Gaps:
Gaps between insulation layers or between insulation and other building components can also compromise insulation effectiveness:
Q_gap = k \* A \* (T1 - T2) / d_gap
where Q_gap is the heat flux through the gap, k is the thermal conductivity, A is the surface area, and d_gap is the thickness of the gap.
The presence of gaps can lead to a significant reduction in insulation effectiveness, compromising the performance of insulation systems.
Air leaks and gaps are a critical problem in building envelopes, compromising the effectiveness of insulation systems. The underlying physics of heat transfer, including conduction, convection, and
radiation, play a crucial role in understanding the impact of these imperfections on insulation efficiency. By recognizing the consequences of air leaks and gaps, designers and builders can take
steps to mitigate their effects, improving the performance and energy efficiency of buildings.
1. Conduct thorough inspections to identify potential air leaks and gaps.
2. Seal air leaks using suitable materials and techniques.
3. Ensure proper installation and maintenance of insulation systems.
4. Consider using advanced insulation materials and technologies that can minimize heat transfer through conduction, convection, and radiation.
By following these recommendations, designers and builders can improve the performance and energy efficiency of buildings, reducing the impact of air leaks and gaps on insulation effectiveness.
Related articles for ‘insulation efficiency’ :
Calculators for ‘insulation efficiency’
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KCPE Past Papers Mathematics 2011
K.C.P.E 2011
The Kenya National Examinations Council
K.C.P.E 20 11
Time: 2 hours
Instructions to candidates (Please read these instruction: carefully)
1. You have been given this question booklet and a separate answer sheet. The question booklet contains 50 questions.
2. Do any necessary rough work in this booklet.
3. When you You have chosen your answer, mark it on the Answer sheet. not in the question booklet
How To Use The Answer Sheet
4. Use only an culinary pencil.
5. Make sure that you have written on the answer sheet
Your index number
Your name
Name of your school
6. By drawing a dark line inside the correct numbered boxes mark your full index numbe (ie school code numbered the three figure candidate’s number) in the grid near the top of the answer sheet
7. Do not make any mark outside the boxes
8. Keep the sheet as clean as possible and do not fold it
9. For each of the questions 1-50 four answers are given. The answers are labeled A, B, C and D. In each case only one of the four answers is correct. Choose the correct answer. 10. On the answer
sheet the correct answer is to be shown by drawing a dark line inside the box in which the letter you have chosen is written
In the question booklet 11. what is the value of 6(24 – 18)+6 x 4/6? A. 30
B. 25
C. 10
D. 28
The correct answer is C (10)
On the answer sheet
4. [A] [B] [C] [D] 14. [A] [B] [C] [D] 24. [A] [B] [C] [D]
34. [A] [B] [C] [D] 44. [A] [B] [C] [D]
11. Your dark line must be within the box
12. For each question only one box is to be marked in each set of four boxes
1. What is 9301854 written in words?
A. Nine million three thousand and one, eight hundred and fifty four.
B. Ninety three and one thousand, eight hundred and fifty four.
C. Nine million three hundred and one thousand eight hundred and fifty four.
D. Nine hundred and thirty thousand eighteen hundred and fifty four.
2. What is the value of
A. 2 B. 14 C. 18 D. 24
3. What is 4.59954 written correct to three decimal places?
A. 4.599
B. 4.6
C. 4.60
D. 4.600
4. Wm is the L.C.M of 30, 45 and 50? A. 15
B. 135
C. 180
D. 540
5. What is the place value of digit 2 in the product of the total value of digit 4 multiplied by the total value of digit 3 in the number 57438?
A. Ones
B. Tens
C. Hundreds
D. Thousands
6. Jebet bought the following items: 3 packets of maize flour at sh.90 each 2 kg of beans for sh 170 1 1/2 kg of potatoes at sh 40 per kg 2 loaves of bread at sh 34 each If she had sh 800,how much
money was she left with?
A. sh 62
B. sh 232
C. sh 466
D. sh 568
7. What is the value of x in the equation
2(x + 1)/3-4 = 6
A. 14
B. 10
C. 8
D. 4
8. The area of a square is 3844 cm”. What is the length of each side of the square?
A. 1922 cm
B. 961 cm
67 cm
62 cm
9. Which is the correct order of writing the fractions 2/5, 4/15, 1/6, 1/2, 2/3 starting from the smallest to the largest‘?
A. 4/15, 2/5, 2/3, 1/6, 1/2 B. 2/3, 1/2, 2/5, 4/15, 1/6 C. 1/2, 2/3, 2/5, 1/6, 4/15 D. 1/6, 4/15, 2/5, 1/2, 2/3
10. In the triangle PQR below, construct the bisectaor of angle PQR to cut line PR at M and the bisector of angle QPR to cut line QR at N‘ The two bisectmjs intersect at point X. Join RX.
What is the size of angle RXM?
A. 58°
B. 60°
C. 65°
D. 117°
11. How many fencing posts, spaced Sm apart, are required to fence a rectangular plot measuring 745 m by 230 m?
A. 391
B. 390
C. 195
D. 196
12. Awinja bought a pair of shoes for sh 810 after getting a discount of 10%. What was the marked price of the pair of shoes?
A. sh 81
B. sh 729
C. sh 891
D. sh 900
13. The table below shows the amount of milk delivered by a farmer to the dairy in 6 days.
What was the median sale of milk, in litres, for the 6 days?
A. 18
B. 19 1/3
c. 20 1/2
D. 21
14. Mutiso and Oluoch shared the profit of their business such that Mutiso got gr of the profit. What was the ratio of Mutiso’s share to Oluoch’s share’?
A. 3: 2
B. 5: 3
C. 3: 5
D. 2: 3
15. What is the value of 0.5 +0.2 +0.25/0.2
A. 14
B. 6.5
C. 4. 5
D. 2.75
16. Mulwa had 5 one thousand shillings notes, 7 five hundred shillings notes, 10 two hundred shilling notes and 6 one hundred shillings notes. He then changed the money into filly shillings .
notes. How many notes altogether did he get?
A. 555000
B. ll 100
C. 2 220
D. 222
17. The figure below is a map of a village drawn to the scale 1:250 000
What is the perimeter of the village in kilometers?
A. 6000
B. 600
C. 60
D. 6
18. A cylindrical container has a circumference of 176 cm and a height of 40 cm. What is the volume of the container in urn’? (Take n = 22/7})
A. 394240
B. 98560
C. 7040
D. 3 520
19. What is 1/2 (3x + 4y) + 1/5 (2x + 7y)- 1 1/4x – 1/2y in a simplified form
A. 13/20x + 2 9/10y
B. 13/20x + 10 1/2y
C. 3 3/20x + 3 9/10y
D. 4 1/4x + 2 9/10y
20. The figure below is a sketch of a triangle XYZ in which angle ZXY=50°, angle YZX=70° and line ZX= 6cm
Which one of the statements below leads to the correct construction of the triangle?
A. Use a ruler to draw line ZX=6cm long and drop a perpendicular from Y to ZX. Then join Y to X and to Z.
B. Use a ruler to draw line ZX=6cm long and a pair of compasses to construct angle ZXY=50° and YZX=7O°.
C. Measure and draw the angles ZXY=50° and YZX=70° using a protractor and draw line ZX =6cm long.
D. Use a ruler to draw line ZX=6cm long. Use a protractor to mark off an angle 70° at Z and angle of 50° at X Let the lines formed by the angles meet at Y.
21. Each of the diagonals of a rectangular flower garden is 65m. If one side of the garden measures 25m, what is the measurement” of the other side?
A. 90m
B. 60m
C. 40m
D. 20m
22. A meeting was attended by 150 people. Out of these. 0.14 were men, 0.2 were women and the rest were children. How many more children than women were these?
A. 59
B. 78
C. 99
D. 129
23. The triangle PQR shown below has been drawn accurately.
what is the size of angle QPR?
A. 95°
B. 85°
C. 50°
D. 45°
24. Mwaruwa is paid sh 3750 after working for 25 days. How much money would he be paid if he does not work for 4 days?
A. sh 600
B. sh 4464
C. sh 4350
D. sh 3150
25. A family uses decilitres of milk each day. How many litres of milk altogether would the family use in the months of June and July?
A. 305
B. 30.5
C. 30.0 D. 3.05
26. In the triangle ABC below, construct a perpendicular from A to meet line BC at N.
Which one of the following statements is correct?
A. Line AN bisects line BC
B. Angle BAN is equal to angle CAN
C. Angle ANB is equal to angle ANC
D. Line AB is equal to line BN.
27. The graph below shows the journey made by a social worker on a certain day
Between which two places was his speed the highest?
A. Home and school
B. School and health centre
C. Health centre and the market
D. Market and home.
28. What is the value of p(2r + q) —r/q, where p=3,q—p = 4 and r = p+q/2?
A. 8 5/7
B. 6 4/7
C. 2 2/7
D. 2/7
29. What is the surface area of a cylindrical rod of height 17 cm and diameter 14 cm?
(Take n = 22/7)
A. 748 cm
B. 902 cm
C. 1056 cm’
D. 2728 cm‘
30. What is the value of 2 1/2 – 2/3 +7/8 x 5/7- 1 2/5 of 5/6?
A. 3 5/24,
B. 2 23/168
C. 1 97/336
D. 1 1/8
31. Halima bought 50 bananas @ sh 3 each. She spent sh 75 for transportation. During transportation 5 bananas got spoilt but she sold the rest making a 20% profit. For how much did she sell each
A. sh 4.00
B. sh 5.40
C. sh 5.60
D. sh 6.00
32. Two sides of a parallelogram EFGH have been dfawn beluw. Colnpiae the parallelogram EFGH. Draw diagonals EG and FH to intersect at J.
What is the length of line FJ? A. 2.76 cm
B. 3.50 cm
C. 4.4 cm
D. 6.5 cm
33. Mutuma leftMombasa on Tuesday at 6.30 pm. and took 8 hours 45 minutes to reach his home. On what day and at what time in a 24 hour system did he reach home? A. Wednesday 0315 h
B. Wednesday 1515 h
C. Tuesday 1515 h
D. Tuesday 0315 h
34 The pie chart below represents the population Working Space of 1800 animals in a farm.
How many more chickens than goats are there in the farm?
A. 300
B. 900
C. 1200
D. 180
35. A car traveled 216 km at an average speed of 48km/h. On the return journey the avarage speed increased to 72km/h. Calculate the average speed, in km/h for the whole journey?
A. 57.6
B. 60
C. 28.8
D. 68.6
36. Which one of the statements below is a property of a right angled triangle‘?
A. All sides are equal.
B. Adjacent angles are supplementary.
C. Two of its sides are perpendicular.
D. The longest side of the triangle is opposite the smallest angle.
37. A mathematics text book has 97 sheets of paper and a cover. Each sheet of paper has a mass of 4 grams and the cover has a mass of 20 g. Find the mass of the book in kilograms.
A. 0.408
B. 4.08
C. 40.8
D. 408
38.The diagram below is a trapezium MNPQ.Line MQ is parallel to line NP.The length of line MQ = 8cm and that of line NR = 7cm The perpendicular line MR-=12cm.
If the area of the trapezium is 198cm‘, what is the length of RP‘?
A. 15cm
B. 18cm
C. 25cm
D. 32cm
39. Ali is now two yuan older than Martha. If Mania’s age is represented by x, what will be their total age after 10 years? A. 2x+22
B. 3x+20
C. x+22
D. 2x+18
40.A football match was attended by 42000 men. The number of women who attended was 27000 less than the number of men and 12000 more than than the number of children. The entrance fee for adults was
sh 100 and for children was sh 50. How much money was collected all together?
A. sh 11700000
B. sh 7500000
C. sh 7050000
D. sh 5850000
41. In the figure below EFG is a straight line. Working Space Lines GH and FH are equal and lines HI and Fl are also equal. Angle GHF is a light angle and angle HIF is 32°. What is the size of angle
A. 61
B. 45°
C. 74°
D. 103°
42. The table below shows pan of Tariff for Ordinary Money Order and Postapay.
Karimi has two children in one school. To pay for their school fees he sent sh 8900 by Ordinary Money Order and sh 15100 by Postapay. How much money would he have saved had he bought one Ordinary
Money Order to pay for all the fees?
A. sh 125
B. sh 400
C. sh 525
D. sh 925
43. Nina is mid a basic salary of sh 8000 as a sales agent. In addition she is paid a 5% commission for goods sold above sh 15000. In one month she earned sh 12000 altogether. What was the total
sales? A. sh 255 000
B. sh 95 000
C. Sh 80 000
D. sh 65 000
44. What is the next number in the pattern 4, 9, 25, 49, 121, 169, -—-‘.7 A. 289
B. 256
C. 225
D. 196
45. The marked price of a motorcycle was sh 30000 but a discount of 5% was allowed for cash payment. Taabu bought the motorcycle an hire purchase terms by paying a deposit of sh 8500 followed by ten
equal monthly instalments of sh 2400 each. I-low much money would Taabu have saved had she bought it for cash?
A. sh 4 000
B. sh 2 500
C. sh 1 500
D. sh 28 500
46. Figure ABCDE below represents a vegetable garden in which AE=12m, AB=36m and CD=24m. Angle DEA is a right angle. The distance from A to D is 15m. A perpendicular dlstsnee fromC to AB is 10m.
What is the area of the garden?
A. 474 m‘
B. 390 m
C. 354 m’
D. 300 m
47. A tailor made 48″pieces of uniform. Half of the number of the uniforms was each made using 1 1/4 meters of material.A quatre of the remainder was each made using 1/2 metres a of the material and
the rest was made using 1 3/4 meters 13 metres of material. The tailor also fixed a logo made using 1/16 meters of material on each uniform.
How many metres of material did the tailor use?
A.72 1/2 meters
B. 72 meters
c. 70 9/6 meters
D. 70 1/2 meters
48. The table below shows the train fares for Nairobi—Mombasa route.
The following passengers travelled in the train:
23 pupils of age 12 years and above
12 pupils aged between 7 and 10 years
2 children below 3 years
3 parents
5 teachers
1 headteacher
The passengers occupied the following classes in the train:
1st class: Headteacher; 1 parent
2nd class: 5 teachers, 2 parents and all pupils and children
How much money did they pay for the journey to Mombasa?
A sh 119 560
B. sh 151 300
C. sh 156 100
D. sh 64080
49. A man deposited sh 50000 in 3 bank for 2 Working Sum years. The hank paid compound interest at the me of 10% per annum. How much money was in his account at the end of the two years?
A. 51110500
B. sh 55 500
C. sh 60000
D. sh 60 500
50. The figures below represent a pattern
Which one of the following is the next shape in the pattern above’?
Free KNEC KCPE Past Papers Mathematics 2011 Answers
KCPE 2011 ANSWERS
# Math
1 C
2 C
3 D
4 C
5 D
6 B
7 A
8 D
9 D
10 C
11 B
12 D
13 C
14 A
15 B
16 D
17 C
18 B
19 A
20 D
21 B
22 A
23 B
24 D
25 B
26 C
27 B
28 C
29 C
30 A
31 D
32 C
33 A
34 B
35 A
36 C
37 A
38 B
39 A
40 D
41 B
42 C
43 B
44 A
45 A
46 C
47 A
48 B
49 D
50 D
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What is Sudoku? - T-Chertz
What is Sudoku?
Sudoku is a logic-based, number-placement puzzle. The objective is to fill a 9×9 grid with digits so that each column, each row, and each of the nine 3×3 boxes (also called cages) contain the
The puzzle initially starts with some of the digits already placed in the grid. The object is to fill in the rest of the grid according to the following rules:
All of the digits in a row or column must be unique.
No two digits can occur in the same box.
Digits cannot be repeated within a row or column.
The puzzles can be solved using logic alone. No calculations are necessary.
Sudoku puzzles are popular because they are easy to learn but can be challenging to solve.
Sudoku Rules and Sudoku History
Sudoku is a logic-based, number-placement puzzle. The game is played on a 9×9 grid, divided into nine 3×3 boxes. In the game, the player must fill each row, column, and box with the numbers 1 through
The game Sudoku was invented in 1988 by Howard Garns, an Electrical Engineer from the United States. The game was patented in the United States in 1989. The game appeared in the Japanese puzzle
magazine Nikoli, where it was known as Sudoku, meaning “single number”. The game became popular in Japan, and in 2004, it appeared in the United States.
The basic rules of Sudoku are simple. The player must fill each row, column, and box with the numbers 1 through 9. However, the challenge of the game comes from the fact that each number can only
appear once in each row, column, and box.
There are many strategies that can be used to solve Sudoku puzzles. Some people use pencil and paper to solve the puzzles, while others use software or online tools. There are also websites that
offer online Sudoku puzzles, with different levels of difficulty.
Sudoku puzzles are a type of logic puzzle that is popular around the world. The puzzles are usually a 9×9 grid with a total of 81 squares. The goal of the puzzle is to fill in the squares with the
numbers 1-9 so that each column, row and 3×3 box has the same number of instances of each number.
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Time Complexity of Algorithms Explained with Examples
There are indeed multiple ways of solving any problem that we might face, but the question comes down to which one among them is the best and should be chosen. But let’s focus only on algorithms, the
best way to find the right solution for a specific problem is by comparing the performances of each available solution. Here Time complexity of algorithms plays a crucial role with Space Complexity
as well, but let’s keep it for some other time.
In this blog, we will see what is time complexity, how to calculate it, and how many common types of time complexities are there.
Let’s begin…
1. What is Time Complexity
2. Big O Notation
3. How to calculate Time Complexity
4. Short Hand Rule to calculate Time complexity — Drop the constants and Remove all non-dominant terms
5. Example Time
6. Types of Time Complexities — Constant, Linear, Quadratic, Polynomial, Logarithmic, linaerithmic and Exponential, Time Complexity.
7. Conclusion
8. FAQ’s
What is Time Complexity of algorithms?
Time complexity is the amount of time taken by an algorithm to run, as a function of the length of the input. Here, the length of input indicates the number of operations to be performed by the
It depends on lots of things like hardware, operating system, processors, etc, and not just on the length of the input. However, we don’t consider any of these factors while analyzing the algorithm.
We will only consider the execution time of an algorithm.
From this, we can conclude that if we had an algorithm with a fixed length or size of the input, the time taken by that algorithm will always remain the same.
Let’s take an example to understand this better -
The above statement is only printed once as no input value was provided (number of times it should run), thus the time taken by the algorithm is constant.
But for the above code, the time taken by the algorithm will not be constant as the above code contains a ‘for loop’ iterating the algorithm equal to the size of the input. In the above code, the
size of the input is taken as 5, thus the algorithm is executed 5 times.
From this, we can conclude that if the statement of an algorithm has only been executed once, the time taken will always remain constant, but if the statement is in a for loop, the time taken by an
algorithm to execute the statement increases as the size of the input increases.
But if the algorithm contains nested loops or a combination of both, a single executed statement and a loop statement, the time is taken by the algorithm will increase according to the number of
times each statement is executed as shown
From this, we can draw a graph between the size of the input and the number of operations performed by the algorithm which will give us a clear picture of different types of algorithms and the time
taken by them or the number of operations performed by that algorithm of the given size of the input.
From the above graph, we can say that there exists a relationship between the size of the input and the number of operations performed by an algorithm, and this relation is known as the order of
growth and is denoted by Big Oh (O) notation which is an asymptotic notation.
Big O Notation
It is used to express the upper limit of an algorithm’s running time, or we can also say that it tells us the maximum time an algorithm will take to execute completely. It is also used to determine
the worst-case scenario of an algorithm.
Asymptotic Notation is used to describe the running time of an algorithm with a given input. There are mainly three types of asymptotic notations -
Big Oh (O) — used to calculate the maximum time taken by an algorithm to execute completely.
Big Theta (Θ) — used to calculate the average time taken by an algorithm to execute completely.
Big Omega (Ω) — used to calculate the minimum time taken by an algorithm to execute completely.
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Computer-Assisted Proofs and Verification Methods
Computer-Assisted Proofs and Verification Methods
• Time: 18.9.2011 - 22.9.2011
Japanese-German Workshop
September 18 - 22, 2011
Location — Organizers — Schedule — Speakers
The concept of computer-assisted proofs can be regarded as a special approach to constructive mathematics. In recent years, various mathematical problems have been solved by computer-assisted proofs,
among them the Kepler conjecture (a 3-dimensional sphere packing problem), the existence of chaos, the existence of the Lorenz attractor, and more.
In the SIAM News 1-2/2002, Lloyd N. Trefethen (Oxford) proposed a 10x10 digit Challenge. These are ten difficult numerical problems with a single number as result. The problems vary from global
optimization, numerical integration, partial differential equations to probabilistic problems and more. The challenge was to produce 10 correct digits of each solution. Various researchers
participated in the contest, but only few of them could give correct answers. An outstanding approach, winning the contest, was the result of effort by a group of four researchers from four different
countries. Remarkably, five out of the ten problems were solved using verification methods and the MATLAB toolbox INTLAB for reliable computing developed by S.M. Rump (TU Hamburg-Harburg).
Many problems involving partial differential equations, with a huge number of applications in science and engineering, allow very stable numerical computations of approximate solutions, but are still
lacking analytical existence and multiplicity proofs. In recent years, methods have been developed which supplement purely analytical arguments by computer-assisted approaches, and these methods have
turned out to be successful in various examples where purely analytical methods have failed. One of the organizers (Plum) and the Japanese scientists M.T. Nakao and S. Oishi have made internationally
recognized contributions by computer-assisted proofs in the field of partial differential equations and their applications, like fluid dynamics (Navier-Stokes problems, Orr-Sommerfeld equation,
Rayleigh-Benard heat convection), electro-dynamics (photonic crystals, scattering problems described by the Lippmann Schwinger equation), magneto-hydrodynamics (plasma flow), nonlinear waves (for
example, travelling waves in a beam), elasticity/plasticity, variational inequalities, obstacle problems, and semilinear elliptic boundary and eigenvalue problems in general. In this context, various
kinds of constants appearing e.g. in finite element error bounds or in Sobolev embeddings are needed explicitly, and have been determined by computations of analytical and computer-assisted means.
S.M. Rump and S. Oishi have extensively developed arithmetic and linear algebra tools for computer assisted proofs based on numerical computations with guaranteed accuracy. For example, they have
developed fast and ultra-fast methods for solving numerical linear algebraic problems with guaranteed accuracy, adaptive methods for solving ill-posed problems in numerical linear algebra, accurate
summation and dot product algorithms for floating point vectors. These methods are also provided in the aforementioned MATLAB interval toolbox INTLAB for verification algorithms, which is widely used
all over the world. As an application of these tools, Oishi and Rump have also developed algorithms in computational geometry, which always give mathematically correct results even if they are
executed by floating point calculations.
The particular expertise of Japanese and German scientists indicates that a workshop with participants primarily from Japan and Germany and from all the described subareas will be very promising. We
are convinced that our workshop has high potential for very fruitful interactions between these areas. Moreover, it will lead to significant developments in several inter-related fields like
numerical functional analysis, ordinary and partial differential equations, ill-posed and ill-conditioned problems, constrained programming, reliable algorithms, and topics related to computer
Kaiserstraße 93, Rooms 1C-04, 1C-01 - Please see map!
G. Alefeld (Karlsruhe), M. Plum (Karlsruhe), S. M. Rump (Hamburg-Harburg)
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Finding cliques using few probes
I will talk about algorithms (with unlimited computational power) which adaptively probe pairs of vertices of a graph to learn the presence or absence of edges and whose goal is to output a large
clique. I will focus on the case of the random graph G(n,1/2), in which case the size of the largest clique is roughly 2\log(n). Our main result shows that if the number of pairs queried is linear in
n and adaptivity is restricted to finitely many rounds, then the largest clique cannot be found; more precisely, no algorithm can find a clique larger than c\log(n) where c < 2 is an explicit
constant. This is joint work with Uriel Feige, David Gamarnik, Joe Neeman, and Prasad Tetali.
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Science:Math Exam Resources/Courses/MATH101/April 2014/Question 07 (b)
MATH101 April 2014
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q2 • Q3 • Q4 • Q5 • Q6 • Q7 (a) • Q7 (b) • Q8 • Q9 (a) • Q9 (b) •
Question 07 (b)
Long Problem. Show your work. No credit will be given for the answer without the correct accompanying work.
Determine, with explanation, whether the following series converges or diverges.
${\displaystyle \sum _{n=1}^{\infty }{\frac {n\cos(n\pi )}{2^{n}}}}$
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is
correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Evaluate ${\displaystyle \cos(n\pi )}$ for the first few values for ${\displaystyle n}$ to see a pattern.
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
• If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you
are stuck or if you want to check your work.
• If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem
and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies.
First, if we evaluate a few terms we see that ${\displaystyle \cos(\pi )=-1}$, ${\displaystyle \cos(2\pi )=1}$, ${\displaystyle \cos(3\pi )=-1}$, ${\displaystyle \cdots }$, ${\displaystyle \cos(n\pi
)=(-1)^{n}}$ and therefore the terms alternate. Thus, the given series is alternating series,
{\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }(-1)^{n}{\frac {n}{2^{n}}}.\end{aligned}}}
Since ${\displaystyle \lim _{n\to \infty }{\frac {n}{2^{n}}}=0}$ and ${\displaystyle a_{n}}$ is decreasing for ${\displaystyle n\geq 3}$, by the alternating series test, the series ${\displaystyle \
sum _{n=1}^{\infty }{\frac {n\cos(n\pi )}{2^{n}}}}$ converges.
Click here for similar questions
MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Alternating series test, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag
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Lecture 025
Definition of Tree and Forest
Definition of Cycles
Cycle: a path from a vertex to itself
Simple Cycle: with at least one edge and without repeated edges
Definition of Trees
Tree: a connected graph without simple cycles
• a connected graph with no simple cycles
• a vertex, or two trees connected by an edge (rec)
• a vertex, or a tree connected to a vertex by an edge (rec)
• a connected graph with v vertices and v-1 edges
• a connected graph with exactly 1 path between any two vertices
Definition of Forests
Forest: a graph where each connected component is a tree (notice a tree is a forest)
• a graph with (connected or not) no simple cycles
• a graph with at most one path between any two vertices
• a forest with v vertices has at most v-1 edges
DFS, BSF complexity of a tree: O(v) reduced from O(v+e) unlike graph
Spanning Tree
Binary Search Tree
BST: a tree where every vertex has at most 3 edges
Spanning Tree
Superimposing a tree
subgraph: a subgraph of graph G is a graph with the same vertices and a subset of its edges
spannning tree: a spannning tree for G is a subgraph that
spanning forest: a spanning forest for G is a subgraph that (a bunch of spanning trees)
Computing a Spanning Tree
Edge-Centric algorithm (O(ev)?O(elog(v)))
Start with a spanning forest of singleton trees and add edges from the graph as long as they don't form a cycles (def.B)
Adding Edges
Initialize T with isolated vertices of G (O(1)=graph_new)
For each edge (u, v) in G: (O(e))
if connected(u, v): (O(v) where v=v_in_tree_so_far by DFS/BFS)
discard the edge
add it to T (O(1))
stop once T has v-1 edges
costs O(ev)
- O(v^2) for sparse graphs
(it can create spanning forests) (Edge-Centric algorithm is greedy, DFS/BFS is not since they have worklist)
Making a Maze
Using graph
Vertex-Centric algorithm (O(e))
Start with a single vertex in the tree and add edges to vertices not in the tree (def.C)
Adding Vertex
Actual Algorithm
(Complexity: O(e) as DFS/BFS)
Minimum Spanning Tree
Minimum spanning tree: one with the least total weight of its edges.
• a graph may have several minimum spanning trees
Kruskal's Algorithm (O(ev)->O(elog(e)))
We sort first
• total is O(ev)
• sort is O(e log(e))
□ O(elog(e) + ev) above
□ since log(e) \in O(v)
☆ because e \in O(v^2), so log e \in O(log v)
☆ and log v \in O(v)
Finished Visualization
Union Find: A way to check connectivity by grouping O(ev)
Union Find Concept: checking connectivity and connect
Thinking in terms of relation
Building Graph using list
• check if two point to the same representative (O(v))
• merge representative, always have 2 choice (don't merge node) O(1)
• stop once T has v-1 edges
Height Tracking: Trying to build balanced tree by tracking height (O(elog(e)))
Height Tracking Array
• store length in root as negative
• merge shorter trees into taller trees
• initialize array into "-1"s
• A tree T of height h has at least $2^{h-1}$ vertices
• A tree T with v vertices has height at most $log(v)+1$ (balanced tree = O(log v), for loop O(e), total O(eloge + elog v), assume sparse O(eloge))
Sample Code
Path Compression (O(eAck^{-1}(v)) amortized)
Path Compression: Ackermann Function
Prim's Algorithm (O(e log e))
Using priority queue
Complexity Summary
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Question ID - 155365 | SaraNextGen Top Answer
For $f(x)=|x|,$ with $\frac{d f}{d x}$ denoting the derivative, the mean value theorem is not
applicable because
(A) $f(x)$ is not continuous at $x=0$
(B) $f(x)=0$ at $x=0$
(C) $\quad \frac{d f}{d x}$ is not defined at $x=0$
(D) $\quad \frac{d f}{d x}=0$ at $x=0$
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This is my first cautious approach to C#. I wrote this to get a idea how programming in C# feels like. The program hence makes no use of the object-oriented paradigm and implements a fully static
Yet, I wanted my first C# program to be something more interesting than
What came to my mind was a prime number generator:
Domains like shared-key cryptography require the generation of large prime numbers, which is computationally expensive using the naive approach. A more effective way to test, whether a given number
is prime was suggested by
as an iterative test that trades completeness for efficiency. Using Fermat's little theorem, the algorithm randomly picks
numbers in the range {1,..,n-1} as potential whitnesses for
not being prime. If no whitness is found,
is assumed to be prime. The higher
the lower is the chance that
is falsely assumed to be prime (false positive).
The implementation at hand allows the fast generation of all prime numbers within a given range. To lower the risk of false positives, the number of iterations (
) can be passed as a parameter (default=10).
The archive below contains the C# source files and a shell-based .NET executable. For use in your own projects, please refer to the !Readme.txt in the archive.
This program is a .NET based application. You may need the .NET runtime Framework to make it run on older Windows Systems.
You can get the runtime distributable from
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How to Sum If Less Than in Excel
To sum values in Excel if they are less than a specific number, you can use the SUMIF function. SUMIF allows you to add up the values in a range that meet a certain criterion.
Here's a step-by-step guide on how to use SUMIF to sum values less than a specific number:
1. Prepare your data in Excel. Organize your data in columns, with one column containing the values you want to sum and another column containing the criteria you want to use (in this case, the
specific number).
2. In an empty cell, type the following formula:
=SUMIF(range, "<" & criteria, [sum_range])
• range: The range of cells you want to apply the criteria to.
• criteria: The specific number you want to compare the values to.
• sum_range: The range of cells containing the values you want to sum. If not provided, Excel will assume the range contains the values to sum.
3. Replace the range, criteria, and sum_range in the formula with the appropriate cell references or values.
4. Press Enter to get the result.
Let's say you have the following data in Excel:
A B
You want to sum the values in column A that are less than 20. Here's how to do it using the SUMIF function:
1. In an empty cell, type the following formula:
=SUMIF(A1:A4, "<20")
2. Press Enter. The result (25) will be displayed in the cell, as 10 + 15 = 25, and both 10 and 15 are less than 20.
If you want to sum the values in column B based on the criteria in column A, you can use the following formula:
=SUMIF(A1:A4, "<20", B1:B4)
Press Enter. The result (26) will be displayed in the cell, as 24 + 2 = 26, since 10 and 15 from column A are less than 20, and their corresponding values in column B are 24 and 2, respectively.
Did you find this useful?
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Universal quantum operations and ancilla-based read-out for tweezer clocks - Rin
Experimental considerations and constraints
Our experimental set-up has been detailed in previous work^13,16,58. Here, we discuss the experimental requirements specific to executing quantum circuits on a quantum processor with optical clock
qubits. In particular, several trade-offs need to be balanced in the choice of magnetic field, trap depth and interatomic spacing.
Due to laser frequency noise, high-fidelity single-qubit rotations benefit from a large Rabi frequency on the clock transition (^1S[0]â â â ^3P[0]). The clock-transition Rabi frequency scales
linearly with the magnetic field, and we achieved Ωâ =â 2Ï â à â 2.1â kHz at 450â G. On the other hand, the Rydberg interaction strength varies with the magnetic field due to admixing with
other Rydberg states^61. Specifically, a numerical calculation (using the â Pairinteractionâ package^61 and limiting the considered Rydberg states to nâ ±â 5 for faster convergence) shows that
the interaction energy peaked around 380â G and decreases for higher magnetic fields (Extended Data Fig. 1a). Our experimental measurements for several magnetic fields are consistent with this
overall trend. Therefore, we operate at a magnetic field of 450â G, which provides a balance between a sufficiently high clock Rabi frequency and sufficiently strong Rydberg interactions.
Our nominal tweezer trap depth (U[0]â â â 450â μK) is chosen to ensure efficient atom loading and high-survival, high-fidelity imaging^62, as well as efficient driving of the carrier transition
for clock qubits. However, we find that the Rydberg gate performs slightly better when the trap is turned off. In our case, this is predominantly due to beating between adjacent tweezers, which
results in trap-depth fluctuations at a frequency of 650â kHz (equal to the tone separation on our tweezer-creating acousto-optic deflector). As the Rydberg transition (^3P[0]â â â 61^3S[1]) is
not under magic trap conditions, this results in detuning noise for the gate operation. For an optimized CZ gate with traps kept on at 0.2U[0], we find that the two-qubit gate fidelity is lower by
approximately 3â Ã â 10^â 3 (not shown). By placing an acousto-optic modulator in the tweezer optical path, we implement a fast switch-off of the trapping light (rise/fall time of the order of 50â
ns). At this timescale, we find that switching traps off from a shallower trap 0.2U[0] is preferable as this imparts minimal heating with no observed loss for clock qubits. Furthermore, in MCR,
efficient motional shelving relies on strong sideband coupling^44, which becomes stronger as the Lambâ Dicke factor increases (trap depth decreases). On the other hand, array reconfiguration
benefits from sufficiently deep traps.
The experimental sequence, thus, involves adiabatic ramps of trap depth as well as fast switch-off and -on. After loading the atoms into the tweezers at U[0], we lower the tweezer depth to 0.5U[0] to
perform erasure-cooling^44. For gate operations, we drive coherent clock rotations at 0.2U[0] and switch the trap off for about 500â ns to perform the Rydberg entangling pulse. When selective local
MCR is applied, we adiabatically ramp to deeper traps of U[0] for the ancilla qubits, while holding the clock qubits at fixed depth.
We now discuss how we perform the dynamical array reconfiguration, shown in Fig. 1 as part of a full quantum operation toolbox. We coherently transport an atom across several sites by performing a
minimal-jerk trajectory that follows x(t)â =â 6t^5â â â 15t^4â +â 10t^3 for tâ â â [0, 1]. For this trajectory, the acceleration is zero at the two end points, which avoids the sudden jump
in the acceleration profile and minimizes the associated jerk. The aim is to achieve minimal heating, which is especially important for driving optical clock transitions in the sideband-resolved
With this trajectory, we find no significant temperature increase for atoms transported over four sites (equivalent to 13.26â μm) in 160â μs at trap depth U[0] (Extended Data Fig. 1b). This is
the typical distance applied in dynamical array reconfiguration. Another interatomic spacing choice is 13.26â μmâ â â 19â à â 698â nm (corresponding to the clock-transition wavelength), which
would ensure an effective zero displacement-induced phase shift^16.
Single-qubit (clock) error model
We characterize our ability to perform coherent single-qubit rotations with a global addressing beam and test our error model by driving the clock transition on atoms with an average motional
occupation of \(\bar{n}\approx 0.01\), following erasure-cooling along the optical axis^44. We drive Rabi oscillations with a nominal Rabi frequency Ωâ =â 2.1â kHz and observe 52.2(8) coherent
cycles (Extended Data Fig. 2a). Applying a train of Ï /2 pulses along the X axis, we find a per-pulse fidelity of 0.9988(2). Note that in such sequences, the effect of slow frequency variations is
suppressed. We, thus, characterize the Ï /2 pulse fidelity by applying a pulse train of Ï /2 pulses with random rotation axes ±X and ±Y (ref. ^63). The resulting Ï /2 pulse fidelity is measured to
be 0.9978(4) (Extended Data Fig. 2c).
The dominant error source in the single-qubit operation is laser frequency noise, which is characterized by the frequency power spectral density (PSD) function S[ν](f). We characterize this with
Ramsey and spin-lock^64 sequences.
Ramsey sequence
The Ramsey sequence is sensitive to the low-frequency component of the laser frequency PSD up to the inverse of the Ramsey interrogation time (approximately 100â Hz). In our experimental set-up, we
observe day-to-day fluctuations in the Ramsey coherence time (Extended Data Fig. 2d). We use an effective model of the PSD at low frequencies (Extended Data Fig. 2f) to account for the fluctuations
of the Ramsey coherence time. We set the PSD to be a constant H at low frequencies up to some frequency of interest (approximately 200â Hz) and find numerically that the Ramsey coherence time was
inversely proportional to H.
Spin lock
To probe and quantify fast frequency noise up to our Rabi frequency, we perform a spin-lock sequence. We initialize all atoms in an eigenstate of \(\widehat{X}\) and turn on a continuous drive along
the X axis for a variable time. Then we apply a Ï /2 pulse along the Y axis, which transfers all atoms into state \(| 1\rangle \) in the absence of errors. The probability of returning to \(| 1\
rangle \) decays over time (Extended Data Fig. 2e), and the decay rate is predominantly sensitive to frequency noise at this Rabi frequency^64. By varying the Rabi frequency of the continuous drive
field and measuring the decay of the probability in \(| 1\rangle \), we determine the frequency PSD by using the linear relation between the decay rate and the frequency PSD S[ν](f) at the Rabi
frequency (Extended Data Fig. 2f).
To account for both the fast frequency noise measured by the spin-lock experiment and the slow frequency noise that determines the Ramsey coherence time, we interpolate the laser frequency PSD with a
power-law function \({S}_{\nu }(f)={h}_{0}+{({h}_{\alpha }/f)}^{\alpha }\) upper-bounded by H at low frequencies. The model parameters h[0], h[α] and α are obtained by fitting the spin-lock data
(Extended Data Fig. 2e,f). The upper bound H is flexible within a range (shown as the shaded area in Extended Data Fig. 2f) and can effectively describe the day-to-day fluctuations of the Ramsey
coherence time. This range is reflected in the uncertainties of the error model predictions quoted throughout this work.
In addition to laser frequency noise, note that although the single-qubit operations are sensitive to the finite temperature, we perform erasure-cooling^44 to prepare atoms close to their motional
ground state (\(\bar{n}\approx 0.01\)). This has a negligible impact (approximately 1â Ã â 10^â 4) on the clock Ï /2 pulse fidelity, as predicted by our error model.
In addition to the error sources described above, we also include laser intensity noise, pulse shape imperfection, spatial Rabi frequency inhomogeneity and Raman scattering induced by the tweezer
light. All these error sources result in an aggregate of approximately 1â Ã â 10^â 4 infidelity for the clock Ï /2 pulse.
With all described error sources included, the error model predicts an average Ï /2 fidelity of 0.9981(8) (Extended Data Fig. 2c), which is in good agreement with the experimental value of 0.9978(4).
Two-qubit gate fidelity benchmarking
Here, we give more details about the randomized circuit (Fig. 2), which is used to benchmark the CZ gate fidelity. We first apply a randomized circuit like the one proposed and used in ref. ^17,
which includes echo pulses (Ï pulses along X) interleaved with random single-qubit rotations and CZ gates (Extended Data Fig. 3a). For this circuit, we observe that both two-qubit (Rydberg) errors
and single-qubit (clock) errors contribute to the inferred infidelity (Extended Data Fig. 3b,c). That such a circuit is sensitive to single-qubit gate errors, although the number of single-qubit
gates was kept fixed, is because the probability distribution of two-qubit states before each single-qubit gate changes as a function of the number of CZ gates applied. Note that as errors affect
entangled states and non-entangled states differently, changing the probability distribution would result in a non-unity return probability, even if the fidelity of CZ gates were perfect, in the
presence of single-qubit gate errors. In this context, the sequence used in Extended Data Fig. 6 of ref. ^17 also showed sensitivity to single-qubit errors, as the probability distribution between
entangled and non-entangled states was not fixed as a function of the number of CZ gates.
To mitigate this effect, we design a randomized circuit (Extended Data Fig. 3a) such that the probability of finding any one of the 12 two-qubit symmetric stabilizer states would be uniform,
irrespective of the number of CZ gates, at each stage of the circuit^47. We term this circuit the symmetric stabilizer benchmarking (SSB) circuit. Specifically, the probabilities of finding an
entangled or separable state are equal throughout the circuit.
Using an interleaved experimental comparison, we find a difference of about 3â Ã â 10^â 3 between benchmarking methods in the fidelity directly inferred from the slope of the return probability
(Extended Data Fig. 3b). This difference stems from the higher sensitivity of the echo circuit benchmarking to single-qubit gate errors. This observation is in good agreement with a full error model
that accounts for both clock and Rydberg excitation imperfections (Extended Data Fig. 3c). This model confirms that the fidelity inferred from the symmetric stabilizer benchmarking circuit is an
accurate proxy of the gate fidelity averaged over all two-qubit symmetric stabilizer states. We confirm that this observation holds over a wide range of error rates by rescaling the strength of
individual error sources in the numerical model (Extended Data Fig. 3c). These include incoherent and coherent errors. However, note that coherent errors or gate miscalibration of larger magnitude
would result in an increased error in estimating the gate fidelity, which is a common issue across various benchmarking techniques. Also note that the gate fidelity averaged over all two-qubit
symmetric stabilizer states is equal to the gate fidelity averaged over two-qubit symmetric input states. This can be seen because these symmetric stabilizer states form a quantum state two-design on
the symmetric subspace^47.
Correcting for the false contribution from leakage errors
We read out the return probability for the randomized circuit benchmarking by pushing out ground-state atoms and pumping clock-state atoms to the ground state for imaging. As part of this optical
pumping, any population in the ^3P[2] state would be pumped and identified as bright, which is the clock-state population. We, thus, correct for leakage from the Rydberg state into the state ^3P[2]
identified as bright. We separately measure the decay into ^3P[2] per gate by repeating the benchmarking sequence, which is followed by pushing out the atoms in the qubit subspace and repumping the ^
3P[2] state for imaging. At a Rydberg Rabi frequency of 5.4â MHz, the false contribution to the CZ fidelity is measured to be 1.8(4)â Ã â 10^â 4 per gate, in good agreement with numerical
predictions. The CZ fidelity quoted throughout this work has been corrected downwards for this effect.
Two-qubit gate (Rydberg) error model
Our model for the two-qubit gate accounting for Rydberg errors is based on previous modelling of errors during Rydberg entangling operations^58,65. We adapt it to model the dynamics of a three-level
system with ground (\(| 0\rangle \)), clock (\(| 1\rangle \)), and Rydberg states (\(| r\rangle \)). Following the optimization of the gate parameters for a time-optimal pulse^17,18 in the error-free
case (Extended Data Fig. 4a), we fix these parameters and simulated noisy dynamics with the Monte Carlo wavefunction approach. The model includes Rydberg laser intensity noise, Rydberg laser
frequency noise, Rydberg decay (quantum jumps) and atomic motion. The predicted contribution of each error source to the CZ gate infidelity is shown in Extended Data Fig. 4c. For the analysis shown
in Extended Data Fig. 3c, we repeat the numerical simulation several times and change the magnitude of one of the error model parameters in each run. For example, we rescale the overall magnitude of
the noise PSD for frequency or intensity noise or the Rydberg decay rate.
Data-taking and analysis
Data-taking and clock laser feedback
Here, we discuss the general data-taking procedure for all experiments described in the main text. Typically, each experimental repetition takes approximately 1â s. To collect enough statistics, we
perform the same sequence for from several hours up to several days (for the randomized circuit two-qubit gate characterization). However, on this timescale, the clock laser reference cavity
experiences environmental fluctuations, resulting in clock laser frequency drifts from approximately 10 to approximately 100â Hz over a timescale of approximately 10â min.
We, thus, interleave data-taking with calibration and feedback runs^16,65. To measure the clock laser detuning from atomic resonance, we perform Rabi spectroscopy with the same nominal power and Ï
pulse time as used in the experiment. The laser frequency is then shifted accordingly by an acousto-optic modulator. Such feedback is performed every 5â 10â min, depending on the details of the
experimental sequence. We reccord the applied laser frequency shifts, which can serve as an indicator of the clock laser stability during the experimental runs. To compare the stability from
experiment to experiment, we take the standard deviation of the feedback values. In the main text, the gate benchmarking (Fig. 2a,b) and the simultaneous preparation of a cascade of GHZ states (Fig.
3) have feedback standard deviations of 73 and 68â Hz, respectively. During the data-taking of the optical-clock-transition Bell-state generation experiment (Fig. 2c,d), the feedback standard
deviation is 203â Hz, significantly higher than other experiments. To ensure the consistency of clock laser conditions among all experiments, we select the Bell-state generation experimental runs
with associated clock laser feedbacks of less than 100â Hz. After applying this cutoff, the standard deviation of the feedback frequencies is 67â Hz, comparable with the other experiments.
To study the effects of the short-term clock stability, we analyse the Bell-state parity experimental runs with associated clock feedback frequencies less than a certain cutoff. With the cutoff
frequency increasing from 100â Hz (results are presented in Fig. 2d) to 400â Hz (all data included), the parity contrast shows a clear decreasing trend (Extended Data Fig. 6a). This is consistent
with our Bell-pair generation fidelity being limited by clock laser phase noise.
In contrast, note that using the randomized symmetric stabilizer benchmarking circuit to characterize the CZ gate itself, our results are consistent run to run and day to day within our experimental
error bars. This further attests to the largely reduced sensitivity of this sequence to single-qubit gate errors stemming from clock laser drift and clock laser phase noise.
Error bars and fitting
Error bars on individual data points throughout this work represent 68% confidence intervals for the standard error of the mean. If not visible, error bars are smaller than the markers. The
randomized circuit return probability shown in Fig. 2b and the parity signal shown in Fig. 2d are fitted using the maximum-likelihood method^58 (see details in the next section). Error bars on fitted
parameters represent one standard deviation. Fitting for all other experimental data is done using the weighted least squares method.
Data analysis of Bell-state fidelity
We analyse the results of the Bell-state experiments as in our previous work^58. We use a beta distribution to assess the underlying probabilities. For the parity signal shown in Fig. 2d, we fit the
data with a sine function with four free parameters: offset, contrast, phase and frequency. We find that using the maximum-likelihood method while taking the underlying beta distribution of each data
point into account is necessary, as the standard Gaussian fit typically overestimate the contrast by approximately 0.015. This is because the beta distribution deviates from a Gaussian distribution
when the two-atom parity is close to ±1, which breaks the underlying assumption of a Gaussian fit. From this, we obtain a parity contrast of \(0.96{3}_{-10}^{+7}\) (\(0.98{3}_{-10}^{+7}\) SPAM
corrected). Together with the measured population overlap P[00]â +â P[11]â =â \(0.98{8}_{-7}^{+5}\) (\(0.99{4}_{-7}^{+5}\) SPAM corrected) (not shown), we obtain a Bell-state generation fidelity
of \(0.97{6}_{-6}^{+4}\) (\(0.98{9}_{-6}^{+4}\) SPAM corrected). These results are obtained by analysing experimental runs with associated clock feedback frequencies of less than 100â Hz.
SPAM correction
The dominant measurement error stems from the long tails in a typical fluorescence imaging scheme. In our experiment, we infer the imaging true negative and true positive rates as F[0] = 0.99997 and
F[1]â =â 0.99995, respectively, from experimental measurements through a model-free calculation (ref. ^66, section 2.6.7). Note that these are not state detection fidelities in a circuit, as state
detection in a circuit would require a further push-out before imaging^62. The probability of successfully expelling the ground-state atom from the trap for state discrimination is Bâ =â 0.9989(1).
Taking these into account, the single-atom measurement-corrected values \({P}_{0}^{{\rm{m}}}\) and \({P}_{1}^{{\rm{m}}}\) have the following relation with the raw values \({P}_{0}^{{\rm{r}}}\) and \
$$\left[\begin{array}{c}{P}_{0}^{{\rm{m}}}\\ {P}_{1}^{{\rm{m}}}\end{array}\right]=\left[\begin{array}{cc}1-C & 1-A-C\\ C & A+C\end{array}\right]\,\left[\begin{array}{c}{P}_{0}^{{\rm{r}}}\\ {P}_{1}^
where \(A={[B({F}_{0}+{F}_{1}-1)]}^{-1}=1.0012\) and \(C=1-{F}_{1}{[B({F}_{0}+{F}_{1}-1)]}^{-1}=-0.0011\). By assuming that a measurement was independent among the atoms, we extend this correction to
multi-qubit measurements by taking the Kronecker product of the above matrix.
After correcting the measurements, we then correct for state preparation errors for the Bell-pair generation circuit (Extended Data Fig. 8). At the circuit initialization (state preparation) stage,
we implement an erasure-cooling scheme^44 and analysed the results conditioned on no erasure detected. We identify that the dominant imperfections in this state preparation stage are (1) atom loss
(with probability ε[l] = 0.0027 for a single atom) and (2) decay from \(| 1\rangle \) to \(| 0\rangle \) (with probability ε[d] = 0.0037 for a single atom). We keep track of how all two-qubit
initial states (\(| 11\rangle ,| 10\rangle ,| 01\rangle ,| 1,\,{\rm{lost}}\rangle ,| {\rm{lost}},1\rangle ,\) â ¦) contribute to the population distribution and the coherence at the measurement
For the population distribution, apart from the ideal initial state \(| 11\rangle \), we keep track of how the erroneous initial states evolve under a perfect circuit execution and contribute to the
final population distribution (Extended Data Fig. 8). We correct the bit-string populations to the first order of ε[d] and ε[l]. Following the probability tree, we can write:
$$\begin{array}{l}{P}_{00}^{{\rm{m}}}\,=\,(1-2{\varepsilon }_{{\rm{l}}}-2{\varepsilon }_{{\rm{d}}}){P}_{00}^{{\rm{c}}}+\frac{1}{4}\times 2{\varepsilon }_{{\rm{d}}}{+\cos }^{2}\frac{{\rm{\pi }}}{8}\
times 2{\varepsilon }_{{\rm{l}}},\\ {P}_{11}^{{\rm{m}}}\,=\,(1-2{\varepsilon }_{{\rm{l}}}-2{\varepsilon }_{{\rm{d}}}){P}_{11}^{{\rm{c}}}+\frac{1}{4}\times 2{\varepsilon }_{{\rm{d}}},\end{array}$$
where the bit-string probabilities \({P}_{{\rm{b}}}^{{\rm{m}}}\) are measurement-corrected and \({P}_{{\rm{b}}}^{{\rm{c}}}\) (the SPAM-corrected population) are issued from perfect initial state
preparation, inherent to the quantum circuit execution errors.
For the coherence measurement, we keep track of how the different erroneous initial states contribute to the observed parity contrast. The error channel with one lost atom (initial state being \(| 1,
\,\text{lost}\rangle \) or \(| \text{lost},\,1\rangle \)) does not affect the contrast due to having a different oscillation frequency. On the other hand, if an atom has decayed to the ground state (
\(| 01\rangle \) or \(| 10\rangle \)), its parity oscillation frequency remains the same but with a Ï phase shift and a contrast of 0.5. This contributes negatively to the observed parity contrast.
Hence, the measured parity oscillation contrast C^m (after measurement correction), in terms of the SPAM-corrected contrast C^c, to the first order of error probabilities, is
$${C}^{{\rm{m}}}=(1-2{\varepsilon }_{{\rm{l}}}-2{\varepsilon }_{{\rm{d}}}){C}^{{\rm{c}}}-2{\varepsilon }_{{\rm{d}}}\times \frac{1}{2}.$$
Erasure conversion for motional qubit initialization
In the main text, there are several results where the analysis of a final image is conditioned on a preceding fast image that verified the state preparation (after erasure-cooling) or motional qubit
initialization (after shelving). First, note that erasure-cooling is needed strictly only for the motional qubit initialization in mid-circuit measurements. Additionally, we find improved
single-qubit (clock) gate fidelities following erasure-cooling, and this improvement becomes significant for shallow tweezers. In contrast, the improvement in CZ gate fidelity following
erasure-cooling is insignificant. In the full error model, we find only a 2â Ã â 10^â 5 increase when cooling the radial degree of freedom to its motional ground state.
For experiments in which MCR is applied (Figs. 4 and 5), we report the results conditioned on not detecting atoms in the ground state after a shelving pulse^44. We provide the results here with no
conditioning for completeness. For measurement-based Bell-state generation (Fig. 5d), without erasure excision, the contrast is 0.39(3), and the population overlap would be 0.64(2), yielding a raw
Bell-state fidelity of 0.52(2). For ancilla-based \(\widehat{X}\) measurement (Fig. 4c), the contrast, conditioned on the ancilla result \(| 0\rangle \) (\(| 1\rangle \)), is 0.60(3) (0.45(3)).
We attribute the limited shelving fidelity mostly to the limited Rabi frequency on the sideband transition, compared with typical frequency variations of the addressing laser or the trap frequency.
Further limitations may arise from the uniformity of the trap waists (and depths) of different tweezers across the array. These limitations can be overcome with a more stable clock laser or by
employing more advanced pulse sequences designed to be insensitive to such inhomogeneities^67.
Effects of clock error on four-qubit GHZ-state preparation
As discussed in the main text, during the preparation of the four-qubit GHZ state, the entangled state is vulnerable to finite atomic temperature and clock laser noise. Entangled states dephase
during the array reconfiguration time, which is considered idle time in the quantum circuit, due to laser frequency noise. To quantitatively study the effect of laser frequency noise, we perform the
experiment with different idle times. We then measure the parity oscillation contrast and the population overlap of the four-qubit GHZ state and compared with our error model predictions, assuming
perfect CZ gates (Extended Data Fig. 9). The experimental pulse sequence is shown in Extended Data Fig. 9a. We increase the total idle time from 280 to 840â μs per arm and observe a decrease in
both the overlap of the GHZ-state population and the parity oscillation contrast (Extended Data Fig. 9c,e). We also observe similar trends in a numerical simulation with our error model, which
assumes perfect CZ gates, a finite temperature of \(\bar{n}=0.24\) and the calibrated clock laser frequency PSD. With the actual reconfiguration time (280â μs), this error model predicts the parity
oscillation contrast to be 0.66 and the state fidelity to be 0.75, consistent with our experimental realization (contrast being 0.68(3) and fidelity being 0.71(2)).
Error model with a 26-mHz clock laser system
The experimental results with the variable idle time show that the generation fidelity of four-qubit GHZ states is limited by the clock laser frequency noise. This motivates us to simulate this state
generation circuit with our clock error model and the frequency PSD of a 26-mHz laser^53. Keeping a finite temperature of \(\bar{n}=0.24\) and assuming perfect CZ gates, we find a contrast of 0.79
and a fidelity of 0.84 (Extended Data Fig. 6b). With this reduced frequency noise, we find the four-qubit GHZ generation fidelity less sensitive to the idle time. For the simultaneous generation of a
cascade of GHZ states, we find that the four-qubit GHZ fidelity is consistent with the shorter idle time sequence. Furthermore, with zero temperature (\(\bar{n}=0\)), the clock error model predicts
near-unity state fidelity (over 0.999). In this low-temperature and 26-mHz clock laser scenario, the state fidelity is limited by the entangling gate fidelity. With the high-fidelity entangling gate
demonstrated in this work, we estimate the generation fidelity of four-qubit GHZ states to be approximately 0.97. This improvement of the atomic temperature could readily be achieved by
erasure-cooling^44 or other methods^68. Note that erasure-cooling is not applied during this particular experiment to speed up the data-taking on a four-atom register.
Projected metrological gain
We analyse the experimental fidelities required to obtain a metrological gain in phase estimation. The metrological gain g is defined as the ratio of posterior variances^2. If we consider the gain of
a protocol with N entangled atoms over the interrogation of N uncorrelated atoms, it can be written as \(g={(\Delta {\phi }_{{\rm{UC}}})}^{2}/{(\Delta {\phi }_{{\rm{C}}})}^{2}\), where \({(\Delta {\
phi }_{{\rm{C}}})}^{2}\) and \({(\Delta {\phi }_{{\rm{UC}}})}^{2}\) are the posterior variances for the entangled case and the uncorrelated case, respectively. For both cases, we assume a
dual-quadrature read-out^5,16,51. We first describe the expected metrological gain with perfect state preparation and then consider the case of imperfect state preparation.
There are two distinct regimes for phase estimation. Local phase estimation corresponds to the limit of a vanishing prior phase width or equivalently to short interrogation times in atomic clocks.
This limit holds only if the prior phase width is smaller than the dynamic range of the quantum state. In this limit, the optimal probe state is an N-atom GHZ state, and the gain is^2 gâ â â N.
For a large phase prior distribution width or equivalently for long interrogation times in atomic clocks, GHZ states do not provide a metrological gain due to their limited dynamic range, so that new
protocols are needed. In the main text, we consider the protocol proposed in refs. ^1,4 and demonstrate a scheme to generate the required input state and read out the phase in both quadratures (Fig.
3). The protocol uses N atoms divided into M groups of GHZ states with Kâ =â 2^j atoms each, where jâ =â 0, â ¦, Mâ â â 1. The number of atoms in the largest GHZ state is, thus, \({K}_{\max }=
{2}^{M-1}.\) The projected metrological gain with ideal state preparation was predicted to be \(g\approx {{\rm{\pi }}}^{2}N/(64\log (N))\) (ref. ^4). The gain can be understood by considering phase
estimation at the Heisenberg limit for the largest GHZ state, which contained N/2 atoms. To exponentially suppress rounding errors in phase estimation, one needs to use n[0] copies of each GHZ state
with \({n}_{0}=(16/{{\rm{\pi }}}^{2})\log ({\rm{N}})\). Note that these expressions hold only in the limit of large N.
We now consider a limited number of atoms N and analyse the effect of finite state preparation fidelity on the projected metrological gain. We assume that the interrogation time is long and perform
numerical Bayesian phase estimation. We use a Gaussian prior distribution to model the prior knowledge of the laser phase. For Bayesian phase estimation, the posterior variance of a given protocol
depends on the variance of the prior distribution. The figure of merit of phase estimation is typically given by^6,37 Râ =â Î Ï [C]/Î´Ï , where Î´Ï is the prior phase distribution width. R
quantifies how much information was obtained in the measurement compared to our initial knowledge of the parameter. It has been shown numerically that for relevant values of N (Nâ â â 100), the
optimal performance, quantified by R, is obtained for a phase prior width around Î´Ï â =â 0.7â rad for a large class of states^6,37. We, therefore, choose to work with this prior width.
Experimentally, the prior width is set by the Ramsey interrogation time and can be tuned to this optimal value.
Using the protocol in ref. ^4 for GHZ states with one, two or four atoms, the minimal number of copies per GHZ size is n[0]â =â 6, resulting in Nâ =â 42 atoms (Extended Data Fig. 7). Using
optimal Bayesian estimators^37, we find numerically a metrological gain of 1.627 (2.114â dB) with perfect state preparation. Imperfect preparation fidelity results in a parity signal with limited
contrast C(K) for a K-atom GHZ state. The probabilities for the outcome of a parity measurement are then modified to \(P(\pm )=(1\pm C(K)\cos (K\phi ))/2\), where {+, â } denote even and odd parity,
To estimate the effect of such limited contrast on the metrological gain, we consider two characteristic scenarios, motivated by our experimental results. First, we look at a case with perfect state
preparation for the one-atom and two-atom GHZ states (C(1)â =â C(2)â =â 1), whereas the four-atom GHZ state has a finite parity signal contrast C(4). In this case, we find numerically that the
minimal contrast to obtain a gain gâ â ¥â 1 is C(4)â =â 0.656. This is higher than the threshold contrast for a narrow prior width given the same state, that is \(C\ge 1/\sqrt{{K}_{\max }}=0.5\)
(refs. ^2,52).
Second, we look at a case where the contrast of each GHZ state scales as \(C(K)={F}_{0}^{K}\), where F[0] is the effective fidelity per qubit. We repeat the calculation and find that the threshold
fidelity to obtain a gain was F[0]â =â 0.969, or equivalently a contrast of C(4)â =â 0.969^4â =â 0.883 for the four-atom GHZ state. Numerically, this threshold seems to be robust to the
introduction of more copies of the one-atom and two-atom GHZ states: if n[0]â =â 12 only for these states (keeping n[0]â =â 6 for the four-atom GHZ states), the threshold is only slightly reduced
to F[0]â â ³â 0.965. Finally, we show the projected metrological gain for various F[0] with respect to the number of atoms in the largest GHZ state used in the protocol of ref. ^4 in Extended Data
Fig. 7.
Our experimental values for C(K) from the simultaneous GHZ-state generation scheme (C(1)â =â 0.82, C(2)â =â 0.68 and C(4)â =â 0.52) fall between the two cases considered here: the best case
scenario of C(K)â =â 1 for \(K < {K}_{\max }\) and the worst case scenario of \(C(K)={F}_{0}^{K}\). Hence, we expect the threshold for metrological gain to lie between the values obtained from
these two cases.
Note that the observed parity oscillation contrasts in our current experimental demonstration are below these thresholds. However, the contrast reduction is entirely dominated by clock laser noise
(Extended Data Fig. 6). With the high-fidelity CZ gates obtained in this work, where F[CZ]â â â 0.996, combined with a reduced clock laser noise (achieved, for example, by a laser with frequency
PSD as in ref. ^53 as discussed above), we numerically project a performance superior to the same number of uncorrelated atoms. Specifically, if we assume F[0]â â â 0.996, the predicted
metrological gain with six copies of GHZ states with one, two or four atoms each is calculated to be 1.519 (1.815â dB). If eight-atom GHZ states are included, the predicted metrological gain with
eight copies of GHZ states with one, two, four or eight atoms each is calculated to be 1.893 (2.772â dB). Note that the above gain analysis assumes zero dead time. Introducing dead time would
degrade the gain, and we defer the analysis of this effect to future work.
Repeated ancilla detection with ancilla reuse
For the MCR illustrated in Fig. 4, fast 18â μs imaging^58 is applied with a fidelity of approximately 0.96 at a tweezer spacing of 3.3â μm. The strong driving on the ^1S[0]â â â ^1P[1]
transition without cooling results in the low survival of detected atoms. Therefore, for experiments where repeated ancilla measurements are needed, we refill the original ancilla position with
another atom through array reconfiguration. Alternatively, we recool and reuse the ancilla atoms instead of refilling them. In this section, we describe a proof-of-concept experiment with different
imaging parameters for the ancilla atoms (Extended Data Fig. 10).
The new imaging scheme is based on the standard high-fidelity, high-survival imaging with cooling light (^1S[0]â â â ^3P[1] intercombination line)^62. We increase the imaging power to collect more
photons over 10â ms and apply the cooling light on one of the axes for another 10â ms, which is shorter than the motional shelving coherence time of approximately 100â ms (ref. ^44). This imaging
scheme allows us to obtain an imaging fidelity of 0.98 with 0.965(2) survival (Extended Data Fig. 10a).
We then check whether we could coherently apply single-qubit rotations after this 10â ms imaging by applying a Ï /2 pulse and a second Ï /2 pulse with a variable phase (Extended Data Fig. 10b). The
measured coherence after imaging 0.94(1) (Extended Data Fig. 10c) is mainly limited by survival, which could readily be improved with further optimizations on cooling during the imaging. Once added
to the complete MCR (Extended Data Fig. 10d), we see a similar coherence for the detected atoms in the ground state (blue) and a slightly lower coherence for the undetected atoms in the clock state
(red), due to the decay (time constant approximately 300â ms) during the 10â ms cooling. These decayed atoms contribute doubly to the loss of coherence of approximately 0.07. With this coherent
driving, we can see when the ancilla atoms are ready to be reused. In the same experiment, we also measure a coherence of approximately 0.73 on the shelved atoms after unshelving them (not shown),
matching the numbers for motional coherence in our previous work^44.
Weight-2 ancilla-based parity read-out
Here, we give the fitted parameters of the plot presented in Fig. 5b. For the direct read-out on the Bell pair, we fit an oscillation of 16.5(1)â kHz with a phase 3.16(7)â rad and a contrast of
0.77(3). The ancilla read-out gives a Ramsey oscillation of 16.5(2)â kHz with a phase of 3.19(9)â rad and a contrast of 0.59(4). In a separate experiment interleaved with this experiment, we
measure the single-atom detuning to be 8.26(5)â kHz.
With perfect gates, the quantum circuit in Fig. 5a yields an oscillating state between \(| {\varPhi }^{+}\rangle \otimes {| 0\rangle }_{{\rm{ancilla}}}\) and \(| {\varPhi }^{-}\rangle \otimes {| 1\
rangle }_{{\rm{ancilla}}}\). For either state, the pair would be measured in \(| 00\rangle \) or \(| 11\rangle \) in this ideal case. Given this information, one can post-select on the experimental
repetitions where the pair is measured in the expected state, either \(| 00\rangle \) or \(| 11\rangle \). This post-selection could identify errors in the execution of the circuit. Looking at the
oscillating state, this post-selection should not bias either of the ancilla measurement outcomes. Performing this post-selection analysis (not shown), we see a Ramsey oscillation of 16.5(3)â kHz
and a contrast of 0.71(8).
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Posted: Tuesday 02nd of Jan 08:18
You seem to be stuck on that I had some time back. I too thought of getting a paid tutor to work it out for me. But they are so costly that I just could not afford them. So I turned to the internet
and found so many software that can help with algebra homework on adding matrices, least common denominator and quadratic formula. After some trials I found that Algebrator is the best of the lot. I
haven’t found a algebra assignment that I can’t get done through Algebrator. It is absolutely amazing. Best part is, the software gives you a step-by-step break-up on how to do it yourself. So you
actually learn how to work it out yourself. Isn’t it cool?
Posted: Thursday 04th of Jan 10:45
Hi Dude, Algebrator assisted me with my learning sessions last week. I got the Algebrator from https://graph-inequality.com/inequalities-1.html. Go ahead, check that and let us know your opinion. I
have even suggested Algebrator to a couple of my friends at school .
Posted: Friday 05th of Jan 16:04
Is it really true that a program can do that? I don’t really know much anything about this Algebrator but I am really seeking for some help so would you mind sharing me where could I find that
software ? Is it downloadable over the internet ? I’m hoping for your fast reply because I really need assistance desperately.
Posted: Saturday 06th of Jan 08:33
Sure. It is quite effortless to access the program as it is just a click away. Go to: https://graph-inequality.com/graphing-systems-of-linear-equat.html. Go through the site and read what the program
offers you. Also note that there is a money back promise if you are not happy . I am sure you will find it as good as I did. Good luck to you.
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CPM Homework Help
Roger is in an Algebra 2 class and has found the radian setting on his calculator. He has not learned about radians but would like to know.
a. Explain to Roger how a radian is defined.
1 radian is the angle measure found by wrapping one radius around a circle.
$\frac{\pi\text{ radians}}{180\degree}$
b. Show him how to change 240° to radians.
Use a Giant One of $240\degree\cdot\frac{\pi\text{ radians}}{180\degree}$.
c. Show him how to change $\frac { 5 \pi } { 6 }$ radians to degrees.
$\frac{5\pi}{6}\cdot \frac{180\degree}{\pi}=?$
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differential equations
I was recently asked about how to spot which direction field corresponds to which differential equation. I hope that by working through a few examples here we will get a reasonable intuition as to
how to do this.
Remember that a direction field is a method for getting the general behaviour of a first order differential equation. Given an equation of the form:
For any function of x and y, the solution to this differential equation must be some function (or indeed family of functions) where the gradient of the function satisfies the above relationship.
The first such equation that we looked at was the equation:
We are trying to find some function, or indeed family of functions y(x) which satisfy this equation. We need to find a function whose derivative (y'(x)) at each point x is equal to the value of the
function (ie. y(x)), plus that value of x.…
Chaos from differential equations
In all of this talk about differential equations, we haven’t spoken all that much about the uses of them, apart from a little about population dynamics, nor indeed about their amazing properties.
Part of the reason for this is that in general (though of course not exclusively), the most interesting differential equations are a single step beyond what we have been looking at. They are
differential equations in more than one variable. For instance, rather than just having a $y$ be a function of $x$ or $t$, they have $y$ a function of both $x$ and $t$. It turns out that this little
change makes all the difference in the world. All of a sudden we can see how things change in both space and time. We can look at real dynamics of systems which are not local to a single place.
This is a topic for another time, and comes under the term partial differential equation.…
UCT MAM1000 lecture notes part 37 – differential equations part vi – second order differential equations
Second Order differential equations
We are only going to look at a particular subset of all possible second order differential equations (that is, equations which contain at most second derivatives) but these particular equations are
absolutely ubiquitous across every field of science. The particular subset we are going to look at are linear, homogenous second order differential equations with constant coefficients. These can be
written in general as:
$\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c y=0$
It is linear because it contains at most (and in this case at least) a single power of $y$ in each term. It is homogenous because there is no term which has no powers of $y$ (ie. the right hand side
is not a constant), and the coefficients $b$ and $c$ are any real numbers (though you can extend this to having complex numbers very easily). We will see that depending on the relationship between
these numbers ($b$ and $c$) we can have very different behaviour of the equation.…
UCT MAM1000 lecture notes part 36 – differential equations part v – first order differential equations
First order linear differential equations
We are now going to deal with another subset of first order differential equations which in some ways are easier than the previous and in other ways more complicated. These are linear first order
differential equations. The general form of a first order linear differential equation is:
where $P(x)$ and $Q(x)$ are any functions of $x$.
Very importantly, I’m leaving off the fact that $y$ is dependent on $x$ in the notation, but you should remember that this is really $y(x)$ and that is the function you are trying to solve for.
Sometimes you will be given an equation which is not obviously in this form but it can be transformed to this form. For instance:
$\frac{1}{y}\frac{dy}{dx}=x^2+\frac{\sin x}{y}$
This can easily be transformed into the canonical form for a linear first order DE. We are going to try and rewrite the left hand side of the equation in a form which will mean that we can solve the
differential equation very easily.…
UCT MAM1000 lecture notes part 35- differential equations part iv – separable differential equations
Separable differential equations
In some ways these are the easiest differential equations to solve in theory, though in practice the final step (that of integrating) may be difficult or impossible. A separable differential equation
is one of the form:
where $f(x)$ and $g(y)$ are any functions of $x$ and $y$ respectively. For instance:
$\frac{dy}{dx}=x y$
is of this form where $f(x)=x$ and $f(y)=\frac{1}{y}$. The reason that these equations are simple in theory is because we can rearrange them to be:
ie. we have all the $x$ stuff on one side and all the $y$ stuff on the other and then we can integrate both sides:
$\int g(y)dy=\int f(x)dx$
and that’s it. As long as you can do the integrals, you can get a function $y$ in terms of $x$. let’s look at some examples:
$\frac{dy}{dx}=x y$
gives the following integral:
$\int\frac{1}{y}dy=\int x dx$
and so:
$\ln |y|+c_1=\frac{x^2}{2}+c_2$
here we have one constant of integration from each side of the interval, but because they are just constants, we can put them into one constant and call it $c$:
$\ln |y|=\frac{x^2}{2}+c$
we can then rearrange this to give:
We can then call $e^c$ just a constant, and let’s call it $y_0$ because we can see that when $x$ is zero, $|y|$ is just going to be given by this constant:
This has two solutions (one where $y$ is positive, and one where it is negative), so we can choose one of them, depending on the initial condition for $y$.…
UCT MAM1000 lecture notes part 34 – differential equations part iii – Direction flows and Euler’s method
We haven’t yet studied any general ways to solve differential equations. In the first case of exponential growth we found an easy way to solve the equation, but for the logistic equation we just gave
the solution and showed that it indeed satisfied the equation. Here we are going to look at some methods for finding not the exact solution, but approximations of the solutions. The first method is
the method of Direction Fields and it will give us a good idea of what the solutions are going to look like. The second method, Euler’s method will give us an approximation to a single solution and
we will be able to improve it to get arbitrarily good solutions to any differential equation (so long as there aren’t particularly nasty pathologies in the differential equation).
Direction Fields
Let’s take a differential equation:
Note that sometimes we will say explicitly that $y(x)$ and sometimes we will leave it implicit, because the equation has a derivative of $y$ with respect to $x$.…
UCT MAM1000 lecture notes part 32 – differential equations and rabbits moving at the speed of light
Up to now if I gave you an equation, and asked you to solve it for $x$ you would be, in general, looking for a value of $x$ which solved the equation. Given:
You can solve this equation to find two values of $x$.
I could also give you an equation which linked $x$ and $y$ explicitly, and you could find a relationship between the two which, given a value of $x$ would give you a value of $y$. You’ve been doing
this now for many years. Now we’re going to add a hugely powerful tool to our mathematical arsenal. We’re going to allow our equations to include information about gradients of the function…let’s see
what this means…
We’re going to take everything that you learnt about integration and turn it into a way to model and understand the world around us. This is a very powerful statement and indeed differential
equations are without a doubt the most powerful mathematical tool we have to understand the behaviour of everything from fundamental particles to populations, economies, weather, flow of wealth,
heat, fluids, the motion of planets, the life of stars, the flight of an aircraft, the trajectory of a meteor, the way a pendulum swings, the way a ponytail swings (see paper on this here), the way
fish move, the way algae grow, the way a neuron fires, the way a fire spreads…and so much more.…
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Q&A: What are some of the most logical and rational Hindu text?
What are some of the most logical and rational Hindu text?
In any logical/mathematical sysem, we have certain Axioms, whose validity are taken for granted within that system.
For example, a mathematical system that tries to introduce the concept of real/complex numbers, the three Axioms - field, order and completeness Axioms - are taken for granted. That is they are not
required to be proved (within that particular system). All the remaining propositions/theorems of that system are then proved using those Axioms.
The only Hindu scriptures which share this pattern to an extent are the Darshana Shastras (the 6 Philosophical scriptures - Samkhya, Nyaya etc).
In the Darshanas, we have the proofs/standards (known as Pramana) which serve the same purpose as does an Axiom in a mathematical system.
For example, for the Samkhya Philosophy, we have:
Drishtamanumanamaptavachanancha sarvapramanasiddhatvat |
Trividham pramanamishtam prameyasiddhih pramanAddhi ||
Pratyaksha (direct perception), anumAna (inferrence) and Apta Vakya (i.e words of the Rishis or scriptures) - these three standards are accepted in Samkhya. All other standards are accomplished/
established by these three only. By using these three pramanas the propositions are established.
Samkhya Karika 4
So, just like all the theorems/propositions in a logical system, are established using the validity of the Axioms which are accepted in that system, in this Samkhya doctrine, all the propositions are
similarly established using the three Pramanas which are accepted as valid in the doctrine.
Note: “The question: What are some of the most logical and rational Hindu text?” is licensed by Stack Exchange Inc (
user contributions licensed under CC BY-SA.
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American Mathematical Society
Invariants, Boolean algebras and ACA$_{0}^{+}$
HTML articles powered by AMS MathViewer
Trans. Amer. Math. Soc. 358 (2006), 989-1014
The sentences asserting the existence of invariants for mathematical structures are usually third order ones. We develop a general approach to analyzing the strength of such statements in second
order arithmetic in the spirit of reverse mathematics. We discuss a number of simple examples that are equivalent to ACA$_{0}$. Our major results are that the existence of elementary equivalence
invariants for Boolean algebras and isomorphism invariants for dense Boolean algebras are both of the same strength as ACA$_{0}^{+}$. This system corresponds to the assertion that $X^{(\omega )}$
(the arithmetic jump of $X$) exists for every set $X$. These are essentially the first theorems known to be of this proof theoretic strength. The proof begins with an analogous result about these
invariants on recursive (dense) Boolean algebras coding $0^{(\omega )}$. References
• Andreas R. Blass, Jeffry L. Hirst, and Stephen G. Simpson, Logical analysis of some theorems of combinatorics and topological dynamics, Logic and combinatorics (Arcata, Calif., 1985) Contemp.
Math., vol. 65, Amer. Math. Soc., Providence, RI, 1987, pp. 125–156. MR 891245, DOI 10.1090/conm/065/891245
• Csima, B., Montalban, A. and Shore, R.A., Boolean algebras, Tarski invariants and index sets, Notre Dame Journal of Formal Logic, to appear.
• C. C. Chang and H. J. Keisler, Model theory, 2nd ed., Studies in Logic and the Foundations of Mathematics, vol. 73, North-Holland Publishing Co., Amsterdam-New York-Oxford, 1977. MR 0532927
• Ju. L. Eršov, Decidability of the elementary theory of relatively complemented lattices and of the theory of filters, Algebra i Logika Sem. 3 (1964), no. 3, 17–38 (Russian). MR 0180490
• Yu. L. Ershov, S. S. Goncharov, A. Nerode, J. B. Remmel, and V. W. Marek (eds.), Handbook of recursive mathematics. Vol. 1, Studies in Logic and the Foundations of Mathematics, vol. 138,
North-Holland, Amsterdam, 1998. Recursive model theory. MR 1673617
• Harvey M. Friedman, Stephen G. Simpson, and Rick L. Smith, Countable algebra and set existence axioms, Ann. Pure Appl. Logic 25 (1983), no. 2, 141–181. MR 725732, DOI 10.1016/0168-0072(83)90012-X
• Sergeĭ S. Goncharov, Schetnye bulevy algebry i razreshimost′, Sibirskaya Shkola Algebry i Logiki. [Siberian School of Algebra and Logic], Nauchnaya Kniga (NII MIOONGU), Novosibirsk, 1996
(Russian, with Russian summary). MR 1469495
• John Doner and Wilfrid Hodges, Alfred Tarski and decidable theories, J. Symbolic Logic 53 (1988), no. 1, 20–35. MR 929372, DOI 10.2307/2274425
• R. Björn Jensen, The fine structure of the constructible hierarchy, Ann. Math. Logic 4 (1972), 229–308; erratum, ibid. 4 (1972), 443. With a section by Jack Silver. MR 309729, DOI 10.1016/
• Sabine Koppelberg, Handbook of Boolean algebras. Vol. 1, North-Holland Publishing Co., Amsterdam, 1989. Edited by J. Donald Monk and Robert Bonnet. MR 991565
• A. S. Morozov, Strong constructivizability of countable saturated Boolean algebras, Algebra i Logika 21 (1982), no. 2, 193–203 (Russian). MR 700992
• Stephen G. Simpson, Subsystems of second order arithmetic, Perspectives in Mathematical Logic, Springer-Verlag, Berlin, 1999. MR 1723993, DOI 10.1007/978-3-642-59971-2
• Robert I. Soare, Recursively enumerable sets and degrees, Perspectives in Mathematical Logic, Springer-Verlag, Berlin, 1987. A study of computable functions and computably generated sets. MR
882921, DOI 10.1007/978-3-662-02460-7
• Tarski, A., Arithmetical classes and types of Boolean algebras, Bulletin of the American Mathematical Society 55, 63.
• Vaught, R. L., Topics in the Theory of Arithmetical Classes and Boolean Algebras, Ph.D. Thesis, University of California, Berkeley.
• White, W. M., Characterizations for Computable Structures, Ph.D. Thesis, Cornell University.
Similar Articles
• Retrieve articles in Transactions of the American Mathematical Society with MSC (2000): 03B25, 03B30, 03C57, 03D28, 03D35, 03D45, 03F35, 06E05
• Retrieve articles in all journals with MSC (2000): 03B25, 03B30, 03C57, 03D28, 03D35, 03D45, 03F35, 06E05
Additional Information
• Richard A. Shore
• Affiliation: Department of Mathematics, Cornell University, Ithaca, New York 14853
• MR Author ID: 161135
• Email: shore@math.cornell.edu
• Received by editor(s): March 22, 2004
• Published electronically: April 13, 2005
• Additional Notes: The author was partially supported by NSF Grant DMS-0100035.
• © Copyright 2005 Richard A. Shore
• Journal: Trans. Amer. Math. Soc. 358 (2006), 989-1014
• MSC (2000): Primary 03B25, 03B30, 03C57, 03D28, 03D35, 03D45, 03F35, 06E05
• DOI: https://doi.org/10.1090/S0002-9947-05-03802-X
• MathSciNet review: 2187642
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A Sharp Threshold for Random Graphs with a Monochromatic Triangle in Every Edge Coloringsearch
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A Sharp Threshold for Random Graphs with a Monochromatic Triangle in Every Edge Coloring
eBook ISBN: 978-1-4704-0446-8
Product Code: MEMO/179/845.E
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A Sharp Threshold for Random Graphs with a Monochromatic Triangle in Every Edge Coloring
eBook ISBN: 978-1-4704-0446-8
Product Code: MEMO/179/845.E
List Price: $59.00
MAA Member Price: $53.10
AMS Member Price: $35.40
• Memoirs of the American Mathematical Society
Volume: 179; 2006; 66 pp
MSC: Primary 05
Let \(\mathcal{R}\) be the set of all finite graphs \(G\) with the Ramsey property that every coloring of the edges of \(G\) by two colors yields a monochromatic triangle. In this paper we
establish a sharp threshold for random graphs with this property. Let \(G(n,p)\) be the random graph on \(n\) vertices with edge probability \(p\). We prove that there exists a function \(\
widehat c=\widehat c(n)=\Theta(1)\) such that for any \(\varepsilon > 0\), as \(n\) tends to infinity, \(Pr\left[G(n,(1-\varepsilon)\widehat c/\sqrt{n}) \in \mathcal{R} \right] \rightarrow 0\)
and \(Pr \left[ G(n,(1+\varepsilon)\widehat c/\sqrt{n}) \in \mathcal{R}\ \right] \rightarrow 1\). A crucial tool that is used in the proof and is of independent interest is a generalization of
Szemerédi's Regularity Lemma to a certain hypergraph setting.
□ Chapters
□ 1. Introduction
□ 2. Outline of the proof
□ 3. Tepees and constellations
□ 4. Regularity
□ 5. The core section (proof of Lemma 2.4)
□ 6. Random graphs
□ 7. Summary, further remarks, glossary
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Requests
Volume: 179; 2006; 66 pp
MSC: Primary 05
Let \(\mathcal{R}\) be the set of all finite graphs \(G\) with the Ramsey property that every coloring of the edges of \(G\) by two colors yields a monochromatic triangle. In this paper we establish
a sharp threshold for random graphs with this property. Let \(G(n,p)\) be the random graph on \(n\) vertices with edge probability \(p\). We prove that there exists a function \(\widehat c=\widehat c
(n)=\Theta(1)\) such that for any \(\varepsilon > 0\), as \(n\) tends to infinity, \(Pr\left[G(n,(1-\varepsilon)\widehat c/\sqrt{n}) \in \mathcal{R} \right] \rightarrow 0\) and \(Pr \left[ G(n,(1+\
varepsilon)\widehat c/\sqrt{n}) \in \mathcal{R}\ \right] \rightarrow 1\). A crucial tool that is used in the proof and is of independent interest is a generalization of Szemerédi's Regularity Lemma
to a certain hypergraph setting.
• Chapters
• 1. Introduction
• 2. Outline of the proof
• 3. Tepees and constellations
• 4. Regularity
• 5. The core section (proof of Lemma 2.4)
• 6. Random graphs
• 7. Summary, further remarks, glossary
Permission – for use of book, eBook, or Journal content
Please select which format for which you are requesting permissions.
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A probabilistic model for flood forecasting based on hydrological data in the state of Maranhão, Brazil
1. Introduction
Floods can have a devastating impact on any region where they occur. Nonetheless, the frequency and severity of these events have increased significantly due to climate change in recent years (
Waseem, Rana 2023).
In Brazil, it is estimated that there are approximately 3,000 km^2 of areas prone to extreme weather events. At least 825 municipalities are considered highly vulnerable to disasters such as
landslides and flash floods (Alvalá et al. 2019; Dias et al. 2020). Floods are the most common type of natural disaster worldwide and pose substantial risks to populations (Mishra et al. 2022).
Floods caused by heavy rains have had a drastic impact on the municipalities in the state of Maranhão, located in the northeastern region of Brazil, as indicated by official decrees issued by the
Municipal Governments of Açailândia, Arame, Buriticupu, and Santa Luzia (Diário Oficial 2023a, 2023b, 2023c, 2023d).
Some recent research employs boxplots to identify anomalies in precipitation time series (Gogien et al. 2023; Moreira et al. 2023); stage-discharge curves to describe the behavior of rivers (
Vishwakarma et al. 2023); flow duration curves (FDC) for optimized water resources management (Ridolfi et al. 2020); and the Mann-Kendall test for detecting trends in rainfall time series (Penereiro
et al. 2018; Zhang et al. 2022).
The investigation of intense rainfall phenomena associated with flooding can be conducted through predictive models and proper hydrological data collection (Lima et al. 2019, Lima, Scofield 2021;
Alves et al. 2022). In this study, all these tools were combined to develop a probabilistic model that estimates the likelihood of new flood events occurring in four municipalities in Maranhão:
Açailândia, Arame, Buriticupu, and Santa Luzia.
2. Material and methods
2.1. Description of the study area
The study area encompasses the municipalities of Açailândia (5806 km^2), Arame (2976 km^2), Buriticupu (2545 km^2), and Santa Luzia (5462 km^2), in the state of Maranhão, in northeastern Brazil.
These municipalities are in a state of emergency because of extreme precipitation events, as reported by the Ministry of Social Development (Brasil 2023). In the research reported here, hydrological
time series data were investigated to develop a probabilistic model for predicting the occurrence of new flood events using data from the National Water and Sanitation Agency (Agência Nacional de
Águas e Saneamento Básico – ANA). Figure 1 displays the municipalities and the selected hydrological stations.
The selection of stations was based on data availability in the Hydrological Information System (HIDROWEB). If the downloaded file contained data, the sample's consistency was verified through
precipitation graphs (mm/day) over time. The goal was to identify the time interval during which the data were available and suitable for analysis; any station that did not provide valid data was
discarded. Table 1 presents the stations that met this criterion and were selected for further investigation.
Table 1.
Municipality Time series Station code Latitude Longitude River Drainage area (km^2)
Açailândia (1996-2023) 447004^1 –4.9308 –47.4967 - -
(1979-2023) 330250^2 –4.6969 –46.9347 Pindaré 5480
Arame (1983-2023) 445008^1 –4.8611 –46.0078 - -
(2003-2023) 333330^2 –5.1447 –45.7953 Grajaú 11400
Buriticupu (2004-2023) 644019^1 –4.2175 –46.4906 - -
(2004-2023) 330700^2 –4.2008 –46.4872 Pindaré 10400
Santa Luzia (1982-2023) 445001^1 –4.0303 –45.771 - -
(1979-2023) 330750^2 –4.2989 –46.4944 Buriticupu 4750
2.2. Boxplot
The box plot is an effective graphical tool for detecting atypical precipitation events. It displays the data distribution and allows for the visual identification of outlier values that represent
exceptionally high or low precipitation events. The construction of boxplots in this study involved the calculation of several measures.
The median (Quartile 2) is the value that divides the data set in half as follows:
where n – total number of observations.
In addition to Q[2], Q[1] (Quartil 1) and Q[3] (Quartil 3) are also required:
The interquartile range (IQR) is used to determine the whiskers in the boxplots, calculated as follows:
lQR = Q[3] − Q[1] (4)
Rainfall values beyond the boundaries represent the outliers. After plotting the graph, the coincidence of these points with flooding events in the municipalities of Maranhão was verified.
2.3. Stage-discharge rating curve
Stage-discharge curves relate the discharge of a river to the water level. These curves are essential in hydrology to understand the behavior of rivers and calculate discharge (m^3 s^-1) at different
times. In this study, rating curves were used to assist in predicting a probabilistic model for the recurrence of discharges with magnitude equal to or greater than those of the initial days of
flooding in the municipalities (March 18-20, 2023).
The relationship between discharge (Q) and water level (h) in a stage-discharge curve can be represented by an exponential equation in the following form:
where: Q – discharge (m^3 s^-1); h – stage (m), a and b are rating curve constants, h[0] – the stage corresponding to zero discharge (m) (Ramírez et al. 2018).
2.4. Flow Duration Curve (FDC)
The FDC provides a comprehensive graphical representation of the relationship between discharge and frequency (Ma et al. 2023). It was used to indicate the percentage of time when discharges equal to
or greater than the reference discharge were observed. Initially, monthly historical discharge data from the stations were collected, and the probability of exceedance was calculated by associating
these data with percentiles ranging from 0.05 to 0.95. This range was adopted to emphasize the absence of zero discharge (0%) or absolute discharge (100%) records.
The percentage of time when a specific discharge is equaled or exceeded can be calculated as:
where: f(x) – cumulative frequency for discharge x; n – total number of observations; i – position of discharge x on the y-axis.
2.5. Mann-Kendall test
2.5.1. Missing Data Index (MDI)
Data consistency is a critical aspect of any hydrological analysis (Becker et al. 2023; Peixoto et al. 2023; Pereira 2023; Tsuha 2023). Therefore, before proceeding with this step, the sample quality
in the time series of precipitation was rechecked, adopting a threshold of ≤10% of failures (Holender, Santos 2023).
The data provided in HIDROWEB are categorized for quality in each measurement, assigning a number to their category: (0) – blank data (unmeasured), (1) – actual data (measured and verified), (2) –
estimated data, (3) – doubtful data (instrumental failures), and (4) – accumulated data (ANA 2002).
The missing data index was calculated as follows:
where: n[0] – number of blank data points; n[3] – number of questionable data points; n[t] – sample space.
2.5.2. Sequential test
After qualitative investigation of the data, the non-parametric Mann (1945) and Kendall (1975) tests, as sequenced by Sneyers (1990) and Onoz, Bayazit (2003), were applied in this study to test the
significance of a trend present in the pluviometric series.
A time series of a variable y[i] consisting of n data points, where 1 ≤ i ≤ n, was considered. The procedure involved calculating the sum t[n] = ∑i=1nmi, where m[i] is the number of terms preceding y
[i] and the preceding values y[j] are less than y[i] (y[j] < y[i]). This procedure was applied to time series with a large number of data points under the null hypothesis H[0] (absence of significant
Based on this premise, it was found that t[n] follows a normal distribution, with the mean and variance parameters defined by equations 8 and 9, respectively:
Evaluating the statistical significance of t[n] with respect to H[0] through a two-tailed test, significance is rejected for high values of U(t[n]), a standardized test statistic, defined as:
Subsequently, using a standardized normal distribution, the calculation of the probability value (a[1]) is done as follows:
Acceptance of H[0] occurs when a[1] > a[0], with a[0] equal to the significance level of the test. If H[0] is rejected, it implies the presence of a significant trend in the series: U(t[n]) < 0
indicates a decreasing trend, while U(t[n]) > 0 indicates an increasing trend.
In the sequential version, U(t[n]) is obtained in the forward direction of the series, starting from i = 1 to i = n. This results in the statistic –1.65 < U(t[n]) < 1.96, where the values of the
two-sided intervals –1.65 to 1.65 and –1.96 to 1.96 are associated with significance levels a[0] = 0.10 (10%) and a[0] = 0.05 (5%), respectively (Mortatti et al. 2004).
The inflection point in the series can be identified following the same approach as with the inverse series U^*(t[n]). The point where U(t[n]) and U^*(t[n]) intersect provides an approximate estimate
of the location of the transition point in the trend. However, this conclusion holds statistical significance only if this change takes place within the two-sided significance interval (Back 2001).
The simplified execution of the steps to obtain the estimated probabilistic model in the present study is shown in Figure 2.
3. Results and discussion
Inaccurate or inconsistent information can lead to erroneous conclusions and negatively impact decisions made based on this data. To ensure the reliability of the data used in this research,
consistency checks were performed at the beginning of the analysis and before the Mann-Kendall test.
The first step involved verifying the availability of valid data from the pluviometric stations from 1983 to 2023. The results obtained show that Arame (Fig. 3b) and Santa Luzia (Fig. 3d) had data
available for the entire series, Açailândia (Fig. 3a) from 1996 onward; in Buriticupu, the station provides data from 2004 forward (Fig. 3c).
Figure 4 shows the box plots for all stations, plotted for each year from 2004 onward. Recent atypical precipitation events were observed, indicated by the outliers (red points).
In the municipality of Açailândia (Fig. 4a), extreme precipitation events were recorded in March 2008 and 2023 (515.80 mm, 577.40 mm), coinciding with the onset of flooding events (Mar. 18-20 2023).
In Arame (Fig. 4b), the outliers correspond to precipitation in February 2007 (264.30 mm); April 2010 (332.30 mm); and March 2019 (376.70 mm). In Buriticupu (Fig. 4c), rainfall anomalies were
identified at various times over the years: in February 2007 (411.40 mm); March 2008, 2012, 2022, and 2023 (537.70 mm, 146.70 mm, 475.70 mm, 301.70 mm, respectively); and April 2010 and 2023 (245.30
mm, 358.80 mm). These records confirm the coin-cidence of atypical rainfall with flooding events in the municipality. In Santa Luzia (Fig. 4d), the records indicate critical precipitation in March
2006, 2012, and 2022 (430.70 mm, 363.40 mm, 703.30 mm); and April 2006 (450.20 mm). The results reaffirm that March is typically the rainiest month in the state of Maranhão (Cerqueira, Cerqueira 2023
3.1. Probabilistic model
Figure 5 shows the stage-discharge curves obtained for each municipality, using streamflow data from the respective station. The exponential equations estimating the relationship between discharge (Q
[m^3 s^-1]) and water level (h [m]) are provided in the caption.
These equations (Fig. 5) were used to estimate the discharge (Q [m^3 s^-1]) during the first days of flooding in the municipalities (Table 2), at fixed station measurement times (7:00 am and 5:00
Table 2.
Date Time Municipality h[0] (m) h (m) Q (m^3 s^-1)
03/18/2023 7:00 a.m. 3.93 72.22
5:00 pm. 3.97 75.21
03/19/2023 7:00 am. Açailândia –0.49 4.10 85.59
5:00 pm. 4.28 101.76
03/20/2023 7:00 am. 4.20 94.30
5:00 pm. 4.15 89.86
03/18/2023 7:00 am. 3.96 59.20
5:00 pm. 4.22 65.28
03/19/2023 7:00 am. Arame 0.53 4.49 71.76
5:00 pm. 4.49 71.76
03/20/2023 7:00 am. 4.46 71.04
5:00 pm. 4.48 71.52
03/18/2023 7:00 a.m. Buriticupu –1.03 4.72 89.58
5:00 pm. 4.70 88.21
03/19/2023 7:00 am. 4.70 88.21
5:00 pm. 4.69 87.53
03/20/2023 7:00 am. 4.74 90.96
5:00 pm. 4.76 92.36
03/18/2023 7:00 a.m. 4.19 19.27
5:00 pm. 4.19 19.27
03/19/2023 7:00 am. Santa Luzia 2.95 4.20 20.51
5:00 pm. 4.20 20.51
03/20/2023 7:00 am. 4.18 20.67
5:00 pm. 4.18 20.67
Using the estimated discharges on flood days as a parameter, the probability of recurring events of equal or greater magnitude was examined through the monthly FDC of the rivers passing through the
municipalities in Maranhão, before (solid lines) and after (dashed lines) the onset of floods, as illustrated in Figure 6.
The probability of exceedance in the municipalities can be derived from the data in Table 2 and the FDCs. In the municipality of Açailândia (Fig. 6a), the average probability of exceedance is 10%;
about 32% for Buriticupu (Fig. 6b); 15% for Arame (Fig. 6c); and less than 5% for Santa Luzia (Fig. 6d).
3.1.1. Statistical significance
Having established the connection between extreme precipitation events and flooding, data quality was confirmed through the Missing Data Index (MDI) (Table 3), which demonstrated that gaps in the
series were within the adopted limit (≤10%). Subsequently, a non-parametric test was applied to the monthly precipitation series.
Table 3.
Time series Data Missing Estimated Doubtful Accumulated Missing Data Index
Açailândia (1996-2022) 14353 307 0.00 29 0 2.34%
Arame (1983-2022) 22,134 470 0.00 93 2 2.29%
Buriticupu (2004-2022) 6602 78 0.00 0 2 1.18%
Santa Luzia (1982-2022) 26,288 903 0.00 1 8 3.42%
The results of the Mann-Kendall test are presented in Table 4, focusing on two different significance levels: α[0] = 0.10 (10%) and α[0] = 0.05 (5%). α[1] values equal to or lower than the
significance levels confirm a significant trend, depending on the sign of U(t[n]) for increasing (+) or decreasing (-) trend.
Table 4.
Month Açailândia (t[n]) Arame
Var (t[n]) α[1] Var (t[n]) U (t[n]) α [1]
Jan 2301.00 –1.42 0.14 6833.67 –2.32 <0.05
Feb 2301.00 0.08 0.93 6833.67 –1.16 0.24
Mar 2058.33 –0.62 0.51 6833.67 –1.61 0.10
Apr 2301.00 –0.69 0.47 7366.67 –1.20 0.22
May 2301.00 –1.00 0.30 6833.67 –1.88 <0.10
Jun 1257.67 1.52 0.13 1833.33 –1.99 <0.05
Jul 125.00 2.77 <0.05 1257.67 0.23 0.82
Aug 165.00 0.93 0.35 333.67 0.44 0.66
Sep 950.00 0.19 0.85 817.00 –0.21 0.78
Oct 1833.33 0.82 0.41 4165.33 –1.86 <0.10
Nov 2058.33 0.40 0.69 4958.33 –0.16 0.85
Dec 2058.33 -0.44 0.63 7366.67 –0.80 0.41
Month Buriticupu Santa Luzia
Var (t[n]) U (t[n]) α [1] Var (t[n]) U (t[n]) α 1
Jan 816.00 3.47 <0.10 12658.67 –0.15 0.88
Feb 815.00 0.84 0.40 13458.67 0.34 0.74
Mar 816.00 2.91 <0.10 13458.67 1.13 0.26
Apr 813.33 0.74 0.46 13458.67 1.16 0.24
May 812.33 0.49 0.62 6833.67 –1.90 <0.10
Jun 788.67 –1.82 <0.10 12658.67 0.55 0.58
Jul 788.67 2.31 <0.05 10450.00 0.16 0.88
Aug 800.33 –1.84 <0.10 7366.67 –0.86 0.39
Sep 812.33 –3.65 <0.05 6833.67 –1.19 0.24
Oct 816.00 1.79 <0.10 10450.00 –1.39 0.16
Nov 817.00 1.40 0.16 11891.00 0.55 0.58
Dec 816.00 1.72 <0.10 12658.67 –0.22 0.82
The values highlighted in bold in Table 4 indicate that α[1]< α[0], demonstrating that the null hypothesis of no trend (h[0]) can be rejected for these months. The municipalities that showed trends
in March were Arame (negative) and Buriticupu (positive).
When U(t[n]) exceeds the confidence interval, the trend can be considered significant, and the points of intersection between U(t[n]) and U^*(t[n]) represent the onset of this trend in the time
series, if it occurs within the confidence intervals (Fig. 7).
In the municipality of Açailândia (Fig. 7a), there is a possibility of an increasing trend, becoming significant in July. In contrast, Arame (Fig. 7b) indicates decreasing trends that are significant
in March, June, and October. Meanwhile, in Buriticupu (Fig. 7c), there are potential positive trends, with a notable emphasis on March and July. In Figure 7d, Santa Luzia exhibited a significant
propensity for growth, starting in February and May.
4. Conclusions
The results obtained for the municipalities of Açailândia, Arame, Buriticupu, and Santa Luzia highlighted the direct influence of heavy rainfall on the occurrence of flooding events, particularly
during February, March, and April. In Açailândia and Buriticupu, extreme precipitation events were recorded in March 2023, with rainfall volumes of 577.40 mm and 301.70 mm, respectively, coinciding
with the floods that occurred in March 2023.
The stage-discharge curves obtained provided equations that describe the discharge behavior as a function of water level in each municipality, providing parameters for flood prevention. This
relationship was evident through the Flow Duration Curves (FDC), which indicated the probability of events of equal or greater magnitude at different times before and after the onset of floods.
Furthermore, the Mann-Kendall test revealed significant trends in some monthly precipitation series, emphasizing the presence of ascending or descending patterns in rainfall during specific months.
These results underscore the critical importance of monitoring precipitation patterns to ensure water resource management and mitigate the impacts of floods in areas susceptible to extreme events.
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Denotational World-indexed Logical Relations and Friends
As part of the long-standing drive for mathematical machinery to reason about
computer programs, we use the technique of denotational logical relations to prove
contextual equivalence of stateful programs. We propose the notion of approximate
locations to solve the non-trivial problem of existence and solve the fundamental
type-worlds circularity by metric-space theory. This approach scales to state-of-theart
step-indexed techniques and permits unrestricted relational reasoning by the use
of so-called Bohr relations.
Along the way, we develop auxiliary theory; most notably a generalized version
of a classical xed-point theorem for functors on certain metric spaces by America
and Rutten. Also we investigate the use of recursively dened metric worlds in an
operational setting and arrive at constructions akin to step-indexed models.
On a dierent, though related, note, we explore a relational reading of separation
logic with assertion variables. In particular, we give criteria for when standard,
unary separation logic proofs lift to the binary setting. Phrased dierently, given
a module-dependent client and a standard separation logic proof of its correctness,
we ponder the default question of representation-independence: is the client able to
{ or unable to { observe implementation specic details about its module.
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Statistical Decisions ModelStatistical Decisions Model - Essay Typing
Inferential statistics refers to a branch of statistics that is used to assist researchers in testing hypotheses and making inferences from sample data to a larger sample or population. It consists
of procedures used to make inferences about population characteristics from information contained in a sample drawn from the same population. Statistical tests such as t test, analysis of variance or
f test, or chi square test are used to see whether two or more groups of participants tend to differ on some variable of interest. However, due to the complexity and number of inferential statistical
tests available, it becomes a difficult decision for a researcher to decide on which statistical tool to employ for a particular study. Therefore, there arises a need for a matrix or model that
facilitates the decision making by a researcher. This matrix can take the form of a decision tree, outline or model that encompasses a series of questions that guide a researcher to decide on which
statistical test in most appropriate for a particular study. This paper presents such as model.
The importance of inferential statistics in research cannot be over emphasized. It facilitates the acquisition of information that would otherwise not have been obtainable about a particular
population. It is less costly, practical, and saves on time and labor. However, and most importantly, inferential statistics provides reliable information which is accurate and of high quality and
whose margin of error can be specified. This paper presents a tabulated statistical decision model that can guide a researcher on deciding on which statistical test to use for a particular research
Steps Involved In the Model and the Easiest and Most Difficult Parts of the Process
When trying to make a decision with regards to which statistical test to use, one needs to ask oneself questions such as:
1. Am I interested in….?
2. Description (Association)- Factor Analysis, Correlation, Path Analysis
3. Intervention (Group Differences)- T-Test, ANOVA, MANOVA, Chi-Square
4. Explanation (Prediction)- Regression, Discriminant Analysis, Logistic Regression
5. Is my Dependent Variable Nominal, Ordinal, Interval, or Ratio…?
6. Nominal- Chi-Square
7. Dichotomous-Logistic Regression
8. Ordinal-Chi-Square
9. Interval/Ratio – Correlation, Multiple Regression, T-Test, ANOVA, MANOVA,
10. Other questions a researcher needs to ask is: Do differences exists between groups?; Do the differences exists between 2 groups or more on one DV, or on multiple DVs?; How strongly and in what
direction are the IV and DV related?; and finally, What is the likelihood of the dependent variable occurring as the values of the independent variables change?.
These guiding questions, together with considerations such: what form the research question took such as to investigate group differences, degree of relation ship and predictions of group membership;
number and type of the dependent variables and independent variables; covariate; and finally, the goal of the research question
Statistical Decisions Model
Research Question? Number And Type Of DV? Number and Type Of IV? Covariates Test Goal Of Analysis
Nominal Or Higher 1 Nominal Or Higher Chi Square To Determine Whether there are Differences between Groups
1 Dichotomous T-Test
1 Categorical One-Way ANOVA
Continuous 1+ One-Way ANCOVA To Determine Significance of Mean Group Differences
Group Differences 2+ Categorical Factorial ANOVA
1+ Factorial ANCOVA
1 Categorical One-Way MANOVA
2+ Continuous 1+ One-Way MANCOVA To Create Linear Combo of DVs to Maximize Mean Group Differences
2+ Categorical Factorial MANOVA
1+ Factorial MANCOVA
Continuous 1 Continuous Correlation Determine Relationship/ Prediction
Degree Of Relationship 2+ Continuous Regression Linear Combination to Predict The DV
1+ Continuous 2+ Continuous Path Analysis Estimate Causal Relations among Variables
Prediction Of Group Membership Dichotomous 2+ Nominal Or Higher Regression Create Linear Combo of IVs of the Log Odds of Being in one Group
The most difficult and challengingpart in the construction of the statistical decision tree was in finding the broad and umbrella differentiating characteristics of the statistical tools so as to
broadly group the tests, as well as finding the unique distinguishing attributes of each tool which makes them distinct from the other tools. This required keen scrutiny and examination of each
statistical tool and necessitated the spending of a lot of time noting down the differences.
Research Question 1: What is the impactof contemporary police strategies in reducing crime rates?
The research study wants to investigate whichcontemporary police strategy between Geographic Policing and Geographic Profiling reduces community crime rates more. The following are the null
hypothesis and alternative hypothesis for the research study:
H[0]: µ[Geographic Policing] = µ[Geographic Profiling]
H[1]: µ[Geographic Policing] ≠ µ[Geographic Profiling]
The null hypothesis for this research question supposes that the two police approaches to addressing crime; geographic policing and geographic profiling, have equal impact in reducing crime levels.
The alternative hypothesis supposes that at least one of the police approach to reducing crime is more effective than the other
The contemporary police strategies that include Geographic Policing and Geographic Profiling form the independent variables of the study, while level of crime rates form the dependent variable.
The contemporary police strategies,the effectiveness of geographic policing and effectiveness of geographic profiling can be measured by use of proxies such as the perception of the law enforcement
officers and the community towards their effectiveness, and or through the use of official statistics that reflect a change in crime levels since the introduction of geographic policing or geographic
profiling into the community. They are categorical variablesthatare measurable using ordinal scale.Crime levels can be measured using the number of reported cases to law enforcement establishments.
It is a continuous variable.
Utilization of the statistical decisions model to make a decision
Using the statistical decision model to decide on an appropriate statistical test to use in the study, I would first look at the umbrella characteristics of the study such as the nature and number of
independent variables as well as number of dependent variables, their attributes and scales of measurement and finally the goal of the analysis.The research study has one independent variable which
is categorical in nature and one dependent variable which is continuous in nature.Additionally, the two variables in the research question do not covary.Lastly, the study aims at determining
significance of mean group differences. Therefore, using the decision model, the study will employ the use of a One-Way ANOVA. The model guidedme to the correct statistical tool, this is because,
aOne-Way ANOVAwould be the most appropriate tool for drawing conclusions for this study.
Research Question2: What is the correlation between choice of forensic tools employed and the number of solved criminal cases (clearances) and crime rates?
The objective of the study is to establish whether there exists any relationship between the choice of forensic investigation tool by the police and the number of solved criminal cases and crime
rates. Following are the statistical notations of the null hypothesis and alternative hypothesis;
H[0]: µ[(x)] = µ[(y)]
H[1]: µ[(x)] ≠ µ[(y)]
The null hypothesis supposes no significant relationships exists between the choice of forensic investigation tool (X) and the number of solved criminal cases and crime rates (Y), while the
alternative hypothesis supposes that significant relationships exists between the choice of forensic investigation tool and the number of solved criminal cases and crime rates.
The variables for the study were the different forensic investigation tools available on one hand and the number of solved criminal cases (Clearances) and crime rates. The forensic investigation
tools are nominal variables, and so are the number of solved criminal cases and crime rates. Both of these variables are categorical or nominal variables since they can take two or more categories
that have no intrinsic ordering.
Utilization of the Statistical Decisions Model to Make Decisions
To choose an appropriate tool for the study, we look at the broad characteristics of the research question such nature and number of variables, their attributes and scales of measurement, goal of the
analysis and nature of research question. This research question has two variables, both of which are continuous variables and do not covary, and its goal is to establish whether there are
differences between groups. Based on the statistical decision model, the statistical tool to use is a chi-square. As a researcher, I would also have used the chi-square statistical tool. Therefore, I
consider the decision model to have given chosen the correct tool.
Limitations and Usefulness of Statistical Decision Models in Statistics and Research Methods
Statistical decision models play an important role in statistics and research methods. They are useful as they provide a direct and often simpler approach to deciding which statistics tool to employ
for a particular research question. Statistical decision models also have the advantage that they are self-explanatory and easy to follow and can easilybe grasped by non-professional users.
Additionally, they can handle both nominal and numeric input attributes thereby making them more accepting to statistical tools with various differentiating attributes.However, these decision models
have disadvantages, the key ones being; firstly, they predominantlyemploy the use the “divide and conquer” method, thus they tend to perform well only when a small number of highly relevant
characteristics exist, but less so if numerous composite correlations are present. Secondly, considerable time is necessary to make a summary of the various statistical tools characteristics and
attributes to create the statistical decision model, thus it is tiresome in its construction.
What Was Learnt From Creating the Model and Applying It to the Study of Interest and How It Might Be Used the Future
Creating this statistical decision model necessitated me to keenly examine the characteristics and attributes of each of the statistical tools so as to note the unique differentiating qualities and
elements of each tool. Therefore, I can consider myself to be more conversant with the various statistical tools more than before, and I can now easily identify a particular tool to use for a
particular research question without referring back to a textbook or to the decision model. The model will be important in future especially in times when a decision with regards to which statistical
tool to use is needed. Indeed, it was an eye-opening exercise that summarized all that I have learnt in this class.
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Navigation: Realtest Script Language > Syntax Element Details >
Daily Test Statistics
The number of this stat period in a test
In the context of the Graphs Section, S.Number is also the count of stat periods so far as of this test date.
In the context of the Results Section, S.Number is also the total number of stat periods in the test.
The size of each stat period in a test will always be the smallest bar size used in any strategy.
For example, if a script includes a daily bar strategy and a weekly bar strategy, and is run for one year, there will be 252 stat periods for both strategies, and S.Number will be 252 for the
final period of either strategy.
If a script only includes weekly bar strategies and is run for one year, there will be 52 stat periods and S.Number will be 52 for the final period.
Because this is a test-level statistic, the value returned will be the same for every strategy (hence no need for "Combined").
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