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http://encyclopedia2.thefreedictionary.com/Wronskian | 1,537,879,564,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00110.warc.gz | 80,034,128 | 11,836 | # Wronskian
Also found in: Wikipedia.
## Wronskian
[′vrän·skē·ən]
(mathematics)
An n × n matrix whose i th row is a list of the (i - 1)st derivatives of a set of functions f1, …, fn ; ordinarily used to determine linear independence of solutions of linear homogeneous differential equations.
## Wronskian
a functional determinant composed of n functions f1(x), f2(x)....,fn(x) and their derivatives up to the order n - 1 inclusive:
The vanishment of the Wrońskian [W(x) = 0] is a necessary and, under certain additional assumptions, a sufficient condition for the linear dependence between the given n functions, differentiated n - 1 times. Based on this, the Wrońskian is used in the theory of linear differential equations. The Wrońskian was introduced by J. Wroński in 1812.
References in periodicals archive ?
Thus a dual Grothendieck polynomial is still a discrete Wronskian that one identifies with a multiSchur function (in the case of an increasing or decreasing sequence of alphabets, one also uses the term flagged Schur function cf.
where <y, z> := yz' - y'z is the Wronskian of y and z.
If X, Y [member of] D, then we define the Wronskian matrix of X(t) and Y(t) by
21) ensures that the Wronskian matrix of the linearly independent functions ([v.
are linearly independent by computing its Wronskian at t = 0.
W is the Wronskian of the two independent solutions, which is constant, being the layer matrix M unimodular, and:
4) are linearly independent if and only if their Wronskian is different from zero.
i] are linearly independent over F, then there exists a generalized Wronskian of the (3), which does not vanish.
The accuracy of the results can be checked using the Wronskians for Mathieu functions, as the algorithms for computing Mathieu functions are not based on the Wronskian of the functions.
Since the Wronskian of any two solution of (8) is independent of t, while t [right arrow] [infinity] in (11) we also get
Marini, On intermediate solutions and the Wronskian for half-linear differential equations, J.
This generalization is obtained by considering the Wronskian of functions over F[x].
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https://abbaskets.com/2023/07/09/616-kg-to-lbs.html | 1,695,382,761,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00757.warc.gz | 92,792,186 | 8,954 | Intro text, can be displayed through an additional field
## What is 61.6 Kg To Lbs?
When it comes to converting units of measurement, it is essential to have a clear understanding of the values involved. In this article, we will explore the conversion of 61.6 kilograms (kg) to pounds (lbs). This conversion is commonly needed in various scenarios, such as international travel, fitness tracking, or scientific calculations.
### The Basics: Understanding Kilograms and Pounds
Before diving into the conversion process, it is important to grasp the concept of kilograms and pounds individually. Let's start by understanding what each unit represents:
#### Kilograms (kg)
Kilograms are a unit of mass commonly used in the metric system. It is the base unit of mass in the International System of Units (SI). One kilogram is equal to 1000 grams and is primarily used to measure the weight of objects or individuals.
#### Pounds (lbs)
Pounds, on the other hand, are a unit of weight widely used in the United States and a few other countries. It is commonly used in everyday life for measuring the weight of groceries, people, or other objects. One pound is equivalent to 0.45359237 kilograms.
### Converting 61.6 Kg to Lbs
Now that we have a clear understanding of kilograms and pounds, let's proceed with the conversion of 61.6 kg to lbs:
To convert kilograms to pounds, we multiply the given value by the conversion factor. In this case, the conversion factor is 2.20462 pounds per kilogram. So, by multiplying 61.6 kg by 2.20462, we get:
61.6 kg * 2.20462 lbs/kg = 135.58632 lbs
Therefore, 61.6 kilograms is equal to 135.58632 pounds.
### FAQs about 61.6 Kg To Lbs
#### Q: How many pounds are there in 61.6 kilograms?
A: There are 135.58632 pounds in 61.6 kilograms.
#### Q: Why is it important to convert kilograms to pounds?
A: Converting kilograms to pounds is crucial when dealing with different measurement systems, especially when communicating or calculating values related to weight in countries that predominantly use pounds as a unit of measurement.
#### Q: Is there a simple method to convert kilograms to pounds?
A: Yes, converting kilograms to pounds is a straightforward process. You can multiply the kilogram value by 2.20462 to obtain the equivalent weight in pounds.
### Conclusion
In conclusion, converting 61.6 kg to lbs is a simple process that involves multiplying the kilogram value by the conversion factor of 2.20462. The resulting value is 135.58632 pounds. Understanding these conversions is essential for various practical situations, whether in daily life or in more specialized fields. By having a clear understanding of the conversion process, you can easily navigate between different systems of measurement and ensure accurate and meaningful communication of weight-related information.
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https://www.mapleprimes.com/users/strokebow/replies | 1,685,681,469,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00170.warc.gz | 938,210,286 | 27,390 | ## 15 Reputation
11 years, 90 days
## The solution maple gives is: QUESTION...
The solution maple gives is:
QUESTION:
Given a particular value for a,q, and omega.
How would I plot x(t) against time in Maple?
* This may be really easy to do (or not) but I am still getting used to maple and would appreciate any help anyone can offer.
Thanks!
## The solution maple gives is: QUESTION...
The solution maple gives is:
QUESTION:
Given a particular value for a,q, and omega.
How would I plot x(t) against time in Maple?
* This may be really easy to do (or not) but I am still getting used to maple and would appreciate any help anyone can offer.
Thanks!
## @ecterrab Thanks for your post!!So...
So there is no way to plot those transition curves inbetween a0, a1, a2,... b1, b2, b3, etc. That is a shame!
## @ecterrab Thanks for your post!!So...
So there is no way to plot those transition curves inbetween a0, a1, a2,... b1, b2, b3, etc. That is a shame!
## Its been almost a year and still no one ...
Its been almost a year and still no one has even attempted an answer.
Is it impossible to do this in Maple or is just that those select people who know how to do this haven't seen the post?
## @Markiyan Hirnyk Ah! I see...You ...
Ah! I see...You are correct. The trem beta does not appear in the link. But the stability diagrm is there none the less.
Like I said, I am no expert at all.
I have some further information in a document but I am unable to attach it... :-/
I have found this online:
Hopefully that helps.
Thanks :-)
## @Markiyan Hirnyk Fair comment. My a...
@Markiyan Hirnyk Fair comment. My apologies. By the way, nor am I :-)
After some digging, from the solution to the equation, one is able to obtain a parameter beta which is a function of a and q and is used to define the stability boundary. The stability boundary is marked by beta = 0 and beta = 1 curves. The stability diagram is plotted in a,q space.
Hope this helps.
thanks :-)
Page 1 of 1
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https://documen.tv/question/a-flat-roof-is-very-susceptible-to-wind-damage-during-a-thunderstorm-and-or-tornado-if-a-flat-ro-16929378-41/ | 1,723,404,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00071.warc.gz | 162,177,455 | 17,060 | ## A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and winds of speed
Question
A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and winds of speed 39.0 m/s blow across it, determine the magnitude of the force exerted on the roof. The density of air is 1.29 kg/m3.
in progress 0
3 years 2021-08-25T09:46:59+00:00 1 Answers 43 views 0
1. Answer: The magnitude of the force exerted on the roof is 490522.5 N.
Explanation:
The given data is as follows.
Below the roof, = 0 m/s
At top of the roof, = 39 m/s
We assume that is the pressure at lower surface of the roof and be the pressure at upper surface of the roof.
Now, according to Bernoulli’s theorem,
=
=
= 981.045 Pa
Formula for net upward force of air exerted on the roof is as follows.
F =
=
= 490522.5 N
Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N. | 285 | 1,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.924422 |
blog.turlockinsurance.com | 1,542,739,099,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746528.84/warc/CC-MAIN-20181120171153-20181120192426-00037.warc.gz | 51,004,839 | 16,934 | How Much Life Insurance Do I Need?
The biggest problem I’ve noticed in the insurance industry is people want life insurance, but with all of the variables and unexpected costs, they have no idea how much coverage they actually need. Having the correct amount of coverage is extremely important. With not enough coverage you could leave your family in debt, and with too much coverage you’ll be paying too much of your hard earned money on monthly premiums! That’s why with help from our friends at Nationwide, I’ve created an easy way for you to calculate exactly how much coverage you and your family needs.
Let’s use Jane Smith as an example. She’s a single mother with a 13-year-old daughter named Sara.
• Jane has a mortgage on her home and she still owes \$180,000.
• She makes \$40,000 a year, and if something were to happen to her, she wants her daughter to have half of her yearly income for the next 5 years. \$80,000 divided by 2 multiplied by 5 years = \$100,000.
• Sara dreams of going to college in California, and Jane wants to make sure if she’s not around to pay tuition Sara can still go to college. Let’s assume that tuition will continue to steadily rise and when Sara goes to college it will cost \$30,000 a year. \$30,000 a year for 4 years = \$120,000.
• Jane has \$4,500 in credit card debt that she does not want to leave behind for Sara to have to deal with.
• Jane wants to leave Sara an emergency fund for unexpected expenses such as a new roof for the house or new engine for the car. Experts suggest 3 months of income for your emergency fund. \$40,000 divided by 12 multiplied by 3 = \$10,000.
• Jane also wants her funeral to be paid for by her life insurance. The average cost of a funeral is just over \$8,000 let’s round it to \$9,000 to be safe.
Now we’ll add all of this up to determine her family’s need which = \$423,500
Now we’ll subtract her assets from the total to determine how much of her family’s need is not covered.
• Jane has a life insurance policy through her job which is worth \$100,000.
• Jane also has \$11,000 in her savings account.
\$423,500 – \$111,000 = \$312,500
Now we have done all of the math and we know Jane needs a \$312,500 life insurance policy to fully protect her family. Of course not many people want to do all of this math for themselves, so that’s where I come in! I’m more than happy to sit down with anyone and go over their finances to make sure their family is protected. If you have any questions or concerns give me a call today: (209)410-0270.
4 Businesses That Have Life Insurance To Thank For Their Success
Everyone knows that life insurance will pay out a benefit when the insured passes on, but most people don’t know the other benefits of having life insurance. One of the greatest things about life insurance is the ability to borrow money from the policy.
Foster Farms Everyone knows and loves Foster Farms chicken, but did you know that Max and Verda Foster payed for their first farm with money borrowed from a life insurance policy? In 1939 the Fosters borrowed \$1,000 from their life insurance to purchase 80 acres of land in Modesto and began raising chickens. This investment led to the creation of Foster Farms.
JC Penney James Penney incorporated his clothing store in 1913, 16 years before the great depression. When the depression struck it destroyed his wealth and almost ended his franchise. He was able to borrow from his Life insurance policy to pay his employees and keep his business afloat. Had he not had life insurance we most likely would not be able to shop at JC Penney today.
McDonalds Ray Krok bought out the McDonald’s brothers and took over the franchise in 1961. During the early years of the company he constantly had cash shortages. On 2 occasions he had to borrow money from his life insurance to cover the salaries of employees.
Disneyland Walt Disney had a dream to create a nice clean amusement park where families could come together and enjoy themselves. He was unable to get the financing he needed to open Disneyland, so he decided to finance it himself. Most of the money used to create the first Disneyland came from the cash value of his life insurance policy. If Walt had not had life insurance, Disneyland may have never been built.
by Dylan Delhart, Financial Advisor @insureCAL
I didn’t buy a Powerball lottery ticket and here’s why:
Yesterday, a friend of mine was telling me that he was going to buy a Powerball lottery ticket on his lunch break. I thought about it and figured I’ll go spend \$2 to be a part of the fun,”why not”. Then I began thinking about the consequences of publicly winning a billion dollars and here’s WHY NOT:
You can’t remain anonymous.
There are only 6 states that the laws allow you to remain anonymous. Since California is not one of those states, that is my 1st why not. The 6 States that allow you to remain anonymous are Delaware, Kansas, Maryland, North Dakota, Ohio, and South Carolina.
Your life will never be the same.
I love my life. Just the way it is. It might not be “perfect”, I might not have the money I want (and sometimes “need”) and there could be fewer bills, less cleaning and less laundry, but when you look at the big picture we all have awesome lives. Great friends of our choosing and great family-not always of our choosing. The facts, almost 70% of lottery winners end up broke after 7 years. Even worse, some have committed suicide after their lives were turned upside down.
You will never know who to trust.
People will come from everywhere to be your best friend. Yes, you have the friends you talked to everyday before you became a billionaire, but after a while of fending off all of those long lost high school friends you do not remember, you will begin questioning your closest friends. Are you certain they don’t like you even a little more now that you can take them to one of your three houses on Lake Como?
Safety.
I enjoy walking to the grocery store while on the phone with a friend and not being bothered by anyone at that grocery store. I also enjoy that the craziest person posting pictures of my 3 month old baby on the Internet is my Mother. Most lottery winners have been robbed or attempted to be robbed. I am paranoid enough walking around with \$122 cash in a \$400 wallet. Let alone everyone in the country knowing I just won a BILLION dollars. You would need a compound, 24/7 body guards, and probably a moat around it all.
I wouldn’t trade all of those things for every rare, vintage Ferrari and Alfa Romeo, all of the houses in Italy, Greece, Switzerland, or for the court side seats at Wimbledon (or even the press box with John McEnroe commentating in it) Now, let’s all get out there and work towards earning the money we will feel good about spending and enjoying the awesome lives we get to live!
Basics of Your Personal Auto Insurance Coverages
Unsure of exactly what all of those numbers and coverage’s mean on your Auto Insurance policy?
Here’s a quick list and explanation of the basics:
BI/Bodily Injury Liability:
Covers bodily injuries that are caused by you/your vehicle to others
PD/Property Damage:
Covers the damage that you/your vehicle causes to other parties vehicle or property
UMBI/Uninsured motorist Bodily Injury:
Covers you and those in your vehicle if injured by an uninsured or under-insured motorist
UMPD/Uninsured Motorist Property Damage: (this coverage is when you do NOT have physical damage coverage)
Covers the damage done to your vehicle by an uninsured or under-insured motorist (only covers up to \$3500 in damage)
CDW/Collision Deductible Waiver: (this coverage is when you DO have physical damage coverage)
Waives your deductible when damage was caused by an uninsured or under-insured motorist
Medical Expense:
Covers yourself and those in your vehicle when the accident is AT FAULT.
Comprehensive Deductible:
Amount that insured is responsible for if the vehicle is vandalized, stolen, etc (anything other than collision)
Collision Deductible:
Amount that insured is responsible for if the vehicle is involved in a collision
Insurance: When Cheap Isn’t Always Best
by Patrick Ramsay
While shopping for home or automotive insurance, it’s easy to become overwhelmed by the waves of insurance companies all vying for your business. The word “cheap” is used so often in the insurance market, it’s hard to recall its meaning as you peruse the various options online and hear commercials boasting about the newest, cheapest rates being offered.
There’s a psychological phenomenon known as semantic satiation, in which a person repeats a word or phrase so many times that it temporarily loses its meaning and begins sounding like repeated meaningless sounds. It could be argued that’s what has happened in the insurance market with the word “cheap.” We hear more about cheap insurance than about quality insurance, so naturally we tend to believe the cheaper the deal on the insurance, the more desirable it is. However, what we forget to realize is that cheaper isn’t always better. In fact, when we’re talking about insurance, cheaper is hardly ever better.
Less coverage means less protection. When it comes to insurance, you’re going to get what you pay for. Are lower monthly payments worth potentially being hung out to dry when you need quality coverage most?
Raising your deductible means more out-of-pocket. While you may see a decrease in the cost of a monthly payment by raising your deductibles, when the time comes to use your policy you will have to pay a large sum of money out-of-pocket to cover your repair.
Low cost oftentimes means low quality. When it comes to insurance, it’s tempting to choose the cheapest option. After buying a home or a car, it’s only natural to seek out ways to save, but your insurance coverage is not the place. A quality insurance company will offer attentive customer service and go the extra mile to take care of their customers and help them save money without cutting corners.
insureCAL offers combined Home and Auto Insurance packages to utilize your insurance policies for the best coverage possible. By combining your home and auto insurance policies, you may be eligible for various discounts, a guaranteed full-year policy term for your car, On Your Side® insurance protection, 24-hour Customer and Claim service, and more. You deserve insurance that brings you peace of mind. Contact insureCAL to find out more about our Home and Auto Insurance packages.
9 Reasons Employers Should Offer Life Insurance
Positive employees make for a positive business. So, how can you ensure your employees start each Monday with a positive foot forward? Offer them peace of mind. In an increasingly hectic world, you can’t put a price on peace of mind, and it’s even harder to achieve. However, you can ensure peace of mind in your employees with something as simple and cheap as offering life insurance. So, why should you offer life insurance to your employees?
1. Your employees want it! According to recent surveys within the private sector 97% of employees offered Life insurance choose to enroll.
1. Less payroll and income taxes. If all the requirements are met, the cost of the premiums for the first \$50,000 of group-term life insurance isn’t included in the employee’s gross income.
1. Employee retention. Employees with quality life insurance are less likely to take seasonal jobs or leave the company altogether, saving the employer turnover expenses.
1. Increased morale. For many workers, good benefits are the difference between a job and a career. Employees with life insurance generally feel more essential to their workplace and may feel the desire to go above and beyond expectations.
1. More appealing job offers. Any employer will tell you how important good employees are. Quality employees are hard to come by these days. By simply offering life insurance, your business can compete with the power houses for the best employees.
1. You could be a hero. The difficulties surrounding the death of their primary wage earner oftentimes extend beyond the initial mourning and emotional distress of the death. Without life insurance the family could be subject to serious financial hardships. The monthly premiums are minuscule compared to the immense benefits. Although no amount of money could replace a loved one, it’s much easier to recover emotionally when you’re not worried about coming up with thousands of dollars for funeral expenses.
1. It’s inexpensive. Term life insurance costs less than your daily cup of coffee! If someone could offer you peace of mind for a price that low, wouldn’t you be interested? Life insurance is one of the cheapest forms of insurance to purchase, which is marvelous considering how important it is.
1. It’s easy to get! Many employers are intimidated by the amount of paperwork involved with getting life insurance. While the hassle may seem daunting, our life insurance professionals are prepared to streamline the process and take care of the tedious tasks that surround getting enrolled in a life insurance program.
1. Increased productivity. A recent study by economists at Warwick University, discovered that happy employees saw an increase in productivity of 12%, while unhappy workers were 10% less productive. The happier, more appreciated your employees feel, the more inclined they will be to bring that positive attitude to work with them. Every employer knows that both positivity and negativity are infectious in the workplace. By offering life insurance, you’re contributing to a more positive work environment because every employee knows that happiness in the workplace starts with their employer.
Contact Dylan Delhart at InsureCAL to talk about the best life insurance options for your employees.
by Dylan Delhart and Patrick Ramsay
Nut Theft in California
By Patrick Ramsay
If you’re unfamiliar with the phrase ”nut heist,” it might bring to mind images of something like the movie Oceans 11 playing out on a pistachio farm. However, if you’re one of the hardworking nut farmers of California’s Central Valley, you know a nut heist is far more serious than that. A nut heist is a complex, organized, large-scale act of crime, and in recent years, California’s tree nut industry has suffered millions of dollars in losses due to these organized nut heists.
In one common technique, thieves fabricate shipping documents to pose as truck drivers and pick up cargo loads of tree nuts that can be worth upwards of \$500,000 depending on the type of nut. While some of the stolen nuts have been tracked down within storefronts, bakeries, and farmer’s markets around the US, law enforcement officials believe organized criminal enterprises are sending cargo loads of nuts to the export market as well.
In response to a recent rash of these nut heists, the California’s tree nut industry is taking action. On Dec. 3, growers, processors and law enforcement gathered at the Emergency Nut Theft Summit in Visalia, California, to discuss the best ways to combat nut theft.
Nut growers have been using various methods to prevent loss of their produce and safeguard their protocols. These methods include fingerprinting and doing a more thorough, in-person check of the truck drivers who come to pick up the loads of tree nuts. Some nut processors are also using GPS in the cargo loads to track the shipments if they go off course. A nut heist is very hard to recognize and harder to stop. Growers, processors, and truck drivers are encouraged to be vigilant, as this is not the kind of crime that can be recognized outright and immediately most of the time. What are some solutions you can think of to stop these nut thieves?
The Bittersweet Relief of Godzilla El Niño in California
By Patrick Ramsay
The infamous weather pattern, El Niño, meaning “the child” in Spanish, first got its name because it was discovered on Christmas. This year, El Niño is bringing a much needed gift to Californians: rain. We aren’t talking about a light drizzle either, we’re talking about a weather pattern so epic, it has earned the title “Godzilla.” Essentially, it is the slight warming of the Pacific Ocean causing a change in the weather patterns every few years. A powerful El Niño is described by a 1.5 degree Celsius change, but this year we were seeing temperatures in the Pacific already 3 degrees C above normal by September.
Drought Relief
California has been suffering from a drought for the last four years and will be the greatest beneficiary of this year’s El Niño. We’re already seeing signs of relief from the drought in the Sierra Nevada snowpack. Roughly 30 percent of California’s water supply comes from the runoff of this snowpack. During the first manual survey of the winter snowpack, officials with the California Cooperative Snow Surveys Program for the Department of Water Resources (DWR) say it’s currently at 136 percent of normal. Around the same time last year, the snowpack was only at 45 percent of the historical average, so officials with the DWR are saying it’s a very promising sign.
1997 El Niño Twins
According to NASA, this year’s El Niño bears an uncanny resemblance to that of the powerful 1997 El Niño, and it’s continuing to grow. Forecasters expect California to start seeing the effects of Godzilla El Niño in early 2016. Bill Patzert, Climatologist for Jet Propulsion Laboratories, said “Reservoir levels have fallen to record or near-record lows, while groundwater tables have dropped dangerously in many areas. Now we’re preparing to see the flip side of nature’s water cycle — the arrival of steady, heavy rains and snowfall.”
Side Effects of Godzilla El Niño
In 1997, the strong El Niño brought twice the average amount of rainfall to Southern California. While the El Niño weather cycle brings watery relief to the drought-stricken state of California, it has also been known to bring mudslides, floods, high winds, lightning strikes and high surfs along with it. How are you preparing for Godzilla El Niño’s intensity?
Save | 3,785 | 18,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | longest | en | 0.976727 |
https://www.unitconverters.net/length/handbreadth-to-mil.htm | 1,560,978,928,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00248.warc.gz | 947,549,935 | 4,513 | Home / Length Conversion / Convert Handbreadth to Mil
# Convert Handbreadth to Mil
Please provide values below to convert handbreadth to mil [mil, thou], or vice versa.
From: handbreadth To: mil
### Handbreadth to Mil Conversion Table
0.01 handbreadth30 mil, thou
0.1 handbreadth300 mil, thou
1 handbreadth3000 mil, thou
2 handbreadth6000 mil, thou
3 handbreadth9000 mil, thou
5 handbreadth15000 mil, thou
10 handbreadth30000 mil, thou
20 handbreadth60000 mil, thou
50 handbreadth150000 mil, thou
100 handbreadth300000 mil, thou
1000 handbreadth3000000 mil, thou
### How to Convert Handbreadth to Mil
1 handbreadth = 3000 mil, thou
1 mil, thou = 0.0003333333 handbreadth
Example: convert 15 handbreadth to mil, thou:
15 handbreadth = 15 × 3000 mil, thou = 45000 mil, thou | 245 | 780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-26 | longest | en | 0.46155 |
https://gradebuddy.com/doc/3428123/hw2-45c-sp15/ | 1,621,214,671,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991921.61/warc/CC-MAIN-20210516232554-20210517022554-00225.warc.gz | 306,168,283 | 10,270 | # UCI CBEMS 45C - HW2_45C_Sp15 (2 pages)
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## HW2_45C_Sp15
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## HW2_45C_Sp15
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Cbems 45c - Chem Eng Thermodyns
##### Chem Eng Thermodyns Documents
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CBEMS 45C Chemical Engineering Thermodynamics Spring 2015 Homework 2 Due at the BEGINNING of lecture on 04 17 2015 1 The Clausius equation of state describes the behavior of a certain fluid P V b RT with b 10 5 m3 mol For this fluid CP 25 4 10 2 T J mol K a Derive an explicit algebraic expression for the CP of the fluid valid at any pressure b The pressure of the fluid must be reduced from 4 MPa to 800 kPa The initial temperature of the fluid is 300 K Your colleagues are considering two different ways of reducing the pressure either by passing the fluid through a throttling valve or by passing it through a small gas turbine that runs reversibly and adiabatically What will the final temperature of the fluid be in each case 2 Assume that nitrogen can be described by the virial equation of state B C PV 1 2 RT V V with B 10 5 m3 mol and C 1 5 10 9 m6 mol2 In an industrial process 40 moles of nitrogen is being isothermally compressed at 20 C from 1 bar to 100 bar in a piston and cylinder device Calculate the work required for this compression the amount of heat that must be removed to keep the process isothermal and the amount of entropy generated in the gas during this process 3 Assuming that the van der Waals equation of state is applicable for methane answer the following questions a Calculate H S and U in going from 300 K and 10 MPa to 200 K and 4 MPa and compare your answers to those obtained from Figure 3 3 2 Comment on the accuracy of the van der Waals equation of state b One mole of methane initially at 100 kPa and 300 K is isothermally compressed to 10 MPa Determine the molar volume at the initial and final conditions and the amount of work necessary to carry out the isothermal compression 4 The following data is available from the steam tables for a temperature of 350 C this data will be available on the course website 1 Specific Volume m3 kg 0 0013118 0 0013538 0 001397 0 001442 0 0014894 0 0015405 0 0015975 0 0016657 0 0022098 0 002782 0 0035023 0 0044092 0 0055508 0 0069881 0 010029 0 015953 0 022442 0 030243 0 039859 0 051834 0 066818 0 085617 0 10924 0 13894 0 17630 0 22332 0 28249 0 35698 0 45074 0 56878 0 71737 0 90443 Pressure bar 1000 0 794 33 630 96 501 19 398 11 316 23 251 19 199 53 165 29 165 29 165 29 165 29 165 29 165 29 158 49 125 89 100 00 79 433 63 096 50 119 39 811 31 623 25 119 19 953 15 849 12 589 10 000 7 9433 6 3096 5 0119 3 9811 3 1623 a Using your choice of software e g Excel Matlab Scilab etc plot the above isotherm data on a P V plot P on y axis specific volume on x axis Plot both axes on a log scale b On the same axes plot the corresponding isotherm predicted by the idea gas law c On the same axes plot the corresponding isotherm predicted by the van der Waals equation of state d Comment on the range of accuracy for the two equations of state Notes use the appropriate vdW constants for water and watch units Include a legend and axes labels 2
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Unlocking... | 996 | 3,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-21 | latest | en | 0.853987 |
https://vlab.amrita.edu/?sub=1&brch=192&sim=346&cnt=3 | 1,642,893,703,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00101.warc.gz | 639,995,346 | 7,001 | . .
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Temperature Coefficient of Resistance
.
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1) What is the principle behind Carey Foster Bridge? Wheatstone's principle Bernoulli's principle Both (a) and (b) None of these
2) What is the relation connecting resistance and resistivity? R=ρl/A R=ρA/l ρ=RAl ρ=Al/R
3) Carey Foster Bridge is used to measure High resistance Low resistance E.M.F All the above
4) As the temperature increases, resistance Decreases Increases No change First increases then decreases
5) In Wheatstone's bridge, P = 10 ohm, Q = 20 ohm R = 5 ohm, then value of X is 5 15 20 10
Cite this Simulator:
.....
..... .....
Copyright @ 2022 Under the NME ICT initiative of MHRD | 186 | 660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-05 | latest | en | 0.83668 |
https://www.physicsoverflow.org/8913/first-and-second-fundamental-forms | 1,716,513,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00186.warc.gz | 831,037,401 | 26,030 | # First and second fundamental forms
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I'm writing notes about the 3+1 formalism in general relativity, for myself. Inevitably I came across the notions of first and second fundamental forms. Mathematically, it is clear how these objects are defined: ($M$ is a 4-dim manifold with metric $g$, $\Sigma$ a hypersurface of $M$)
The first fundamental form is the induced metric on $\Sigma$, also given by the pullback of the spacetime metric $g$.
The second fundamental form $K: T_{p}(\Sigma)\times T_{p}(\Sigma)\rightarrow \mathbb{R}$ is given through the Weingarten map $\chi$, i.e. $(u,v)\mapsto -u\cdot \chi(v)$.
Now, I'm having difficulties with underlying physcial intuition for these two objects (especially the second fundamental form). Is there a way for a physicist to "visualise" them in a way? What kind of objects are these forms exactly?
This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user ConciseAndClear
retagged Mar 25, 2014
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For the first fundamental form - if you've got two vectors tangent to $\Sigma$, and $\Sigma$ is embedded in $M$, and $M$ has a metric, just use the embedding to consider the vectors as living tangent to $M$ and use $M$'s metric to compute their inner product.
For the second fundamental form, basically, if you imagine a two surface $\Sigma$ embedded in $\mathbb{R}^3$, and you imagine the normals as arrows orthogonal to $\Sigma$ sticking out like a hedgehog's spines. Then if the surface is gently curved, the normals dont change much as you go from a point $x$ on $\Sigma$ to another point $x'$ on $\Sigma$ infinitesimally separated from $x$ by a vector tangent to $\Sigma$. Conversely if the embedding is highly curved, the normals change a lot when you do this small displacement.
The Weingarten map, your $\chi$, is just the map $$u\rightarrow \nabla_un$$ where $n$ is the normal to $\Sigma$ and $u$ is tangent to $\Sigma$, and this encodes how much the normals are changing when you nudge by $u$ along $\Sigma$.
The 2nd FF, or "extrinsic curvature" is just another way of representing the information in the Weingarten map. Explicitly, it's a bilinear form which maps a pair of vectors $u,v$ tangent to $\Sigma$ to a number $-u.\chi(v)$ ($\chi(v)$ is also tangent to $\Sigma$ so this makes sense).
BTW, since you're studying 3+1, this reference (which came up in one of Alex Nelson's posts), is really informative and steers you through the minefield of conflicting approaches.
This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59
answered Apr 30, 2013 by (2,500 points)
thank you very much twistor59! that's exactly the kind of explanation I wanted!
This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user ConciseAndClear
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To add a bit of fluff to twistor59's answer, let's take a bird's eye view of Riemannian geometry.
The Riemannian metric gives us the notion of lengths and angles as well as the concept of straight lines (geodesics).
Any submanifold inherits these notions from the ambient space, made explicit via the first fundamental form, which makes the submanifold is a Riemannian manifold in its own right.
As such, it comes with the notion of intrinsic curvature, eg manifest in the sum of the angles of a triagle formed by straight lines, which is independent of the embedding into any larger space.
However, there's a second notion of curvature, the extrinsic one, which does make use of this embedding, eg via normal vectors, osculating circles or approximation by paraboloids. The second fundamental form is of this type.
These different notions of curvature are of course related: You can get the intrinsic curvatue of a submanifold $N\hookrightarrow M$ (measured by the Riemann curvature tensor $R_N$) from its extrinsic curvature (measured by the second fundamental form $\mathrm{II}$) and the intrinsic curvature $R_M$ of the ambient space via $$\langle R_N(u,v)w,z\rangle = \langle R_M(u,v)w,z\rangle+\langle \mathrm I\!\mathrm I(u,z),\mathrm I\!\mathrm I(v,w)\rangle-\langle \mathrm I\!\mathrm I(u,w),\mathrm I\!\mathrm I(v,z)\rangle$$ (The formula was taken from this Wikipedia article).
This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Christoph
answered Apr 30, 2013 by (210 points)
Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor) Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification | 1,358 | 5,464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.874068 |
https://en.wikipedia.org/wiki/Monoclinic | 1,639,005,291,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363598.57/warc/CC-MAIN-20211208205849-20211208235849-00083.warc.gz | 297,245,133 | 17,782 | # Monoclinic crystal system
(Redirected from Monoclinic)
Monoclinic crystal
An example of the monoclinic crystal orthoclase
In crystallography, the monoclinic crystal system is one of the seven crystal systems. A crystal system is described by three vectors. In the monoclinic system, the crystal is described by vectors of unequal lengths, as in the orthorhombic system. They form a rectangular prism with a parallelogram as its base. Hence two pairs of vectors are perpendicular (meet at right angles), while the third pair makes an angle other than 90°.
## Bravais lattices
### Two-dimensional
There is only one monoclinic Bravais lattice in two dimensions: the oblique lattice.
Bravais lattice Oblique
Pearson symbol mp
Unit cell
### Three-dimensional
Two monoclinic Bravais lattices exist: the primitive monoclinic and the base-centered monoclinic lattices.
Rectangular vs rhombic unit cells for the 2D base layers of the monoclinic lattice. The two lattices swap in centering type when the axis setting is changed.
Bravais lattice Primitive
monoclinic
Base-centered
monoclinic
Pearson symbol mP mS
Standard unit cell
Oblique rhombic prism
unit cell
In the monoclinic system there is a rarely used second choice of crystal axes that results in a unit cell with the shape of an oblique rhombic prism;[1] it can be constructed because the rectangular two-dimensional base layer can also be described with rhombic axes. In this axis setting, the primitive and base-centered lattices swap in centering type.
## Crystal classes
The table below organizes the space groups of the monoclinic crystal system by crystal class. It lists the International Tables for Crystallography space group numbers,[2] followed by the crystal class name, its point group in Schoenflies notation, Hermann–Mauguin (international) notation, orbifold notation, and Coxeter notation, type descriptors, mineral examples, and the notation for the space groups.
# Point group Type Example Space groups
Name[3] Schön. Intl Orb. Cox. Primitive Base-centered
3–5 Sphenoidal C2 2 22 [2]+ enantiomorphic polar halotrichite P2, P21 C2
6–9 Domatic Cs (C1h) m *11 [ ] polar hilgardite Pm, Pc Cm, Cc
10–12 Prismatic C2h 2/m 2* [2,2+] centrosymmetric gypsum P2/m, P21/m C2/m
13–15 P2/c, P21/c C2/c
Sphenoidal is also called monoclinic hemimorphic, domatic is also called monoclinic hemihedral, and prismatic is also called monoclinic normal.
The three monoclinic hemimorphic space groups are as follows:
• a prism with as cross-section wallpaper group p2
• ditto with screw axes instead of axes
• ditto with screw axes as well as axes, parallel, in between; in this case an additional translation vector is one half of a translation vector in the base plane plus one half of a perpendicular vector between the base planes.
The four monoclinic hemihedral space groups include
• those with pure reflection at the base of the prism and halfway
• those with glide planes instead of pure reflection planes; the glide is one half of a translation vector in the base plane
• those with both in between each other; in this case an additional translation vector is this glide plus one half of a perpendicular vector between the base planes. | 791 | 3,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-49 | latest | en | 0.799167 |
https://math.answers.com/Q/How_far_is_it_between_each_base | 1,716,115,561,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00159.warc.gz | 350,638,972 | 49,637 | 0
# How far is it between each base?
Updated: 9/27/2023
Wiki User
16y ago
360 feet in high school college and professional Baseball 240 feet in little league and softball
Wiki User
16y ago
Wiki User
16y ago
When running the bases in Major League Baseball, each base is 90 feet from the next base.
Wiki User
14y ago
Each base and home plate are 90 ft. from each other.
Wiki User
14y ago
it depends on the age group...
Wiki User
13y ago
90 feet.
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60 feet.
### How far is it between the bases in Major League baseball?
There are 90 feet between each base in Major League baseball.
### How far is the 1st base line to the 2nd first base line?
every base is 90 feet away from each other
### How far is it from first base to second base in college?
It is 90 feet from each base from high school on up through college and pro baseball.
### Are stars far or close to each other?
Far Very large gaps between
hydrogen bond
90 feet
### How far is second base to first?
The distance between first base and second base is 90 feet.
10 yards
90 feet
### Are fgura and marsascala far from each other?
The distance between Fgura to Marsascala is not far. Going from Fgura to Marsascala will take about nine minutes.
### How do you figure the distance between first base and third base in a straight line?
This involves a bit of geometry. First of all, a baseball diamond including the four bases, first base, second base, third base and home plate is a square. To find the exact distance in a straight line between first base and third base (or home plate to second base, for that matter), simply use half the diamond by using the right triangle formed by two of the base lines (each of which is 90 feet in a straight line between the bases. That leaves the hypotenuse of your right triangle (a straight line between first base and third base) to be determined. Geometry will give you your answer. Square each of the base lines (multiply each by itself), add them together, and then determine the square root of your total. That amount will be your hypotenuse, or the straight line distance between first base over to third base (the answer is the same from home plate to second base). I'm not going to tell you what the exact answer is, but you will find that it's slightly more than 127 feet. | 566 | 2,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-22 | latest | en | 0.954692 |
http://www.lwfree.cn/zidonghua/20190413/31987.html | 1,560,980,281,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00463.warc.gz | 281,975,842 | 6,785 | ## ANSYS上引连铸铜镁合金杆温度场模拟研究
Research on Temperature field simulation of copper magnesium alloy rod in upward continuous casting
Abstract: A temperature field model of the copper magnesium alloy rod is established in this paper, and the temperature field of the copper magnesium alloy rod is simulated by ANSYS.. The influence of pouring temperature, upward speed and top temperature on the temperature field of copper magnesium alloy rod is investigated, and the thermal equilibrium of the mould is calculated according to the principle of thermal equilibrium.. The simulation results show that the cited in the speed and the top rod at the same temperature, 1185℃ when the liquid phase zone depth were greater than 1160℃ when the depth of liquid hole; in the same on approach speed and pouring temperature, the top rod temperature equal to 50 DEG C when the liquid phase hole depth minimum; under the same roof bar temperature and pouring temperature, drawing speed is equal to the 225mm/min liquid hole depth of minimum. Finally, when t was 1160, 50,v was 225mm/min, the minimum of the liquid phase region was 6.69mm. At this point lead continuous casting copper magnesium alloy rod doesn’t easily produce shrinkage hole.
Key words:Temperature field simulation; Upward continuous casting; Pouring temperature; Casting speed
1.前言 1
1.1上引连铸工作原理及机组产品简介 1
1.2上引连铸影响因素分析 1
1.3对上引连铸生产铜镁合金铸造缺陷分析与控制 1
1.3.1. .镁含量波动控制 3
1.3.2. 合金杆断杆分析及对策 3
1.3.3. 合金杆裂纹分析及对策 4
1.4连铸过程中温度场数值模拟 4
1.4.1导热方程 4
1.4.2初试条件与边界条件 5
1.4.3平面温度场的有限差分求解 6
1.5选题的目的和意义 6
1.6课题的研究内容和目标 7
1.6.1研究内容 7
1.6.2研究目标 7
2.模拟 8
2.1 模拟准备 8
2.2 模拟数据 9
2.2.1初始条件 9
2.2.2边界条件 9
2.2.3铜镁合金凝固过程中焓变的确定 13
2.2.4铜镁合金原始参数 14
2.3模拟步骤 14
2.3.1 模型的建立 15
2.3.2 网格划分 16
2.3.3 添加边界条件 18
2.3.4 瞬态时间步长设置 21
2.3.5 求解结果与显示云图 21
3.数学模拟结果及讨论分析 22
3.1结晶器的热平衡计算 22
3.2模拟结果 23
3.3浇注温度对铜镁合金杆温度场的影响 ….40 源`自*六)维[论*文'网www.lwfree.cn
3.4铸顶温度对铜镁合金杆温度场的影响 41 ANSYS上引连铸铜镁合金杆温度场模拟研究:http://www.lwfree.cn/zidonghua/20190413/31987.html
------分隔线---------------------------- | 859 | 2,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-26 | latest | en | 0.408683 |
https://www.geeksforgeeks.org/how-to-begin-with-competitive-programming/?ref=rp | 1,590,655,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00154.warc.gz | 739,223,171 | 29,075 | # How to begin with Competitive Programming?
At the very beginning to competitive programming, barely anyone knows the coding style to be followed. Below is an example to help you understand how problems are crafted in competitive programming.
Let us consider below problem statement as an example.
Problem Statement:
Linear Search: Given an integer array and an element x, find if element is present in array or not. If element is present, then print index of its first occurrence. Else print -1.
Input:
First line contains an integer, the number of test cases ‘T’. Each test case should be an integer. Size of the array ‘N’ in the second line. In the third line, input the integer elements of the array in a single line separated by space. Element X should be inputted in the fourth line, i.e., after entering the elements of array. Repeat the above steps second line onwards for multiple test cases.
Output:
Print the output in a separate line returning the index of the element X. If the element is not present, then print -1.
Constraints:
1 <= T <= 100
1 <= N <= 100
1 <= Arr[i] <= 100
Example Input and Output for Your Program:
```Input:
2
4
1 2 3 4
3
5
10 90 20 30 40
40
Output:
2
4```
Explanation:
```There are 2 test cases (Note 2 at the beginning of input)
Test Case 1: Input: arr[] = {1, 2, 3, 4},
Element to be searched = 3.
Output: 2
Explanation: 3 is present at index 2.
Test Case 2: Input: arr[] = {10, 90, 20, 30, 40},
Element to be searched = 40.
Output: 4
Explanation: 40 is present at index 4.```
## C
`// A Sample C program for beginners with Competitive Programming ` `#include ` ` ` `// This function returns index of element x in arr[] ` `int` `search(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` ` ``int` `i; ` ` ``for` `(i = 0; i < n; i++) ` ` ``{ ` ` ``// Return the index of the element if the element ` ` ``// is found ` ` ``if` `(arr[i] == x) ` ` ``return` `i; ` ` ``} ` ` ` ` ``//return -1 if the element is not found ` ` ``return` `-1; ` `} ` ` ` `int` `main() ` `{ ` ` ``// Note that size of arr[] is considered 100 according to ` ` ``// the constraints mentioned in problem statement. ` ` ``int` `arr[100], x, t, n, i; ` ` ` ` ``// Input the number of test cases you want to run ` ` ``scanf``(``"%d"``, &t); ` ` ` ` ``// One by one run for all input test cases ` ` ``while` `(t--) ` ` ``{ ` ` ``// Input the size of the array ` ` ``scanf``(``"%d"``, &n); ` ` ` ` ``// Input the array ` ` ``for` `(i=0; i
## C++
`// A Sample C++ program for beginners with Competitive Programming ` `#include ` `using` `namespace` `std; ` ` ` `// This function returns index of element x in arr[] ` `int` `search(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ``// Return the index of the element if the element ` ` ``// is found ` ` ``if` `(arr[i] == x) ` ` ``return` `i; ` ` ``} ` ` ` ` ``// return -1 if the element is not found ` ` ``return` `-1; ` `} ` ` ` `int` `main() ` `{ ` ` ``// Note that size of arr[] is considered 100 according to ` ` ``// the constraints mentioned in problem statement. ` ` ``int` `arr[100], x, t, n; ` ` ` ` ``// Input the number of test cases you want to run ` ` ``cin >> t; ` ` ` ` ``// One by one run for all input test cases ` ` ``while` `(t--) ` ` ``{ ` ` ``// Input the size of the array ` ` ``cin >> n; ` ` ` ` ``// Input the array ` ` ``for` `(``int` `i=0; i> arr[i]; ` ` ` ` ``// Input the element to be searched ` ` ``cin >> x; ` ` ` ` ``// Compute and print result ` ` ``cout << search(arr, n, x) << endl; ` ` ``} ` ` ``return` `0; ` `} `
## Python
`# A Sample Python program for beginners with Competitive Programming ` ` ` `# Returns index of x in arr if it is present, ` `# else returns -1 ` `def` `search(arr, x): ` ` ``n ``=` `len``(arr) ` ` ``for` `j ``in` `range``(``0``,n): ` ` ``if` `(x ``=``=` `arr[j]): ` ` ``return` `j ` ` ``return` `-``1` ` ` `# Input number of test cases ` `t ``=` `int``(``raw_input``()) ` ` ` `# One by one run for all input test cases ` `for` `i ``in` `range``(``0``,t): ` ` ` ` ``# Input the size of the array ` ` ``n ``=` `int``(``raw_input``()) ` ` ` ` ``# Input the array ` ` ``arr ``=` `map``(``int``, ``raw_input``().split()) ` ` ` ` ``# Input the element to be searched ` ` ``x ``=` `int``(``raw_input``()) ` ` ` ` ``print``(search(arr, x)) ` ` ` ` ``# The element can also be searched by index method ` ` ``# But you need to handle the exception when element is not found ` ` ``# Uncomment the below line to get that working. ` ` ``# arr.index(x) `
## Java
`// A Sample Java program for beginners with Competitive Programming ` `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` ` `class` `LinearSearch ` `{ ` ` ``// This function returns index of element x in arr[] ` ` ``static` `int` `search(``int` `arr[], ``int` `n, ``int` `x) ` ` ``{ ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``{ ` ` ``// Return the index of the element if the element ` ` ``// is found ` ` ``if` `(arr[i] == x) ` ` ``return` `i; ` ` ``} ` ` ` ` ``// return -1 if the element is not found ` ` ``return` `-``1``; ` ` ``} ` ` ` ` ``public` `static` `void` `main (String[] args) ``throws` `IOException ` ` ``{ ` ` ``// Note that size of arr[] is considered 100 according to ` ` ``// the constraints mentioned in problem statement. ` ` ``int``[] arr = ``new` `int``[``100``]; ` ` ` ` ``// Using BufferedReader class to take input ` ` ``BufferedReader br = ``new` `BufferedReader(``new` `InputStreamReader(System.in)); ` ` ` ` ``int` `t = Integer.parseInt(br.readLine()); ` ` ` ` ``// String Buffer to store answer ` ` ``StringBuffer sb = ``new` `StringBuffer(); ` ` ` ` ``// One by one run for all input test cases ` ` ``while` `(t > ``0``) ` ` ``{ ` ` ``// Input the size of the array ` ` ``int` `n = Integer.parseInt(br.readLine()); ` ` ` ` ``// to read multiple integers line ` ` ``String line = br.readLine(); ` ` ``String[] strs = line.trim().split(``"\\s+"``); ` ` ` ` ``// Input the array ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``arr[i] = Integer.parseInt(strs[i]); ` ` ` ` ``// Input the element to be searched ` ` ``int` `x = Integer.parseInt(br.readLine()); ` ` ` ` ``// Compute and print result ` ` ``sb.append(search(arr, n, x)+``"\n"``); ` ` ` ` ``t--; ` ` ``} ` ` ` ` ``System.out.print(sb); ` ` ``} ` `} `
Common mistakes by beginners
1. Program should not print any extra character. Writing a statement like `printf("Enter value of n")` would cause rejection on any platform.
2. Input and output format specifications must be read carefully. For example, most of the problems expect a new line after every output. So if we don’t write printf(“\n”) or equivalent statement in a loop that runs for all test cases, the program would be rejected.
How to Begin Practice?
You can begin with above problem itself. Try submitting one of the above solutions here. It is recommended solve problems on Practice for cracking any coding interview.
Now you know how to write your first program in Competitive Programming Environment, you can start with School Practice Problems for Competitive Programming or Basic Practice Problems for Competitive Programming.
How to Begin Study?
Top 10 Algorithms and Data Structures for Competitive Programming.
Visit here to decide which category suits you more.
See this for more FAQs for beginners.
How to prepare for ACM – ICPC?
My Personal Notes arrow_drop_up
Improved By : Studentmotivation
Article Tags :
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 2,559 | 8,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | longest | en | 0.789856 |
https://hackage-origin.haskell.org/package/clash-prelude-0.99.2/candidate/docs/Clash-Sized-Vector.html | 1,696,364,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00594.warc.gz | 310,687,444 | 23,714 | clash-prelude-0.99.2: CAES Language for Synchronous Hardware - Prelude library
Clash.Sized.Vector
Description
Synopsis
# Vector data type
data Vec :: Nat -> * -> * where Source #
Fixed size vectors.
• Lists with their length encoded in their type
• Vector elements have an ASCENDING subscript starting from 0 and ending at length - 1.
Constructors
Nil :: Vec 0 a
Bundled Patterns
pattern (:>) :: a -> Vec n a -> Vec (n + 1) a infixr 5 Add an element to the head of a vector.>>> 3:>4:>5:>Nil <3,4,5> >>> let x = 3:>4:>5:>Nil >>> :t x x :: Num a => Vec 3 a Can be used as a pattern:>>> let f (x :> y :> _) = x + y >>> :t f f :: Num a => Vec ((n + 1) + 1) a -> a >>> f (3:>4:>5:>6:>7:>Nil) 7 Also in conjunctions with (:<):>>> let g (a :> b :> (_ :< y :< x)) = a + b + x + y >>> :t g g :: Num a => Vec ((((n + 1) + 1) + 1) + 1) a -> a >>> g (1:>2:>3:>4:>5:>Nil) 12 pattern (:<) :: Vec n a -> a -> Vec (n + 1) a infixl 5 Add an element to the tail of a vector.>>> (3:>4:>5:>Nil) :< 1 <3,4,5,1> >>> let x = (3:>4:>5:>Nil) :< 1 >>> :t x x :: Num a => Vec 4 a Can be used as a pattern:>>> let f (_ :< y :< x) = y + x >>> :t f f :: Num a => Vec ((n + 1) + 1) a -> a >>> f (3:>4:>5:>6:>7:>Nil) 13 Also in conjunctions with (:>):>>> let g (a :> b :> (_ :< y :< x)) = a + b + x + y >>> :t g g :: Num a => Vec ((((n + 1) + 1) + 1) + 1) a -> a >>> g (1:>2:>3:>4:>5:>Nil) 12
Instances
Functor (Vec n) Source # Instance detailsDefined in Clash.Sized.Vector Methodsfmap :: (a -> b) -> Vec n a -> Vec n b #(<$) :: a -> Vec n b -> Vec n a # KnownNat n => Applicative (Vec n) Source # Instance detailsDefined in Clash.Sized.Vector Methodspure :: a -> Vec n a #(<*>) :: Vec n (a -> b) -> Vec n a -> Vec n b #liftA2 :: (a -> b -> c) -> Vec n a -> Vec n b -> Vec n c #(*>) :: Vec n a -> Vec n b -> Vec n b #(<*) :: Vec n a -> Vec n b -> Vec n a # (KnownNat n, 1 <= n) => Foldable (Vec n) Source # Instance detailsDefined in Clash.Sized.Vector Methodsfold :: Monoid m => Vec n m -> m #foldMap :: Monoid m => (a -> m) -> Vec n a -> m #foldr :: (a -> b -> b) -> b -> Vec n a -> b #foldr' :: (a -> b -> b) -> b -> Vec n a -> b #foldl :: (b -> a -> b) -> b -> Vec n a -> b #foldl' :: (b -> a -> b) -> b -> Vec n a -> b #foldr1 :: (a -> a -> a) -> Vec n a -> a #foldl1 :: (a -> a -> a) -> Vec n a -> a #toList :: Vec n a -> [a] #null :: Vec n a -> Bool #length :: Vec n a -> Int #elem :: Eq a => a -> Vec n a -> Bool #maximum :: Ord a => Vec n a -> a #minimum :: Ord a => Vec n a -> a #sum :: Num a => Vec n a -> a #product :: Num a => Vec n a -> a # (KnownNat n, 1 <= n) => Traversable (Vec n) Source # Instance detailsDefined in Clash.Sized.Vector Methodstraverse :: Applicative f => (a -> f b) -> Vec n a -> f (Vec n b) #sequenceA :: Applicative f => Vec n (f a) -> f (Vec n a) #mapM :: Monad m => (a -> m b) -> Vec n a -> m (Vec n b) #sequence :: Monad m => Vec n (m a) -> m (Vec n a) # (KnownNat n, Eq a) => Eq (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methods(==) :: Vec n a -> Vec n a -> Bool #(/=) :: Vec n a -> Vec n a -> Bool # (KnownNat n, Ord a) => Ord (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodscompare :: Vec n a -> Vec n a -> Ordering #(<) :: Vec n a -> Vec n a -> Bool #(<=) :: Vec n a -> Vec n a -> Bool #(>) :: Vec n a -> Vec n a -> Bool #(>=) :: Vec n a -> Vec n a -> Bool #max :: Vec n a -> Vec n a -> Vec n a #min :: Vec n a -> Vec n a -> Vec n a # Show a => Show (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector MethodsshowsPrec :: Int -> Vec n a -> ShowS #show :: Vec n a -> String #showList :: [Vec n a] -> ShowS # Lift a => Lift (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodslift :: Vec n a -> Q Exp # NFData a => NFData (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodsrnf :: Vec n a -> () # ShowX a => ShowX (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector MethodsshowsPrecX :: Int -> Vec n a -> ShowS Source #showX :: Vec n a -> String Source #showListX :: [Vec n a] -> ShowS Source # KnownNat n => Ixed (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodsix :: Index (Vec n a) -> Traversal' (Vec n a) (IxValue (Vec n a)) (Default a, KnownNat n) => Default (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodsdef :: Vec n a (KnownNat n, Arbitrary a) => Arbitrary (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodsarbitrary :: Gen (Vec n a)shrink :: Vec n a -> [Vec n a] CoArbitrary a => CoArbitrary (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Methodscoarbitrary :: Vec n a -> Gen b -> Gen b (KnownNat n, KnownNat (BitSize a), BitPack a) => BitPack (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector Associated Typestype BitSize (Vec n a) :: Nat Source # Methodspack :: Vec n a -> BitVector (BitSize (Vec n a)) Source #unpack :: BitVector (BitSize (Vec n a)) -> Vec n a Source # KnownNat n => Bundle (Vec n a) Source # Instance detailsDefined in Clash.Signal.Bundle Associated Typestype Unbundled domain (Vec n a) = (res :: *) Source # Methodsbundle :: Unbundled domain (Vec n a) -> Signal domain (Vec n a) Source #unbundle :: Signal domain (Vec n a) -> Unbundled domain (Vec n a) Source # (LockStep en a, KnownNat n) => LockStep (Vec n en) (Vec n a) Source # Instance detailsDefined in Clash.Prelude.DataFlow MethodslockStep :: DataFlow dom (Vec n en) Bool (Vec n a) (Vec n a) Source #stepLock :: DataFlow dom Bool (Vec n en) (Vec n a) (Vec n a) Source # type Unbundled t (Vec n a) Source # Instance detailsDefined in Clash.Signal.Bundle type Unbundled t (Vec n a) = Vec n (Signal t a) type Index (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector type Index (Vec n a) = Index n type IxValue (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector type IxValue (Vec n a) = a type BitSize (Vec n a) Source # Instance detailsDefined in Clash.Sized.Vector type BitSize (Vec n a) = n * BitSize a # Accessors ## Length information length :: KnownNat n => Vec n a -> Int Source # The length of a Vector as an Int value. >>> length (6 :> 7 :> 8 :> Nil) 3 lengthS :: KnownNat n => Vec n a -> SNat n Source # Length of a Vector as an SNat value ## Indexing (!!) :: (KnownNat n, Enum i) => Vec n a -> i -> a Source # "xs !! n" returns the n'th element of xs. NB: vector elements have an ASCENDING subscript starting from 0 and ending at length - 1. >>> (1:>2:>3:>4:>5:>Nil) !! 4 5 >>> (1:>2:>3:>4:>5:>Nil) !! (length (1:>2:>3:>4:>5:>Nil) - 1) 5 >>> (1:>2:>3:>4:>5:>Nil) !! 1 2 >>> (1:>2:>3:>4:>5:>Nil) !! 14 *** Exception: Clash.Sized.Vector.(!!): index 14 is larger than maximum index 4 ... head :: Vec (n + 1) a -> a Source # Extract the first element of a vector >>> head (1:>2:>3:>Nil) 1 >>> head Nil <interactive>:... • Couldn't match type ‘1’ with ‘0’ Expected type: Vec (0 + 1) a Actual type: Vec 0 a • In the first argument of ‘head’, namely ‘Nil’ In the expression: head Nil In an equation for ‘it’: it = head Nil last :: Vec (n + 1) a -> a Source # Extract the last element of a vector >>> last (1:>2:>3:>Nil) 3 >>> last Nil <interactive>:... • Couldn't match type ‘1’ with ‘0’ Expected type: Vec (0 + 1) a Actual type: Vec 0 a • In the first argument of ‘last’, namely ‘Nil’ In the expression: last Nil In an equation for ‘it’: it = last Nil at :: SNat m -> Vec (m + (n + 1)) a -> a Source # "at n xs" returns n'th element of xs NB: vector elements have an ASCENDING subscript starting from 0 and ending at length - 1. >>> at (SNat :: SNat 1) (1:>2:>3:>4:>5:>Nil) 2 >>> at d1 (1:>2:>3:>4:>5:>Nil) 2 indices :: KnownNat n => SNat n -> Vec n (Index n) Source # Generate a vector of indices. >>> indices d4 <0,1,2,3> indicesI :: KnownNat n => Vec n (Index n) Source # Generate a vector of indices, where the length of the vector is determined by the context. >>> indicesI :: Vec 4 (Index 4) <0,1,2,3> findIndex :: KnownNat n => (a -> Bool) -> Vec n a -> Maybe (Index n) Source # "findIndex p xs" returns the index of the first element of xs satisfying the predicate p, or Nothing if there is no such element. >>> findIndex (> 3) (1:>3:>2:>4:>3:>5:>6:>Nil) Just 3 >>> findIndex (> 8) (1:>3:>2:>4:>3:>5:>6:>Nil) Nothing elemIndex :: (KnownNat n, Eq a) => a -> Vec n a -> Maybe (Index n) Source # "elemIndex a xs" returns the index of the first element which is equal (by ==) to the query element a, or Nothing if there is no such element. >>> elemIndex 3 (1:>3:>2:>4:>3:>5:>6:>Nil) Just 1 >>> elemIndex 8 (1:>3:>2:>4:>3:>5:>6:>Nil) Nothing ## Extracting sub-vectors (slicing) tail :: Vec (n + 1) a -> Vec n a Source # Extract the elements after the head of a vector >>> tail (1:>2:>3:>Nil) <2,3> >>> tail Nil <interactive>:... • Couldn't match type ‘1’ with ‘0’ Expected type: Vec (0 + 1) a Actual type: Vec 0 a • In the first argument of ‘tail’, namely ‘Nil’ In the expression: tail Nil In an equation for ‘it’: it = tail Nil init :: Vec (n + 1) a -> Vec n a Source # Extract all the elements of a vector except the last element >>> init (1:>2:>3:>Nil) <1,2> >>> init Nil <interactive>:... • Couldn't match type ‘1’ with ‘0’ Expected type: Vec (0 + 1) a Actual type: Vec 0 a • In the first argument of ‘init’, namely ‘Nil’ In the expression: init Nil In an equation for ‘it’: it = init Nil take :: SNat m -> Vec (m + n) a -> Vec m a Source # "take n xs" returns the n-length prefix of xs. >>> take (SNat :: SNat 3) (1:>2:>3:>4:>5:>Nil) <1,2,3> >>> take d3 (1:>2:>3:>4:>5:>Nil) <1,2,3> >>> take d0 (1:>2:>Nil) <> >>> take d4 (1:>2:>Nil) <interactive>:... • Couldn't match type ‘4 + n0’ with ‘2’ Expected type: Vec (4 + n0) a Actual type: Vec (1 + 1) a The type variable ‘n0’ is ambiguous • In the second argument of ‘take’, namely ‘(1 :> 2 :> Nil)’ In the expression: take d4 (1 :> 2 :> Nil) In an equation for ‘it’: it = take d4 (1 :> 2 :> Nil) takeI :: KnownNat m => Vec (m + n) a -> Vec m a Source # "takeI xs" returns the prefix of xs as demanded by the context. >>> takeI (1:>2:>3:>4:>5:>Nil) :: Vec 2 Int <1,2> drop :: SNat m -> Vec (m + n) a -> Vec n a Source # "drop n xs" returns the suffix of xs after the first n elements. >>> drop (SNat :: SNat 3) (1:>2:>3:>4:>5:>Nil) <4,5> >>> drop d3 (1:>2:>3:>4:>5:>Nil) <4,5> >>> drop d0 (1:>2:>Nil) <1,2> >>> drop d4 (1:>2:>Nil) <interactive>:... • Couldn't match expected type ‘2’ with actual type ‘4 + n0’ The type variable ‘n0’ is ambiguous • In the first argument of ‘print’, namely ‘it’ In a stmt of an interactive GHCi command: print it dropI :: KnownNat m => Vec (m + n) a -> Vec n a Source # "dropI xs" returns the suffix of xs as demanded by the context. >>> dropI (1:>2:>3:>4:>5:>Nil) :: Vec 2 Int <4,5> select :: CmpNat (i + s) (s * n) ~ GT => SNat f -> SNat s -> SNat n -> Vec (f + i) a -> Vec n a Source # "select f s n xs" selects n elements with step-size s and offset f from xs. >>> select (SNat :: SNat 1) (SNat :: SNat 2) (SNat :: SNat 3) (1:>2:>3:>4:>5:>6:>7:>8:>Nil) <2,4,6> >>> select d1 d2 d3 (1:>2:>3:>4:>5:>6:>7:>8:>Nil) <2,4,6> selectI :: (CmpNat (i + s) (s * n) ~ GT, KnownNat n) => SNat f -> SNat s -> Vec (f + i) a -> Vec n a Source # "selectI f s xs" selects as many elements as demanded by the context with step-size s and offset f from xs. >>> selectI d1 d2 (1:>2:>3:>4:>5:>6:>7:>8:>Nil) :: Vec 2 Int <2,4> ### Splitting splitAt :: SNat m -> Vec (m + n) a -> (Vec m a, Vec n a) Source # Split a vector into two vectors at the given point. >>> splitAt (SNat :: SNat 3) (1:>2:>3:>7:>8:>Nil) (<1,2,3>,<7,8>) >>> splitAt d3 (1:>2:>3:>7:>8:>Nil) (<1,2,3>,<7,8>) splitAtI :: KnownNat m => Vec (m + n) a -> (Vec m a, Vec n a) Source # Split a vector into two vectors where the length of the two is determined by the context. >>> splitAtI (1:>2:>3:>7:>8:>Nil) :: (Vec 2 Int, Vec 3 Int) (<1,2>,<3,7,8>) unconcat :: KnownNat n => SNat m -> Vec (n * m) a -> Vec n (Vec m a) Source # Split a vector of (n * m) elements into a vector of "vectors of length m", where the length m is given. >>> unconcat d4 (1:>2:>3:>4:>5:>6:>7:>8:>9:>10:>11:>12:>Nil) <<1,2,3,4>,<5,6,7,8>,<9,10,11,12>> unconcatI :: (KnownNat n, KnownNat m) => Vec (n * m) a -> Vec n (Vec m a) Source # Split a vector of (n * m) elements into a vector of "vectors of length m", where the length m is determined by the context. >>> unconcatI (1:>2:>3:>4:>5:>6:>7:>8:>9:>10:>11:>12:>Nil) :: Vec 2 (Vec 6 Int) <<1,2,3,4,5,6>,<7,8,9,10,11,12>> # Construction ## Initialisation singleton :: a -> Vec 1 a Source # Create a vector of one element >>> singleton 5 <5> replicate :: SNat n -> a -> Vec n a Source # "replicate n a" returns a vector that has n copies of a. >>> replicate (SNat :: SNat 3) 6 <6,6,6> >>> replicate d3 6 <6,6,6> repeat :: KnownNat n => a -> Vec n a Source # "repeat a" creates a vector with as many copies of a as demanded by the context. >>> repeat 6 :: Vec 5 Int <6,6,6,6,6> iterate :: SNat n -> (a -> a) -> a -> Vec n a Source # "iterate n f x" returns a vector starting with x followed by n repeated applications of f to x. iterate (SNat :: SNat 4) f x == (x :> f x :> f (f x) :> f (f (f x)) :> Nil) iterate d4 f x == (x :> f x :> f (f x) :> f (f (f x)) :> Nil) >>> iterate d4 (+1) 1 <1,2,3,4> "interate n f z" corresponds to the following circuit layout: iterateI :: KnownNat n => (a -> a) -> a -> Vec n a Source # "iterate f x" returns a vector starting with x followed by n repeated applications of f to x, where n is determined by the context. iterateI f x :: Vec 3 a == (x :> f x :> f (f x) :> Nil) >>> iterateI (+1) 1 :: Vec 3 Int <1,2,3> "interateI f z" corresponds to the following circuit layout: generate :: SNat n -> (a -> a) -> a -> Vec n a Source # "generate n f x" returns a vector with n repeated applications of f to x. generate (SNat :: SNat 4) f x == (f x :> f (f x) :> f (f (f x)) :> f (f (f (f x))) :> Nil) generate d4 f x == (f x :> f (f x) :> f (f (f x)) :> f (f (f (f x))) :> Nil) >>> generate d4 (+1) 1 <2,3,4,5> "generate n f z" corresponds to the following circuit layout: generateI :: KnownNat n => (a -> a) -> a -> Vec n a Source # "generateI f x" returns a vector with n repeated applications of f to x, where n is determined by the context. generateI f x :: Vec 3 a == (f x :> f (f x) :> f (f (f x)) :> Nil) >>> generateI (+1) 1 :: Vec 3 Int <2,3,4> "generateI f z" corresponds to the following circuit layout: ### Initialisation from a list listToVecTH :: Lift a => [a] -> ExpQ Source # Create a vector literal from a list literal. $(listToVecTH [1::Signed 8,2,3,4,5]) == (8:>2:>3:>4:>5:>Nil) :: Vec 5 (Signed 8)
>>> [1 :: Signed 8,2,3,4,5]
[1,2,3,4,5]
>>> \$(listToVecTH [1::Signed 8,2,3,4,5])
<1,2,3,4,5>
## Concatenation
(++) :: Vec n a -> Vec m a -> Vec (n + m) a infixr 5 Source #
Append two vectors.
>>> (1:>2:>3:>Nil) ++ (7:>8:>Nil)
<1,2,3,7,8>
(+>>) :: KnownNat n => a -> Vec n a -> Vec n a infixr 4 Source #
Add an element to the head of a vector, and extract all but the last element.
>>> 1 +>> (3:>4:>5:>Nil)
<1,3,4>
>>> 1 +>> Nil
<>
(<<+) :: Vec n a -> a -> Vec n a infixl 4 Source #
Add an element to the tail of a vector, and extract all but the first element.
>>> (3:>4:>5:>Nil) <<+ 1
<4,5,1>
>>> Nil <<+ 1
<>
concat :: Vec n (Vec m a) -> Vec (n * m) a Source #
Concatenate a vector of vectors.
>>> concat ((1:>2:>3:>Nil) :> (4:>5:>6:>Nil) :> (7:>8:>9:>Nil) :> (10:>11:>12:>Nil) :> Nil)
<1,2,3,4,5,6,7,8,9,10,11,12>
Arguments
:: KnownNat n => Vec n a The old vector -> Vec m a The elements to shift in at the head -> (Vec n a, Vec m a) (The new vector, shifted out elements)
Shift in elements to the head of a vector, bumping out elements at the tail. The result is a tuple containing:
• The new vector
• The shifted out elements
>>> shiftInAt0 (1 :> 2 :> 3 :> 4 :> Nil) ((-1) :> 0 :> Nil)
(<-1,0,1,2>,<3,4>)
>>> shiftInAt0 (1 :> Nil) ((-1) :> 0 :> Nil)
(<-1>,<0,1>)
Arguments
:: KnownNat m => Vec n a The old vector -> Vec m a The elements to shift in at the tail -> (Vec n a, Vec m a) (The new vector, shifted out elements)
Shift in element to the tail of a vector, bumping out elements at the head. The result is a tuple containing:
• The new vector
• The shifted out elements
>>> shiftInAtN (1 :> 2 :> 3 :> 4 :> Nil) (5 :> 6 :> Nil)
(<3,4,5,6>,<1,2>)
>>> shiftInAtN (1 :> Nil) (2 :> 3 :> Nil)
(<3>,<1,2>)
Arguments
:: (Default a, KnownNat m) => SNat m m, the number of elements to shift out -> Vec (m + n) a The old vector -> (Vec (m + n) a, Vec m a) (The new vector, shifted out elements)
Shift m elements out from the head of a vector, filling up the tail with Default values. The result is a tuple containing:
• The new vector
• The shifted out values
>>> shiftOutFrom0 d2 ((1 :> 2 :> 3 :> 4 :> 5 :> Nil) :: Vec 5 Integer)
(<3,4,5,0,0>,<1,2>)
Arguments
:: (Default a, KnownNat n) => SNat m m, the number of elements to shift out -> Vec (m + n) a The old vector -> (Vec (m + n) a, Vec m a) (The new vector, shifted out elements)
Shift m elements out from the tail of a vector, filling up the head with Default values. The result is a tuple containing:
• The new vector
• The shifted out values
>>> shiftOutFromN d2 ((1 :> 2 :> 3 :> 4 :> 5 :> Nil) :: Vec 5 Integer)
(<0,0,1,2,3>,<4,5>)
merge :: KnownNat n => Vec n a -> Vec n a -> Vec (2 * n) a Source #
Merge two vectors, alternating their elements, i.e.,
>>> merge (1 :> 2 :> 3 :> 4 :> Nil) (5 :> 6 :> 7 :> 8 :> Nil)
<1,5,2,6,3,7,4,8>
# Modifying vectors
replace :: (KnownNat n, Enum i) => i -> a -> Vec n a -> Vec n a Source #
"replace n a xs" returns the vector xs where the n'th element is replaced by a.
NB: vector elements have an ASCENDING subscript starting from 0 and ending at length - 1.
>>> replace 3 7 (1:>2:>3:>4:>5:>Nil)
<1,2,3,7,5>
>>> replace 0 7 (1:>2:>3:>4:>5:>Nil)
<7,2,3,4,5>
>>> replace 9 7 (1:>2:>3:>4:>5:>Nil)
<1,2,3,4,*** Exception: Clash.Sized.Vector.replace: index 9 is larger than maximum index 4
...
## Permutations
Arguments
:: (Enum i, KnownNat n, KnownNat m) => (a -> a -> a) Combination function, f -> Vec n a Default values, def -> Vec m i Index mapping, is -> Vec (m + k) a Vector to be permuted, xs -> Vec n a
Forward permutation specified by an index mapping, ix. The result vector is initialised by the given defaults, def, and an further values that are permuted into the result are added to the current value using the given combination function, f.
The combination function must be associative and commutative.
Arguments
:: (Enum i, KnownNat n) => Vec n a Source vector, xs -> Vec m i Index mapping, is -> Vec m a
Backwards permutation specified by an index mapping, is, from the destination vector specifying which element of the source vector xs to read.
"backpermute xs is" is equivalent to "map (xs !!) is".
For example:
>>> let input = 1:>9:>6:>4:>4:>2:>0:>1:>2:>Nil
>>> let from = 1:>3:>7:>2:>5:>3:>Nil
>>> backpermute input from
<9,4,1,6,2,4>
Arguments
:: (Enum i, KnownNat n, KnownNat m) => Vec n a Default values, def -> Vec m i Index mapping, is -> Vec (m + k) a Vector to be scattered, xs -> Vec n a
Copy elements from the source vector, xs, to the destination vector according to an index mapping is. This is a forward permute operation where a to vector encodes an input to output index mapping. Output elements for indices that are not mapped assume the value in the default vector def.
For example:
>>> let defVec = 0:>0:>0:>0:>0:>0:>0:>0:>0:>Nil
>>> let to = 1:>3:>7:>2:>5:>8:>Nil
>>> let input = 1:>9:>6:>4:>4:>2:>5:>Nil
>>> scatter defVec to input
<0,1,4,9,0,4,0,6,2>
NB: If the same index appears in the index mapping more than once, the latest mapping is chosen.
Arguments
:: (Enum i, KnownNat n) => Vec n a Source vector, xs -> Vec m i Index mapping, is -> Vec m a
Backwards permutation specified by an index mapping, is, from the destination vector specifying which element of the source vector xs to read.
"gather xs is" is equivalent to "map (xs !!) is".
For example:
>>> let input = 1:>9:>6:>4:>4:>2:>0:>1:>2:>Nil
>>> let from = 1:>3:>7:>2:>5:>3:>Nil
>>> gather input from
<9,4,1,6,2,4>
### Specialised permutations
reverse :: Vec n a -> Vec n a Source #
The elements in a vector in reverse order.
>>> reverse (1:>2:>3:>4:>Nil)
<4,3,2,1>
transpose :: KnownNat n => Vec m (Vec n a) -> Vec n (Vec m a) Source #
Transpose a matrix: go from row-major to column-major
>>> let xss = (1:>2:>Nil):>(3:>4:>Nil):>(5:>6:>Nil):>Nil
>>> xss
<<1,2>,<3,4>,<5,6>>
>>> transpose xss
<<1,3,5>,<2,4,6>>
Arguments
:: (KnownNat n, KnownNat d) => SNat d Interleave step, d -> Vec (n * d) a -> Vec (d * n) a
"interleave d xs" creates a vector:
<x_0,x_d,x_(2d),...,x_1,x_(d+1),x_(2d+1),...,x_(d-1),x_(2d-1),x_(3d-1)>
>>> let xs = 1 :> 2 :> 3 :> 4 :> 5 :> 6 :> 7 :> 8 :> 9 :> Nil
>>> interleave d3 xs
<1,4,7,2,5,8,3,6,9>
rotateLeft :: (Enum i, KnownNat n) => Vec n a -> i -> Vec n a Source #
Dynamically rotate a Vector to the left:
>>> let xs = 1 :> 2 :> 3 :> 4 :> Nil
>>> rotateLeft xs 1
<2,3,4,1>
>>> rotateLeft xs 2
<3,4,1,2>
>>> rotateLeft xs (-1)
<4,1,2,3>
NB: use rotateLeftS if you want to rotate left by a static amount.
rotateRight :: (Enum i, KnownNat n) => Vec n a -> i -> Vec n a Source #
Dynamically rotate a Vector to the right:
>>> let xs = 1 :> 2 :> 3 :> 4 :> Nil
>>> rotateRight xs 1
<4,1,2,3>
>>> rotateRight xs 2
<3,4,1,2>
>>> rotateRight xs (-1)
<2,3,4,1>
NB: use rotateRightS if you want to rotate right by a static amount.
rotateLeftS :: KnownNat n => Vec n a -> SNat d -> Vec n a Source #
Statically rotate a Vector to the left:
>>> let xs = 1 :> 2 :> 3 :> 4 :> Nil
>>> rotateLeftS xs d1
<2,3,4,1>
NB: use rotateLeft if you want to rotate left by a dynamic amount.
rotateRightS :: KnownNat n => Vec n a -> SNat d -> Vec n a Source #
Statically rotate a Vector to the right:
>>> let xs = 1 :> 2 :> 3 :> 4 :> Nil
>>> rotateRightS xs d1
<4,1,2,3>
NB: use rotateRight if you want to rotate right by a dynamic amount.
# Element-wise operations
## Mapping
map :: (a -> b) -> Vec n a -> Vec n b Source #
"map f xs" is the vector obtained by applying f to each element of xs, i.e.,
map f (x1 :> x2 :> ... :> xn :> Nil) == (f x1 :> f x2 :> ... :> f xn :> Nil)
and corresponds to the following circuit layout:
imap :: forall n a b. KnownNat n => (Index n -> a -> b) -> Vec n a -> Vec n b Source #
Apply a function of every element of a vector and its index.
>>> :t imap (+) (2 :> 2 :> 2 :> 2 :> Nil)
imap (+) (2 :> 2 :> 2 :> 2 :> Nil) :: Vec 4 (Index 4)
>>> imap (+) (2 :> 2 :> 2 :> 2 :> Nil)
<2,3,*** Exception: Clash.Sized.Index: result 4 is out of bounds: [0..3]
...
>>> imap (\i a -> fromIntegral i + a) (2 :> 2 :> 2 :> 2 :> Nil) :: Vec 4 (Unsigned 8)
<2,3,4,5>
"imap f xs" corresponds to the following circuit layout:
smap :: forall k a b. KnownNat k => (forall l. SNat l -> a -> b) -> Vec k a -> Vec k b Source #
Apply a function to every element of a vector and the element's position (as an SNat value) in the vector.
>>> let rotateMatrix = smap (flip rotateRightS)
>>> let xss = (1:>2:>3:>Nil):>(1:>2:>3:>Nil):>(1:>2:>3:>Nil):>Nil
>>> xss
<<1,2,3>,<1,2,3>,<1,2,3>>
>>> rotateMatrix xss
<<1,2,3>,<3,1,2>,<2,3,1>>
## Zipping
zipWith :: (a -> b -> c) -> Vec n a -> Vec n b -> Vec n c Source #
zipWith generalises zip by zipping with the function given as the first argument, instead of a tupling function. For example, "zipWith (+)" applied to two vectors produces the vector of corresponding sums.
zipWith f (x1 :> x2 :> ... xn :> Nil) (y1 :> y2 :> ... :> yn :> Nil) == (f x1 y1 :> f x2 y2 :> ... :> f xn yn :> Nil)
"zipWith f xs ys" corresponds to the following circuit layout:
NB: zipWith is strict in its second argument, and lazy in its third. This matters when zipWith is used in a recursive setting. See lazyV for more information.
zipWith3 :: (a -> b -> c -> d) -> Vec n a -> Vec n b -> Vec n c -> Vec n d Source #
zipWith3 generalises zip3 by zipping with the function given as the first argument, instead of a tupling function.
zipWith3 f (x1 :> x2 :> ... xn :> Nil) (y1 :> y2 :> ... :> yn :> Nil) (z1 :> z2 :> ... :> zn :> Nil) == (f x1 y1 z1 :> f x2 y2 z2 :> ... :> f xn yn zn :> Nil)
"zipWith3 f xs ys zs" corresponds to the following circuit layout:
NB: zipWith3 is strict in its second argument, and lazy in its third and fourth. This matters when zipWith3 is used in a recursive setting. See lazyV for more information.
zip :: Vec n a -> Vec n b -> Vec n (a, b) Source #
zip takes two vectors and returns a vector of corresponding pairs.
>>> zip (1:>2:>3:>4:>Nil) (4:>3:>2:>1:>Nil)
<(1,4),(2,3),(3,2),(4,1)>
zip3 :: Vec n a -> Vec n b -> Vec n c -> Vec n (a, b, c) Source #
zip takes three vectors and returns a vector of corresponding triplets.
>>> zip3 (1:>2:>3:>4:>Nil) (4:>3:>2:>1:>Nil) (5:>6:>7:>8:>Nil)
<(1,4,5),(2,3,6),(3,2,7),(4,1,8)>
izipWith :: KnownNat n => (Index n -> a -> b -> c) -> Vec n a -> Vec n b -> Vec n c Source #
Zip two vectors with a functions that also takes the elements' indices.
>>> izipWith (\i a b -> i + a + b) (2 :> 2 :> Nil) (3 :> 3:> Nil)
<*** Exception: Clash.Sized.Index: result 3 is out of bounds: [0..1]
...
>>> izipWith (\i a b -> fromIntegral i + a + b) (2 :> 2 :> Nil) (3 :> 3 :> Nil) :: Vec 2 (Unsigned 8)
<5,6>
"imap f xs" corresponds to the following circuit layout:
NB: izipWith is strict in its second argument, and lazy in its third. This matters when izipWith is used in a recursive setting. See lazyV for more information.
## Unzipping
unzip :: Vec n (a, b) -> (Vec n a, Vec n b) Source #
unzip transforms a vector of pairs into a vector of first components and a vector of second components.
>>> unzip ((1,4):>(2,3):>(3,2):>(4,1):>Nil)
(<1,2,3,4>,<4,3,2,1>)
unzip3 :: Vec n (a, b, c) -> (Vec n a, Vec n b, Vec n c) Source #
unzip3 transforms a vector of triplets into a vector of first components, a vector of second components, and a vector of third components.
>>> unzip3 ((1,4,5):>(2,3,6):>(3,2,7):>(4,1,8):>Nil)
(<1,2,3,4>,<4,3,2,1>,<5,6,7,8>)
# Folding
foldr :: (a -> b -> b) -> b -> Vec n a -> b Source #
foldr, applied to a binary operator, a starting value (typically the right-identity of the operator), and a vector, reduces the vector using the binary operator, from right to left:
foldr f z (x1 :> ... :> xn1 :> xn :> Nil) == x1 f (... (xn1 f (xn f z))...)
foldr r z Nil == z
>>> foldr (/) 1 (5 :> 4 :> 3 :> 2 :> Nil)
1.875
"foldr f z xs" corresponds to the following circuit layout:
NB: "foldr f z xs" produces a linear structure, which has a depth, or delay, of O(length xs). Use fold if your binary operator f is associative, as "fold f xs" produces a structure with a depth of O(log_2(length xs)).
foldl :: (b -> a -> b) -> b -> Vec n a -> b Source #
foldl, applied to a binary operator, a starting value (typically the left-identity of the operator), and a vector, reduces the vector using the binary operator, from left to right:
foldl f z (x1 :> x2 :> ... :> xn :> Nil) == (...((z f x1) f x2) f...) f xn
foldl f z Nil == z
>>> foldl (/) 1 (5 :> 4 :> 3 :> 2 :> Nil)
8.333333333333333e-3
"foldl f z xs" corresponds to the following circuit layout:
NB: "foldl f z xs" produces a linear structure, which has a depth, or delay, of O(length xs). Use fold if your binary operator f is associative, as "fold f xs" produces a structure with a depth of O(log_2(length xs)).
foldr1 :: (a -> a -> a) -> Vec (n + 1) a -> a Source #
foldr1 is a variant of foldr that has no starting value argument, and thus must be applied to non-empty vectors.
foldr1 f (x1 :> ... :> xn2 :> xn1 :> xn :> Nil) == x1 f (... (xn2 f (xn1 f xn))...)
foldr1 f (x1 :> Nil) == x1
foldr1 f Nil == TYPE ERROR
>>> foldr1 (/) (5 :> 4 :> 3 :> 2 :> 1 :> Nil)
1.875
"foldr1 f xs" corresponds to the following circuit layout:
NB: "foldr1 f z xs" produces a linear structure, which has a depth, or delay, of O(length xs). Use fold if your binary operator f is associative, as "fold f xs" produces a structure with a depth of O(log_2(length xs)).
foldl1 :: (a -> a -> a) -> Vec (n + 1) a -> a Source #
foldl1 is a variant of foldl that has no starting value argument, and thus must be applied to non-empty vectors.
foldl1 f (x1 :> x2 :> x3 :> ... :> xn :> Nil) == (...((x1 f x2) f x3) f...) f xn
foldl1 f (x1 :> Nil) == x1
foldl1 f Nil == TYPE ERROR
>>> foldl1 (/) (1 :> 5 :> 4 :> 3 :> 2 :> Nil)
8.333333333333333e-3
"foldl1 f xs" corresponds to the following circuit layout:
NB: "foldl1 f z xs" produces a linear structure, which has a depth, or delay, of O(length xs). Use fold if your binary operator f is associative, as "fold f xs" produces a structure with a depth of O(log_2(length xs)).
fold :: (a -> a -> a) -> Vec (n + 1) a -> a Source #
fold is a variant of foldr1 and foldl1, but instead of reducing from right to left, or left to right, it reduces a vector using a tree-like structure. The depth, or delay, of the structure produced by "fold f xs", is hence O(log_2(length xs)), and not O(length xs).
NB: The binary operator "f" in "fold f xs" must be associative.
fold f (x1 :> x2 :> ... :> xn1 :> xn :> Nil) == ((x1 f x2) f ...) f (... f (xn1 f xn))
fold f (x1 :> Nil) == x1
fold f Nil == TYPE ERROR
>>> fold (+) (5 :> 4 :> 3 :> 2 :> 1 :> Nil)
15
"fold f xs" corresponds to the following circuit layout:
ifoldr :: KnownNat n => (Index n -> a -> b -> b) -> b -> Vec n a -> b Source #
Right fold (function applied to each element and its index)
>>> let findLeftmost x xs = ifoldr (\i a b -> if a == x then Just i else b) Nothing xs
>>> findLeftmost 3 (1:>3:>2:>4:>3:>5:>6:>Nil)
Just 1
>>> findLeftmost 8 (1:>3:>2:>4:>3:>5:>6:>Nil)
Nothing
"ifoldr f z xs" corresponds to the following circuit layout:
ifoldl :: KnownNat n => (a -> Index n -> b -> a) -> a -> Vec n b -> a Source #
Left fold (function applied to each element and its index)
>>> let findRightmost x xs = ifoldl (\a i b -> if b == x then Just i else a) Nothing xs
>>> findRightmost 3 (1:>3:>2:>4:>3:>5:>6:>Nil)
Just 4
>>> findRightmost 8 (1:>3:>2:>4:>3:>5:>6:>Nil)
Nothing
"ifoldl f z xs" corresponds to the following circuit layout:
## Specialised folds
Arguments
:: KnownNat k => Proxy (p :: TyFun Nat * -> *) The motive -> (forall l. SNat l -> a -> (p @@ l) -> p @@ (l + 1)) Function to fold.NB: The SNat l is not the index (see (!!)) to the element a. SNat l is the number of elements that occur to the right of a. -> (p @@ 0) Initial element -> Vec k a Vector to fold over -> p @@ k
A dependently typed fold.
Using lists, we can define append (a.k.a. Data.List.++) in terms of Data.List.foldr:
>>> import qualified Data.List
>>> let append xs ys = Data.List.foldr (:) ys xs
>>> append [1,2] [3,4]
[1,2,3,4]
However, when we try to do the same for Vec, by defining append' in terms of Clash.Sized.Vector.foldr:
append' xs ys = foldr (:>) ys xs
we get a type error:
>>> let append' xs ys = foldr (:>) ys xs
<interactive>:...
• Occurs check: cannot construct the infinite type: ... ~ ... + 1
Expected type: a -> Vec ... a -> Vec ... a
Actual type: a -> Vec ... a -> Vec (... + 1) a
• In the first argument of ‘foldr’, namely ‘(:>)’
In the expression: foldr (:>) ys xs
In an equation for ‘append'’: append' xs ys = foldr (:>) ys xs
• Relevant bindings include
ys :: Vec ... a (bound at ...)
append' :: Vec n a -> Vec ... a -> Vec ... a
(bound at ...)
The reason is that the type of foldr is:
>>> :t foldr
foldr :: (a -> b -> b) -> b -> Vec n a -> b
While the type of (:>) is:
>>> :t (:>)
(:>) :: a -> Vec n a -> Vec (n + 1) a
We thus need a fold function that can handle the growing vector type: dfold. Compared to foldr, dfold takes an extra parameter, called the motive, that allows the folded function to have an argument and result type that depends on the current length of the vector. Using dfold, we can now correctly define append':
import Data.Singletons.Prelude
import Data.Proxy
data Append (m :: Nat) (a :: *) (f :: TyFun Nat *) :: *
type instance Apply (Append m a) l = Vec (l + m) a
append' xs ys = dfold (Proxy :: Proxy (Append m a)) (const (:>)) ys xs
We now see that append' has the appropriate type:
>>> :t append'
append' :: KnownNat k => Vec k a -> Vec m a -> Vec (k + m) a
And that it works:
>>> append' (1 :> 2 :> Nil) (3 :> 4 :> Nil)
<1,2,3,4>
NB: "dfold m f z xs" creates a linear structure, which has a depth, or delay, of O(length xs). Look at dtfold for a dependently typed fold that produces a structure with a depth of O(log_2(length xs)).
Arguments
:: KnownNat k => Proxy (p :: TyFun Nat * -> *) The motive -> (a -> p @@ 0) Function to apply to every element -> (forall l. SNat l -> (p @@ l) -> (p @@ l) -> p @@ (l + 1)) Function to combine results.NB: The SNat l indicates the depth/height of the node in the tree that is created by applying this function. The leafs of the tree have depth/height 0, and the root of the tree has height k. -> Vec (2 ^ k) a Vector to fold over.NB: Must have a length that is a power of 2. -> p @@ k
A combination of dfold and fold: a dependently typed fold that reduces a vector in a tree-like structure.
As an example of when you might want to use dtfold we will build a population counter: a circuit that counts the number of bits set to '1' in a BitVector. Given a vector of n bits, we only need we need a data type that can represent the number n: Index (n+1). Index k has a range of [0 .. k-1] (using ceil(log2(k)) bits), hence we need Index n+1. As an initial attempt we will use sum, because it gives a nice (log2(n)) tree-structure of adders:
populationCount :: (KnownNat (n+1), KnownNat (n+2))
=> BitVector (n+1) -> Index (n+2)
populationCount = sum . map fromIntegral . bv2v
The "problem" with this description is that all adders have the same bit-width, i.e. all adders are of the type:
(+) :: Index (n+2) -> Index (n+2) -> Index (n+2).
This is a "problem" because we could have a more efficient structure: one where each layer of adders is precisely wide enough to count the number of bits at that layer. That is, at height d we want the adder to be of type:
Index ((2^d)+1) -> Index ((2^d)+1) -> Index ((2^(d+1))+1)
We have such an adder in the form of the plus function, as defined in the instance ExtendingNum instance of Index. However, we cannot simply use fold to create a tree-structure of pluses:
>>> :{
let populationCount' :: (KnownNat (n+1), KnownNat (n+2))
=> BitVector (n+1) -> Index (n+2)
populationCount' = fold plus . map fromIntegral . bv2v
:}
<interactive>:...
• Couldn't match type ‘((n + 2) + (n + 2)) - 1’ with ‘n + 2’
Expected type: Index (n + 2) -> Index (n + 2) -> Index (n + 2)
Actual type: Index (n + 2)
-> Index (n + 2) -> AResult (Index (n + 2)) (Index (n + 2))
• In the first argument of ‘fold’, namely ‘plus’
In the first argument of ‘(.)’, namely ‘fold plus’
In the expression: fold plus . map fromIntegral . bv2v
• Relevant bindings include
populationCount' :: BitVector (n + 1) -> Index (n + 2)
(bound at ...)
because fold expects a function of type "a -> a -> a", i.e. a function where the arguments and result all have exactly the same type.
In order to accommodate the type of our plus, where the result is larger than the arguments, we must use a dependently typed fold in the the form of dtfold:
{-# LANGUAGE UndecidableInstances #-}
import Data.Singletons.Prelude
import Data.Proxy
data IIndex (f :: TyFun Nat *) :: *
type instance Apply IIndex l = Index ((2^l)+1)
populationCount' :: (KnownNat k, KnownNat (2^k))
=> BitVector (2^k) -> Index ((2^k)+1)
populationCount' bv = dtfold (Proxy @IIndex)
fromIntegral
(\_ x y -> plus x y)
(bv2v bv)
And we can test that it works:
>>> :t populationCount' (7 :: BitVector 16)
populationCount' (7 :: BitVector 16) :: Index 17
>>> populationCount' (7 :: BitVector 16)
3
Some final remarks:
• By using dtfold instead of fold, we had to restrict our BitVector argument to have bit-width that is a power of 2.
• Even though our original populationCount function specified a structure where all adders had the same width. Most VHDL/(System)Verilog synthesis tools will create a more efficient circuit, i.e. one where the adders have an increasing bit-width for every layer, from the VHDL/(System)Verilog produced by the Clash compiler.
NB: The depth, or delay, of the structure produced by "dtfold m f g xs" is O(log_2(length xs)).
vfold :: forall k a b. KnownNat k => (forall l. SNat l -> a -> Vec l b -> Vec (l + 1) b) -> Vec k a -> Vec k b Source #
Specialised version of dfold that builds a triangular computational structure.
Example:
compareSwap a b = if a > b then (a,b) else (b,a)
insert y xs = let (y',xs') = mapAccumL compareSwap y xs in xs' :< y'
insertionSort = vfold (const insert)
Builds a triangular structure of compare and swaps to sort a row.
>>> insertionSort (7 :> 3 :> 9 :> 1 :> Nil)
<1,3,7,9>
The circuit layout of insertionSort, build using vfold, is:
# Prefix sums (scans)
scanl :: (b -> a -> b) -> b -> Vec n a -> Vec (n + 1) b Source #
scanl is similar to foldl, but returns a vector of successive reduced values from the left:
scanl f z (x1 :> x2 :> ... :> Nil) == z :> (z f x1) :> ((z f x1) f x2) :> ... :> Nil
>>> scanl (+) 0 (5 :> 4 :> 3 :> 2 :> Nil)
<0,5,9,12,14>
"scanl f z xs" corresponds to the following circuit layout:
NB:
last (scanl f z xs) == foldl f z xs
scanr :: (a -> b -> b) -> b -> Vec n a -> Vec (n + 1) b Source #
scanr is similar to foldr, but returns a vector of successive reduced values from the right:
scanr f z (... :> xn1 :> xn :> Nil) == ... :> (xn1 f (xn f z)) :> (xn f z) :> z :> Nil
>>> scanr (+) 0 (5 :> 4 :> 3 :> 2 :> Nil)
<14,9,5,2,0>
"scanr f z xs" corresponds to the following circuit layout:
NB:
head (scanr f z xs) == foldr f z xs
postscanl :: (b -> a -> b) -> b -> Vec n a -> Vec n b Source #
postscanl is a variant of scanl where the first result is dropped:
postscanl f z (x1 :> x2 :> ... :> Nil) == (z f x1) :> ((z f x1) f x2) :> ... :> Nil
>>> postscanl (+) 0 (5 :> 4 :> 3 :> 2 :> Nil)
<5,9,12,14>
"postscanl f z xs" corresponds to the following circuit layout:
postscanr :: (a -> b -> b) -> b -> Vec n a -> Vec n b Source #
postscanr is a variant of scanr that where the last result is dropped:
postscanr f z (... :> xn1 :> xn :> Nil) == ... :> (xn1 f (xn f z)) :> (xn f z) :> Nil
>>> postscanr (+) 0 (5 :> 4 :> 3 :> 2 :> Nil)
<14,9,5,2>
"postscanr f z xs" corresponds to the following circuit layout:
mapAccumL :: (acc -> x -> (acc, y)) -> acc -> Vec n x -> (acc, Vec n y) Source #
The mapAccumL function behaves like a combination of map and foldl; it applies a function to each element of a vector, passing an accumulating parameter from left to right, and returning a final value of this accumulator together with the new vector.
>>> mapAccumL (\acc x -> (acc + x,acc + 1)) 0 (1 :> 2 :> 3 :> 4 :> Nil)
(10,<1,2,4,7>)
"mapAccumL f acc xs" corresponds to the following circuit layout:
mapAccumR :: (acc -> x -> (acc, y)) -> acc -> Vec n x -> (acc, Vec n y) Source #
The mapAccumR function behaves like a combination of map and foldr; it applies a function to each element of a vector, passing an accumulating parameter from right to left, and returning a final value of this accumulator together with the new vector.
>>> mapAccumR (\acc x -> (acc + x,acc + 1)) 0 (1 :> 2 :> 3 :> 4 :> Nil)
(10,<10,8,5,1>)
"mapAccumR f acc xs" corresponds to the following circuit layout:
# Stencil computations
Arguments
:: KnownNat n => SNat (stX + 1) Windows length stX, at least size 1 -> (Vec (stX + 1) a -> b) The stencil (function) -> Vec ((stX + n) + 1) a -> Vec (n + 1) b
1-dimensional stencil computations
"stencil1d stX f xs", where xs has stX + n elements, applies the stencil computation f on: n + 1 overlapping (1D) windows of length stX, drawn from xs. The resulting vector has n + 1 elements.
>>> let xs = (1:>2:>3:>4:>5:>6:>Nil)
>>> :t xs
xs :: Num a => Vec 6 a
>>> :t stencil1d d2 sum xs
stencil1d d2 sum xs :: Num b => Vec 5 b
>>> stencil1d d2 sum xs
<3,5,7,9,11>
Arguments
:: (KnownNat n, KnownNat m) => SNat (stY + 1) Window hight stY, at least size 1 -> SNat (stX + 1) Window width stX, at least size 1 -> (Vec (stY + 1) (Vec (stX + 1) a) -> b) The stencil (function) -> Vec ((stY + m) + 1) (Vec ((stX + n) + 1) a) -> Vec (m + 1) (Vec (n + 1) b)
2-dimensional stencil computations
"stencil2d stY stX f xss", where xss is a matrix of stY + m rows of stX + n elements, applies the stencil computation f on: (m + 1) * (n + 1) overlapping (2D) windows of stY rows of stX elements, drawn from xss. The result matrix has m + 1 rows of n + 1 elements.
>>> let xss = ((1:>2:>3:>4:>Nil):>(5:>6:>7:>8:>Nil):>(9:>10:>11:>12:>Nil):>(13:>14:>15:>16:>Nil):>Nil)
>>> :t xss
xss :: Num a => Vec 4 (Vec 4 a)
>>> :t stencil2d d2 d2 (sum . map sum) xss
stencil2d d2 d2 (sum . map sum) xss :: Num b => Vec 3 (Vec 3 b)
>>> stencil2d d2 d2 (sum . map sum) xss
<<14,18,22>,<30,34,38>,<46,50,54>>
Arguments
:: KnownNat n => SNat (stX + 1) Length of the window, at least size 1 -> Vec ((stX + n) + 1) a -> Vec (n + 1) (Vec (stX + 1) a)
"windows1d stX xs", where the vector xs has stX + n elements, returns a vector of n + 1 overlapping (1D) windows of xs of length stX.
>>> let xs = (1:>2:>3:>4:>5:>6:>Nil)
>>> :t xs
xs :: Num a => Vec 6 a
>>> :t windows1d d2 xs
windows1d d2 xs :: Num a => Vec 5 (Vec 2 a)
>>> windows1d d2 xs
<<1,2>,<2,3>,<3,4>,<4,5>,<5,6>>
Arguments
:: (KnownNat n, KnownNat m) => SNat (stY + 1) Window hight stY, at least size 1 -> SNat (stX + 1) Window width stX, at least size 1 -> Vec ((stY + m) + 1) (Vec ((stX + n) + 1) a) -> Vec (m + 1) (Vec (n + 1) (Vec (stY + 1) (Vec (stX + 1) a)))
"windows2d stY stX xss", where matrix xss has stY + m rows of stX + n, returns a matrix of m+1 rows of n+1 elements. The elements of this new matrix are the overlapping (2D) windows of xss, where every window has stY rows of stX elements.
>>> let xss = ((1:>2:>3:>4:>Nil):>(5:>6:>7:>8:>Nil):>(9:>10:>11:>12:>Nil):>(13:>14:>15:>16:>Nil):>Nil)
>>> :t xss
xss :: Num a => Vec 4 (Vec 4 a)
>>> :t windows2d d2 d2 xss
windows2d d2 d2 xss :: Num a => Vec 3 (Vec 3 (Vec 2 (Vec 2 a)))
>>> windows2d d2 d2 xss
<<<<1,2>,<5,6>>,<<2,3>,<6,7>>,<<3,4>,<7,8>>>,<<<5,6>,<9,10>>,<<6,7>,<10,11>>,<<7,8>,<11,12>>>,<<<9,10>,<13,14>>,<<10,11>,<14,15>>,<<11,12>,<15,16>>>>
# Conversions
toList :: Vec n a -> [a] Source #
Convert a vector to a list.
>>> toList (1:>2:>3:>Nil)
[1,2,3]
bv2v :: KnownNat n => BitVector n -> Vec n Bit Source #
Convert a BitVector to a Vec of Bits.
>>> let x = 6 :: BitVector 8
>>> x
0000_0110
>>> bv2v x
<0,0,0,0,0,1,1,0>
v2bv :: KnownNat n => Vec n Bit -> BitVector n Source #
Convert a Vec of Bits to a BitVector.
>>> let x = (0:>0:>0:>1:>0:>0:>1:>0:>Nil) :: Vec 8 Bit
>>> x
<0,0,0,1,0,0,1,0>
>>> v2bv x
0001_0010
# Misc
lazyV :: KnownNat n => Vec n a -> Vec n a Source #
What you should use when your vector functions are too strict in their arguments.
For example:
-- Bubble sort for 1 iteration
sortV xs = map fst sorted :< (snd (last sorted))
where
lefts = head xs :> map snd (init sorted)
rights = tail xs
sorted = zipWith compareSwapL lefts rights
-- Compare and swap
compareSwapL a b = if a < b then (a,b)
else (b,a)
Will not terminate because zipWith is too strict in its second argument.
In this case, adding lazyV on zipWiths second argument:
sortVL xs = map fst sorted :< (snd (last sorted))
where
lefts = head xs :> map snd (init sorted)
rights = tail xs
sorted = zipWith compareSwapL (lazyV lefts) rights
Results in a successful computation:
>>> sortVL (4 :> 1 :> 2 :> 3 :> Nil)
<1,2,3,4>
NB: There is also a solution using flip, but it slightly obfuscates the meaning of the code:
sortV_flip xs = map fst sorted :< (snd (last sorted))
where
lefts = head xs :> map snd (init sorted)
rights = tail xs
sorted = zipWith (flip compareSwapL) rights lefts
>>> sortV_flip (4 :> 1 :> 2 :> 3 :> Nil)
<1,2,3,4>
data VCons (a :: *) (f :: TyFun Nat *) :: * Source #
To be used as the motive p for dfold, when the f in "dfold p f" is a variation on (:>), e.g.:
map' :: forall n a b . KnownNat n => (a -> b) -> Vec n a -> Vec n b
map' f = dfold (Proxy @(VCons b)) (_ x xs -> f x :> xs)
Instances
type Apply (VCons a :: TyFun Nat * -> *) (l :: Nat) Source # Instance detailsDefined in Clash.Sized.Vector type Apply (VCons a :: TyFun Nat * -> *) (l :: Nat) = Vec l a
asNatProxy :: Vec n a -> Proxy n Source #
Vector as a Proxy for Nat
# Primitives
## Traversable instance
traverse# :: forall a f b n. Applicative f => (a -> f b) -> Vec n a -> f (Vec n b) Source #
## BitPack instance
unconcatBitVector# :: forall n m. (KnownNat n, KnownNat m) => BitVector (n * m) -> Vec n (BitVector m) Source # | 15,257 | 44,774 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-40 | latest | en | 0.619902 |
https://brainmass.com/math/linear-algebra/solving-systems-linear-equations-books-magazines-215965 | 1,495,608,360,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607802.75/warc/CC-MAIN-20170524055048-20170524075048-00408.warc.gz | 726,443,600 | 19,767 | Share
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# Solving Systems of Linear Equations of Books and Magazines
Systems of linear equations
Books and magazines. At Gwen's garage sale, all books were one price, and all magazines were another price.
Harriet bought four books and three magazines for \$1.45,and June bought two books and five magazines for \$1.25.
What was the price of a book and what was the price of a magazine?
Solving by Substitution
Solve each system by substitution. Determine whether the equations are independent, dependent, or inconsistent.
y=-3x+19
y=2x-1
y=-4x-7
y=3x
y=x+4
3y-5x=6
x-2y=-1
-x+5y=4
3x-5y=-11
x-2y=11
#### Solution Preview
Please refer attached file for better clarity of expressions.
Solutions:
Books and magazines. At Gwen's garage sale, all books were one price, and all magazines were another price.
Harriet bought four books and three magazines for \$1.45,and June bought two books and five magazines for \$1.25.
What was the price of a book and what was the price of a magazine?
Let price of each book =x
Price of each magazine=y
Harriet bought four books and three magazines for \$1.45,
4x+3y=1.45 ---------------(1)
June bought two books and five magazines for \$1.25.
2x+5y=1.25 ---------------(2)
Multiply equation (2) by -2, we get
-4x-10y=-2.5 ------------(3)
Adding equation (1) and equation (3), we get
4x+3y=1.45
-4x-10y=-2.5
-7y=-1.05
y=0.15
Put value of y=0.15 in equation (1), we ...
#### Solution Summary
There are 6 problems. Solutions to these problems explain the steps for finding the values of variables by Addition and Substitution method.
\$2.19 | 460 | 1,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2017-22 | longest | en | 0.949588 |
http://blog.51cto.com/computerdragon/1305985 | 1,540,178,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00259.warc.gz | 49,535,689 | 13,389 | package whut.tree;
//二叉排序树的非递归插入，非递归查询，寻找最大值，寻找最小值
class Node {
private int data;
private Node left;
private Node right;
public Node(int data) {
this.data = data;
}
public int getData() {
return data;
}
public Node getLeft() {
return left;
}
public void setLeft(Node left) {
this.left = left;
}
public Node getRight() {
return right;
}
public void setRight(Node right) {
this.right = right;
}
// 展示节点数据
public void displayNode() {
System.out.println(" " + data + " ");
}
}
// 树
public class BinaryTree {
private Node root;// 树根
// 非递归方式插入新的节点数据，
public void insert(int data) {
Node newNode = new Node(data);
if (root == null) {
root = newNode;
} else {
// 子节点，当前节点
Node current = root;
// 父节点
Node parent;
while (true)// 寻找插入的位置
{
parent = current;
// 当小于根节点，插入到左边
if (data < current.getData()) {
current = current.getLeft();
// 跳出循环
if (current == null) {
parent.setLeft(newNode);
return;
}
} else {
current = current.getRight();
if (current == null) {
parent.setRight(newNode);
return;
}
}
}
}
}
// 非递归方式实现查询
public Node find(int data) {
Node current = root;
while (current.getData() != data) {
if (current.getData() < data)
current = current.getLeft();
else
current = current.getRight();
// 当一个元素不在一个二叉树中，肯定为null了
if (current == null)
return null;
}
// 跳出循环说明找到了那个相等的
return current;
}
// 找二叉树最小的节点,一直遍历左孩子，直到其左孩子为null
public Node findMinNode() {
Node current;
Node parent;
//
if (root == null) {
return null;
} else {
parent = root;
current = parent.getLeft();
while (current != null) {
parent = current;
current = current.getLeft();
}
return parent;
}
}
// 找到二叉排序树的最大值，也就是最右边的孩子
public Node findMaxNode() {
Node current;
Node parent;
//
if (root == null) {
return null;
} else {
parent = root;
current = parent.getRight();
while (current != null) {
parent = current;
current = current.getRight();
}
return parent;
}
}
// 先要找到节点，然后根据要删除节点的位置进行删除
// 删除的节点有三种，叶子节点，有一个节点的节点，有两个节点的节点
public boolean delete(int key) {
Node current = root;
Node parent = root;
// 这里主要是为了区分删除的是左孩子还是右孩子
boolean isLeftChild = false;
// 显然，当current.iData == key 时，current就是需要删除的节点
// 在循环中利用parent保存了父类节点
while (current.getData() != key) {
parent = current;
if (key < current.getData()) {
isLeftChild = true;
current = current.getLeft();
} else {
isLeftChild = false;
current = current.getRight();
}
if (current == null)// 找不到key时返回false
return false;
}
// 当节点为叶子节点的时候
if (current.getLeft() == null && current.getRight() == null) {
if (current == root)
root = null;
else if (isLeftChild)
parent.setLeft(null);
else
parent.setRight(null);
}
// 当删除的节点为含有一个子节点的节点
// 删除的节点只有一个左子节点时
// 必须要考虑被删除的节点是左节点还是右节点
else if (current.getRight() == null) {
if (current == root)// 要删除的节点为根节点
root = current.getLeft();
else if (isLeftChild)// 要删除的节点是一个左子节点
parent.setLeft(current.getLeft());
else
parent.setRight(current.getLeft());// 要删除的节点是一个右子节点
}
// 当删除的节点为含有一个子节点的节点
// 删除的节点只有一个右子节点时
// 必须要考虑被删除的节点是左节点还是右节点
else if (current.getLeft() == null) {
if (current == root)// 要删除的节点为根节点
root = current.getRight();
else if (isLeftChild)// 要删除的节点是一个左子节点
parent.setLeft(current.getRight());
else
parent.setRight(current.getRight());// 要删除的节点是一个右子节点
}
// 当要删除的节点是含有两个节点的时候
else
{
//首先要获取被删除节点的后继节点，current
Node successor = getSuccessor(current);
if(current == root)
root = successor ;
//这里已经屏蔽了后继节点是叶子和非叶子节点
else if(isLeftChild)
parent.setLeft(successor);
else
parent.setRight(successor);
successor.setLeft(current.getLeft());
}
return true;
}
// 寻找后记节点，主要是当要删除的节点包含了两个子节点的时候
// 返回后继节点,后继节点就是比要删除的节点的关键值要大的节点集合中的最小值。
//后继节点要么是被删除节点的不包含左子节点的右节点，要么就是包含左子节点的右节点的子节点
private Node getSuccessor(Node delNode)
{
// 后继节点的父节点
Node successorParent = delNode;
// 后继节点
Node successor = delNode.getRight();
//判断后继节点是否有左孩子
Node current = successor.getLeft();
while (current != null) {
successorParent = successor;
successor = current;
current = current.getLeft();
}
//当该后继节点是属于包含左子节点的右节点的子节点
if (successor != delNode.getRight())
{
successorParent.setLeft(successor.getRight());
//连接被删除节点的右孩子
successor.setRight(delNode.getRight());
}
return successor;
}
// 下面三种遍历树
// 三种遍历均是采用递归实现的
// 前序遍历
public void preOrder(Node localRoot) {
if (localRoot != null) {
localRoot.displayNode();// 访问这个节点
preOrder(localRoot.getLeft());// 调用自身来遍历左子树
preOrder(localRoot.getRight());// 调用自身来遍历右子树
}
}
// 中序遍历
public void midOrder(Node localRoot) {
if (localRoot != null) {
preOrder(localRoot.getLeft());// 调用自身来遍历左子树
localRoot.displayNode();// 访问这个节点
preOrder(localRoot.getRight());// 调用自身来遍历右子树
}
}
// 后续遍历
public void lastOrder(Node localRoot) {
if (localRoot != null) {
preOrder(localRoot.getLeft());// 调用自身来遍历左子树
preOrder(localRoot.getRight());// 调用自身来遍历右子树
localRoot.displayNode();// 访问这个节点
}
}
}
package whut.tree;
//二叉树的存储，这里会建立一个二叉排序树,递归方式
public class NodeTree {
public int value;
public NodeTree left;
public NodeTree right;
//递归实现存储
public void store(int value) {
if (value < this.value) {
if (left == null) {
left = new NodeTree();
left.value = value;
} else {
left.store(value);
}
} else if (value > this.value) {
if (right == null)
{
right = new NodeTree();
right.value = value;
} else {
right.store(value);
}
}
}
public boolean find(int value) {
System.out.println("find happen " + this.value);
if (value == this.value) {
return true;
} else if (value > this.value) {
if (right == null)
return false;
return right.find(value);
} else {
if (left == null)
return false;
return left.find(value);
}
}
//前序遍历
public void preList() {
System.out.print(this.value + ",");
if (left != null)
left.preList();
if (right != null)
right.preList();
}
//中序遍历
public void middleList() {
if (left != null)
left.middleList();
System.out.print(this.value + ",");
if (right != null)
right.middleList();
}
//后续遍历
public void afterList() {
if (left != null)
left.afterList();
if (right != null)
right.afterList();
System.out.print(this.value + ",");
}
public static void main(String[] args) {
int[] data = new int[20];
for (int i = 0; i < data.length; i++) {
data[i] = (int) (Math.random() * 100) + 1;
System.out.print(data[i] + ",");
}
System.out.println();
NodeTree root = new NodeTree();
root.value = data[0];
for (int i = 1; i < data.length; i++) {
root.store(data[i]);
}
//查询
root.find(data[19]);
//先序
root.preList();
System.out.println();
//中序
root.middleList();
System.out.println();
//后续
root.afterList();
}
} | 2,040 | 6,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-43 | latest | en | 0.161222 |
https://cran.rstudio.com/web/packages/lolR/vignettes/pca.html | 1,519,482,233,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815812.83/warc/CC-MAIN-20180224132508-20180224152508-00662.warc.gz | 650,198,683 | 311,394 | Principal Component Analysis (PCA)
2018-02-05
Data for this notebook will be n=400 examples of d=30 dimensions.
PCA
Stacked Cigar Simulation
We first visualize the first 2 dimensions:
testdat <- lol.sims.cigar(n, d)
X <- testdat$X Y <- testdat$Y
data <- data.frame(x1=X[,1], x2=X[,2], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, y=x2, color=y)) +
geom_point() +
xlab("x1") +
ylab("x2") +
ggtitle("Simulated Data")
Projecting with PCA to 3 dimensions and visualizing the first 2:
result <- lol.project.pca(X, r)
data <- data.frame(x1=result$Xr[,1], x2=result$Xr[,2], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, y=x2, color=y)) +
geom_point() +
xlab("x1") +
ylab("x2") +
ggtitle("Projected Data using PCA")
Projecting with LDA to K-1=1 dimensions:
liney <- MASS::lda(result$Xr, Y) result <- predict(liney, result$Xr)
lhat <- 1 - sum(result$class == Y)/length(Y) data <- data.frame(x1=result$x[,1], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, fill=y)) +
xlab("$x_1$") +
ylab("Density") +
ggtitle(sprintf("PCA-LDA, L = %.2f", lhat))
Trunk Simulation
We visualize the first 2 dimensions:
testdat <- lol.sims.rtrunk(n, d)
X <- testdat$X Y <- testdat$Y
data <- data.frame(x1=X[,1], x2=X[,2], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, y=x2, color=y)) +
geom_point() +
xlab("x1") +
ylab("x2") +
ggtitle("Simulated Data")
Projecting with PCA to 3 dimensions and visualizing the first 2:
result <- lol.project.pca(X, r)
data <- data.frame(x1=result$Xr[,1], x2=result$Xr[,2], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, y=x2, color=y)) +
geom_point() +
xlab("x1") +
ylab("x2") +
ggtitle("Projected Data using PCA")
Projecting with LDA to K-1=1 dimensions:
liney <- MASS::lda(result$Xr, Y) result <- predict(liney, result$Xr)
lhat <- 1 - sum(result$class == Y)/length(Y) data <- data.frame(x1=result$x[,1], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, fill=y)) +
xlab("x1") +
ylab("Density") +
ggtitle(sprintf("PCA-LDA, L = %.2f", lhat))
Rotated Trunk Simulation
We visualize the first 2 dimensions:
testdat <- lol.sims.rtrunk(n, d, rotate=TRUE)
X <- testdat$X Y <- testdat$Y
data <- data.frame(x1=X[,1], x2=X[,2], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, y=x2, color=y)) +
geom_point() +
xlab("x1") +
ylab("x2") +
ggtitle("Simulated Data")
Projecting with PCA to 3 dimensions and visualizing the first 2:
result <- lol.project.pca(X, r)
data <- data.frame(x1=result$Xr[,1], x2=result$Xr[,2], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, y=x2, color=y)) +
geom_point() +
xlab("x1") +
ylab("x2") +
ggtitle("Projected Data using PCA")
Projecting with LDA to K-1=1 dimensions:
liney <- MASS::lda(result$Xr, Y) result <- predict(liney, result$Xr)
lhat <- 1 - sum(result$class == Y)/length(Y) data <- data.frame(x1=result$x[,1], y=Y)
data$y <- factor(data$y)
ggplot(data, aes(x=x1, fill=y)) +
ggtitle(sprintf("PCA-LDA, L = %.2f", lhat)) | 1,016 | 2,923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-09 | longest | en | 0.236827 |
http://mathhelpforum.com/statistics/281392-combination-events.html | 1,568,896,154,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573519.72/warc/CC-MAIN-20190919122032-20190919144032-00141.warc.gz | 127,596,946 | 12,064 | 1. ## Combination of events
Hello forumites
Ten pairs of shoes are in a closet. Four shoes are selected at random. Author wants me to find the probability that there will be at least one pairs of shoes among the four shoes selected.
Solution:-
Four shoes can be selected out of 10 pairs (20 number) in $\binom{20}{4}$ ways. Now we want to find the probability that there will be at least one pair of shoes among the four shoes selected which is equal to the probability that remains after deducting the probability of no pairs of shoes among the four shoes selected from the total probability.. So it is $1-\frac{\binom{10}{4}}{\binom{20}{4}}=0.956656$
But answer provided is $\frac{99}{323}=\frac{\binom{55}{2}}{\binom{20}{4} }$ Now which is wrong?
2. ## Re: Combination of events
If shoes within a pair are indistinguishable I agree with your answer.
I can't imagine where they came up with $\dbinom{55}{2}$
3. ## Re: Combination of events
Originally Posted by romsek
If shoes within a pair are indistinguishable I agree with your answer.
I can't imagine where they came up with $\dbinom{55}{2}$
Ok I was wrong. They are correct. Though $\dbinom{55}{2}$ is just coincidence I believe
Let's pick shoes so that there are no pairs
first pick is 20/20 shoes.
second pick is 18/19
third is 16/18
fourth is 14/17
We want 1 minus the product of these
$p = 1 - \dfrac{20 \cdot 18 \cdot 16 \cdot 14}{20 \cdot 19 \cdot 18 \cdot 17} = 1 - \dfrac{224}{323} = \dfrac{99}{323}$
4. ## Re: Combination of events
Originally Posted by romsek
Ok I was wrong. They are correct. Though $\dbinom{55}{2}$ is just coincidence I believe
Let's pick shoes so that there are no pairs
first pick is 20/20 shoes.
second pick is 18/19
third is 16/18
fourth is 14/17
We want 1 minus the product of these
$p = 1 - \dfrac{20 \cdot 18 \cdot 16 \cdot 14}{20 \cdot 19 \cdot 18 \cdot 17} = 1 - \dfrac{224}{323} = \dfrac{99}{323}$
Hello,
5. ## Re: Combination of events
Originally Posted by Vinod
Hello,
The problem with your answer is that by picking pairs, you remove the shoe in each pair that you don't pick from the possible shoes to pick.
6. ## Re: Combination of events
Originally Posted by romsek
The problem with your answer is that by picking pairs, you remove the shoe in each pair that you don't pick from the possible shoes to pick.
Hello,
If we want to compute the probability of exactly one pair among the four selected from the total ten pairs of shoes, how it should be computed? Secondly what is the probability of two pairs among the four selected out of ten pairs of shoes?
7. ## Re: Combination of events
Originally Posted by Vinod
Hello,
If we want to compute the probability of exactly one pair among the four selected from the total ten pairs of shoes, how it should be computed?
do you mean by randomly selecting pairs of shoes or by randomly selecting single shoes?
8. ## Re: Combination of events
Originally Posted by romsek
do you mean by randomly selecting pairs of shoes or by randomly selecting single shoes?
Hello,
It means random selection of four single shoes out of twenty shoes which are equal to ten pairs of shoes.
9. ## Re: Combination of events
Originally Posted by romsek
do you mean by randomly selecting pairs of shoes or by randomly selecting single shoes?
$p = \dfrac{20 \cdot 1 \cdot 18 \cdot 16}{20\cdot 19 \cdot 18 \cdot 17} = \dfrac{16}{323}$
10. ## Re: Combination of events
Originally Posted by romsek
$p = \dfrac{20 \cdot 1 \cdot 18 \cdot 16}{20\cdot 19 \cdot 18 \cdot 17} = \dfrac{16}{323}$
Hello,
11. ## Re: Combination of events
From the ten pairs of shoes, choose 3. Among the 3, choose one pair to take both shoes. From the other two pairs, choose one of the two for each. The total number of ways to select the shoes:
$$\dbinom{10}{3}\dbinom{3}{1}\dbinom{2}{1}\dbinom{ 2}{1}$$
Out of the total of 20 shoes where you choose 4.
$$\dbinom{20}{4}$$
So, the probability is:
$$\dfrac{\dbinom{10}{3}\dbinom{3}{1}\dbinom{2}{1} \dbinom{2}{1}}{\dbinom{20}{4}} = \dfrac{96}{323}$$
12. ## Re: Combination of events
Originally Posted by Vinod
Hello forumites
If an insect forum exists, I will greet them as forum mites. | 1,189 | 4,166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-39 | latest | en | 0.928417 |
https://www.jiskha.com/display.cgi?id=1165383357 | 1,516,161,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00669.warc.gz | 929,006,841 | 3,821 | # Pre-Cal
posted by .
Find the 5 roots of x^5+1=0 in polar and Cartesian form. (x^5 means x to the 5th power)
x^5 = -1 = e^[i (2n + 1) pi]
where i is any integer
x = [e^[i (2n + 1) pi]]^(1/5)
= e^(i pi/5)= cos pi/5 + i sin (pi/5)
= e^(3 i pi/5)
= cos (2 pi/5) + i sin (2 pi/5)
etc.
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find the product: (4k to the 5th power)(-2k)to the 3rd power. 5th power is inside the parenthesis and the 3rd is outside the parenthesis. Is it -32k to the 8th power or -32k to the 2nd power? | 465 | 1,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-05 | latest | en | 0.864889 |
https://tbc-python.fossee.in/convert-notebook/RCC_Theory_and_Design_by_M._G._Shah_and_C._M._Kale/Chapter11_0iM0C4n.ipynb | 1,717,051,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00729.warc.gz | 468,279,690 | 38,019 | # Chapter 11:Column Footing Design¶
## Ex11.1:pg-591¶
In [1]:
import math
b=0.2 #column width in m
D=0.3 #column depth in m
fck=15 #in MPa
sigma_cbc=5 #in MPa
sigma_st=230 #in MPa
P1=600 #load on column in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=150 #bearing capacity of soil in kN/sq m
A=P/q #in sq m
L=math.sqrt(A) #assuming footing to be square
L=2.1 #assume, in m
p=P1/L**2 #soil pressure, in kN/sq m
p=136 #assume, in sq m
bc=b/D
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: consider greater width of shaded portion in Fig. 11.3 of textbook
d1=L*(L-b)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 11.4 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+D*p+2*(b+D)*Tc
g=-(L**2-b*D)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.362 #assume, in m
#bending moment consideration, refer Fig. 11.5 of textbook
Mx=1*((L-b)/2)**2/2*p #in kN-m
My=1*((L-D)/2)**2/2*p #in kN-m
d3=math.sqrt(Mx*10**6/0.65/10**3) #<362 mm, hence OK
z=0.9*d2*10**3 #lever arm, in mm
Ast1=(Mx*10**6/sigma_st/z) #in sq mm
Ast=L*Ast1 #steel required for full width of 2.1 m, in sq mm
#provide 12 mm dia bars
dia=12 #in mm
n=Ast/0.785/dia**2 #no. of 12 mm dia bars
n=16 #assume
Tbd=0.84 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=825 #assume, in mm
c=50 #side cover, in mm
La=(L-D)/2*10**3-c #>Ld, hence OK
D=d2*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover=100 mm bottom; 50 mm side\nSteel-",(n)," bars of 12 mm dia both ways"
Summary of design:
Overall depth of footing= 468.0 mm
Cover=100 mm bottom; 50 mm side
Steel- 16 bars of 12 mm dia both ways
## Ex11.2:pg-592¶
In [3]:
import math
b=0.4 #column width, in m
D=0.4 #column depth, in m
fck=15 #in MPa
sigma_cbc=5 #in MPa
sigma_st=140 #in MPa
P1=1000 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=200 #bearing capacity of soil, in kN/sq m
A=P/q #in sq m
L=math.sqrt(A) #assuming footing to be square
L=2.3 #assume, in m
p=P1/L**2 #soil pressure, in kN/sq m
p=189 #assume, in kN/sq m
bc=b/D
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: consider greater width of shaded portion in Fig. 11.7 of textbook
d1=L*(L-b)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 11.8 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+D*p+2*(b+D)*Tc
g=-(L**2-b*D)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.425 #assume, in m
d=max(d1,d2) #in m
#bending moment consideration, refer Fig. 11.9 of textbook
Mx=1*((L-b)/2)**2/2*p #in kN-m
d3=math.sqrt(Mx*10**6/0.87/10**3) #<425 mm, hence OK
z=0.87*d*10**3 #lever arm, in mm
Ast1=(Mx*10**6/sigma_st/z) #in sq mm
Ast=L*Ast1 #steel required for full width of 2.3 m, in sq mm
#provide 18 mm dia bars
dia=18 #in mm
n=Ast/0.785/dia**2 #no. of 18 mm dia bars
n=15 #assume
Tbd=0.6 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
c=50 #side cover, in mm
La=(L-D)/2*10**3-c #in mm
#providing hook at ends
La=La+16*dia #>Ld, hence OK
D=d2*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover=100 mm bottom; 50 mm side\nSteel-",(n)," bars of 18 mm dia both ways"
Summary of design:
Overall depth of footing= 534.0 mm
Cover=100 mm bottom; 50 mm side
Steel- 15 bars of 18 mm dia both ways
## 11.3:pg-593¶
In [6]:
import math
B=0.5 #column diameter, in m
fck=20 #in MPa
sigma_cbc=7 #in MPa
sigma_st=230 #in MPa
P1=1600 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=300 #bearing capacity of soil, in kN/sq m
A=P/q #in sq m
L=math.sqrt(A) #assuming footing to be square
L=2.4 #assume, in m
p=P1/L**2 #soil pressure, in kN/sq m
p=278 #assume, in kN/sq m
bc=1
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I: refer Fig. 11.11 of textbook
d1=L*(L-B)/2*p/(Tc*L+L*p) #in m
#case II: refer Fig. 11.12 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=math.pi/4*p+math.pi*Tc
f=2*math.pi/4*B*p+math.pi*B*Tc
g=-(L**2-math.pi/4*B**2)*p
d2=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d2=0.57 #assume, in m
d=max(d1,d2) #in m
#bending moment consideration, refer Fig. 11.13 of textbook
M=1*((L-B)/2)**2/2*p #in kN-m
d3=math.sqrt(M*10**6/0.88/10**3) #<570 mm, hence OK
z=0.9*d*10**3 #lever arm, in mm
Ast1=(M*10**6/sigma_st/z) #in sq mm
Ast=L*Ast1 #steel required for full width of 2.4 m
#provide 20 mm dia bars
dia=20 #in mm
n=Ast/0.785/dia**2 #no. of 20 mm dia bars
n=9 #assume
Tbd=1.12 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
Ld=1030 #assume, in mm
c=50 #side cover, in mm
La=(L-B)/2*10**3-c #in mm
#bend bar at right angle and provide length, l
l=Ld-La #in mm
D=d*10**3+dia/2+100 #in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover:100 mm bottom; 50 mm side\nSteel:",(n),"-20 mm dia bars both ways"
Summary of design:
Overall depth of footing= 680.0 mm
Cover:100 mm bottom; 50 mm side
Steel: 9 -20 mm dia bars both ways
## Ex11.4:pg-595¶
In [9]:
import math
b=0.3 #column width in m
c1=0.4 #column depth in m
fck=20 #in MPa
sigma_cbc=7 #in MPa
sigma_st=275 #in MPa
P1=1200 #load on column, in kN
P2=0.05*P1 #weight of footing, in kN
P=P1+P2 #in kN
q=200 #bearing capacity of soil, in kN/sq m
A=P/q #in sq m
L1=2 #in m
L2=A/L1 #assuming footing to be square
L2=3.2 #assume, in m
p=P1/L1/L2 #soil pressure, in kN/sq m
bc=b/c1
ks=0.5+bc #>1
ks=1
Tc=0.16*math.sqrt(fck)*10**3 #in kN/sq m
Tv=Tc
#let d be the depth of footing in metres
#case I, refer Fig. 11.15 of textbook
#short direction
d1=L1*(L2-c1)/2*p/(Tc*L1+L1*p) #in m
#long direction
d2=L2*(L1-b)/2*p/(Tc*L2+L2*p) #in m
#case II: refer Fig. 11.16 of textbook; we get a quadratic equation of the form e d**2 + f d + g = 0
e=p+4*Tc
f=b*p+c1*p+2*(b+c1)*Tc
g=-(L1*L2-b*c1)*p
d3=(-f+math.sqrt(f**2-4*e*g))/2/e #in m
d3=0.47 #assume, in m
d=max(d1,d2,d3) #in m
#bending moment consideration, refer Fig. 11.17 of textbook
Mx=1*((L1-b)/2)**2/2*p #in kN-m
My=1*((L2-c1)/2)**2/2*p #in kN-m
d4=math.sqrt(My*10**6/0.8/10**3) #in mm
d4=480 #>470 mm (provided for shear)
d=d4 #in mm
z=0.92*d #lever arm, in mm
#short direction
Ast1=(Mx*10**6/sigma_st/z) #in sq mm
Ast=L2*Ast1 #steel required for full width of 3.2 m, in sq mm
b1=L1 #central band width, in m
beta=L2/L1
Astc=L1/(beta+1)*Ast #in sq mm
#provide 12 mm dia bars
dia=12 #in mm
n1=Astc/0.785/dia**2 #no. of 12 mm dia bars
n1=13 #assume
Astr=Ast-Astc #steel in remaining width, in sq mm
n2=Astr/0.785/dia**2
n2=4 #assume
n2=n2/2 #on each side
Tbd=1.12 #in MPa
Ld=dia*sigma_st/4/Tbd #in mm
c=50 #side cover, in mm
La=(L1-b)/2*10**3-c #>Ld, hence OK
#long direction
Ast1=(My*10**6/sigma_st/z) #in sq mm
Ast=L1*Ast1 #steel required for full width of 2 m, in sq mm
#provide 18 mm dia bars
dia=18 #in mm
n=Ast/0.785/dia**2 #no. of 18 mm dia bars
n=12 #assume
Ld=dia*sigma_st/4/Tbd #in mm
c=50 #side cover, in mm
La=(L2-c1)/2*10**3-c #>Ld, hence OK
D=d+dia/2+100 #in mm
D=590 #assume, in mm
print "Summary of design:\nOverall depth of footing=",(D)," mm\nCover=100 mm bottom; 50 mm side\nSteel-long direction\n",(n)," bars of 18 mm dia in ",(L1)," m width equally spaced\nShort direction\nCentral band ",(L1)," m:",(n1),"-12 mm dia bars equally spaced\nRemaining sides:",(n2),"-12 mm dia bars on each side"
Summary of design:
Overall depth of footing= 590 mm
Cover=100 mm bottom; 50 mm side
Steel-long direction
12 bars of 18 mm dia in 2 m width equally spaced
Short direction
Central band 2 m: 13 -12 mm dia bars equally spaced
Remaining sides: 2 -12 mm dia bars on each side | 3,321 | 7,597 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.360443 |
https://tfresource.org/topics/Smoothing_in_project_level_traffic_forecasting | 1,695,697,309,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510130.53/warc/CC-MAIN-20230926011608-20230926041608-00311.warc.gz | 616,118,430 | 7,649 | # Smoothing in project-level traffic forecasting
## # Objective
The objective of smoothing is to reveal the underlying trend in a time series so that the time series can be more clearly related to explanatory variables.
## # Background
Smoothing is used to stabilize a time series containing considerable variations, but smoothing is used in traffic forecast principally for removing cyclical variations prior to any estimation process. For example, smoothing can be used to eliminate seasonal variations in traffic due to vacations, school sessions, holiday shopping and other effects tied to the time of year. One set of results from smoothing are “seasonal adjustment factors” that can be used to relate smoothed or yearly forecasts to individual time periods (such as months). The preferred method of smoothing of traffic data is central moving average.
## # Guidelines
Smoothing can be helpful when developing a linear trend model or a linear model with explanatory variables. Central moving average takes the average of traffic counts for exactly one complete cycle with ½ cycle before a particular period and ½ cycle after that period, including the period itself. For example, if the moving average of traffic is being calculated for May of 2005, then the average should be taken over the twelve months between November of 2004 and October of 2005. The smoothed data series will terminate about ½ cycle ahead of the unsmoothed data series.
Smoothing is done prior to the statistical analysis step. When dealing with cycles of an even number of periods, the averaging range should be selected such that the last complete smoothed data point is as near to recent as possible. The statistical analysis of the smoothed data is carried out in the same way as unsmoothed data, using linear regression. See Trend models and Linear models with explanatory variables.
A seasonal adjustment factor for a period is average of the ratios of the unsmoothed data series to the smoothed data series. A traffic forecast for a specific time period in the future may be obtained by applying that period’s seasonal adjustment factor to a forecast of the smoothed traffic.
A series of monthly average traffic counts can be statistically stronger than a series of annual traffic counts, given that there are more data points. However, explanatory variables should be available monthly or nearly so for a monthly traffic forecast. Some interpolation to obtain monthly data is acceptable.
Central moving average may be used to smooth any cyclical data series, such a traffic counts across a day or across a week.
Other simple smoothing techniques from the literature, such as “exponential smoothing”, have not been shown to be advantageous for analysis of traffic counts.
Explanatory variables exhibiting cyclical variations should also be smoothed. In such cases, a smoothed forecasted value for the explanatory variable should be used in any forecast. | 547 | 2,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.926677 |
http://www.mathworks.com/help/mpc/examples/dc-servomotor-with-constraint-on-unmeasured-output.html?prodcode=MP&language=en&nocookie=true | 1,419,856,403,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447562966.132/warc/CC-MAIN-20141224185922-00082-ip-10-231-17-201.ec2.internal.warc.gz | 300,371,142 | 7,650 | Accelerating the pace of engineering and science
# Model Predictive Control Toolbox
## DC Servomotor with Constraint on Unmeasured Output
This example shows how to design a model predictive controller for a DC servomechanism under voltage and shaft torque constraints.
Reference
[1] A. Bemporad and E. Mosca, ''Fulfilling hard constraints in uncertain linear systems by reference managing,'' Automatica, vol. 34, no. 4, pp. 451-461, 1998.
Define DC-Servo Motor Model
The linear open-loop dynamic model is defined in "plant". Varible "tau" is the maximum admissible torque to be used as an output constraint.
```[plant, tau] = mpcmotormodel;
```
Design MPC Controller
Specify input and output signal types for the MPC controller. The second output, torque, is unmeasurable.
```plant = setmpcsignals(plant,'MV',1,'MO',1,'UO',2);
```
MV Constraints
The manipulated variable is constrained between +/- 220 volts. Since the plant inputs and outputs are of different orders of magnitude, you also use scale factors to faciliate MPC tuning. Typical choices of scale factor are the upper/lower limit or the operating range.
```MV = struct('Min',-220,'Max',220,'ScaleFactor',440);
```
OV Constraints
Torque constraints are only imposed during the first three prediction steps.
```OV = struct('Min',{Inf, [-tau;-tau;-tau;-Inf]},'Max',{Inf, [tau;tau;tau;Inf]},'ScaleFactor',{2*pi, 2*tau});
```
Weights
The control task is to get zero tracking offset for the angular position. Since you only have one manipulated variable, the shaft torque is allowed to float within its constraint by setting its weight to zero.
```Weights = struct('MV',0,'MVRate',0.1,'OV',[0.1 0]);
```
Construct MPC controller
Create an MPC controller with plant model, sample time and horizons.
```Ts = 0.1; % Sampling time
p = 10; % Prediction horizon
m = 2; % Control horizon
mpcobj = mpc(plant,Ts,p,m,Weights,MV,OV);
```
Simulate Using SIM Command
Use sim command to simulate the closed-loop control of the linear plant model in MATLAB.
```disp('Now simulating nominal closed-loop behavior');
Tstop = 8; % seconds
Tf = round(Tstop/Ts); % simulation iterations
r = [pi*ones(Tf,1) zeros(Tf,1)];% reference signal
[y1,t1,u1] = sim(mpcobj,Tf,r);
```
```Now simulating nominal closed-loop behavior
-->Converting model to discrete time.
-->The "Model.Noise" property of the "mpc" object is empty. Assuming white noise on each measured output channel.
```
Plot results.
```subplot(311)
stairs(t1,y1(:,1));
hold on
stairs(t1,r(:,1));
hold off
title('Angular Position')
subplot(312)
stairs(t1,y1(:,2));
title('Torque')
subplot(313)
stairs(t1,u1);
title('Voltage')
```
To run this example, Simulink® is required.
```if ~mpcchecktoolboxinstalled('simulink')
disp('Simulink(R) is required to run this example.')
return
end
```
Simulate closed-loop control of the linear plant model in Simulink. Controller "mpcobj" is specified in the block dialog.
```mdl = 'mpc_motor';
open_system(mdl)
sim(mdl)
```
The closed-loop response is identical to the simulation result in MATLAB.
```bdclose(mdl)
``` | 811 | 3,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2014-52 | latest | en | 0.809447 |
https://askfilo.com/physics-question-answers/if-the-ke-of-a-particle-becomes-four-times-its-initial-value-then-the-new | 1,723,378,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00053.warc.gz | 82,265,327 | 37,736 | Question
Medium
Solving time: 3 mins
# If the KE of a particle becomes four times its initial value, then the new momentum will be more than its initial momentum by;
A
B
C
D
## Text solutionVerified
The relationship between the kinetic energy (KE) and Linear momentum (p) is -
When the kinetic energy of the body becomes four times of its initial value, then
Hence, option (B) is correct.
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Question Text If the KE of a particle becomes four times its initial value, then the new momentum will be more than its initial momentum by; Updated On Nov 23, 2022 Topic System of Particles and Rotational Motion Subject Physics Class Class 11 Answer Type Text solution:1 Video solution: 1 Upvotes 231 Avg. Video Duration 5 min | 339 | 1,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.84722 |
https://books.google.gr/books?id=z0IXAAAAYAAJ&qtid=877e78ba&dq=editions:HARVARD32044096994090&hl=el&source=gbs_quotes_r&cad=5 | 1,606,416,112,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00703.warc.gz | 229,281,409 | 6,354 | Αναζήτηση Εικόνες Χάρτες Play YouTube Ειδήσεις Gmail Drive Περισσότερα »
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To reduce the fraction of one denomination to the fraction, of another, but less, retaining the same value. RULE. Multiply the...
The American Tutor's Assistant Revised: Or, A Compendious System of ... - Σελίδα 121
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## A Compendium of Practical Arithmetick
John Thomas Hope - 1790 - 387 σελίδες
...•|{• ^ to a f1mple fraction? Anfwer | (6) Reduce -,7j { to a fur.ple fraction ? Anfwer 7ST. CASE IX. To reduce the fraction of one denomination to the fraction of another i bat greater retaining the fame value. RULE. Firft, Reduce the given fraction to a compound fraction,...
## The American Tutor's Guide: Being a Compendium of Arithmetic. In Six Parts ...
James Thompson - 1808 - 172 σελίδες
...fraction of a <xt. Ans. Tfj 24. Reduce 4 of 5*. to the fraction of 21 shillings. Ans. VII. To reduce a fraction of one denomination to the fraction of another,...— Multiply the given numerator by the parts of the denominations between it and that denomination you would reduce it to, for a new numerator, which place...
## A New and Complete System of Arithmetick: Composed for the Use of the ...
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## The Teachers Assistant. Or, A System of Practical Arithmetic: Wherein the ...
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...fraction. Result 1141 — us-' 6. Reduce if of f of | to a single fraction. Result -fgj = -j~. CASEÍ6. To reduce the fraction of one denomination to the fraction of another, but greater, retaining thersame value. RULE. Make the fraction a compound one, by comparing it with all...
## Daboll's Schoolmaster's Assistant: Improved and Enlarged : Being a Plain and ...
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## The Teachers' Assistant, Or, A System of Practical Arithmetic: Wherein the ...
1817 - 198 σελίδες
...a single fraction. Result !Mi=MI6. Reduce jf off of £ to a single fraction. Result M=*fi. CASE 6. To reduce the fraction of one denomination to the fraction of another, but greater, retaining the same value. Make the fraction a compound one, by comparing it with all the denominations...
## Daboll's Schoolmaster's Assistant: Improved and Enlarged. Being a Plain ...
Nathan Daboll - 1818 - 240 σελίδες
...if 4. Reduce -j | f and T9ff to their least common denominator. " " Jlns. T88 || { % T»T CASE VII. To reduce the fraction of one denomination to the fraction , of another, retaining the same value. RULE. Reduce the giveji fraction to such a compound one, as will express...
## The Scholar's Guide to Arithmetic: Being a Collection of the Most Useful ...
Phinehas Merrill - 1819 - 107 σελίδες
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## The Scholar's Arithmetic: Designed for the Use of Schools in the United States
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...fraction of a day. Ans. TjV*' CASE 9. To reduce a fraction of one denomination to a fraction of lother, but less, retaining the same value. RULE. Multiply the given numerator by the parts of the denominations !tweeii it and that to which it is to be reduced, for a new nuer-ttor, ami place it over... | 1,173 | 4,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | latest | en | 0.530457 |
https://statsletters.com/category/uncategorized/ | 1,516,373,576,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888041.33/warc/CC-MAIN-20180119144931-20180119164931-00048.warc.gz | 809,579,763 | 30,814 | ## On types of randomized algorithms
There is more than Monte Carlo when talking about randomized algorithms. It is not uncommon to see the expresions “Monte Carlo Approach” and “randomized approach”
## Basic Statistics with Sympathy – Part 4: Building arbitrary RNGs in Sympathy
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## Basic Statistics with Sympathy – Part 2: Plotting and using the Calculator Node for common functions.
Allow me to introduce you to your new best friend from Sympathy 1.2.x: The improved calculator node. The node takes a list of tables, from | 296 | 1,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-05 | longest | en | 0.888104 |
http://m.dlxedu.com/m/askdetail/3/833ba897430047e870d340af0535ab64.html | 1,553,399,788,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00383.warc.gz | 124,437,228 | 3,965 | I have two matrices that looks like this:
A
``ColumnA ColumnB ColumnC ColumnDA D N FDF N A SP F K lqw AS O Wn H QD E``
B
`` ColumnA ColumnB ColumnC ColumnDA DH K FSnp N A SAS Q O lmP n N WEAS PV QNQ E``
I would like a third matrix C containing the common elements column by column between the two matrices.
I tried to do this work by using R but it seems impossible since the two matrices are too large: ~5000 rows and 1500 columns. The two matrices have the same number of columns ad the same column names.
Can anyone help me please?
Best
Desired output:
C
``ColumnA ColumnB ColumnC ColumnDA N N SP AS A Qn K EO``
You could try
``````library(stringi)
#Here `A` and `B` are "data.frames"
m1 <- stri_list2matrix( Map(`intersect`, A, B), fill='')
C <- setNames(as.data.frame(m1, stringsAsFactors=FALSE), colnames(A))
C
# ColumnA ColumnB ColumnC ColumnD
# 1 A N N S
# 2 P AS A Q
# 3 n K E
# 4 O
``````
Or
``````lst <- lapply(rbind(A,B), function(x) x[duplicated(x)& x!=''] )
m2 <- sapply(lst, `length<-`, max(sapply(lst, length)))
m2[is.na(m2)] <- ''
as.data.frame(m2, stringsAsFactors=FALSE)
# ColumnA ColumnB ColumnC ColumnD
#1 A N K S
#2 P n A Q
#3 AS O E
#4 N
``````
Do you know how to use sqlite?
In sqlite you could try something like
``````SELECT DISTINCT newtable
FROM A
WHERE newtable Not IN (SELECT DISTINCT newtable FROM B)
``````
it shoulnt be too much hassle to create a .db file
Note: if you're running linux you have sqlite or sqlite3 installed already
Top | 488 | 1,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-13 | latest | en | 0.74983 |
http://agam.github.io/post/2015/03/19/euler-38-pandigital-multiples/ | 1,537,300,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155676.21/warc/CC-MAIN-20180918185612-20180918205612-00332.warc.gz | 8,105,845 | 2,976 | # euler 38 pandigital multiples
Mar 19, 2015
I’m embarassed at how ridiculously naive my solution is. In hindsight, it’s so obvious that the solution has to be a four-digit number and its double, and that the first digit is going to be a 9.
Still, C++ makes you lazy. This took `24 milliseconds`, and I didn’t feel the need to speed it up :)
Statutory Warning: Spoilers ahead
``````static const long kMaxNumbers = 10000;
int main() {
long largestPanDigital = 0;
for (long num = 2; num < kMaxNumbers; ++num) {
vector<int> pandigits;
int o = 1;
while (o <= 20) {
long product = num * o;
vector<int> pd;
while (product > 0) {
int d = product % 10;
if (d == 0 ||
(std::find(pandigits.begin(), pandigits.end(), d) != pandigits.end()) ||
(std::find(pd.begin(), pd.end(), d) != pd.end())) {
break;
}
pd.push_back(d);
product /= 10;
}
if (product > 0) {
// Found an existing digit
break;
}
// We formed a new product; keep going!
for (auto it = pd.rbegin(); it != pd.rend(); ++it) {
pandigits.push_back(*it);
}
++o;
}
if (pandigits.size() == 9 && o > 1) {
long panDigital = 0L;
std::cout << "Found: ";
for (auto& d : pandigits) {
std::cout << d;
panDigital = panDigital * 10 + d;
}
std::cout << " = " << num << " * " << o-1 << std::endl;
if (panDigital > largestPanDigital) {
largestPanDigital = panDigital;
}
}
}
std::cout << "The largest one is " << largestPanDigital << std::endl;
}
``````
It runs as follows:
``````\$ ~/cpp/Test
Found: 918273645 = 9 * 5
Found: 192384576 = 192 * 3
Found: 219438657 = 219 * 3
Found: 273546819 = 273 * 3
Found: 327654981 = 327 * 3
Found: 672913458 = 6729 * 2
Found: 679213584 = 6792 * 2
Found: 692713854 = 6927 * 2
Found: 726914538 = 7269 * 2
Found: 729314586 = 7293 * 2
Found: 732914658 = 7329 * 2
Found: 769215384 = 7692 * 2
Found: 792315846 = 7923 * 2
Found: 793215864 = 7932 * 2
Found: 926718534 = 9267 * 2
Found: 927318546 = 9273 * 2
Found: 932718654 = 9327 * 2
The largest one is <redacted>
`````` | 690 | 1,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-39 | latest | en | 0.555257 |
https://www.unitconverters.net/time/octennial-to-month.htm | 1,715,990,728,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00003.warc.gz | 948,930,001 | 3,176 | Home / Time Conversion / Convert Octennial to Month
# Convert Octennial to Month
Please provide values below to convert octennial to month, or vice versa.
From: octennial To: month
### Octennial to Month Conversion Table
OctennialMonth
0.01 octennial0.96 month
0.1 octennial9.6 month
1 octennial96 month
2 octennial192 month
3 octennial288 month
5 octennial480 month
10 octennial960 month
20 octennial1920 month
50 octennial4800 month
100 octennial9600 month
1000 octennial96000 month
### How to Convert Octennial to Month
1 octennial = 96 month
1 month = 0.0104166667 octennial
Example: convert 15 octennial to month:
15 octennial = 15 × 96 month = 1440 month | 188 | 669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.437395 |
https://www.physicsforums.com/threads/algebra-ii-tossed-up-with-physics.3017/ | 1,669,986,945,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00117.warc.gz | 983,867,977 | 15,072 | # Algebra II tossed up with Physics
Two motorized toy boats (they are 5 meters apart)
are released in a pool at time t = 0. Boat 1 travels 35 degree North of east at a rate of 0.65 meter per second. Boat 2 travels due east at a rate of 0.4 meter per second. Good grief, I forgot the size of the pool, can we say it is AxB or assign any number such as 10m X 20m?
> a. Write a set of parametric equations to describe the path of each boat.
> b. At what point will each boat hit the east edge of the pool?
> c. At what point do the paths of the two boats cross?
STAii
First of all, welcome to the forums.
In the homework help forum you are asked to show us where you got stuck in order to help you.
But anyway, your question needs a lot more info to be solved (at least this is what i see).
First, we need to know more about the "5 meters apart" fact, you need to know in which direction (relative to each other) the boats are places initially.
Second, you need to give coordination points of the boats (or at least a clue about that) initially relatively to any of the edge of the pool, otherwise we cannot know when it will hit the pool (you see, if the boats started to move at the middle of the pool, they will not hit the east edge like if they started at the west edge of the pool, i think this is obvious).
I am sure you will be helped as soon as u provide the information, and where you got stuck in the question.
Thanks.
a. Write a set of parametric equations to describe the path of
each boat.
Boat 1 has subscript 1 boat 2 has subscript 2
each boat has x and y coordinates which are a function of t
(parametric)
boat 1 has x1(t) and y1(t) boat 2 has x2(t) and y2(t)
x1(t) = 0.65 cos(35)t y1(t) = 0.65 sin(35)t x2(t) =
.4t
y2(t) = constant
This is what my good neighbor told me. Since he was so kind and prompt, I didn't have a heart to ask him where those sin and cos coming from.
Boat 1's coordnate (0,0)
Boat 2's coordinate (0,5)
The width of pool is 15m. I guess the length of the pool won't be the issue.
STAii
Ok, this is better.
First of all, to clear out what where the cos() and sin() comes from.
The velocity of each boat is a vector, vectors can be analyzed into components, in our case horizontal (x) and vertical (y) component.
To find the magnitude of the horizontal component, multiply the magnitude of the vector by Cos() of the angle between the vector and the x-axis (you can figure this out if you draw a triangle with a side having the length of the vector, and another side on the X axis, and remember the definition of Cos() ).
So, the magnitude the horizontal component of the velocity of the fisrt boat (Vx1) = V1*cos(35) .
The magnitude of the vertical component of the velocity is the velocity of the boat multiplied by the Sin() of the angle between the velocity and the X-axis (which is equal to the Cos() of the angle between the velocity and the Y-axis).
So, Vy1 = V1*Sin(35) .
And it is well known that
S = V*t
(where S is the displacement)
So :
x1 = Vx1*t = V1*Cos(35)*t
y1 = Vy1*t = V1*Sin(35)*t
And do the same for the other boat. (remember that the angle between the velocity of boat2 and X-Axis is 0).
For question B, boat1 will hit the edge when x1(t) = the width of the pool, and boat2 will hit it when x2(t2) = the width of the pool.
Now to solve question C, let's change the parametric equations a little so that they refer to the origin point. (add the value of x1 when t=0)
x1(t1) = V1*Cos(35)*t2 + 0
y1(t1) = V1*Sin(35)*t2 + 0
x2(t2) = V2*t2 + 0
y2(t2) = 0 + 5
Now, when the boats paths meet x1(t1)=x2(t2) and y1(t1)=y2(t2) (or in other words, we are trying to find the point that both pathes share).
Note that the time in which each boat will reach this point might not be the same ! (this is why i called them t1 and t2).
Solve those two equations, and you will find the value of t1 or t2, use that to find x and y (from the parametric equations). | 1,099 | 3,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-49 | latest | en | 0.964396 |
https://polytope.miraheze.org/wiki/Parabiaugmented_hexagonal_prism | 1,723,745,250,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00092.warc.gz | 365,594,180 | 12,116 | # Parabiaugmented hexagonal prism
Parabiaugmented hexagonal prism
Rank3
TypeCRF
Notation
Bowers style acronymPabauhip
Coxeter diagramoxxxo oxuxo&#xt
Elements
Faces
Edges2+4+4+8+8
Vertices2+4+8
Vertex figures2 squares, edge length 1
8 irregular tetragons, edge lengths 1, 1, 2, 3
4 isosceles triangles, edge lengths (1+5)/2, 2, 2
Measures (edge length 1)
Volume${\displaystyle {\frac {2{\sqrt {2}}+9{\sqrt {3}}}{6}}\approx 3.06948}$
Dihedral angles3–4: ${\displaystyle \arccos \left(-{\sqrt {\frac {7+2{\sqrt {6}}}{12}}}\right)\approx 174.73561^{\circ }}$
3–6: ${\displaystyle \arccos \left(-{\frac {\sqrt {6}}{3}}\right)\approx 144.73561^{\circ }}$
4–4: 120°
3–3: ${\displaystyle \arccos \left(-{\frac {1}{3}}\right)\approx 109.47122^{\circ }}$
4–6: 90°
Central density1
Number of external pieces14
Level of complexity13
Related polytopes
ArmyPabauhip
RegimentPabauhip
DualParalaterobitruncated hexagonal tegum
ConjugateParabiaugmented hexagonal prism
Abstract & topological properties
Flag count104
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryK3, order 8
Flag orbits13
ConvexYes
NatureTame
The parabiaugmented hexagonal prism (OBSA: pabauhip) is one of the 92 Johnson solids (J55). It consists of 4+4 triangles, 4 squares, and 2 hexagons. It can be constructed by attaching square pyramids to two opposite square faces of the hexagonal prism.
## Vertex coordinates
A parabiaugmented hexagonal prism of edge length 1 has the following vertices:
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {\sqrt {3}}{2}},\,\pm {\frac {1}{2}}\right)}$,
• ${\displaystyle \left(\pm 1,\,0,\,\pm {\frac {1}{2}}\right)}$,
• ${\displaystyle \left(0,\,\pm {\frac {{\sqrt {2}}+{\sqrt {3}}}{2}},\,0\right)}$. | 617 | 1,732 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.543984 |
http://community.fantasyflightgames.com/index.php?/topic/88932-threat-treshold/ | 1,416,861,894,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400381177.56/warc/CC-MAIN-20141119123301-00251-ip-10-235-23-156.ec2.internal.warc.gz | 68,771,713 | 18,763 | # Threat treshold
14 replies to this topic
### #1 wolph42
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Posted 21 August 2013 - 01:11 PM
On page 284 is the Threat threshold introduced to create 'balancing' encounters. What eludes me however is what it actually is. On first glance it looks like that an encounter for e.g. 6 rank 5 PCs should be 12*6 = 72 NPC's but that short of ridiculous.
So how should I interpret it?
for reference:
Threat Threshold
The threat threshold is a number that limits the maximum
number and strength of NPCs that should be included in
an encounter. To determine the threat threshold for an
encounter, use the encounter rank on Table 12–1: Threat
Threshold on page 284 to determine the multiplier for the
total number of PCs in the warband. The result is the threat
threshold, which should be modified as needed should the
Acolytes have NPC allies aiding them or other factors.
Using Threat Threshold
Once the threat threshold has been determined for an
encounter, the GM can start creating the encounter by
he adds together all of their threat values. Ideally, the
combined total of all threat values in the encounter should
equal the threat threshold. However, this is often difficult to
accomplish. If the total threat value of an encounter exceeds
the determined threat threshold, the GM should recalculate
the rank of the encounter.
### #2 MagnusPihl
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Posted 21 August 2013 - 01:19 PM
Each NPC has a threat rating listed in the circle on the top-right of their profile, usually ranging from 5 to 30ish.
You can combine any number of NPCs to reach the threat threshold that matches your party.
### #3 wolph42
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Posted 21 August 2013 - 01:28 PM
You beat me to the edit of my post. Looking further i noticed those obscure black circles and quickly came to the same conclusion.
Where is this explained though? And how are these numbers drived?
### #4 MagnusPihl
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Posted 21 August 2013 - 02:45 PM
You beat me to the edit of my post. Looking further i noticed those obscure black circles and quickly came to the same conclusion.
Where is this explained though? And how are these numbers drived?
Threat Levels are explained on page 282, under NPC Profiles. Maybe it could be clearer.
I think your other point is the real kicker: How are these calculated? I love the idea, but it's not straightforward to make up your own NPC threat levels. A formula of some sort would be neat, but probably isn't really possible (a combat-oriented NPC with a very high Fellowship stat might get a high threat level without being very dangerous...).
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### #5 Bilateralrope
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Posted 22 August 2013 - 03:59 AM
There may not be any way to calculate a threat rating for your own NPCs without either extensive playtesting, or knowing both the game and your players enough that you can just ignore the threat rating.
### #6 Morangias
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Posted 22 August 2013 - 04:41 AM
I really don't think we need a threat calculator mechanic at all. I've yet to see a system that did it well, and this incarnation doesn't seem to stand out as the one that will somehow get it right.
There is no truth in flesh, only betrayal.
There is no strenght in flesh, only weakness.
There is no constancy in flesh, only decay.
There is no certainty in flesh but death.
### #7 MagnusPihl
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Posted 22 August 2013 - 05:10 AM
I really don't think we need a threat calculator mechanic at all. I've yet to see a system that did it well, and this incarnation doesn't seem to stand out as the one that will somehow get it right.
I disagree. Even a basic system like what we have is a huge step forward.
I remember looking at Deathwatch for the first time and trying to set up some balanced encounters. It was hopeless. I had absolutely no idea what a starting group could expect to handle. What use is a Troop/Elite/Master differentiation when there's multiple power levels in each group and none of them are explained?
The current system gives a rough guideline for new GMs. It's not perfect - it's never going to be - but it's a start. From there, you can start getting a feel for the system. At least your first encounter isn't likely to wipe out the group unintentionally.
### #8 Tom Cruise
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Posted 22 August 2013 - 05:17 AM
I think Morangias was more arguing against including a threat calculator for custom made NPCs, rather than the threat system itself. And I agree. Threat calculators perform badly enough in really carefully designed, combat focused games like DnD, I can't imagine it'd be possible to make a particularly useful one for a more open ended system like Dark Heresy
### #9 knasserII
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Posted 22 August 2013 - 05:22 AM
I think Morangias was more arguing against including a threat calculator for custom made NPCs, rather than the threat system itself. And I agree. Threat calculators perform badly enough in really carefully designed, combat focused games like DnD, I can't imagine it'd be possible to make a particularly useful one for a more open ended system like Dark Heresy
Yes. I found the threat ratings hugely useful when I was testing combat. A major help to newer GMs to the system. But a threat calculator for NPCs would be very hard and might lead to a false sense of assurance for GMs. There are sometimes things that are better not done at all than done badly, and risk calculation is one of those things.
Edited by knasserII, 22 August 2013 - 05:22 AM.
I lack credentials.
### #10 Morangias
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Posted 22 August 2013 - 05:49 AM
I think Morangias was more arguing against including a threat calculator for custom made NPCs, rather than the threat system itself. And I agree. Threat calculators perform badly enough in really carefully designed, combat focused games like DnD, I can't imagine it'd be possible to make a particularly useful one for a more open ended system like Dark Heresy
No, I meant the whole idea of threat ratings and setting up encounters in accord with them. It never works, and it encourages all sorts of bad behavior, from GMs not thinking about the kind of encounter they want to run to players feeling entitled about how combat should work out for them. I've seen that happen, and I've seen both sides ending up bitter and disappointed when it became obvious this kind of mechanic will never function properly.
There is no truth in flesh, only betrayal.
There is no strenght in flesh, only weakness.
There is no constancy in flesh, only decay.
There is no certainty in flesh but death.
### #11 Bilateralrope
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Posted 22 August 2013 - 02:54 PM
I've seen what happens when a powergaming GM with a GM vs Players mindset gets hold of a threat rating calculator for his own NPCs. He will optimize the NPCs for the situation where they run into the PCs. Then, when anyone points out how unfair the fight is, he responds by saying that the threat rating number matches what the party should be able to fight. Therefore the fight is balanced.
### #12 Magnus Grendel
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Posted 23 August 2013 - 01:43 AM
A 'threat calculator' is only ever a vague guide at best anyway.
As noted, a group of 6 rank 5 PCs should be encountering 72 'points' of threat to make for a decent encounter.
But In the last two dark heresy games I GM-ed:
One had a Reliquary Retrieval Team, with a preacher, a sage, a tech-savvy scummer, a not-especially combatitive arbites and one augmetic blade-armed assassin*.
The other was essentially an interrogator's personal hit-squad, with an angelus-wielding assassin sniper, a SWAT-equivalent arbitrator, a guardsman, a tech-priest and a scummer who didn't worship khorne (honest).
In theory these should be facing the same quantity of handgun-wielding thugs. In reality there is no way in hell you can create a system which makes a fair fight for both, and cleaving too closely to a system which says you can is asking for disappointment - especially if it tries to tell players there is a certain amount of combatant characters you 'must' have in a party.
It's not bad as a ready-reckoner so someone new to a system can see at a glance who is nasty and who isn't, but Troops/Elites/Master did that fine anyway, as far as I was concerned.
* The team spent a lot of time running.
### #13 knasserII
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Posted 23 August 2013 - 07:04 AM
I've seen what happens when a powergaming GM with a GM vs Players mindset gets hold of a threat rating calculator for his own NPCs. He will optimize the NPCs for the situation where they run into the PCs. Then, when anyone points out how unfair the fight is, he responds by saying that the threat rating number matches what the party should be able to fight. Therefore the fight is balanced.
Well if it is a choice between providing evil GMs with justification for murdering a party which they'll do anyway, or providing guidance to newbie GMs who might do so accidentally, I think the latter is worth more.
Systems like DH are especially hard to guesstimate opposition in. Plus the default encounter level in the book is NOT meant to create a 50:50 chance of survival, you know? If you want dangerous and risky, you give them an encounter of around Rank+2. So if it's a party not optimized for combat, giving them standard encounters for their rank shouldn't be TPKs or anything. They'll just find them harder, as such a party ought to.
I lack credentials.
### #14 Bilateralrope
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Posted 23 August 2013 - 03:41 PM
I've seen what happens when a powergaming GM with a GM vs Players mindset gets hold of a threat rating calculator for his own NPCs. He will optimize the NPCs for the situation where they run into the PCs. Then, when anyone points out how unfair the fight is, he responds by saying that the threat rating number matches what the party should be able to fight. Therefore the fight is balanced.
Well if it is a choice between providing evil GMs with justification for murdering a party which they'll do anyway, or providing guidance to newbie GMs who might do so accidentally, I think the latter is worth more.
That guidance can be provided by pre-written NPCs with their own stat blocks. The problem I suggested only comes in when you give GMs a way to calculate a threat rating on their own custom NPCs.
### #15 Simsum
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Posted 25 August 2013 - 11:56 PM
Perhaps FFG should just stick another little black advice box in one of the first pages of the book, stating that if you can't find mutual ground for a satisfying game by talking to each other like reasonable adults trying to have a good time, you should find other people to play with.
There's no such thing as exploit-proof. I'm sure you can get close, but you just might end up investing more resources in the attempt, than FFG has invested in everything they've done to date. The only sure solution is to not play with people you don't enjoy playing with - and FFG can't do that for you.
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Q
Dingfeng Capacitor -- What are the measurement formulas for capacitor capacity?
A
Q: Dingfeng Capacitor -- What are the measurement formulas for capacitor capacity?
A: If a capacitor is with 1 C quantity of electricity, and the potential difference between the two stages is 1 V, the capacitance of this capacitor is 1 F, namely: C=Q/U.
However, the capacity of the capacitor is not determined by Q(charge) or U (voltage), ie: C = εS / 4πkd. Where ε is a constant, S is the facing area of the capacitor plate, d is the distance of the capacitor plate,and k is the electrostatic force constant. For a common parallel plate capacitor, the capacitance is C=εS/d. (ε is the dielectric constant of the dielectric between the plates, S is the plate area, and d is the distance between the plates.)
Multi-capacitor parallel calculation formula: C=C1+C2+C3+...+Cn.
Multi-capacitor Tandem calculation formula: 1/C=1/C1+1/C2+...+1/Cn. | 251 | 943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.856823 |
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However, the relationship Next, I will demonstrate how to run If you don’t see Data Analysis, install that ToolPak. Multiple linear regression is somewhat more complicated than simple linear regression, because there are more parameters than will fit on a two-dimensional plot. Example of Multiple Linear Regression in Python In the following example, we will use multiple linear regression to predict the stock index price (i.e., the dependent variable) of a fictitious economy by using 2 independent For more information, check out this post on why you should not use multiple linear regression for Key Driver Analysis with example data for multiple linear regression examples. We know that the Linear Regression technique has only one dependent variable and one independent variable. Learn how to install it in my post about using Excel to perform t-tests.It’s free! We need to also include in CarType to our model. Learn ways of fitting models here! 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The results of a stepwise multiple regression, with P-to-enter and P-to-leave both equal to 0.15, is that acreage, nitrate, and maximum depth contribute to the multiple regression equation. New York: Guilford Press. Q. Open How to Run a Multiple Regression in Excel. For instance, say that one stoplight backing up can prevent traffic from passing Example of Interpreting and Applying a Multiple Regression Model We'll use the same data set as for the bivariate correlation example -- the criterion is 1 st year graduate grade point average and the predictors are the program they are in and the three GRE scores. Additionally, examples and applications will be We could have used as little or as many variables we wanted in our regression model(s) — up to all the 13! It is used to discover the relationship and assumes the linearity between target and predictors. Intuitively, I assume that higher IQ, motivation and social support are associated with better job performance. Multiple Regression - Example I run a company and I want to know how my employees’ job performance relates to their IQ, their motivation and the amount of social support they receive. An example of a linear regression … Excel is a great option for running multiple regressions when a user doesn't have access to advanced statistical software. Video created by Johns Hopkins University for the course "Multiple Regression Analysis in Public Health ". In statistical modeling, regression analysis is a set of statistical processes for estimating the relationships between a dependent variable (often called the 'outcome variable') and one or more independent variables (often called 'predictors', 'covariates', or 'features'). Definition 1: We use the same terminology as in Definition 3 of Regression Analysis, except that the degrees of freedom df Res and df Reg are modified to account for the number k of independent variables. The topics below are provided in order of increasing complexity. Example: The simplest multiple regression model for two predictor variables is y = β 0 +β 1 x 1 +β 2 x 2 + The surface that corresponds to the model y =50+10x 1 +7x 2 looks like this. Four Tips on How to Perform a Regression Analysis that Avoids Common Problems: Keep these tips in mind through out all stages of this tutorial to ensure a top-quality regression analysis. This tutorial has covered basics of multiple regression analysis. Regression analysis for categorical moderators. Multiple Regression Analysis Examples A. Here the blood pressure is the dependent These days, website management requires certain tactics to increase traffic from Google. A description of each variable A simple linear regression equation for this would be $$\hat{Price} = b_0 + b_1 * Mileage$$. What if you have more than one independent variable? This is particularly useful to predict the price for gold in the six months from now. With the example of multiple regression, you can predict the blood pressure of an individual by considering his height, weight, and age. The following example illustrates XLMiner's Multiple Linear Regression method using the Boston Housing data set to predict the median house prices in housing tracts. For a thorough analysis, however, we want to make sure we satisfy the iii. We are dealing with a more complicated example in this case though. Regression analysis, when used in business, is often associated with break even analysis which is mainly concerned on determining the safety threshold for a business in … I’ll also explain the Multiple Regression Analysis using this site econoshift.com as an example. A complete example of regression analysis. Multiple (Linear) Regression R provides comprehensive support for multiple linear regression. In this video we review the very basics of Multiple Regression. Multiple Linear Regression is one of the regression methods and falls under predictive mining techniques. Fitting the Model # Multiple Linear Regression Example … Multiple regression analysis, a term first used by Karl Pearson (1908), is an extremely useful extension of simple linear regression in that we use several quantitative (metric) or dichotomous variables in - Work placement salaries analysis through multiple linear regression and their occurrence based on qualifications and work experience. For the calculation of Multiple Regression, go to the Data tab in excel, and then select the data analysis option. Perform a linear regression analysis of PIQ on Brain, Height, and Weight. Upon completion of this tutorial, you should understand the following: Example: Think of SEO with Multiple Regression Analysis. Returning to the Benetton example, we can include year variable in the regression, which gives the result that Sales = 323 + 14 Advertising + 47 Year. We will ignore this violation of the assumption for now, and conduct the multiple linear regression analysis. In this tutorial, We are going to understand Multiple Regression which is used as a predictive analysis tool in Machine Learning and see the example in Python. For the further procedure and calculation refers to the given article here – Analysis ToolPak in Excel The regression formula for the above example will be SPSS Multiple Regression Analysis Tutorial By Ruben Geert van den Berg under Regression Running a basic multiple regression analysis in SPSS is simple. Regression analysis is used to model the relationship between a response variable and one or more predictor variables. Other statistical tools can equally be used to easily predict the outcome of … You can however create non-linear terms in the model. However, there are ways to display your results that include the effects of multiple independent variables on the dependent variable, even though only one independent variable can actually be plotted on the x-axis. Here, we will be citing a scenario that serves as an example of the implementation of simple regression analysis. PhotoDisc, Inc./Getty Images A Within this module, an overview of multiple regression will be provided. Please Note: The purpose of this page is to show how to use various data analysis commands. Example Multiple regression analysis can be performed using Microsoft Excel and IBM’s SPSS. Let’s look at an example. The multiple regression model itself is only capable of being linear, which is a limitation. Let us assume the average speed when 2 highway patrols are deployed is 75 mph, or 35 mph when 10 highway patrols are deployed. Sample size tables for correlation analysis with applications in partial correlation and multiple regression analysis. Multiple linear regression is found in SPSS in Analyze/Regression/Linear… In our example, we need to enter the variable “murder rate” as the dependent variable and the population, burglary, larceny, and vehicle theft variables as independent variables. Multiple linear regression analysis is also used to predict trends and future values.
## multiple regression analysis example
Land With Barn For Sale Washington, Century Pool Pump Manual, Niacin For Muscle Growth, Marigolds Falling Over, Affordable Housing Near Me, | 1,824 | 9,535 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-17 | latest | en | 0.865321 |
http://www.numbersaplenty.com/10001101101011 | 1,582,478,398,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00257.warc.gz | 233,375,896 | 3,337 | Search a number
10001101101011 is a prime number
BaseRepresentation
bin1001000110001001000000…
…0101000001011111010011
31022102002120221010222110022
42101202100011001133103
52302324223340213021
633134235321254055
72051361415504316
oct221422005013723
938362527128408
1010001101101011
113206496930559
12115634463432b
1357713b303397
151252419451ab
hex918901417d3
10001101101011 has 2 divisors, whose sum is σ = 10001101101012. Its totient is φ = 10001101101010.
The previous prime is 10001101101007. The next prime is 10001101101067. The reversal of 10001101101011 is 11010110110001.
Adding to 10001101101011 its reverse (11010110110001), we get a palindrome (21011211211012).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 10001101101011 - 22 = 10001101101007 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 10001101101011.
It is not a weakly prime, because it can be changed into another prime (10001101101071) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5000550550505 + 5000550550506.
It is an arithmetic number, because the mean of its divisors is an integer number (5000550550506).
Almost surely, 210001101101011 is an apocalyptic number.
10001101101011 is a deficient number, since it is larger than the sum of its proper divisors (1).
10001101101011 is an equidigital number, since it uses as much as digits as its factorization.
10001101101011 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1, while the sum is 8.
The spelling of 10001101101011 in words is "ten trillion, one billion, one hundred one million, one hundred one thousand, eleven". | 552 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-10 | latest | en | 0.8359 |
https://mltocups.net/440-ml-to-cups/ | 1,701,922,372,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00808.warc.gz | 449,844,029 | 19,966 | # How much is 440 ml in cups – Convert 440 ml to cups
Converting 440 ml to cups is a common task that can be accomplished with a little bit of math. By using the conversion rate of 1 ml to 0.0042268 US customary cups, it can be determined that 440 ml is equal to 1.8597712486 US cups, which is also known as 1 2/3 cups.
To further illustrate how to convert 440 cup to ml, it can be noted that 100 ml divided by 237 ml (the equivalent of 1 cup) equals approximately 2/5 cup or 0.422675283773 US cups. This is an essential tool for cooks and bakers who need to accurately measure ingredients for their recipes in terms of both milliliters and cups.
It’s important to remember that measurements should always be done with precision and accuracy, as even small discrepancies can have a huge impact on the outcome of any recipe!
## Convert 440 milliliters water to cups
Ml to cups formula: 440 ml / 236.5883953392 =1.85977 Cups
ml
=
ml
### 440 Ml liquid, water to Cups Conversion table / chart
cup cups american Cups canada 439.1 1.8559659250001 1.931808183018 439.2 1.8563886000001 1.9322481302244 439.3 1.8568112750001 1.9326880774307 439.4 1.8572339500001 1.933128024637 439.5 1.8576566250001 1.9335679718434 439.6 1.8580793000001 1.9340079190497 439.7 1.8585019750001 1.934447866256 439.8 1.8589246500001 1.9348878134624 439.9 1.8593473250001 1.9353277606687 440 1.8597700000001 1.9357677078751 440.1 1.8601926750001 1.9362076550814 440.2 1.8606153500001 1.9366476022877 440.3 1.8610380250001 1.9370875494941 440.4 1.8614607000001 1.9375274967004 440.5 1.8618833750001 1.9379674439067 440.6 1.8623060500001 1.9384073911131 440.7 1.8627287250001 1.9388473383194 440.8 1.8631514000001 1.9392872855257 440.9 1.8635740750001 1.9397272327321
See also Convert 250 ml to cups. 250 ml to cups conversion table
## Convert 440 milliliters milk to cups
ml
=
ml
### 440 Ml milk to Cups Conversion table / chart
Ml cups american Cups canada 439.1 1.8559659250001 1.931808183018 439.2 1.8563886000001 1.9322481302244 439.3 1.8568112750001 1.9326880774307 439.4 1.8572339500001 1.933128024637 439.5 1.8576566250001 1.9335679718434 439.6 1.8580793000001 1.9340079190497 439.7 1.8585019750001 1.934447866256 439.8 1.8589246500001 1.9348878134624 439.9 1.8593473250001 1.9353277606687 440 1.8597700000001 1.9357677078751 440.1 1.8601926750001 1.9362076550814 440.2 1.8606153500001 1.9366476022877 440.3 1.8610380250001 1.9370875494941 440.4 1.8614607000001 1.9375274967004 440.5 1.8618833750001 1.9379674439067 440.6 1.8623060500001 1.9384073911131 440.7 1.8627287250001 1.9388473383194 440.8 1.8631514000001 1.9392872855257 440.9 1.8635740750001 1.9397272327321
## Convert 440 milliliters oil to cups
ml
=
ml
### 440 Ml oil to Cups Conversion table / chart
Ml cups american Cups canada 439.1 1.8559659250001 1.931808183018 439.2 1.8563886000001 1.9322481302244 439.3 1.8568112750001 1.9326880774307 439.4 1.8572339500001 1.933128024637 439.5 1.8576566250001 1.9335679718434 439.6 1.8580793000001 1.9340079190497 439.7 1.8585019750001 1.934447866256 439.8 1.8589246500001 1.9348878134624 439.9 1.8593473250001 1.9353277606687 440 1.8597700000001 1.9357677078751 440.1 1.8601926750001 1.9362076550814 440.2 1.8606153500001 1.9366476022877 440.3 1.8610380250001 1.9370875494941 440.4 1.8614607000001 1.9375274967004 440.5 1.8618833750001 1.9379674439067 440.6 1.8623060500001 1.9384073911131 440.7 1.8627287250001 1.9388473383194 440.8 1.8631514000001 1.9392872855257 440.9 1.8635740750001 1.9397272327321
## How to convert 440 ml to cups
Imagine if you had 440 milliliters of something and wanted to measure it in cups…well, wonder no more! All you have to do is divide 440 ml by 237 – presto: that’s your amount in cups!
Without a doubt, consulting the article on convert how to convert ml to cups is highly recommended.
## Exam:
440 mL equals how many cups?
• 440 ml milk to cups =1.85977 cups
• 440 ml water to cups = 1.85977 cups
• 440 ml sugar to cups =1.85977 cups
## F.A.Q 440 ml to cups
### 440 ML IS EQUIVALENT TO HOW MANY CUPS?
440 milliliters convert into 1.85977 cups | 1,653 | 4,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-50 | latest | en | 0.753737 |
https://www.coursehero.com/file/6566044/Current-and-Drift-velocity/ | 1,498,370,046,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00131.warc.gz | 856,305,698 | 46,671 | Current and Drift velocity
# Current and Drift velocity - When a battery or power supply...
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Current and Drift velocity An electric current, which is a flow of charge, occurs when there is a potential difference. For a current to flow also requires a complete circuit, which means the flowing charge has to be able to get back to where it starts. Current (I) is measured in amperes (A), and is the amount of charge flowing per second. current : I = q / t, with units of A = C / s When current flows through wires in a circuit, the moving charges are electrons. For historical reasons, however, when analyzing circuits the direction of the current is taken to be the direction of the flow of positive charge, opposite to the direction the electrons go. We can blame Benjamin Franklin for this. It amounts to the same thing, because the flow of positive charge in one direction is equivalent to the flow of negative charge in the opposite direction.
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Unformatted text preview: When a battery or power supply sets up a difference in potential between two parts of a wire, an electric field is created and the electrons respond to that field. In a current-carrying conductor, however, the electrons do not all flow in the same direction. In fact, even when there is no potential difference (and therefore no field), the electrons are moving around randomly. This random motion continues when there is a field, but the field superimposes onto this random motion a small net velocity, the drift velocity. Because electrons are negative charges, the direction of the drift velocity is opposite to the electric field. In a typical case, the drift velocity of electrons is about 1 mm / s. The electric field,on the other hand, propagates much faster than this, more like 10 8 m / s....
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## This note was uploaded on 11/22/2011 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.
Ask a homework question - tutors are online | 453 | 2,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-26 | longest | en | 0.938877 |
https://thibaultlanxade.com/general/how-much-horsepower-would-a-challenger-srt-hellcat-redeye-widebody-need-to-tow-the-ss-delphine-up-the-industry-standard-davis-dam-grade-climb-pro-tips-davis-dam-assumed-grade-7-air-density-is-sea-level-conditions-002377-slugs-f-t3-w-weight | 1,722,861,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00073.warc.gz | 447,379,500 | 7,904 | Q:
# how much horsepower would a Challenger SRT® Hellcat Redeye Widebody need to tow the SS Delphine up the industry standard: Davis Dam Grade Climb? Pro Tips:Davis Dam assumed grade = 7% Air density is sea level conditions (.002377 slugs/f^t3) w(weight) of the SS Delphine + trailer + Redeye = 3,922,000 + 150,000 + 4451 Lbs Crr = coefficient of rolling resistance = .015 Assume Combined CDA is 1555 ft^2 SAE J2807 (Davis Dam Grande Climb) - min speed is 40mpg (59 ft/s) F(drag) = 1/2p x V^2 x C(d)A F(rolling resistance) = W x cos x Crr F(weight) = W x sin F(sum) = F(drag) + F(rr) + F(weight) P = F(sum)XV convert to horsepower = P(1hp/550ft lbf/sec)
Accepted Solution
A:
Answer: I am getting a result around the 54,800 area, so I am selecting the 55,000 HP answer. | 251 | 771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-33 | latest | en | 0.806118 |
http://research.stlouisfed.org/fred2/series/NYSLIND | 1,432,844,265,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929561.98/warc/CC-MAIN-20150521113209-00128-ip-10-180-206-219.ec2.internal.warc.gz | 193,797,614 | 19,555 | Need Help?
2015-03: 1.80 Percent (+ see more)
Monthly, Seasonally Adjusted, NYSLIND, Updated: 2015-04-28 10:06 AM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
The leading index for each state predicts the six-month growth rate of the state's coincident index. In addition to the coincident index, the models include other variables that lead the economy: state-level housing permits (1 to 4 units), state initial unemployment insurance claims, delivery times from the Institute for Supply Management (ISM) manufacturing survey, and the interest rate spread between the 10-year Treasury bond and the 3-month Treasury bill.
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The leading index for each state predicts the six-month growth rate of the state's coincident index. In addition to the coincident index, the models include other variables that lead the economy: state-level housing permits (1 to 4 units), state initial unemployment insurance claims, delivery times from the Institute for Supply Management (ISM) manufacturing survey, and the interest rate spread between the 10-year Treasury bond and the 3-month Treasury bill.
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
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Suggested Citation
``` Federal Reserve Bank of Philadelphia, Leading Index for New York [NYSLIND], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/NYSLIND/, May 28, 2015. ```
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Name: Email: | 578 | 2,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2015-22 | longest | en | 0.838532 |
https://www.physicsforums.com/threads/binomial-distribution.570283/ | 1,511,604,115,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809746.91/warc/CC-MAIN-20171125090503-20171125110503-00724.warc.gz | 827,738,883 | 14,193 | # Binomial Distribution
1. Jan 23, 2012
### planauts
1. The problem statement, all variables and given/known data
http://puu.sh/epl6 [Broken]
http://puu.sh/eplm [Broken]
2. Relevant equations
3. The attempt at a solution
No clue on how to attempt this problem. Any help would be appreciated, thanks!
Last edited by a moderator: May 5, 2017
2. Jan 24, 2012
### Ray Vickson
(a) What is the formula for $f(k) \equiv \Pr\{X = k|X \geq 2 \},$ for k = 2, 3, ...,n? Then $E(X | X \geq 2) = \sum_{k=2}^n k f(k),$ and you ought to be able to get the cited formula from this.
(b) What is the formula for σ in terms of n and p? Look at it carefully.
RGV
Last edited by a moderator: May 5, 2017 | 232 | 692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-47 | longest | en | 0.876119 |
https://www.majortests.com/essay/Linear-Programm-549945.html | 1,632,539,386,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00510.warc.gz | 876,208,009 | 5,362 | # Linear Programm Essay
Submitted By mclopez05
Words: 1160
Pages: 5
EMBA – LP: Using Excel; Transportation & Assignment; Integer Programming
Learning goals
Modeling in Excel using Solver • Understand the importance of optimization (linear programming) via Excel’s Solver • Understand how to model simple linear programs using Excel and Solver • Understand the power of Solver to perform nonlinear and integer programming
The Transportation Model • Understand the structure and assumptions of the transportation model • Understand the relationship between the transportation model and linear programming • Understand the advantage of using a more general model (linear programming) rather than a specific model (transportation)
The Assignment Model • Understand the structure and assumptions of the assignment model • Understand the relationship between the assignment model and the transportation model and thereby linear programming • Understand the advantage of using a more general model (transportation or linear programming) rather than a specific model (transportation)
Integer Programming • Understand the slight difference between modeling linear programs and integer linear programs • Understand the major difference between the solution of linear programs and integer programs • Understand the additional modeling we can do using 0/1 variables • Understand some standard integer programming models (capital budgeting, set covering)
EMBA – LP: Using Excel; Transportation & Assignment; Integer Programming
Agenda
FRIDAY
1. Interfaces Presentations
|Guess, Sharon |Against Your Better Judgement? How Organizations Can Improve Their Use of Management Judgement in Forecasting |
|JACOBSON, RANDOLPH S |Contract Optimization at the Texas Children's Hospital, |
|REYES, CESAR A |Warner Robins Air Logistics Center Streamlines. Aircraft Repair and Overhaul |
2. Material we did not get to last time - We will use our QM for Windows Lego.lin file for the following • Graph • Possible outcomes o Unique solution (original example) o Multiple solutions (change profit from 17 to 20) o Unbounded solution (change = constraints) o No feasible solution (add constraint # tables >=10)
3. Lecture – (File: lp.excel.trans.assign.xls) Part 1 – Creating linear programming models in Excel (Worksheet: Lego) Software: Excel including Excel’s Solver add-in (see Tools menu or Data tab) Excel – recreate the Lego problem in Excel – examine Solver’s reports – interpretation of the dual in more detail (Worksheet: Lego – revisited) – demonstration of using Solver for nonlinear problems (Worksheet: errors– revisited) Part 2 – The Transportation and Assignment Problems Software: QM for Windows and Excel QM Transportation Model Description (Example 1 – page 4) Model Supplies, demands, shipping costs Assumptions – proportionality, additivity Solution Starting method Shipments Marginal costs (improvement values) Unbalanced problems - (Example 2) Suppose that the demand at Houston drops by 10 units More modeling – (Example 3) Suppose that shipping from Austin to Ft. Worth is not permitted QM for Windows, Excel QM Formulation as linear program (see page 5) Results, Reduced costs, Dual Values Assignment Description | 636 | 3,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-39 | latest | en | 0.80373 |
http://politicalcalculations.blogspot.com/2006/07/whos-in-debt-in-us.html | 1,477,441,284,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720471.17/warc/CC-MAIN-20161020183840-00493-ip-10-171-6-4.ec2.internal.warc.gz | 208,571,447 | 118,286 | Unexpectedly Intriguing!
July 17, 2006
Recently, Political Calculations asked and answered the question of how well American families are handling their debt using data from the Federal Reserve's 2004 Survey of Consumer Finances, which was published in February 2006. Today, we're going back to that source to answer a different question - who's more likely to be in debt: the rich or the poor?
Using net worth as our metric for determining wealth status, we'll first look at who's in debt by seeing how many people in each of the report's net worth percentile brackets (profiled here) have any debt of any kind. The following table has been extracted from Table 11 (p A28) of the survey report:
Debt Holding Status by Net Worth Percentiles
Net Worth Percentile BracketPercentage of Families in Bracket with DebtPercentage of Families in Bracket Without Debt
0 - 2564.9%35.1%
25 - 5083.8%16.2%
50 - 7583.2%16.8%
75 - 9074.6%25.4%
90 - 10072.7%27.3%
Remarkably, we see that the poorest group, defined as those in the 0 to 25th net worth percentile bracket, are the least likely to even have debt, with 35.1% of this group being debt free! Instead, we find that the group most likely to have debt is in the 25th to 50th percentile bracket, whose net worth ranges from \$13,300 at the low end to \$93,100 at the high end. The second most likely group to have debt, following closely behind the 25-50th percentile group, is found in the 50th to 75th percentile bracket, where the net worth of families at the top end of the bracket is \$328,500.
Who might have guessed that those with the lowest net worth would be the least likely to have debt?!
### Previously on Political Calculations
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The tools on this site are built using JavaScript. If you would like to learn more, one of the best free resources on the web is available at W3Schools.com. | 615 | 2,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2016-44 | longest | en | 0.957197 |
http://intranet.math.vt.edu/netmaps/dynamic/deg37/deg37port86_Main.output | 1,579,512,555,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00511.warc.gz | 85,603,155 | 1,567 | These Thurston maps are NET maps for every choice of translation term. They are primitive and have degree 37. ALL THURSTON MULTIPLIERS c/d IN UNREDUCED FORM 0/1, 1/37, 1/1, 2/1, 3/1, 4/1, 5/1, 7/1, 8/1, 9/1, 11/1, 13/1, 14/1, 16/1 18/1, 19/1, 21/1, 25/1, 29/1, 31/1 EXCLUDED INTERVALS FOR THE HALF-SPACE COMPUTATION (-58.640427,58.640427) The half-space computation does not determine rationality. EXCLUDED INTERVALS FOR JUST THE SUPPLEMENTAL HALF-SPACE COMPUTATION INTERVAL COMPUTED FOR HST OR EXTENDED HST -6.082031)(6.083008 infinity EXTENDED HST The supplemental half-space computation shows that these NET maps are rational. SLOPE FUNCTION INFORMATION There are no slope function fixed points because every loop multiplier of the mod 2 slope correspondence graph is at least 1 and the map is rational. No nontrivial cycles were found. The slope function maps some slope to the nonslope. The slope function orbit of every slope p/q with |p| <= 50 and |q| <= 50 ends in the nonslope. If the slope function maps slope p/q to slope p'/q', then |p'| <= |p| for every slope p/q with |p| <= 50 and |q| <= 50. FUNDAMENTAL GROUP WREATH RECURSIONS When the translation term of the affine map is 0: NewSphereMachine( "a=<1,b*c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "b=<1,1,d,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "c=<1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,c>(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "d=(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "a*b*c*d"); When the translation term of the affine map is lambda1: NewSphereMachine( "a=<1,a^-1,d*a,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "b=<1,b,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "c=(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "d=(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "a*b*c*d"); When the translation term of the affine map is lambda2: NewSphereMachine( "a=<1,a,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,1>(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "b=<1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,c>(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "c=<1,1,d,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "d=<1,a*b,a^-1,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "a*b*c*d"); When the translation term of the affine map is lambda1+lambda2: NewSphereMachine( "a=<1,1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c>(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "b=(1,2)(3,4)(5,6)(7,8)(9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22)(23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)", "c=<1,b,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "d=<1,c^-1,c*d,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c,c^-1,c>(2,3)(4,5)(6,7)(8,9)(10,11)(12,13)(14,15)(16,17)(18,19)(20,21)(22,23)(24,25)(26,27)(28,29)(30,31)(32,33)(34,35)(36,37)", "a*b*c*d"); | 2,537 | 4,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-05 | latest | en | 0.518854 |
https://brilliant.org/problems/the-piggy-banks-problem-001/ | 1,490,220,557,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186353.38/warc/CC-MAIN-20170322212946-00303-ip-10-233-31-227.ec2.internal.warc.gz | 782,909,998 | 12,462 | # The Piggy Banks Problem - 001
Two piggy banks contains some coins. The first piggy bank contains 4 fair coins and 2 biased coins in favor of heads. The second piggy bank contains 2 fair coins and 4 biased coins in favor of tails. The biased coins follow the odds 2:1. A piggy bank is selected at random, a coin is obtained from it and is found to be biased, what is the probability that the coin is from the first piggy bank if each piggy bank has an equal chance to be selected?
× | 120 | 485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-13 | longest | en | 0.947179 |
https://www.teacherspayteachers.com/Browse/Search:angles%20coloring | 1,547,783,942,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659677.17/warc/CC-MAIN-20190118025529-20190118051529-00314.warc.gz | 960,768,618 | 51,737 | showing 1-24 of 3,079 results
Parallel Lines Cut by a Transversal Coloring Activity This is a fun way for students to practice solving problems related to parallel lines cut by a transversal, all while coloring a vibrant beach scene! Students must have an understanding of alternate interior, alternate exterior, corresponding, c
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\$2.00
407 Ratings
4.0
PDF (2.37 MB)
Common Core 4MD06 Measure angles in whole-number degrees using a protractor. Students must use a protractor to measure 11 angles. Distractor answers (common mistakes that students make when measuring) are present. Each answer will inform the students what color to use on the coloring page for t
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FREE
150 Ratings
4.0
PDF (381.5 KB)
Use this handout to teach students the confusing vocabulary around angles: alternate interior; alternate exterior; vertical; corresponding; supplementary; complementary
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FREE
90 Ratings
4.0
PDF (83.71 KB)
This is a coloring activity for 10 problems on parallel line angles. Algebra 1 is reinforced for ALL problems. Posted: 5/3/15 so 50% off through 5/6/15 This product is included FREE in the Parallel Line activities BUNDLE.
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\$1.50
19 Ratings
4.0
ZIP (234.83 KB)
This is a color by numbers worksheet of 12 missing angles problems that include supplementary, complementary, vertical, and adjacent angles. Students solve the problems, find the answer on the table, and color the butterfly based on the color code given. Missing Angles Coloring Activity by Suzie's
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Types:
CCSS:
\$3.00
40 Ratings
4.0
PDF (720.14 KB)
In this activity students will solve for x by using their knowledge of complementary and supplementary angles in order to complete the coloring page. There are a total of 10 problems. This resource is grouped with three other activities in the Complementary and Supplementary Angles Activity Pack
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\$1.50
40 Ratings
3.9
PDF (792.55 KB)
Circles - Central and Inscribed Angles Color-By-Number Worksheet This color-by-number worksheet covers the concepts Central and Inscribed Angles in Circles. Students are given multiple situations and types of central and inscribed angles. When they find their answer, they look in the solution box
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\$2.00
34 Ratings
4.0
PDF (1.11 MB)
This is a coloring activity for a set of 10 problems all on vertical angles. Algebra 1 is reinforced in this product. This product is included for FREE in the Angle Pairs BUNDLE. There are several coloring activities on angles. Posted: 10/27/16 so 50% off through 10/30/16
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\$1.50
9 Ratings
4.0
PDF (198.68 KB)
This is a coloring activity for a set of 14 problems on classifying angles as acute, right, obtuse, or straight. All problems have pictures of angles.Posted: 8/5/15 so 50% off through 8/8/15
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\$1.50
32 Ratings
4.0
PDF (199.5 KB)
In this activity, students will classify angles (graphic and numerical representations) as obtuse, acute, straight, or right. Then, they will color a pattern according to their answers to reveal a beautiful, colorful mandala! This no-prep activity is an excellent resource for sub plans, enrichment
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Also included in: Geometry Skills Color By Number Bundle 1: 10 Essential Skills
\$2.00
16 Ratings
4.0
PDF (659.58 KB)
This is a coloring activity for a set of 14 problems on naming angles using 3 letters. There is also a naming angles maze product for \$1.00. Posted: 2/26/16 so 50% off through 2/29/16
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\$1.50
16 Ratings
3.9
PDF (238.99 KB)
In this activity students will solve for x by using their knowledge of vertical angles in order to complete the coloring page. There are a total of 8 problems. This resource is grouped with three other activities in the Vertical and Adjacent Angles Activity Pack You can also find this resource i
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\$1.50
16 Ratings
4.0
PDF (957.75 KB)
Valentine's Day Math Color By Number pictures are a fun and easy way to review measuring angles. Keep students learning and having fun with this no-prep Valentine's Day math option. Save over 25% by buying this resource in the Winter Color by Number Bundle. Save over 50% by buying this resourc
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\$3.00
15 Ratings
3.9
PDF (9.97 MB)
This is a coloring design for a set of 10 problems on inscribed angles. I used 5 colors in the design. There are 2 versions to choose from. Easy version and hard version where Algebra 1 is reinforced in most of the problems. Posted: 5/10/15 so 50% off through 5/13/15
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\$1.50
14 Ratings
4.0
ZIP (204.42 KB)
This is a coloring activity for a set of 10 problems on finding angle measures in parallelograms. Algebra is reinforced in this activity. Posted: 9/16/15 so 50% off through 9/19/15
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Types:
\$1.50
9 Ratings
4.0
ZIP (298.64 KB)
Color by numbers are a favorite in math class! In this activity, students answer 20 questions about angle measures — for 16 questions, they measure the angles; for 4 questions, they find an angle measure when a larger angle is decomposed into 2 angles. The angle measure answers are no closer tha
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Types:
CCSS:
Also included in: Math Color by Number Bundle (#3) - Grades 4-5
\$2.25
7 Ratings
4.0
PDF (2.31 MB)
This is a coloring activity for a set of 12 problems. Problems 1 - 6 are list the angles in order from least to greatest and problems 7 - 12 are on ordering the sides from least to greatest. Students need to apply triangle sum theorem to find the missing angle first for problems 7 - 12. Posted:
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Types:
\$1.50
6 Ratings
4.0
PDF (223.54 KB)
This is a coloring activity for a set of 12 problems on angles in kites and trapezoids. (6 of each) I used 5 colors in the design and a lot of trapezoids. Algebra 1 is needed to solve several problems. Posted: 5/14/15 so 50% off through 5/17/15 This product is included free in the QUAD SUM Bu
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\$1.50
5 Ratings
4.0
ZIP (127.79 KB)
Fun and engaging coloring activity. Students have to find the missing angle of a triangle and select the correct answer. Each problem has an answer with a color next to it. On the back (print double sided) students will color the corresponding question number to the number on the back. For example,
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Types:
\$1.00
4 Ratings
4.0
PDF (458.64 KB)
Hello Happy Teachers! This is a great product for your INB (Interactive Notebooks); or it can be used for enrichment, stations, assessment, or review. I used this in my standard 8th grade math classroom, but it could be used in any grade covering "Angle Relationships". Directions for Teacher: 1.
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CCSS:
\$2.50
2 Ratings
4.0
PDF (1.18 MB)
Use circles with secant lines and tangent lines to complete this coloring activity. This activity uses secant and tangent segments to solve for x in relation to angle measurements. It includes both secant-secant, secants-tangent, and tangent-tangent knowledge. Students use their answers to color a
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\$1.75
2 Ratings
3.9
PDF (927.16 KB)
This complementary and supplementary angles resource is a coloring activity that allows students to practice finding the missing angle. There are ten questions and a coloring worksheet that are a perfect activity to supplement your lesson on complementary and supplementary angles. It can be used wit
Subjects:
Types:
CCSS:
\$1.50
1 Rating
4.0
PDF (1.71 MB)
This is a color by answer that focuses on finding the missing angle measurement in parallel lines cut by a transversal, in triangles, and the exterior angle of a triangle.
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Types:
CCSS:
\$3.00
1 Rating
4.0
PDF (278.83 KB)
Your students will have fun practicing their CLASSIFYING ANGLES skills with this engaging color-by-code math puzzle. It is truly no-prep! Just print and go!I’ve included a color answer key for fast visual checking. Students can even check their own work!Use the coloring tool of your choice. I recomm
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Types:
CCSS:
\$1.00
1 Rating
3.5
PDF (1.01 MB)
showing 1-24 of 3,079 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,080 | 8,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-04 | latest | en | 0.852574 |
https://ycszkj.top/2022/05/28/how-november-23-the-lottery-7-quick-methods-to-boost-your-lottery-win-chances/ | 1,675,339,568,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00778.warc.gz | 1,104,656,752 | 9,084 | # How November 23 The Lottery – 7 Quick Methods To Boost Your Lottery Win Chances
“Can I win the lottery?” Right here is the question when i often asked myself. Besides this question, every lottery player comes with a question about lottery that bothers him/her.
The last good thing of playing online lottery quite simply can get free lotto entry pass. Lotteries online don’t have food with caffeine . issues like of those small town lotteries like electricity bills and room maintenance. Longer than you effortlessly find the ticket, you would get free vacation.
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How November 23 The Lottery – 7 Quick Methods To Boost Your Lottery Win Chances
Scroll to top | 510 | 2,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.964165 |
https://www.motilaloswal.com/blog-details/how-to-calculate-initial-margin-for-futures-contracts/1452 | 1,701,531,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00422.warc.gz | 987,430,221 | 105,114 | How to Calculate Initial Margin for Futures Contracts| Motilal Oswal
# How to Calculate Initial Margin for Futures Contracts
Whether you are a seasoned broker or a novice to the trading game, and searching for ways to make huge gains, the chances are that you may have come across the word “margin”. The concept of a margin is crucial for all those who venture into trading in commodity futures or derivatives in all classes. The futures margin can be referred to as a deposit (a sum of money) made in good faith placed in a trading account in order to control and maintain a futures contract. This is not like a margin that you pay for stocks, where a margin is more akin to a down payment. Instead, a futures margin is like a performance bond, designed to make sure that traders fulfill their obligations of a financial nature.
## What is the margin in futures trading? - Get Your Basics Right
It is important to get one thing right from the start. Trading on a margin and futures margin are two different concepts. In simple words, trading on a margin translates to a way of investing on credit. You may take a loan from your broker so that you can invest in shares or any securities. On the other hand, the futures margin represents a figure which is a sum of money you have to maintain in your trading account, so you are able to enter a position for a futures contract. This is a percentage of the complete value of the futures contract.
## The Importance of Futures Margins
When you buy or sell a futures contract, why are you required to pay margins? The answer is that futures trading involve risk because the price movement could go against you. When you buy futures of the Nifty at a level of 10,300 and if the Nifty goes down to Rs.10,200 there is a loss you are incurring and that is your risk. Markets are by nature volatile and these margins are essentially collected to cover this volatility risk. So, how does futures margining work? Broadly, there are 2 types of margins that are normally collected. At the time of taking the position you are required to pay the Initial Margin on the position (SPAN + Exposure margin).
The SPAN margin is based on a statistical concept called VAR (Value at Risk). It basically means that the initial margin should be large enough to cover the loss of your position in 99% of the cases. There will be Black Swan days but that is a different issue. Initial margin is based on the potential maximum loss in a single day on the portfolio. Greater the volatility of the stock, greater the risk and therefore greater is the initial margin. The second type of margin is the mark-to-market (MTM) margin which is collected for daily volatility in the price of the futures. The initial margin only looks at single-day risk. If the stock continues to move against you (falling when you are long / rising when you are short), then on each subsequent day the MTM will be collected. So how does futures margin work in practice? Let us understand all about margins on futures contracts through a live example of Initial Margins and MTM margins..
## How Initial Margins are calculated?
The chart above calculates the initial margin as a sum of the SPAN and the Exposure margins. Let us use Margin Calculator for our referenceand try to analyse our calculations The minimum margins required for each specific position are defined by the stock exchange. Brokers are free to collect more than this margin but they are not permitted to collect less than this margin. In the above table, the Nov, Dec and Jan contracts of ACC are considered. You will find that as the contract goes farther into the future the margins are higher due to higher risk. But what is important here are the 3 classes of initial margins that are charged by the broker on your futures position. Let us understand the nuances of this futures margin example..
### Carry Forward (Normal Margin):
This is the normal margin that will have to be charged when you propose to carry forwards your futures position beyond the day. Normally, in case of Carry Forward trade the initial margin varies from 10% to 15% of the notional value of the contract depending on the risk and volatility of the stock. In the above case, for the Nov 2017 contract, the notional value of the futures contract is Rs.708,580/- (1771.45 X 400). On that notional value, the initial margin is collected at Rs.89,338/- per lot, which works out to 12.61% of the notional value. As mentioned earlier, this percentage of initial margin for the futures position will depend on the volatility and risk of the stock.
The normal margin pertains to a futures position that you propose to carry forward to the following day. However, what if you want to square off the position intraday? Since the risk is lower, the initial margins (MIS) will be lower. For Intraday index futures the initial margin is set at 40% of the normal initial margin while in case of intraday stock futures, the initial margin is set at 50% of the normal initial margin. In the above case, the margin will be 50% of the normal margin which is Rs.44,669/-. This margin is referred to as Margin Intraday Square-off (MIS) margin.
### CO / BO order Margin:
The third category which is even lower than the MIS margin is the BO/CO margin. These are the Bracket Orders and Cover Orders. In a cover order, the intraday trade is necessarily set with an in-built stop loss. The Bracket Order goes a step further and defines the stop loss and also a profit target making it a closed bracket order. The margin, in this case, will be 30-33% of the normal margins and in the case of ACC it is Rs.28,343/-
One thing you need to remember is that in case of MIS orders, CO orders and BO orders all open positions will normally be closed out by your broker’s risk management system (RMS) 15-30 minutes before the close of stock market trading on the same day.
### Mark to Market (MTM) margining:
Mark to Market (MTM) margin is an accounting adjustment. If you have bought the futures of Tata Motors at Rs.409 and as long as the price is above Rs.409, you really do not have much to worry. The MTM problem will come when come when the market price of Tata Motors goes below Rs.409. Here again there are 2 situations. Firstly, if the price goes down to Rs.407, most brokers will check if your margin balance is sufficient to cover the SPAN margin. (Remember Initial Margin is SPAN + Exposure Margin). That is still OK. However, if the stock price goes below say Rs.395, then your margin balance is likely to fall below the SPAN Margin. Then the broker will make a Margin Call asking you to fill up the deficit in margin and if you are unable to pay the margin then your position will be closed out by the RMS. Remember, MTM margins are only applicable to carry forward positions and not to intraday, BO or CO positions.
So, that is all about margins in futures contracts. The futures margins example above clearly highlights how do futures margins work in practice. In a nutshell, it is a risk management measure!
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https://community.qlik.com/thread/264756 | 1,534,761,624,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216333.66/warc/CC-MAIN-20180820101554-20180820121554-00339.warc.gz | 613,942,713 | 25,569 | 12 Replies Latest reply: Jun 22, 2017 1:03 PM by David Forest
# Hi All, I'm new to Qlik and could use some help.
I have pivot table that looks like the one below. I'm trying to take the difference between price and prior price (for same Lot# and plant), if increase in price, then multiply it by quantity sold at price increase. If no increase then 0. Thank you.
Rows Measures Lot# Customer Plant Month Price Quantity Sold Impact 0021 Johnnies 5 8 \$ 4.35 2 0021 Johnnies 5 9 \$ 5.01 25 =if(5.01>4.35 for same Lot# and plant, then Impact =(5.01-4.35) *25 0021 Johnnies 5 10 \$ 5.10 21 0021 Johnnies 5 11 \$ 5.10 23 0021 Johnnies 5 12 \$ 5.25 10 0035 Joney 7 8 \$ 2.67 6 0035 Joney 7 9 \$ 2.80 4 0035 Joney 7 10 \$ 2.85 53 0035 Joney 7 11 \$ 2.90 5
• ###### Re: Hi All, I'm new to Qlik and could use some help.
[price temp]:
INLINE
[
"Lot#", "Customer", "Plant", "Month", "Price", "Quantity Sold"
'0021', 'Johnnies', 5, 8, 4.35, 2
'0021', 'Johnnies', 5, 9, 5.01, 25
'0021', 'Johnnies', 5, 10, 5.10, 21
'0021', 'Johnnies', 5, 11, 5.10, 23
'0021', 'Johnnies', 5, 12, 5.25, 10
'0035', 'Joney', 7, 8, 2.67, 6
'0035', 'Joney', 7, 9, 2.80, 4
'0035', 'Joney', 7, 10, 2.85, 53
'0035', 'Joney', 7, 11, 2.90, 5
];
[price]:
NoConcatenate
if(peek("Lot#")="Lot#" and peek(Plant)=Plant and Peek(Price)<Price, Price-Peek(Price)*"Quantity Sold",0) AS Impact
RESIDENT [price temp]
ORDER BY "Lot#",Month;
DROP TABLE [price temp];
!
• ###### Re: Hi All, I'm new to Qlik and could use some help.
sample app
• ###### Re: Hi All, I'm new to Qlik and could use some help.
There was one error in your solution which I fixed. It was yielding negative values
So I updated it to fix that ( now shows 16.50 instead of -103.74 for impact on line 2 ) -- and changed up the script to eliminate the temp table.
[Price Updated]:
Load *, if(peek("Lot#")="Lot#" and peek(Plant)=Plant and Peek(Price)<Price, (Price-Peek(Price))*"Quantity Sold",0) AS Impact
INLINE
[
"Lot#", "Customer", "Plant", "Month", "Price", "Quantity Sold"
'0021', 'Johnnies', 5, 8, 4.35, 2
'0021', 'Johnnies', 5, 9, 5.01, 25
'0021', 'Johnnies', 5, 10, 5.10, 21
'0021', 'Johnnies', 5, 11, 5.10, 23
'0021', 'Johnnies', 5, 12, 5.25, 10
'0035', 'Joney', 7, 8, 2.67, 6
'0035', 'Joney', 7, 9, 2.80, 4
'0035', 'Joney', 7, 10, 2.85, 53
'0035', 'Joney', 7, 11, 2.90, 5
];
I'm new too so I wanted to try out your solution and experiment with preceding load. Thank you.
• ###### Re: Hi All, I'm new to Qlik and could use some help.
Thank you for the quick response David. I will try this once IT gets me the proper rights.
• ###### Re: Hi All, I'm new to Qlik and could use some help.
James - I updated David's solution to correct a small bug. Being new also - I posted it as a reply to him when perhaps maybe I needed to put as a reply on the original thread. I did add a change to eliminate the temp table - since I was trying out the preceding load .
• ###### Re: Hi All, I'm new to Qlik and could use some help.
Thanks Mary Jo, I have a question about the [ ]; at the end below INLINE with the column headings and the rows of data. That was a table I made up so you could see the structure of the table I was creating. My data table before the pivot is approx. 750,000 rows. I was trying to get the script to enter in the App script for the pivot. How do I just enter the script to perform the added calculation and add the Impact column to table? Like I said I'm very new to this.
• ###### Re: Hi All, I'm new to Qlik and could use some help.
James,
I am new to this too. It sounds like you have a qvf started - is it possible to upload at least the script for the table you are referring to? In the meanwhile - let me look for the example I created which doesn't use an inline table - that might answer your question.
• ###### Re: Hi All, I'm new to Qlik and could use some help.
INLINE is a way to quickly add some data, the Load*, if(peek("Lot#")="Lot#" and peek(Plant)=Plant and Peek(Price)<Price, (Price-Peek(Price))*"Quantity Sold",0) AS Impact could be added above any other SQL SELECT or LOAD statement
this could also be done in the UI as a measure using Above() instead of Peek()
• ###### Re: Hi All, I'm new to Qlik and could use some help.
Thank you David for letting me know I could use Above() in UI. I did this and it worked perfectly. I can't thank you and Mary Jo enough for all your help. I have a cookbook on Qlik coming today and can't wait to dig into it.
Best regards,
Jim
• ###### Re: Hi All, I'm new to Qlik and could use some help.
I just noticed that the formula only works when price changes in same month. If month changes in the next row with a higher price for the same part, it won't calculate. ??
• ###### Re: Hi All, I'm new to Qlik and could use some help.
yes, above and Peek, not sure which method you're referring to, by default go back one row in the data. if you have more than one row per month, you'd have to summarize the data at the month level, avg or max or FirstSortedValue using -Date to get last. | 1,663 | 5,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-34 | latest | en | 0.766293 |
https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2008-July/035058.html | 1,568,863,038,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573415.58/warc/CC-MAIN-20190919015534-20190919041534-00091.warc.gz | 562,440,724 | 2,981 | # [gmx-users] rot_diff
Xavier Periole X.Periole at rug.nl
Tue Jul 8 09:31:10 CEST 2008
```Have a look at those papers.
JMB-(1994)239:629-636.
JPC/B-(2008)112:6013-6024.
XAvier
On Mon, 7 Jul 2008 18:35:57 -0400
"rams rams" <rams.crux at gmail.com> wrote:
> Dear XAvier,
>
> Thanks for your reply and for the explanation. I am not an NMR guy so I
> would like to know little bit of more about the way we can calculate the
> rotational diffusion. The way I understood is the following and let me know
> if I am wrong.
>
> After obtaining the rotaional correlation function using Gromacs tools
> (g_rotacf), I need to calculate the correlation time I suppose.
> The obtained correlation time is related with the local diffusion constant
> (d) by the relation
> d = 1/l(l+1) t
> t is the correlation time obtained above and l = 1 or 2 depends upon the
> order of the legendre polynomial we will use in the g_rotacf and the
> experimental results with which we are comparing.
>
> then by solving the following relation
> d=n'Qn (n is a unit vector lies along the vector connecting the two spins),
> we can obtain "Q" which inturn is in relation with D (the diffusion tensor).
>
>
> Thats the overall idea I have but I am sure I need to worry alot of finer
> other details while I start putting my hands into it. If the overall idea is
> alright I could put the things in a more detailed way.
>
> Ram.
>
> On Sat, Jul 5, 2008 at 12:51 PM, Xavier Periole <X.Periole at rug.nl> wrote:
>
>> On Sat, 5 Jul 2008 10:40:21 -0400
>> "rams rams" <rams.crux at gmail.com> wrote:
>>
>>> Dear users,
>>>
>>> Is it possible to evaluate the rotational diffusion of proteins using
>>> gromacs tools ??
>>>
>> No directly. However you can use g_rotacf to generate the autocorrelation
>> function of vectors (option -d). By defining vectors representing your
>> molecule/protein you can access the rotational correlation time of your
>> representative vector. You can imagine different way to get a statistically
>> significant value. One would be to define many vectors between backbone
>> atoms and average your results. Another would be to again define many
>> vectors but this time between the center of mass of the protein and each
>> Ca atoms and average ...
>>
>> You can also hack the g_rms code to extract the rotational matrix during
>> the overlay of your protein to a reference structure and apply it to
>> a unit vector from whose trajectory you can again use g_rotacf to get
>> the autocorrelation function of that vector ...
>>
>> An important point in the comparison of your result to experimental
>> values is the way the rotational correlation time is extracted
>> experimentally. They select different mode of relaxation (1 or 2) and thus
>> you have to use the corresponding Legendre polynomial when calculating the
>> autocorrelation function. From NMR relaxation l=2.
>>
>> XAvier.
>>
>> -----------------------------------------------------
>> XAvier Periole - PhD
>>
>> Molecular Dynamics Group
>> - NMR and Computation -
>> University of Groningen
>> The Netherlands
>> -----------------------------------------------------
>> _______________________________________________
>> gmx-users mailing list gmx-users at gromacs.org
>> http://www.gromacs.org/mailman/listinfo/gmx-users
>> Please search the archive at http://www.gromacs.org/search before posting!
>> Please don't post (un)subscribe requests to the list. Use the www interface
>> or send it to gmx-users-request at gromacs.org.
>>
-----------------------------------------------------
XAvier Periole - PhD
Molecular Dynamics Group / NMR and Computation
University of Groningen
The Netherlands
-----------------------------------------------------
``` | 911 | 3,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-39 | latest | en | 0.882324 |
https://www.atilim.edu.tr/en/ects/site-courses/202/18047/detail | 1,656,584,093,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00340.warc.gz | 702,472,468 | 20,645 | # Calculus I (MATH151) Course Detail
Course Name Course Code Season Lecture Hours Application Hours Lab Hours Credit ECTS
Calculus I MATH151 4 2 0 5 7
Pre-requisite Course(s)
N/A
Course Language English N/A Bachelor’s Degree (First Cycle) Face To Face Lecture, Question and Answer, Problem Solving. The course is designed to fill the gaps in students knowledge that they have in their pre-college education and then to give them computational skills in one-variable differential and integral calculus to handle engineering problems The students who succeeded in this course; understand, define and use functions, and represent them by means of graphs understand fundamental concepts of limit and continuity understand the meaning of derivative and calculate derivatives of one-variable functions use derivatives to solve problems involving maxima, minima, and related rates understand integration, know integration techniques, use them to solve area, volume and other problems Preliminaries, limits and continuity, differentiation, applications of derivatives, L`Hopital's Rule, integration, applications of integrals, integrals and transcendental functions, integration techniques and improper integrals, squences.
### Weekly Subjects and Releated Preparation Studies
Week Subjects Preparation
1 P.1 Real Numbers and the Real Line, P.2 Cartesian Coordinates in the Plane, P.3 Graphs of Quadratic Equations, P.4 Functions and Their Graphs, pp:3-33
2 P.5 Combining Functions to Make New Functions, P.6 Polynomials and Rational Functions, P.7 Trigonometric Functions, pp:33-57
3 1.1 Examples of Velocity, Growth Rate, and Area, 1.2 Limits of Functions, 1.3 Limits at Infinity and Infinite Limits, 1.4 Continuity, pp:58-87
4 1.5 The Formal Definition of Limit, 2.1 Tangent Lines and Their Slopes, 2.2 The Derivative, 2.3 Differentiation Rules, pp:87-114
5 2.4 The Chain Rule, 2.5 Derivatives of Trigonometric Functions, 2.6 Higher-Order Derivatives, pp:115-129
6 2.7 Using Differentials and Derivatives, 2.8 The Mean Value Theorem, 2.9 Implicit Differentiation, 3.1 Inverse Functions, pp:129-147 pp:163-169
7 Midterm
8 3.2 Exponential and Logarithmic Functions, 3.3 The Natural Logarithm and Exponential, 3.4 Growth and Decay (Theorem 4, Theorem 5, Theorem 6 and Examples for these theorems), 3.5 The Inverse Trigonometric Functions, pp:169-187 pp:190-197
9 3.6 Hyperbolic Functions (only their definition and derivatives), 4.1 Related Rates, 4.3 Indeterminate Forms, pp:198-203 pp:213-219 pp:227-232
10 4.4 Extreme Values, 4.5 Concavity and Inflections, 4.6 Sketching the Graph of a Function, pp:232-252
11 4.8 Extreme-Value Problems, 4.9 Linear Approximations, 2.10 Antiderivatives and Initial Value Problems (Antiderivatives, The Indefinite Integral), 5.1 Sums and Sigma Notation, pp:258-271 pp:147-150 pp:288-293
12 5.2 Areas as Limits of Sums, 5.3 The Definite Integral, 5.4 Properties of the Definite Integral, 5.5 The Fundamental Theorem of Calculus, pp:293-316
13 5.6 The Method of Substitution, 5.7 Areas of Plane Regions, 6.1 Integration by Parts, pp:316-337
14 6.2 Integrals of Rational Functions, 6.3 Inverse Substitutions, 6.5 Improper Integrals, pp:337-353 pp:359-367
15 7.1 Volumes by Slicing – Solids of Revolution, 7.2 More Volumes by Slicing, 7.3 Arc Length and Surface Area (only Arc Length), Review, pp:390-407
16 Final Exam
### Sources
Course Book 1. Calculus: A complete Course, R. A. Adams, C. Essex, 7th Edition; Pearson Addison Wesley 2. Thomas’ Calculus Early Transcendentals, 11th Edition.( Revised by M. D. Weir, J.Hass and F. R. Giardano; Pearson , Addison Wesley) 3. Calculus: A new horizon, Anton Howard, 6th Edition; John Wiley & Sons 4. Calculus with Analytic Geometry, C. H. Edwards; Prentice Hall 5. Calculus with Analytic Geometry, R. A. Silverman; Prentice Hall
### Evaluation System
Requirements Number Percentage of Grade
Attendance/Participation - -
Laboratory - -
Application - -
Field Work - -
Special Course Internship - -
Quizzes/Studio Critics - -
Homework Assignments - -
Presentation - -
Project - -
Report - -
Seminar - -
Midterms Exams/Midterms Jury 2 60
Final Exam/Final Jury 1 40
Toplam 3 100
Percentage of Semester Work 60 40 100
### Course Category
Core Courses X
### The Relation Between Course Learning Competencies and Program Qualifications
# Program Qualifications / Competencies Level of Contribution
1 2 3 4 5
1 An ability to apply knowledge in mathematics and basic sciences and computational skills to solve manufacturing engineering problems
2 An ability to define and analyze issues related with manufacturing technologies
3 An ability to develop a solution based approach and a model for an engineering problem and design and manage an experiment
4 An ability to design a comprehensive manufacturing system based on creative utilization of fundamental engineering principles while fulfilling sustainability in environment and manufacturability and economic constraints
5 An ability to chose and use modern technologies and engineering tools for manufacturing engineering applications
6 An ability to utilize information technologies efficiently to acquire datum and analyze critically, articulate the outcome and make decision accordingly
7 An ability to attain self-confidence and necessary organizational work skills to participate in multi-diciplinary and interdiciplinary teams as well as act individually
8 An ability to attain efficient communication skills in Turkish and English both verbally and orally
9 An ability to reach knowledge and to attain life-long learning and self-improvement skills, to follow recent advances in science and technology
10 An awareness and responsibility about professional, legal, ethical and social issues in manufacturing engineering
11 An awareness about solution focused project and risk management, enterpreneurship, innovative and sustainable development
12 An understanding on the effects of engineering applications on health, social and legal aspects at universal and local level during decision making process
### ECTS/Workload Table
Activities Number Duration (Hours) Total Workload
Course Hours (Including Exam Week: 16 x Total Hours) 16 4 64
Laboratory
Application 16 2 32
Special Course Internship
Field Work
Study Hours Out of Class 14 3 42
Presentation/Seminar Prepration
Project
Report
Homework Assignments
Quizzes/Studio Critics
Prepration of Midterm Exams/Midterm Jury
Prepration of Final Exams/Final Jury 1 18 18
Total Workload 156 | 1,602 | 6,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-27 | latest | en | 0.84821 |
https://unimath.github.io/agda-unimath/order-theory.meet-suplattices.html | 1,726,839,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00415.warc.gz | 541,287,857 | 39,415 | # Meet-suplattices
Content created by Egbert Rijke, Fredrik Bakke, Julian KG, fernabnor and louismntnu.
Created on 2023-05-08.
module order-theory.meet-suplattices where
Imports
open import foundation.binary-relations
open import foundation.dependent-pair-types
open import foundation.propositions
open import foundation.sets
open import foundation.universe-levels
open import order-theory.meet-semilattices
open import order-theory.posets
open import order-theory.suplattices
## Idea
An l-meet-suplattice is a meet-semilattice L which has least upper bounds for all families of elements x : I → L indexed by a type I : UU l.
Note that meet-suplattices are not required to satisfy a distributive law. Such meet-suplattices are called frames.
## Definitions
### The predicate on meet-semilattices of being a meet-suplattice
module _
{l1 : Level} (l2 : Level) (X : Meet-Semilattice l1)
where
is-meet-suplattice-Meet-Semilattice-Prop : Prop (l1 ⊔ lsuc l2)
is-meet-suplattice-Meet-Semilattice-Prop =
is-suplattice-Poset-Prop l2 (poset-Meet-Semilattice X)
is-meet-suplattice-Meet-Semilattice : UU (l1 ⊔ lsuc l2)
is-meet-suplattice-Meet-Semilattice =
type-Prop is-meet-suplattice-Meet-Semilattice-Prop
is-prop-is-meet-suplattice-Meet-Semilattice :
is-prop is-meet-suplattice-Meet-Semilattice
is-prop-is-meet-suplattice-Meet-Semilattice =
is-prop-type-Prop is-meet-suplattice-Meet-Semilattice-Prop
### Meet-suplattices
Meet-Suplattice : (l1 l2 : Level) → UU (lsuc l1 ⊔ lsuc l2)
Meet-Suplattice l1 l2 =
Σ (Meet-Semilattice l1) (is-meet-suplattice-Meet-Semilattice l2)
module _
{l1 l2 : Level} (A : Meet-Suplattice l1 l2)
where
meet-semilattice-Meet-Suplattice : Meet-Semilattice l1
meet-semilattice-Meet-Suplattice = pr1 A
poset-Meet-Suplattice : Poset l1 l1
poset-Meet-Suplattice =
poset-Meet-Semilattice meet-semilattice-Meet-Suplattice
type-Meet-Suplattice : UU l1
type-Meet-Suplattice =
type-Poset poset-Meet-Suplattice
leq-meet-suplattice-Prop : (x y : type-Meet-Suplattice) → Prop l1
leq-meet-suplattice-Prop = leq-Poset-Prop poset-Meet-Suplattice
leq-Meet-Suplattice : (x y : type-Meet-Suplattice) → UU l1
leq-Meet-Suplattice = leq-Poset poset-Meet-Suplattice
is-prop-leq-Meet-Suplattice :
(x y : type-Meet-Suplattice) → is-prop (leq-Meet-Suplattice x y)
is-prop-leq-Meet-Suplattice = is-prop-leq-Poset poset-Meet-Suplattice
refl-leq-Meet-Suplattice : is-reflexive leq-Meet-Suplattice
refl-leq-Meet-Suplattice = refl-leq-Poset poset-Meet-Suplattice
antisymmetric-leq-Meet-Suplattice : is-antisymmetric leq-Meet-Suplattice
antisymmetric-leq-Meet-Suplattice =
antisymmetric-leq-Poset poset-Meet-Suplattice
transitive-leq-Meet-Suplattice : is-transitive leq-Meet-Suplattice
transitive-leq-Meet-Suplattice = transitive-leq-Poset poset-Meet-Suplattice
is-set-type-Meet-Suplattice : is-set type-Meet-Suplattice
is-set-type-Meet-Suplattice = is-set-type-Poset poset-Meet-Suplattice
set-Meet-Suplattice : Set l1
set-Meet-Suplattice = set-Poset poset-Meet-Suplattice
is-suplattice-Meet-Suplattice :
is-suplattice-Poset l2 poset-Meet-Suplattice
is-suplattice-Meet-Suplattice = pr2 A
suplattice-Meet-Suplattice : Suplattice l1 l1 l2
suplattice-Meet-Suplattice =
( poset-Meet-Suplattice , is-suplattice-Meet-Suplattice)
meet-Meet-Suplattice :
(x y : type-Meet-Suplattice) →
type-Meet-Suplattice
meet-Meet-Suplattice =
meet-Meet-Semilattice meet-semilattice-Meet-Suplattice
sup-Meet-Suplattice :
{I : UU l2} → (I → type-Meet-Suplattice) →
type-Meet-Suplattice
sup-Meet-Suplattice {I} f = pr1 (is-suplattice-Meet-Suplattice I f) | 1,311 | 3,547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.405241 |
https://nextreleaseunknown.com/qa/what-is-the-frequency-symbol.html | 1,638,152,132,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00017.warc.gz | 510,767,276 | 21,529 | # What Is The Frequency Symbol?
## How many types of frequency are there?
Wavelength:ELFEXTREMELY LOW FREQUENCY Frequency: 3 KHz to 30 KHz Wavelength: 100 km to 10 kmHFHIGH FREQUENCY Frequency: 3 MHz to 30 MHz Wavelength: 100 m to 10 mVHFVERY HIGH FREQUENCY Frequency: 30 MHz to 300 MHz Wavelength: 10 m to 1 mUHFULTRA HIGH FREQUENCY Frequency: 300 MHz to 3 GHz Wavelength: 1 m to 100 mm4 more rows•Jul 10, 2019.
## How do you type a frequency symbol?
The symbols most often used for frequency are f and the Greek letters nu (ν) and omega (ω). Nu is used more often when specifying electromagnetic waves, such as light, X-rays, and gamma rays.
## What is Hz equal to?
It is measured in hertz (Hz), an international unit of measure where 1 hertz is equal to 1 cycle per second. Hertz (Hz) = One hertz is equal to one cycle per second.
## What is the meaning of frequency?
noun, plural fre·quen·cies. rate of occurrence: The doctor has increased the frequency of his visits. Physics. the number of periods or regularly occurring events of any given kind in unit of time, usually in one second. the number of cycles or completed alternations per unit time of a wave or oscillation.
## What is the formula for Hz?
The formula for time is: T (period) = 1 / f (frequency). λ = c / f = wave speed c (m/s) / frequency f (Hz). The unit hertz (Hz) was once called cps = cycles per second….Centimeters per period / div.cmTimebase Yms↓Frequency f = 1/THz4 more rows
## What is frequency and its types?
Definition: The frequency is the number of oscillation per unit time. It is used for defining the cyclic process like rotation, oscillation, wave etc. The completion of the cyclic process at particular interval of time is known as the frequency. Frequency is the total number of oscillations per unit time. …
## How do u find the frequency?
How to calculate frequencyDetermine the action. Decide what action you want to use to determine the frequency. … Select the length of time. Select the length of time over which you will measure the frequency. … Divide the numbers. To calculate frequency, divide the number of times the event occurs by the length of time.Feb 22, 2021
## How do you measure human frequency?
A first real time spectrum analyzer is connected to a broad frequency range antenna placed in contact with the skin, electromagnetic fields emitted by a particular condition of the subject human body are measured to reveal peaks of power, and a second narrow frequency antenna is used to measure the peak more accurately …
## What are the three units of frequency?
The units of frequency are thus cycles per second, or Hertz (Hz). Radio stations have frequencies. They are usually equal to the station number times 1,000,000 Hz. For instance, – the local Washington, DC station HFS has a frequency of 99.1 million Hz in the FM radio band.
## What is the law of frequency?
The law of frequency suggests that the more a person practices the correct and desired behavior, the more a person will use and make the desired behavior a personal habit.
## Is Hz and FPS same?
No; they are two separate things. Remember that FPS is how many frames your gaming computer is producing or drawing, while the refresh rate is how many times the monitor is refreshing the image on the screen. The refresh rate (Hz) of your monitor does not affect the frame rate (FPS) your GPU will be outputting.
## How many watts are in a Hz?
Hz stands for hertzs and W*s stands for watt-seconds. The formula used in hertzs to watt-seconds conversion is 1 Hertz = 6.62606957030801E-34 Watt-Second. In other words, 1 hertz is 1.509190311676E+33 times smaller than a watt-second.
## What is the unit for frequency?
hertz (Hz)The SI unit for frequency is the hertz (Hz). One hertz is the same as one cycle per second.
## How do you find the class frequency?
The relative frequency of a class is found by dividing the frequency by the number of values in the data sample – this gives the proportion that fall into that class. The cumulative relative frequency is found by dividing the relative frequency by the number in the sample.
## What is the frequency of this wave?
Wave frequency is the number of waves that pass a fixed point in a given amount of time. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave passing a fixed point in 1 second. A higher-frequency wave has more energy than a lower-frequency wave with the same amplitude.
## What is frequency and what is its symbol?
The SI unit for frequency is the Hertz (Hz), which is equivalent to the older unit cycles per second (cps). Frequency is also known as cycles per second or temporal frequency. The usual symbols for frequency are the Latin letter f or the Greek letter ν (nu).
## What is frequency and example?
Frequency describes the number of waves that pass a fixed place in a given amount of time. … The hertz measurement, abbreviated Hz, is the number of waves that pass by per second. For example, an “A” note on a violin string vibrates at about 440 Hz (440 vibrations per second).
## How do you find percent frequency?
Percentage is calculated by taking the frequency in the category divided by the total number of participants and multiplying by 100%. To calculate the percentage of males in Table 3, take the frequency for males (80) divided by the total number in the sample (200). Then take this number times 100%, resulting in 40%. | 1,263 | 5,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-49 | latest | en | 0.914281 |
http://perplexus.info/show.php?pid=4772&cid=33739 | 1,537,808,140,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160620.68/warc/CC-MAIN-20180924165426-20180924185826-00292.warc.gz | 191,017,538 | 4,336 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
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An Identity Problem (Posted on 2006-08-17)
Four inhabitants of the island make the statements as given below. Each of these inhabitants are either Liars or Knights. Little is known regarding the identity of the four inhabitants except for the fact that Q and R are not both Knights.
P's Statement: If asked, Q would say that he and S belong to the same group.
Q's Statement: If asked, S would say that he and P do not belong to the same group.
R's Statement: P and myself belong to the same group..
S's Statement: If asked, P would say that he and R do not belong to the same group.
To what group or groups do the four speakers belong?
See The Solution Submitted by K Sengupta Rating: 3.5000 (2 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re: Quick | Comment 4 of 11 |
(In reply to Quick by Federico Kereki)
If you start supposing Q is a Liar, you get the same solution. P must be a Knight, no matter what R is. Then, Q would lie saying that S is the same as he, so S would be a Knight,a nd that would imply R is not the same as P, so he is a Liar.
Posted by Old Original Oskar! on 2006-08-17 22:22:56
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Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information | 420 | 1,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-39 | latest | en | 0.919486 |
https://blog.udemy.com/how-to-alphabetize-in-excel/ | 1,670,167,542,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710974.36/warc/CC-MAIN-20221204140455-20221204170455-00645.warc.gz | 176,237,119 | 17,449 | Excel is an awesome tool for keeping long lists of sortable, easily accessible data. This can include anything from book collections, to-do lists, budgets, invoices, roll call sheets, grocery lists, and more.
In some cases, you may find yourself with a sprawling record of names, titles, and tasks that need to be placed in alphabetical order to really be useful. Luckily for you, this is a pretty basic function, though not without its quirks. In this guide, we’ll go over how to alphabetize in Excel, and what to be aware of when sorting long lists of data.
For extra help, take an Excel training course aimed at beginners here!
## Sorting a Basic List
The first thing you’re going to need is a list. In Excel, a spreadsheet consists of endless cells of data, organized into columns and rows. You could have a spreadsheet that consists of just one column, with multiple rows of data. This would just be a standard list, such as a list of guest names for an upcoming birthday party.
Let’s say you input a list of names into the spreadsheet as they popped into your head, but by the end you’d like to sort them out alphabetically just to make your life easier.
The first thing to do is select the data you want to be sorted, a simple task in this first example because our data is located in just one column. You can click on any row within your column of data – in this case, A1 through A15 – to select that column, or you can click the top or bottom most cell and drag your selection to highlight all the cells in the column.
In the image above, you can see “15R x 1C” displayed in the name box, to the left of the formula bar. This indicates that you have fifteen rows (15R) and one column (1C) highlighted.
Once your list is selected, all you have to do next is click on the Data tab at the top, also shown in the image above, and click the AZ icon with the down pointing arrow. This will sort your list in alphabetical order, from A to Z.
Done! But what if your list has more than one column?
## Sorting a List with Multiple Columns
In some cases, you may have a list made up of several columns. The data in these other columns may or may not correspond to the column(s) you want to sort, which can make things a bit more complicated.
Let’s start out simple, and say you just have a basic birthday guest list again, only you want to use the same spreadsheet to keep track of a list of things you need to buy for the party. These items will have nothing to do with the guests, they’re just a separate list in a separate column. You’ve ordered them according to their location in the grocery store, so that it’ll be easier for you to find them when you go shopping.
You want to sort the names in Column A alphabetically, but keep the list in Column B in the same order in which you input it.
To do so, highlight the cells in Column A just as we did in the first example, select the Data tab, and click the AZ button with the down pointing arrow. This time, you’ll receive the following prompt:
Excel has detected the data in Column B, and wants to know if this data corresponds with the data in Column A. Because it doesn’t, click the option that says “Continue with the current selection” and press OK.
The names in Column A will be sorted alphabetically, and the list in Column B will remain in the order you originally input.
## Sorting Multiple Lists With Corresponding Data
There may be times when your additional columns do contain information relevant to the one you want to sort. Sticking with the birthday guest example, let’s say you wrote out the names in the order the guests called you. This is your original unalphabetized list. The guests also let you know what food item they were bringing to the party.
You want to alphabetize the list of guests, but you want the food item they’re bringing to remain next to their name. You don’t want the food list to be alphabetized, because that will jumble everything up.
To do this, just highlight the data in both columns, and press the AZ button with the down pointing arrow. The two lists will sort themselves, keeping the guest names next to their corresponding food item. Or, you can highlight just Column A, click the AZ sort button, and select the “Expand the selection” option once you’re prompted.
If you highlight Column B – the list of food items – and “Expand the selection” after sorting, you’ll see the food items will alphabetize instead! The corresponding names will remain next to each item in Column A, though.
## Sorting Parts of a List
Let’s say the last five people to call you – Alli, Leslie, Carston, Jose, and Ayumi – are all a family, and they had made plans before you invited them which may or may not fall through. Their attendance is iffy.
You want to keep them on the list as “maybes,” but you don’t want to get them mixed into the bigger one. You still need to alphabetize that guest list, though! To keep the family of five separate, bold their names (just as a visual aid), and highlight the cells in both columns excluding theirs.
Then, simply click the AZ sort button, and voila! The guest list is alphabetized, except for the last five names.
## Sorting Horizontally
You can also sort cells horizontally, if you happen to have a list confined to one row and multiple columns, rather than multiple rows and one (or more) columns in the previous examples.
For this one, let’s stay simple. I’m getting tired of that birthday example, aren’t you?
To sort these columns alphabetically, and horizontally, highlight each cell in the list, and instead of clicking the AZ button with the down pointing arrow, click the box that says Sort in the Data tab.
Once you’re prompted with this box, click “Options…” and the following pop-up will appear:
Select the option that says “Sort left to right” and click OK.
In the original Sort pop-up, select Row 1 from the drop-down menu next to “Sort By,” make sure it’s sorting by “Values” and in “A to Z” order, and click OK!
This has been a basic tutorial for an extremely simple, yet versatile function in Microsoft Excel. To learn more about this feature, and more, consider taking some of the following Excel courses:
Page Last Updated: January 2014
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Bestseller | 1,576 | 6,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.930887 |
https://www.audifans.com/archives/1993/08/msg00045.html | 1,618,421,073,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077843.17/warc/CC-MAIN-20210414155517-20210414185517-00397.warc.gz | 757,254,041 | 2,498 | # [ramiro: Questions about tire stuff....]
```I forgot to tell you, all my calculations were based on the assumption
that you were going to go to a 50 series tire (225/50-15). This
e-mail might be too late to stop the flames.
Ramiro
----------------------------------------------------------------------
Date: Tue, 10 Aug 93 11:12:05 EST
From: ramiro
To: mlr@critter.cv.com
CC: quattro@aries.east.sun.com
In-reply-to: Mike LaRosa 's message of Tue, 10 Aug 93 08:07:08 EDT <9308101207.AA01844@critter.cv.com>
>
> Date: Tue, 10 Aug 93 08:07:08 EDT
> From: mlr@critter.cv.com (Mike LaRosa )
>
> HI everybody.....
>
> Got a question regarding tire sizes, I have an 89 100, with 205/60/15
> Michelin mxv2 tires that are quickly approaching the surface texture of
> a boloney skin.....:-) Some of you may remember me asking questions
> last year about which tire to replace these with, well the time has
> come. I've decided to go with either P500 or COMP T/A HR4's but my
> one nagging question is I'd like to go a little wider, to maybe
> 225.... But what's this gonna do to my speedo ? How much faster do
> you think I'll really be going when the speedo say's 60 mph ?
>
> Mike L.
>
The question is "How much slower do you think you'll really be going
when the speedo says 60 mph ?"
The answer is 57.9 mph. The 225 wheels are going to be 3.5% smaller
than the 205 wheels that you currently have. Therefore, the
speedometer will show higher speeds than the actual speed that you're
traveling. Your speedometer is calibrated to show 60 mph at 14.20
rev/sec. But with the smaller wheels, at 60 mph (actual speed) your
wheels would be rotating at 14.75 rev/sec.
So when the smaller wheels are rotating at 14.20 rev/sec, the
speedometer will read 60 mph (it's been calibrated for 14.20 rev/sec =
60 mph), but the actual speed is only 57.9 mph. When your actual
speed is 60 mph, the wheels will be rotating at 14.70 rev/sec, and the
speedometer will show 62.1 mph.
These are the basis for my calculations; please correct them if
I'm wrong. Diameter of wheels:
205/60-15 = 621 mm (this is what you have now)
225/50-15 = 600 mm (is this what you want ?)
225/55-15 = 622.5 mm (bingo, this is what you need)
225/50-16 = 625 mm (you can use this too; need new rims though)
``` | 685 | 2,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-17 | latest | en | 0.925083 |
https://mersenneforum.org/showthread.php?s=701d57a9e795a719c865cd3bb165e520&t=20372&goto=nextnewest | 1,611,422,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00183.warc.gz | 448,531,919 | 9,066 | mersenneforum.org > Data discrepancy in credit calculator for LL tests?
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2016-03-04, 04:09 #1 ixfd64 Bemusing Prompter "Danny" Dec 2002 California 24×3×72 Posts discrepancy in credit calculator for LL tests? It is my understanding that double an exponent results in an LL test taking approximately four times longer. From the CPU credit calculator, the 49th known Mersenne prime, M74,207,281, takes around 205 GHz-days. The first Mersenne number of prime index with 100 million digits, M332,192,831, needs around 4,941 GHz-days. This also seems reasonable. M601,248,421, the largest Mersenne number with an LL test to date, requires about 16,584 GHz-days, which closely matches the credit that Never Odd or Even received. However, the numbers become weird after that: the calculator says the first Mersenne number with more than a billion digits, M3,321,928,097, requires just 91,630 GHz-days. The actual value should also be much higher; (3,321,928,097 / 601,248,421)2 ≈ 30.5, and multiplying that by 16,584 gives over 500,000 GHz-days. So does the time complexity for LL tests stop exhibiting quadratic growth after a certain point? Or is there an error in the calculator? Last fiddled with by ixfd64 on 2016-03-04 at 04:16
2016-03-04, 04:25 #2 airsquirrels "David" Jul 2015 Ohio 11·47 Posts The calculation is based on a timing chart for a hard coded list of FFT sizes matched to exponent ranges. For numbers outside of the maximum measured FFT size the credit value is just an extrapolated point from the largest FFT size. http://www.mersenne.ca/credit.php?showsource=1 From a quick glance,the largest FFT size is 33.5M or so and credit is computed linearly for larger exponents than 596M Last fiddled with by airsquirrels on 2016-03-04 at 04:29
2016-03-04, 05:43 #3 axn Jun 2003 3×5×17×19 Posts James should modify the calculator to give an error message for out-of-range exponents rather than give out some made up crap. Or try to do a realistic projection.
2019-08-13, 20:11 #4
kriesel
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
12EF16 Posts
Quote:
Originally Posted by ixfd64 It is my understanding that double an exponent results in an LL test taking approximately four times longer. ... So does the time complexity for LL tests stop exhibiting quadratic growth after a certain point? Or is there an error in the calculator?
The number theory folks tell us it's O(p2 log p log log p) per primality test using the best available technology for this size multiplication or squarings mod Mp, the irrational base discrete weighted transform. Over the mersenne.org range, that works out to around p2.117. Double the exponent, about 4.34 times the effort.
Empirical run time scaling for LL, PRP, or P-1 are around p2.1. Any fixed overhead appears to lower the power on the scaling and have greater effect in lowering the scaling power at small p.
prime95 PRP https://www.mersenneforum.org/showpo...78&postcount=2
prime95 P-1 https://www.mersenneforum.org/showpo...92&postcount=3
CUDALucas LL https://www.mersenneforum.org/showpo...23&postcount=2
CUDAPm1 P-1 https://www.mersenneforum.org/showpo...27&postcount=2
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Science 8
acceleration This is the rate at which velocity changes.
balanced forces This is when forces are combined and there is a zero net force and no change in motion.
motion You can tell that this has occured because an object's position has changed compared to its reference point.
net force This is the resulting force when two or more forces are combined.
speed This is the distance covered in a period of time.
unbalanced forces When forces are combined, this results in a net force and causes a change in motion.
reference point This is an object that seems to stay still.
velocity This is the speed of an object in a given direction.
action force The force that is put on an object.
gravity This is the invisible attaction obetween two objects. Its strength depends on mass and distance.
inertia This is the tendency of objects to resist changes in motion.
momentum This is the measure of an object's motion based on mass and velocity.
weight This measures the force of gravity on an objcet.
reaction force This is the force that an object puts back onto the object putting a force on it.
F = MA This is the formula associtated with Newton's 2nd Law.
Newton's Second Law An example of this law is the harder you kick a soccer ball the farther it will go.
Newton's Third Law An example of this law is when you put a force on a table leg with your toe, the table leg puts a force back on your toe.
Newton's First Law The reason we wear seatbelts can be explained by using this law of motion. | 323 | 1,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-24 | latest | en | 0.950126 |
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So, my initial exercise was to define a struct to represent a point in the Cartesian (xy) coordinate plane and write a client program that asks the user to type in two pairs of coordinates and finds the distance between them using a method. This I did with relative ease.
I am having difficulty updating my point class to use getters, setters and constructors, and changing the client program accordingly to achieve the same purpose.
I can't really figure out how to make a method that can process two different objects without defining additional fields in the point class definition, which rather defeats the purpose of having objects.
Could someone please clarify what information I am missing that is making this so difficult for me? I am reading my textbook over and over and I am not finding anything that clarifies this.
Class Definition:
Java Code:
```public class point {
//fields
public int x;
public int y;
//constructors
public point(){
this(0, 0);
}
public point(int x, int y){
setLocation(x,y);
}
public void setLocation(int x, int y){
this.x = x;
this.y = y;
}
//methods
//returns the distance between this point and another point
public double distanceBetween(double x1, double y1, double x2, double y2) {
double deltaX = (x1-x2);
double deltaY = (y1-y2);
return Math.pow(deltaX, 2) + Math.pow(deltaY, 2);
}
public String toString() {
return "(" + x + ", " + y + ")";
}
}```
Client Program:
Java Code:
```import java.util.*;
import java.awt.*;
public class clientprogram {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
point p1 = new point();
Scanner input1 = new Scanner(input.nextLine());
input1.useDelimiter(",");
p1.x = input1.nextInt();
p1.y = input1.nextInt();
point p2 = new point();
Scanner input2 = new Scanner(input.nextLine());
input2.useDelimiter(",");
p2.x = input2.nextInt();
p2.y = input2.nextInt();
System.out.println("The distance between the two points is " + distanceBetween(p1.x, p1.y, p2.x, p2.y));
}
}```
Step 1: Use standard naming conventions. Classes start with an uppercase letter.
Step 2: Make x and y in your Point class private.
Step 3: Now that x and y are private, you need to add getters so your ClientProgram class can access them.
Step 4: Now that you have the getters, you can modify your ClientProgram class to use them.
3. Member
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Okay, so this is what I changed with my point class. I'm sorry if I'm being colossally, royally stupid, but I don't really understand why my ClientProgram still can't use the variables.
I don't really know how to modify it to use the getters.
It says that the constructor for point() is undefined...when I really don't think it is?
Java Code:
```public class point {
//private fields
private int x;
private int y;
//constructor method
public point(int x, int y){
this.x = x;
this.y = y;
}
// accessor methods
public int getX(){
return x;
}
public int getY(){
return y;
}
//mutator methods
public void setX(int x){
x = this.x;
}
public void setY(int y){
y = this.y;
}
//returns the distance between this point and another point
public double distanceBetween(int x1, int y1, int x2, int y2) {
double deltaX = (x1-x2);
double deltaY = (y1-y2);
return Math.pow(deltaX, 2) + Math.pow(deltaY, 2);
}
public String toString() {
return "(" + this.x + ", " + this.y + ")";
}
}```
Again, I beg of you, use standard naming conventions.
You haven't posted your updated ClientProgram class, so it's hard to tell you what's wrong with it. Sounds like you're trying to use a constructor that doesn't exist though. Compare the constructor you're calling in ClientProgram to the constructor you wrote in your Point class.
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• | 955 | 3,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-50 | latest | en | 0.812769 |
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#### tmyers
##### Active member
I am having various issues with rounding across my db.
A lot of the numbers I deal with are in currency format. Both the controls and the fields within the tables are set to currency, yet it appears it is still rounding to the .0001 place (roughly) and is resulting in some interesting numbers.
I looked in the Round() function, but that rounds to the nearest whole dollar up and down. Per my boss, he wants to round to the nearest tenth and always up like excels RoundUp function.
So items like \$10.01 go to \$10.10 and \$10.11 go to \$10.20 etc. etc.
Would I just use the Int function? And to get to the tenth, would it be like Int(-10 * [myfield]) / -10?
#### Minty
##### AWF VIP
Currency is fixed at 4 decimal places. It will do 4/5ths rounding beyond that, which should mean you never get a problem variance at the cent level, as it's accurate to 1/1000 of a cent
Rounding up to the nearest 10c is not a standard accountancy rounding, so you would need to roll your own - your function looks about right, but test it on the same numbers you are having issues with..
#### Gasman
##### Enthusiastic Amateur
I am having various issues with rounding across my db.
A lot of the numbers I deal with are in currency format. Both the controls and the fields within the tables are set to currency, yet it appears it is still rounding to the .0001 place (roughly) and is resulting in some interesting numbers.
I looked in the Round() function, but that rounds to the nearest whole dollar up and down. Per my boss, he wants to round to the nearest tenth and always up like excels RoundUp function.
So items like \$10.01 go to \$10.10 and \$10.11 go to \$10.20 etc. etc.
Would I just use the Int function? And to get to the tenth, would it be like Int(-10 * [myfield]) / -10?
Where did you get that idea?
Code:
``````? round(10.1234,2)
10.12
? round(10.1256,2)
10.13
? round(10.0111,2)
10.01``````
#### tmyers
##### Active member
Currency is fixed at 4 decimal places. It will do 4/5ths rounding beyond that, which should mean you never get a problem variance at the cent level, as it's accurate to 1/1000 of a cent
Rounding up to the nearest 10c is not a standard accountancy rounding, so you would need to roll your own - your function looks about right, but test it on the same numbers you are having issues with..
I tried adding that to my expressions and am still getting weird results.
Here is a break down:
Unit Price: Int(-10*([Price]+IIf(IsNull([Freight]),0,[Freight])+(IIf(IsNull([LampPrice]),0,[LampPrice])*IIf(IsNull([NumberOfLamps]),0,[NumberOfLamps])))/(1-[GP]))/-10
Then I have:
Extended: Int(-10*([Quantity]*(([Price]+IIf(IsNull([Freight]),0,[Freight])+(IIf(IsNull([LampPrice]),0,[LampPrice])*IIf(IsNull([NumberOfLamps]),0,[NumberOfLamps])))/(1-[gp]))))/-10
The very first line in the report is (79) items @ \$216.00/e (which is correct). The extended for this item somehow came out to \$17,056.90. The amount SHOULD have been \$17,064.00. How in the world did it get to that other number? Not only is it close to being \$10 off, it has .90 when it was multiplied by a whole amount.
#### Minty
##### AWF VIP
I tried adding that to my expressions and am still getting weird results.
Here is a break down:
Unit Price: Int(-10*([Price]+IIf(IsNull([Freight]),0,[Freight])+(IIf(IsNull([LampPrice]),0,[LampPrice])*IIf(IsNull([NumberOfLamps]),0,[NumberOfLamps])))/(1-[GP]))/-10
Then I have:
Extended: Int(-10*([Quantity]*(([Price]+IIf(IsNull([Freight]),0,[Freight])+(IIf(IsNull([LampPrice]),0,[LampPrice])*IIf(IsNull([NumberOfLamps]),0,[NumberOfLamps])))/(1-[gp]))))/-10
The very first line in the report is (79) items @ \$216.00/e (which is correct). The extended for this item somehow came out to \$17,056.90. The amount SHOULD have been \$17,064.00. How in the world did it get to that other number? Not only is it close to being \$10 off, it has .90 when it was multiplied by a whole amount.
I'm not going to try and decipher that second calculation!
Do yourself a favour and create a query that does all the is Null value replacements first.
Then make another query to do all the maths and rounding work from that.
I suspect all will become a lot clearer then.
#### tmyers
##### Active member
I'm not going to try and decipher that second calculation!
Do yourself a favour and create a query that does all the is Null value replacements first.
Then make another query to do all the maths and rounding work from that.
I suspect all will become a lot clearer then.
Never thought of doing that.
Those formulas are a royal pain to look at so I can't blame you.
#### tmyers
##### Active member
Got it. Had to do the Int function against that mess and then multiply by the quantity.
I am still going to do your suggestion and break that up. That expression is a royal pain to read.
#### Minty
##### AWF VIP
Often, even if not strictly necessary, breaking things down into more manageable pieces is the best way.
Good luck with the rest of your project.
#### tmyers
##### Active member
Often, even if not strictly necessary, breaking things down into more manageable pieces is the best way.
Good luck with the rest of your project.
Agreed. It is a simple function from a technical standpoint. It is just setting various nulls to 0 so that the formula can correctly add things together. Downside is it is doing it for 3-4 fields so the expression is quite large.
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643 | 1,470 | 5,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.949463 |
http://www4.geometry.net/detail/math_discover/fuzzy_math.html | 1,686,211,168,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00509.warc.gz | 94,486,560 | 15,657 | Home - Math_Discover - Fuzzy Math
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Fuzzy Math: more detail
1. Fuzzy Math Sets, Quick Tutorial On Fuzzy Logic And Sets.
Fuzzy Logic A type of logic for processing imprecise data founded by LA Zadeh. Fuzzy Sets A type of sets in which elements belong to subsets in some degree.
MENTAL MATH TUTORIALS Data Mining Neural Networks Fuzzy Logic MENU Home Addition Subtraction Multiplication ... ñol TUTORIALS Data Mining Neural Networks Fuzzy Logic and Sets LABS Fuzzy Logic New Artificial Life Fuzzy Logic: A type of logic for processing imprecise data founded by L. A. Zadeh. Elements may have infinite gradation between TRUE and FALSE. Fuzzy Sets: A type of sets in which elements belong to subsets in some degree. (Certain speed may be 0.75 SLOW and 0.25 MEDIUM). Click for more fuzzy TEST A: Enter a new speed B: Speed = C: What is the fuzzy value of MEDIUM ? D: Answer with Keyboard »» KEYBOARD D: Press Enter and View report REPORT Yours: Ours: Result: Time: Home Site Map Suggest a link Send Comments ... Help
2. Fuzzy Math (Main Page)
Paul Krugman. fuzzy math. The Essential Guide to the Bush Tax Plan. Wielding his widely recognized powers of explanation, Paul Krugman lays bare the hidden facts behind the \$2 trillion tax cut.
http://www.wwnorton.com/catalog/fall01/005062.htm
##### Fuzzy Math
The Essential Guide to the Bush Tax Plan Wielding his widely recognized powers of explanation, Paul Krugman lays bare the hidden facts behind the \$2 trillion tax cut.
With huge budget surpluses just ahead, the question of whether to cut taxes has shifted to when? and by how much? With Fuzzy Math , Paul Krugman dissects the Bush tax proposal and shows us who wins, who loses, and how quickly the tax cuts will consume the surplus. Always the equal-opportunity critic when it comes to faulty economics, Krugman also tucks into the Democratic alternatives to the Bush plan.
This little book packs a big wallop. Together with major media appearances, it puts Krugman's wisdom and steely-eyed analysis firmly at the center of the debate about how to spend upwards of \$2 trillion. It may very well change the course of history.
Paul Krugman , who "writes better than any economist since John Maynard Keynes" ( Fortune ), writes the biweekly "Reckonings" column for the New York Times . Winner of the John Bates Clark Medal, Krugman teaches at Princeton University.
3. Fuzzy Math, By Jonathan Rowe
Search Newdream.org. Match ALL words. Match ANY word. fuzzy math Cook the Planet, Cook. the Books, Call it Growth. By Jonathan Rowe You are here home publications enough fuzzy math. what
Search Newdream.org
Match ALL words Match ANY word
##### By Jonathan Rowe
Several months ago a professor at the University of North Carolina published findings that turned beliefs about the economy upside down. Health improves, he said, as the economy goes down. When the economy declines, to a point at least, deaths, smoking, obesity, heavy drinking, heart disease and some kinds of back problems all decline as well. "Sounds unlikely," said The New York Times. And indeed it is, by the standard reckonings at least. We all know that an expanding economy makes us better off - or do we? Another study, this one in England, found that shopping, which is the drive train of the entire economy, and which is supposed to make people feel good, actually can make them depressed. "For significant numbers, dissatisfaction is now part of the shopping process," one of the authors said. (As though we needed a study to tell us that.) What's going on here? How could we feel better when the experts say we should feel worse, and worse when they say we should feel better? Could it be that economists don't know up from down to begin with?
4. Fuzzy Math (Aaron Swartz: The Weblog)
fuzzy math. One of the weirdest things I heard when listening to the Cato Institute debate was an p2p is not a crime. fuzzy math. Tufte. Quickly. curb your celebrity
http://www.aaronsw.com/weblog/000616
##### Fuzzy Math
for free More examples: the thriving shareware market, the Baen Free Library, Janis Ian and others. posted September 25, 2002 06:54 PM ( Politics
##### Nearby
up Feeling Lucky
TRAMP Update
Gary Shapiro: Sharing is Legal and Moral!
...
p2p is not a crime
Fuzzy Math
Tufte
Quickly
quick hack
... me@aaronsw.com
All text above by me is in the public domain.
5. President Clinton's Mandate For Fuzzy Math
THE WALL STREET JOURNAL June 11, 1997 President Clinton s Mandate for fuzzy math. By Lynne V. Cheney. Marianne Jennings, an Arizona
http://www.mathematicallycorrect.com/cheney.htm
THE WALL STREET JOURNAL
June 11, 1997
##### President Clinton's Mandate for Fuzzy Math
By Lynne V. Cheney Marianne Jennings, an Arizona State business professor, has brought enlightenment to the multitudes. With her commentaries on her daughter Sarah's eighth grade math book ("MTV Math" she calls it, for its colorful pictures, disconnected ideas, and generally casual attitude), she has helped parents across the country realize they are not the only ones dismayed by current mathematics education. Kids are writing about "What We Can Do to Save the Earth," and inventing their own strategies for multiplying. They're learning that getting the right answer to a math problem can be much less important than having a good rationale for a wrong one. Sometimes called "whole math" or "fuzzy math," this latest project of the nation's colleges of education has some formidable opponents. In California, where the school system embraced whole math in 1992, parents and dissident teachers have set up a World Wide Web site called Mathematically Correct to point out the follies of whole-math instruction.
6. President Clinton's Mandate For Fuzzy Math
THE WALL STREET JOURNAL. June 11, 1997. President Clinton's Mandate for fuzzy math. By Lynne V. Cheney. Marianne Jennings, an Arizona State business professor, has brought enlightenment to the multitudes. Sometimes called "whole math" or "fuzzy math " this latest project of the nation's colleges of education has some
http://mathematicallycorrect.com/cheney.htm
THE WALL STREET JOURNAL
June 11, 1997
##### President Clinton's Mandate for Fuzzy Math
By Lynne V. Cheney Marianne Jennings, an Arizona State business professor, has brought enlightenment to the multitudes. With her commentaries on her daughter Sarah's eighth grade math book ("MTV Math" she calls it, for its colorful pictures, disconnected ideas, and generally casual attitude), she has helped parents across the country realize they are not the only ones dismayed by current mathematics education. Kids are writing about "What We Can Do to Save the Earth," and inventing their own strategies for multiplying. They're learning that getting the right answer to a math problem can be much less important than having a good rationale for a wrong one. Sometimes called "whole math" or "fuzzy math," this latest project of the nation's colleges of education has some formidable opponents. In California, where the school system embraced whole math in 1992, parents and dissident teachers have set up a World Wide Web site called Mathematically Correct to point out the follies of whole-math instruction.
7. 2 Plus 2: The Home Of Mathematically Correct
The advocates of the new, fuzzy math have practiced their rhetoric well. 2+2=5 fuzzy math Invades Wisconsin Schools, .pdf file 80k by Leah Vukmir.
http://www.mathematicallycorrect.com/
"There is a mathematically correct solution" This web site is devoted to the concerns raised by parents and scientists about the invasion of our schools by the New-New Math and the need to restore basic skills to math education. Mathematically Correct is the informal, nationwide organization that fights the Establishment on behalf of sanity and quality in math education. David Gelernter, NY Post Mathematics achievement in America is far below what we would like it to be. Recent " reform " efforts only aggravate the problem. As a result, our children have less and less exposure to rigorous, content-rich mathematics . The advocates of the new, fuzzy math have practiced their rhetoric well. They speak of higher-order thinking, conceptual understanding and solving problems, but they neglect the systematic mastery of the fundamental building blocks necessary for success in any of these areas. Their focus is on things like calculators, blocks, guesswork, and group activities and they shun things like algorithms and repeated practice. The new programs are shy on fundamentals and they also lack the mathematical depth and rigor that promotes greater achievement. Concerned parents are in a state of dismay and have begun efforts to restore content, rigor, and genuinely high expectations to mathematics education. This site provides relevant background and information for parents, teachers, board members and the public from around the country.
8. Mai Gehrke's Curriculum Vita
New Mexico State University Nonstandard mathematics, operators on boolean algebras, fuzzy mathematics, universal algebra, general topology, posets and lattices.
http://www.math.nmsu.edu/mgehrke/mgehrke.html
##### Mai Gehrke
Department of Mathematical Sciences Phone: (505] 646-4218 New Mexico State University Fax: (505) 646-1064 Las Cruces, NM 88003 mgehrke@nmsu.edu Office Location: Science Hall Room 232 House for rent while on sabbatical 2004-2005 Return to Faculty Page Return to Main Index
##### PROFESSIONAL EXPERIENCE
Professor, New Mexico State University, Las Cruces, New Mexico.
1/99Present Part-time consulting, Physical Science Laboratory, Las Cruces, New Mexico.
1/975/97 Visiting Professor, Vanderbilt University, Nashville, Tennessee.
8/9612/96 Visiting Lektor, University of Copenhagen, Copenhagen, Denmark.
8/93 5/00 Associate Professor, New Mexico State University, Las Cruces, New Mexico.
8/905/93 Assistant Professor, New Mexico State University, Las Cruces, New Mexico.
8/885/90 2-year position as Assistant Professor, Vanderbilt University, Nashville, Tennessee.
1/837/87 Teaching Assistant, University of Houston, Houston, Texas.
##### RESEARCH INTERESTS
Logic and its Applications
Universal Algebra and Lattice Theory
##### Mathematical Research Publications
• A new proof of completeness of S4 with respect to the real line , with G. Bezhanishvili, submitted to
• 9. Fuzzy Math Books To Avoid
Warning against fuzzy math books by the NewNew Math (also called fuzzy math and Mickey-Mouse Math) and the need to restore basic skills to math education. As a homeschooler, you
http://www.homeschoolmath.net/fuzzy_math.htm
##### HOMESCHOOL MATH
Home 123-Math ebooks Worksheets Curriculum guide ... DVD rentals
##### Fuzzy math - programs you are better off NOT using
The website www.mathematicallycorrect.com is devoted to the concerns raised by parents and scientists about the invasion of US schools by the New-New Math ( also called Fuzzy Math and Mickey-Mouse Math) and the need to restore basic skills to math education. As a homeschooler, you will be able to avoid these horrible new math programs used in some public schools. Though you can probably find these books cheap in thrift stores if schools are using them in your area, they might not be a good idea from the mathematical point of view. For more information, see Mathematics Program Reviews and Information at their website. "Mathematics achievement in America is far below what we would like it to be. Recent "reform" efforts only aggravate the problem. As a result, our children have less and less exposure to rigorous, content-rich mathematics. The advocates of the new, fuzzy math have practiced their rhetoric well. They speak of higher-order thinking, conceptual understanding and solving problems, but they neglect the systematic mastery of the fundamental building blocks necessary for success in any of these areas. Their focus is on things like calculators, blocks, guesswork, and group activities and they shun things like algorithms and repeated practice. The new programs are shy on fundamentals and they also lack the mathematical depth and rigor that promotes greater achievement."
10. CBS News | Bush Team's Fuzzy Math | February 24, 2004Â 13:21:59
Bush Team s fuzzy math NEW YORK, Feb. 24, 2004 (Photo CBS/AP) There s nothing subtle about these They just decided to leave a
http://www.cbsnews.com/stories/2004/02/24/politics/main601960.shtml
Home U.S. Iraq World ... FREE CBS News Video February 24, 2004 13:21:59 The Early Show CBS Evening News 48 Hours 60 Minutes ... Section Front E-mail This Story Printable Version Bush Team's Fuzzy Math NEW YORK, Feb. 24, 2004 (Photo: CBS/AP) "There's nothing subtle about these: They just decided to leave a lot of big items out." Robert Bixby, Concord Coalition Differences in estimates of the cost of the Medicare drug benefit are due to different predictions of future demand. (Photo: CBS/The Early Show) The war in Iraq is a large potential cost that the president's budget largely ignores. (Photo: AP) (CBS) In a presidential debate nearly four years ago, George W. Bush accused Al Gore of employing "fuzzy math." But increasingly, it's the White House that's being accused of numerical fuzziness on Medicare, the deficit and jobs. President Bush last year signed a Medicare prescription drug benefit with an estimated price tag of \$395 billion. A month later, the White House said the actual cost was more like \$534 billion. In his State of the Union address, the president pledged to halve the deficit by 2009. But his plan largely excludes the cost of operations in Iraq and Afghanistan and does not reflect the long-term impact of Mr. Bush's proposal to make recent tax cuts permanent.
11. ABCNEWS.com : Why Fuzzy Math Makes Sense In Politics
It's a safe bet that you have never heard the presidential candidates explain how adjusted multiple correlation coefficients work. Self-described fuzzy mathematician John Allen Paulos claims that fuzzy numbers are the only kind Nothing Wrong. With fuzzy mathSometimes Fuzziness Is Required When Analyzing Politics
http://abcnews.go.com/sections/science/WhosCounting/whoscounting001101.html
12. Fuzzy Math
Kerby Anderson Commentary fuzzy math. July 6, 1999. For the moment, it appears that traditional math is on the way out and fuzzy math is on the way in.
http://www.probe.org/docs/c-fuzzy.html
Kerby Anderson Commentary
##### Fuzzy Math
July 6, 1999 Concerned parents and political activists have been complaining about fuzzy math for years, but the battle over this new way of teaching math in Plano, Texas has started to bring the issue into focus. For years I've been reading about these new textbooks that change the way math is taught. These techniques are sometimes called "fuzzy math" or "whole math" or the "new new math." I've even heard of a textbook referred to as rainforest algebra because it spends a good portion of the book talking about ecology and the loss of rainforests instead of teaching algebra. Now you would think that math is math. But of course I used to think that grammar was grammar until educators introduced such things as the whole language approach and changed the way students learned English and grammar. The same seems to be taking place with these new math books. The textbook in question is called Connected Mathematics . Math is taught through group work and through discovery learning. In essence, students are to teach themselves and to work in groups. I well remember when that fad hit my high school my senior year. Since I lived just outside Berkeley, California I got to see some of these educational fads 20 years before most school districts. My math teacher came back from a seminar on group learning and implemented it in our pre-calculus class. Let me say it was pretty much of a disaster. Most of us learned math in spite of the program, not because of it.
13. Frankly...: Fuzzy Math
Franklyn Monk is a goddamn artist freak. This page is devoted to various work including comics, poetry, flash design, 3d graphics and animation. fuzzy math. SciScoop fuzzy math And Fuzzy Thinking By The Numbers
http://fmonk.com/archives/000416.php
Monk's Blog
##### Fuzzy Math
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14. Fuzzy Math
Kerby Anderson Commentary fuzzy math. October 4, 1999. The concept is called fuzzy math, because it reminds critics of what was done years ago with English.
http://www.probe.org/docs/c-fuzzy2.html
Kerby Anderson Commentary
##### Fuzzy Math
October 4, 1999 A month ago, I wrote a commentary about Connected Mathematics, often called "fuzzy math." Since then there has been an article about this issue in the Washington Times. When a major national newspaper does a story about a Plano math book, you know that this local controversy is taking on national implications. The concept is called fuzzy math, because it reminds critics of what was done years ago with English. No longer was phonics taught, but a whole language approach was introduced into the schools changing the way students learned English and grammar. The same seems to be taking place in the textbook called Connected Mathematics. Math is taught through group work and through discovery learning. Math teachers no longer teach. Instead students are to teach themselves and to work in groups. Six families from Plano have filed a class-action lawsuit claiming their children were not learning in the connected math program and were not given the option of a more traditional math course. The Washington Times article points out that other school districts have been testing the connected math program. So what happens in Plano will affect more than just students in the Dallas area.
15. King George The Second: Fuzzy Math
Nothing close to half the country voted for this man. George W. Bush's fuzzy math. Soon after the Supreme Court decision that led Al Gore to concede the Even less surprising is the fuzzy math
http://www.irregulartimes.com/fuzzy.html
##### George W. Bush's Fuzzy Math
Soon after the Supreme Court decision that led Al Gore to concede the presidential election to George W. Bush, I wrote an article entitled Bush Wins, Bush Loses . In the article, I compared George W. Bush to Daffy Duck for the way that he has been duped into accepting a job so difficult that only an idiot would take it. The whole point of the article is that the difficulty of the political situation and President George Junior's inexperience, inarticulateness and general inability will combine to make him the least remarkable person to inhabit the Oval Office in American History. Well, it's not too surprising that a large number of the Republican faithful disagreed with me. Even less surprising is the fuzzy math that they used to justify their angry responses. Typical is the following, from a Republican who goes by the name of poetwarrior. This GOP guy writes, "I dissent. Well-written but not a very honest diatribe. Remember: half the country voted for this man." Now, I ask you: what's not honest about saying 1st that George W. Bush is not very smart and 2nd that he's got a very difficult job ahead of him? Honesty hurts sometimes.
16. FactCheck.org Kerry Ad On Abortion: Fuzzy Math Kerry Ad on Abortion fuzzy math. Kerry claims the Supreme Court is just one vote away from overturning abortion rights for women. Make that TWO votes.http://www.factcheck.org/article.aspx?docID=176
17. How Not To Teach Math By Matthew Clavel
from a pedagogical philosophy that goes by several namesÂÂConstructivist Math, ÂNewNew Math, and, to its detractors, Âfuzzy math. IÂll
http://www.city-journal.org/html/eon_3_7_03mc.html
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18. ABCNEWS.com : Why Fuzzy Math Makes Sense In Politics
(James A. Boyle), Nothing Wrong With fuzzy math. Sometimes Fuzziness Is Required When Analyzing Politics and the Economy. By John Allen Paulos Nov.
http://abcnews.go.com/sections/scitech/WhosCounting/whoscounting001101.html | 4,857 | 21,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-23 | latest | en | 0.840259 |
https://oeis.org/A270709 | 1,540,074,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513441.66/warc/CC-MAIN-20181020205254-20181020230754-00237.warc.gz | 748,471,120 | 3,977 | This site is supported by donations to The OEIS Foundation.
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A270709 a(n) = (n+1)*Sum_{k=0..(n-1)/2} (binomial(k+1,n-2*k-1)*binomial(2*k,k)/(k+1)^2). 2
0, 2, 3, 2, 5, 7, 14, 26, 51, 103, 209, 435, 910, 1930, 4122, 8874, 19227, 41893, 91751, 201839, 445841, 988403, 2198547, 4905147, 10974210, 24615134, 55341636, 124694354, 281525678, 636802626, 1442953404, 3274997130, 7444505615, 16946749249 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 FORMULA G.f.: ((3*x+2)*(1-sqrt(1-4*(x^3+x^2))))/(2*(x^2+x)). a(n) ~ (3*r+2) * sqrt(3-4*r^2) * 2^(2*n+2) * r^(n+3) * (r+1)^(n+1) / (n^(3/2) * sqrt(Pi)), where r = 0.41964337760708... is the real root of the equation 4*r^2*(1+r) = 1. - Vaclav Kotesovec, Mar 22 2016 MATHEMATICA Table[(n+1)*Sum[Binomial[k+1, n-2*k-1] * Binomial[2*k, k] / (k+1)^2, {k, 0, (n-1)/2}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 22 2016 *) PROG (Maxima) makelist(coeff(taylor(((3*x+2)*(1-sqrt(1-4*(x^3+x^2))))/(2*(x^2+x)), x, 0, 15), x, n), n, 0, 15); a(n):=(n+1)*sum((binomial(k+1, n-2*k-1)*binomial(2*k, k))/(k+1)^2, k, 0, (n-1)/2); (PARI) x='x+O('x^200); concat(0, Vec(((3*x+2)*(1-sqrt(1-4*(x^3+x^2))))/(2*(x^2+x)))) \\ Altug Alkan, Mar 22 2016 CROSSREFS Cf. A000108. Sequence in context: A092550 A058977 A085818 * A064939 A248012 A151549 Adjacent sequences: A270706 A270707 A270708 * A270710 A270711 A270712 KEYWORD nonn AUTHOR Vladimir Kruchinin, Mar 22 2016 STATUS approved
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Last modified October 20 18:08 EDT 2018. Contains 316401 sequences. (Running on oeis4.) | 822 | 1,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-43 | latest | en | 0.452818 |
https://www.answers.com/Q/Three_coins_are_flip_at_the_same_time_What_is_the_probability_that_the_outcome_is_at_least_2_heads | 1,611,066,417,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00622.warc.gz | 654,187,395 | 32,558 | Math and Arithmetic
Statistics
Probability
# Three coins are flip at the same time What is the probability that the outcome is at least 2 heads?
###### Wiki User
There are four outcomes possible (not considering order)
• HHH
• HHT
• HTT
• TTT
Only in two of the cases are there two or more heads
The probability is 0.5
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## Related Questions
If both tosses are fair, the probability of that outcome is one in four.
The probability of the first coin landing heads is half (or 1/2). Similarly, the probability of the second and third coins landing heads are also 1/2 in each case. Therefore, the probability of having three heads is: (1/2)(1/2)(1/2) = (1/8)
The probability that both coins are heads is the probability of one coin landing heads multiplied by the probability of the second coin landing heads: (.5) * (.5) = .25 or (1/2) * (1/2) = 1/4
The probability of flipping three heads when flipping three coins is 1 in 8, or 0.125. It does not matter if the coins are flipped sequentially or simultaneously, because they are independent events.
The probability of something NOT happening is the complement of the probability of something happening. Since the probability that you DO have 3 heads is 1/8 (that is, 1/2 cubed), the complement is 1 - 1/8 = 7/8.
An outcome is what actually happens, while the probability of that outcome is how likely that particular thing is to happen. Say I was flipping a coin. The probability of the outcome of heads is 1/2 because there are 2 possible outcomes and heads is only 1 of them. Then when I flip the coin, it lands on tails. The outcome is tails.
lets get some facts odds of head on 1 coin 50% or evens odds of no head 50% or evens the possible results vary from 1 coin to 2 coins. 1 coin has 2 results heads or tails 2 coins have 4 results. heads heads, tails tails, tails heads, heads tails. each outcome has a probability of 25%. for the question we remove the heads and tail probability and we have 2 outcomes with heads and one without. so 2 to 1 chance or 33.3333 recuring chance.
well since the coins have two sides,there is a 50% chance of it landing on heads
There are 8 permutations of three coins. Of these, 3 of them have two heads, so the probability of tossing two heads on three coins is 3 in 8, or 0.375. However, you said, "at least", so that includes the case of three heads, so the probability of throwing at least two heads is 4 in 8, or 0.5. T T T T T H T H T T H H * H T T H T H * H H T * H H H *
Possibilities: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. There are 3 chances out of 8 that there will be two heads and one more that there will be AT LEAST two heads.
Coins do not have numbers, there is only the probability of heads or tails.
you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?
The probability is always 50/50 even if you flipped 100 or 1000000 coins.
The sample space is 23 or 8; which can be listed out as: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 2 of the 8 that have exactly 2 heads; so the probability of exactly two coins landing on heads is 2/8 or 1/4.
Because you are thinking permutations rather than combinations. There are four permutations of two coins, but there are only three combinations, because it does not matter which coin is heads and which coin is tails. As a result, the combination of heads and tails has a 0.5 probability, while two heads or two tails each have a 0.25 probability.
The number of total outcomes on 3 tosses for a coin is 2 3, or 8. Since only 1 outcome is H, H, H, the probability of heads on three consecutive tosses of a coin is 1/8.
###### ProbabilityMath and ArithmeticStatisticsAlgebra
Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 1,049 | 3,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-04 | latest | en | 0.949438 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/403/2/k/c/ | 1,597,340,320,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739048.46/warc/CC-MAIN-20200813161908-20200813191908-00250.warc.gz | 731,588,097 | 50,914 | # Properties
Label 403.2.k.c Level 403 Weight 2 Character orbit 403.k Analytic conductor 3.218 Analytic rank 0 Dimension 4 CM no Inner twists 2
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$403 = 13 \cdot 31$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 403.k (of order $$5$$, degree $$4$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$3.21797120146$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\zeta_{10})$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{5}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{10}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( -1 + 2 \zeta_{10} - \zeta_{10}^{2} ) q^{2} + ( 1 - \zeta_{10}^{3} ) q^{3} + ( -3 \zeta_{10} + 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{4} + ( 2 + \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{5} + ( \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{6} + ( -3 \zeta_{10} + 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{7} + ( 4 - 4 \zeta_{10} + \zeta_{10}^{3} ) q^{8} + ( 1 - \zeta_{10} + \zeta_{10}^{3} ) q^{9} +O(q^{10})$$ $$q + ( -1 + 2 \zeta_{10} - \zeta_{10}^{2} ) q^{2} + ( 1 - \zeta_{10}^{3} ) q^{3} + ( -3 \zeta_{10} + 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{4} + ( 2 + \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{5} + ( \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{6} + ( -3 \zeta_{10} + 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{7} + ( 4 - 4 \zeta_{10} + \zeta_{10}^{3} ) q^{8} + ( 1 - \zeta_{10} + \zeta_{10}^{3} ) q^{9} + \zeta_{10} q^{10} + ( 2 \zeta_{10} + \zeta_{10}^{2} + 2 \zeta_{10}^{3} ) q^{11} -3 \zeta_{10} q^{12} + ( 1 - \zeta_{10} + \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{13} + ( 6 - 6 \zeta_{10} + 3 \zeta_{10}^{3} ) q^{14} + ( 3 - \zeta_{10} + \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{15} + ( -5 + 8 \zeta_{10} - 8 \zeta_{10}^{2} + 5 \zeta_{10}^{3} ) q^{16} -\zeta_{10}^{3} q^{17} + ( -2 + 5 \zeta_{10} - 5 \zeta_{10}^{2} + 2 \zeta_{10}^{3} ) q^{18} + ( -2 - 4 \zeta_{10} - 2 \zeta_{10}^{2} ) q^{19} + ( -3 \zeta_{10} - 3 \zeta_{10}^{3} ) q^{20} -3 \zeta_{10} q^{21} + ( -1 + \zeta_{10} + \zeta_{10}^{3} ) q^{22} + ( 4 - 4 \zeta_{10} ) q^{23} + ( \zeta_{10} - 4 \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{24} + ( 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{25} + ( 1 - \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{26} + ( 4 \zeta_{10} - \zeta_{10}^{2} + 4 \zeta_{10}^{3} ) q^{27} + ( -9 + 18 \zeta_{10} - 18 \zeta_{10}^{2} + 9 \zeta_{10}^{3} ) q^{28} + ( 3 + \zeta_{10} + 3 \zeta_{10}^{2} ) q^{29} + ( 1 + \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{30} + ( 3 - 2 \zeta_{10} + 6 \zeta_{10}^{2} ) q^{31} + ( -6 + 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{32} + ( 3 + 2 \zeta_{10} + 3 \zeta_{10}^{2} ) q^{33} + ( 1 - 2 \zeta_{10} + 2 \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{34} + ( -3 \zeta_{10} - 3 \zeta_{10}^{3} ) q^{35} + ( -3 + 6 \zeta_{10}^{2} - 6 \zeta_{10}^{3} ) q^{36} + ( -7 - \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{37} + ( 2 \zeta_{10} - 6 \zeta_{10}^{2} + 2 \zeta_{10}^{3} ) q^{38} + ( 1 - \zeta_{10} - \zeta_{10}^{3} ) q^{39} + ( 3 - 3 \zeta_{10} - 2 \zeta_{10}^{3} ) q^{40} + ( 1 + 2 \zeta_{10} + \zeta_{10}^{2} ) q^{41} + ( 3 \zeta_{10} - 6 \zeta_{10}^{2} + 3 \zeta_{10}^{3} ) q^{42} + ( 1 - 5 \zeta_{10} + \zeta_{10}^{2} ) q^{43} + ( 6 - 3 \zeta_{10} + 3 \zeta_{10}^{2} - 6 \zeta_{10}^{3} ) q^{44} + \zeta_{10}^{3} q^{45} + ( -4 + 12 \zeta_{10} - 12 \zeta_{10}^{2} + 4 \zeta_{10}^{3} ) q^{46} + ( 7 - 10 \zeta_{10} + 10 \zeta_{10}^{2} - 7 \zeta_{10}^{3} ) q^{47} + ( -5 + 5 \zeta_{10} + 2 \zeta_{10}^{3} ) q^{48} + ( -2 + 11 \zeta_{10} - 11 \zeta_{10}^{2} + 2 \zeta_{10}^{3} ) q^{49} + ( 6 - 9 \zeta_{10} + 6 \zeta_{10}^{2} ) q^{50} + ( -\zeta_{10} - \zeta_{10}^{3} ) q^{51} + ( -3 + 3 \zeta_{10} - 3 \zeta_{10}^{2} ) q^{52} + ( -11 + 11 \zeta_{10} + 5 \zeta_{10}^{3} ) q^{53} + ( -5 + 5 \zeta_{10} - \zeta_{10}^{3} ) q^{54} + ( 5 \zeta_{10} + 3 \zeta_{10}^{2} + 5 \zeta_{10}^{3} ) q^{55} + ( -12 + 15 \zeta_{10}^{2} - 15 \zeta_{10}^{3} ) q^{56} + ( -8 - 6 \zeta_{10}^{2} + 6 \zeta_{10}^{3} ) q^{57} + ( 2 \zeta_{10} - \zeta_{10}^{2} + 2 \zeta_{10}^{3} ) q^{58} + ( 2 - 3 \zeta_{10} + 3 \zeta_{10}^{2} - 2 \zeta_{10}^{3} ) q^{59} + ( -3 - 3 \zeta_{10} - 3 \zeta_{10}^{2} ) q^{60} + 11 q^{61} + ( 3 + 2 \zeta_{10} - 7 \zeta_{10}^{2} + 8 \zeta_{10}^{3} ) q^{62} + ( -3 + 6 \zeta_{10}^{2} - 6 \zeta_{10}^{3} ) q^{63} + ( 6 - 5 \zeta_{10} + 6 \zeta_{10}^{2} ) q^{64} + ( 2 - \zeta_{10} + \zeta_{10}^{2} - 2 \zeta_{10}^{3} ) q^{65} + ( \zeta_{10} + \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{66} + ( -5 - \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{67} + ( -3 \zeta_{10}^{2} + 3 \zeta_{10}^{3} ) q^{68} -4 \zeta_{10}^{2} q^{69} + ( 3 - 3 \zeta_{10} ) q^{70} + ( -2 + 2 \zeta_{10} + 8 \zeta_{10}^{3} ) q^{71} + ( 9 - 14 \zeta_{10} + 9 \zeta_{10}^{2} ) q^{72} + ( 2 \zeta_{10} - 7 \zeta_{10}^{2} + 2 \zeta_{10}^{3} ) q^{73} + ( 5 - 11 \zeta_{10} + 5 \zeta_{10}^{2} ) q^{74} + ( 3 - 3 \zeta_{10} + 3 \zeta_{10}^{2} - 3 \zeta_{10}^{3} ) q^{75} + ( -12 + 12 \zeta_{10} + 6 \zeta_{10}^{3} ) q^{76} + ( 6 - 3 \zeta_{10} + 3 \zeta_{10}^{2} - 6 \zeta_{10}^{3} ) q^{77} + ( \zeta_{10} - \zeta_{10}^{2} ) q^{78} + ( 2 - 2 \zeta_{10} + 3 \zeta_{10}^{3} ) q^{79} + ( -7 + 5 \zeta_{10} - 5 \zeta_{10}^{2} + 7 \zeta_{10}^{3} ) q^{80} + ( 6 - 2 \zeta_{10} + 6 \zeta_{10}^{2} ) q^{81} + ( -\zeta_{10} + 3 \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{82} + ( -6 + 4 \zeta_{10} - 6 \zeta_{10}^{2} ) q^{83} + ( -9 + 9 \zeta_{10} ) q^{84} + ( 1 - \zeta_{10} - 2 \zeta_{10}^{3} ) q^{85} + ( 6 \zeta_{10} - 11 \zeta_{10}^{2} + 6 \zeta_{10}^{3} ) q^{86} + ( 7 + 4 \zeta_{10}^{2} - 4 \zeta_{10}^{3} ) q^{87} + ( 5 + 2 \zeta_{10}^{2} - 2 \zeta_{10}^{3} ) q^{88} + ( -2 \zeta_{10} + 7 \zeta_{10}^{2} - 2 \zeta_{10}^{3} ) q^{89} + ( -1 + 2 \zeta_{10} - 2 \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{90} + ( -3 + 3 \zeta_{10} - 3 \zeta_{10}^{2} ) q^{91} + ( -12 + 12 \zeta_{10}^{2} - 12 \zeta_{10}^{3} ) q^{92} + ( 7 + 4 \zeta_{10}^{2} - \zeta_{10}^{3} ) q^{93} + ( 10 - 13 \zeta_{10}^{2} + 13 \zeta_{10}^{3} ) q^{94} + ( -8 - 6 \zeta_{10} - 8 \zeta_{10}^{2} ) q^{95} + ( -3 - 3 \zeta_{10} + 3 \zeta_{10}^{2} + 3 \zeta_{10}^{3} ) q^{96} + 12 \zeta_{10}^{2} q^{97} + ( -11 + 20 \zeta_{10}^{2} - 20 \zeta_{10}^{3} ) q^{98} + ( -1 - \zeta_{10}^{2} + \zeta_{10}^{3} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q - q^{2} + 3q^{3} - 9q^{4} + 6q^{5} - 2q^{6} - 9q^{7} + 13q^{8} + 4q^{9} + O(q^{10})$$ $$4q - q^{2} + 3q^{3} - 9q^{4} + 6q^{5} - 2q^{6} - 9q^{7} + 13q^{8} + 4q^{9} + q^{10} + 3q^{11} - 3q^{12} + q^{13} + 21q^{14} + 7q^{15} + q^{16} - q^{17} + 4q^{18} - 10q^{19} - 6q^{20} - 3q^{21} - 2q^{22} + 12q^{23} + 6q^{24} - 6q^{25} + 6q^{26} + 9q^{27} + 9q^{28} + 10q^{29} + 2q^{30} + 4q^{31} - 30q^{32} + 11q^{33} - q^{34} - 6q^{35} - 24q^{36} - 26q^{37} + 10q^{38} + 2q^{39} + 7q^{40} + 5q^{41} + 12q^{42} - 2q^{43} + 12q^{44} + q^{45} + 12q^{46} + q^{47} - 13q^{48} + 16q^{49} + 9q^{50} - 2q^{51} - 6q^{52} - 28q^{53} - 16q^{54} + 7q^{55} - 78q^{56} - 20q^{57} + 5q^{58} - 12q^{60} + 44q^{61} + 29q^{62} - 24q^{63} + 13q^{64} + 4q^{65} + q^{66} - 18q^{67} + 6q^{68} + 4q^{69} + 9q^{70} + 2q^{71} + 13q^{72} + 11q^{73} + 4q^{74} + 3q^{75} - 30q^{76} + 12q^{77} + 2q^{78} + 9q^{79} - 11q^{80} + 16q^{81} - 5q^{82} - 14q^{83} - 27q^{84} + q^{85} + 23q^{86} + 20q^{87} + 16q^{88} - 11q^{89} + q^{90} - 6q^{91} - 72q^{92} + 23q^{93} + 66q^{94} - 30q^{95} - 15q^{96} - 12q^{97} - 84q^{98} - 2q^{99} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/403\mathbb{Z}\right)^\times$$.
$$n$$ $$249$$ $$313$$ $$\chi(n)$$ $$1$$ $$-\zeta_{10}^{3}$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
66.1
−0.309017 − 0.951057i 0.809017 + 0.587785i −0.309017 + 0.951057i 0.809017 − 0.587785i
−0.809017 2.48990i 0.190983 0.587785i −3.92705 + 2.85317i 0.381966 −1.61803 −3.92705 + 2.85317i 6.04508 + 4.39201i 2.11803 + 1.53884i −0.309017 0.951057i
157.1 0.309017 + 0.224514i 1.30902 0.951057i −0.572949 1.76336i 2.61803 0.618034 −0.572949 1.76336i 0.454915 1.40008i −0.118034 + 0.363271i 0.809017 + 0.587785i
287.1 −0.809017 + 2.48990i 0.190983 + 0.587785i −3.92705 2.85317i 0.381966 −1.61803 −3.92705 2.85317i 6.04508 4.39201i 2.11803 1.53884i −0.309017 + 0.951057i
326.1 0.309017 0.224514i 1.30902 + 0.951057i −0.572949 + 1.76336i 2.61803 0.618034 −0.572949 + 1.76336i 0.454915 + 1.40008i −0.118034 0.363271i 0.809017 0.587785i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
31.d even 5 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 403.2.k.c 4
31.d even 5 1 inner 403.2.k.c 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
403.2.k.c 4 1.a even 1 1 trivial
403.2.k.c 4 31.d even 5 1 inner
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{4} + T_{2}^{3} + 6 T_{2}^{2} - 4 T_{2} + 1$$ acting on $$S_{2}^{\mathrm{new}}(403, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 + T + 4 T^{2} + 2 T^{3} + 9 T^{4} + 4 T^{5} + 16 T^{6} + 8 T^{7} + 16 T^{8}$$
$3$ $$1 - 3 T + T^{2} + T^{3} + 4 T^{4} + 3 T^{5} + 9 T^{6} - 81 T^{7} + 81 T^{8}$$
$5$ $$( 1 - 3 T + 11 T^{2} - 15 T^{3} + 25 T^{4} )^{2}$$
$7$ $$1 + 9 T + 29 T^{2} + 33 T^{3} + 4 T^{4} + 231 T^{5} + 1421 T^{6} + 3087 T^{7} + 2401 T^{8}$$
$11$ $$1 - 3 T + 8 T^{2} - 51 T^{3} + 265 T^{4} - 561 T^{5} + 968 T^{6} - 3993 T^{7} + 14641 T^{8}$$
$13$ $$1 - T + T^{2} - T^{3} + T^{4}$$
$17$ $$1 + T - 16 T^{2} - 33 T^{3} + 239 T^{4} - 561 T^{5} - 4624 T^{6} + 4913 T^{7} + 83521 T^{8}$$
$19$ $$1 + 10 T + 21 T^{2} - 190 T^{3} - 1519 T^{4} - 3610 T^{5} + 7581 T^{6} + 68590 T^{7} + 130321 T^{8}$$
$23$ $$1 - 12 T + 41 T^{2} - 36 T^{3} + 49 T^{4} - 828 T^{5} + 21689 T^{6} - 146004 T^{7} + 279841 T^{8}$$
$29$ $$1 - 10 T + 11 T^{2} + 120 T^{3} - 439 T^{4} + 3480 T^{5} + 9251 T^{6} - 243890 T^{7} + 707281 T^{8}$$
$31$ $$1 - 4 T + 46 T^{2} - 124 T^{3} + 961 T^{4}$$
$37$ $$( 1 + 13 T + 115 T^{2} + 481 T^{3} + 1369 T^{4} )^{2}$$
$41$ $$1 - 5 T - 31 T^{2} + 205 T^{3} + 476 T^{4} + 8405 T^{5} - 52111 T^{6} - 344605 T^{7} + 2825761 T^{8}$$
$43$ $$1 + 2 T - 19 T^{2} - 254 T^{3} + 619 T^{4} - 10922 T^{5} - 35131 T^{6} + 159014 T^{7} + 3418801 T^{8}$$
$47$ $$1 - T + 59 T^{2} - 177 T^{3} + 2264 T^{4} - 8319 T^{5} + 130331 T^{6} - 103823 T^{7} + 4879681 T^{8}$$
$53$ $$1 + 28 T + 401 T^{2} + 4024 T^{3} + 32129 T^{4} + 213272 T^{5} + 1126409 T^{6} + 4168556 T^{7} + 7890481 T^{8}$$
$59$ $$1 - 49 T^{2} + 270 T^{3} + 3211 T^{4} + 15930 T^{5} - 170569 T^{6} + 12117361 T^{8}$$
$61$ $$( 1 - 11 T + 61 T^{2} )^{4}$$
$67$ $$( 1 + 9 T + 153 T^{2} + 603 T^{3} + 4489 T^{4} )^{2}$$
$71$ $$1 - 2 T - 7 T^{2} + 466 T^{3} + 1865 T^{4} + 33086 T^{5} - 35287 T^{6} - 715822 T^{7} + 25411681 T^{8}$$
$73$ $$1 - 11 T - 22 T^{2} + 845 T^{3} - 5609 T^{4} + 61685 T^{5} - 117238 T^{6} - 4279187 T^{7} + 28398241 T^{8}$$
$79$ $$1 - 9 T - 48 T^{2} + 643 T^{3} - 195 T^{4} + 50797 T^{5} - 299568 T^{6} - 4437351 T^{7} + 38950081 T^{8}$$
$83$ $$1 + 14 T + 53 T^{2} + 870 T^{3} + 14801 T^{4} + 72210 T^{5} + 365117 T^{6} + 8005018 T^{7} + 47458321 T^{8}$$
$89$ $$1 + 11 T - 38 T^{2} - 1037 T^{3} - 5625 T^{4} - 92293 T^{5} - 300998 T^{6} + 7754659 T^{7} + 62742241 T^{8}$$
$97$ $$1 + 12 T + 47 T^{2} - 600 T^{3} - 11759 T^{4} - 58200 T^{5} + 442223 T^{6} + 10952076 T^{7} + 88529281 T^{8}$$ | 6,578 | 11,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-34 | latest | en | 0.38781 |
https://www.dataunitconverter.com/yobibit-to-nibble | 1,716,331,092,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00872.warc.gz | 620,188,503 | 16,252 | # Yibit to Nibble → CONVERT Yobibits to Nibbles
expand_more
info 1 Yibit is equal to 302,231,454,903,657,293,676,544 Nibble
Yobibit (binary) --to--> Nibble
Yibit
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toc Table of Contents
## Yobibit (Yibit) Versus Nibble - Comparison
Yobibits and Nibbles are units of digital information used to measure storage capacity and data transfer rate.
Yobibits is a "binary" unit where as Nibbles is a "basic" unit. One Yobibit is equal to 1024^8 bits. One Nibble is equal to 4 bits. There are 0.0000000000000000000000033087224502121106 Yobibit in one Nibble. Find more details on below table.
Unit Name Yobibit Nibble
Unit Symbol Yib or Yibit
Standard binary basic
Defined Value 2^80 or 1024^8 Bits 4 bits
Value in Bits 1,208,925,819,614,629,174,706,176 4
Value in Bytes 151,115,727,451,828,646,838,272 0.5
## Yobibit (Yibit) to Nibble Conversion - Formula & Steps
The Yibit to Nibble Calculator Tool provides a convenient solution for effortlessly converting data units from Yobibit (Yibit) to Nibble. Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Yobibit) and target (Nibble) data units.
Source Data Unit Target Data Unit
Equal to 1024^8 bits
(Binary Unit)
Equal to 4 bits
(Basic Unit)
The formula for converting the Yobibit (Yibit) to Nibble can be expressed as follows:
diamond CONVERSION FORMULA Nibble = Yibit x 10248 ÷ 4
Now, let's apply the aforementioned formula and explore the manual conversion process from Yobibit (Yibit) to Nibble. To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Nibbles = Yobibits x 10248 ÷ 4
STEP 1
Nibbles = Yobibits x (1024x1024x1024x1024x1024x1024x1024x1024) ÷ 4
STEP 2
Nibbles = Yobibits x 1208925819614629174706176 ÷ 4
STEP 3
Nibbles = Yobibits x 302231454903657293676544
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Example : By applying the previously mentioned formula and steps, the conversion from 1 Yobibit (Yibit) to Nibble can be processed as outlined below.
1. = 1 x 10248 ÷ 4
2. = 1 x (1024x1024x1024x1024x1024x1024x1024x1024) ÷ 4
3. = 1 x 1208925819614629174706176 ÷ 4
4. = 1 x 302231454903657293676544
5. = 302,231,454,903,657,293,676,544
6. i.e. 1 Yibit is equal to 302,231,454,903,657,293,676,544 Nibble.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Yobibits to Nibbles using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Yobibit ?
A yobibit (Yib or Yibit) is a binary unit of digital information that is equal to 1,208,925,819,614,629,174,706,176 bits and is defined by the International Electro technical Commission(IEC). The prefix 'yobi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'yottabit' (Yb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
- Learn more..
arrow_downward
#### What is Nibble ?
A Nibble is a unit of digital information that consists of 4 bits. It is half of a byte and can represent a single hexadecimal digit. It is used in computer memory and data storage and sometimes used as a basic unit of data transfer in certain computer architectures.
- Learn more..
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## Excel Formula to convert from Yobibit (Yibit) to Nibble
Apply the formula as shown below to convert from 1 Yobibit (Yibit) to Nibble.
A B C
1 Yobibit (Yibit) Nibble
2 1 =A2 * 302231454903657293676544
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Yobibit (Yibit) to Nibble Conversion
You can use below code to convert any value in Yobibit (Yibit) to Yobibit (Yibit) in Python.
yobibits = int(input("Enter Yobibits: "))
nibbles = yobibits * (1024*1024*1024*1024*1024*1024*1024*1024) / 4
print("{} Yobibits = {} Nibbles".format(yobibits,nibbles))
The first line of code will prompt the user to enter the Yobibit (Yibit) as an input. The value of Nibble is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Yobibits(Yibit) are there in a Nibble?expand_more
There are 0.0000000000000000000000033087224502121106 Yobibits in a Nibble.
#### What is the formula to convert Nibble to Yobibit(Yibit)?expand_more
Use the formula Yibit = Nibble x 4 / 10248 to convert Nibble to Yobibit.
#### How many Nibbles are there in a Yobibit(Yibit)?expand_more
There are 302231454903657293676544 Nibbles in a Yobibit.
#### What is the formula to convert Yobibit(Yibit) to Nibble?expand_more
Use the formula Nibble = Yibit x 10248 / 4 to convert Yobibit to Nibble.
#### Which is bigger, Yobibit(Yibit) or Nibble?expand_more
Yobibit is bigger than Nibble. One Yobibit contains 302231454903657293676544 Nibbles.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,559 | 5,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-22 | latest | en | 0.767446 |
http://multiwingspan.co.uk/python.php?page=bitwise | 1,716,827,805,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00196.warc.gz | 18,757,529 | 4,768 | # Python For GCSE Bitwise Operators
## Introduction
The screenshot shows the bitwise operators in Python. The are called 'bitwise' because they operate on the numbers at the level of the bit. That means that they are using the binary representation of the number when they perform the operation.
Let's work through each one of these to see if we can make sense of how we are getting the results that we are.
### 5 AND 3
The AND operator is the ampersand (&). To see how it is working, we line up the binary representations of each number, one above the other. Then we perform a logic AND operation with each column. The result is a 1 if the column contains two 1s, otherwise we get a zero.
```101 011 001```
That's how we end up with a result of 1.
### 5 OR 3
The OR operator is the pipe character (|). Again, we line up the numbers and do the logical OR operation on each column. The result is a 1 if either or both of the two numbers in the column are 1s.
```101 011 111```
### 5 XOR 3
The XOR operation is based on a logic gate that is not a part of the GCSE specification. It means 'exclusive OR'. It works in a similar way to the OR function except that two inputs of a 1 give the result of a zero. You can summarise an XOR by saying that it needs one or other of the inputs to be a 1, but not both. The symbol for this operator is the circumflex (^).
```101 011 110```
### NOT 5
The NOT operator is a tilde (~). It flips the bits. When using this, you end up flipping the sign (positive or negative) of the number. You can read up about how the Two's Complement way of representing negative binary integers works to understand this more.
```101 flip all bits, make the number negative and set the leftmost bit to 1 110```
### Left & Right Shifts
The left and right shift operators move all of the bits of a number a specified number of places to the left or right. A left shift of 1 is the equivalent of multiplying a number by 2. A right shift of 2 would divide a number by 2. When we are using binary integers, right shifts can mean that some place values are lost when they are moved right of the units bit.
```1 << 2 001 becomes 100. 128 >> 4 10000000 becomes 1000.```
## Some Observations
### AND
If we do an AND between a number and 1, the result depends on the number we start with. This is useful for reading bits in a number.
If we do an AND between a number and 0, we ignore the bits in the number we start with. This is useful for masking bits.
For example, take an 8 bit number.
```10101001 00001111 00001001```
If we do, number & 15, we can extract the digits from the first nibble. We mask (ignore) the other bits.
### OR
OR is useful for turning on specific bits. If we do an OR with 1, we make a bit equal to 1, no matter what it was,
```1001 1111 1111```
### XOR
XOR is useful for flipping bits. If we do and XOR with 1, we would transform the original bit of 0 to 1 and one of 1 to 0.
These are quite challenging and they are meant to be. The idea behind these tasks is to build up a series of useful functions that you can use when you find the need for one of them in a program you are writing. Follow the guidance for each of the tasks and restrict yourself to using references for Python statements rather than looking for someone's answer.
Many of these can be achieved with only a few lines, some with just a return statement.
Get the number of bits in a number. Find the base 2 logarithm of n+1. Round it up using ceil. Both of these methods are in the math module.
Do an AND with 1. If the rightmost bit is a 1, the number is odd, otherwise even.
The idea for this one is to find out if the bit at position p in the number n is set to 1 or not. It should return True or False. To find out if the bit at place p is on, do an AND between n and a 1 which has been left shifted p places.
To toggle the bit at place p, do an XOR between n and a 1 which has been left shifted p places.
You are setting the bit at place p to 1 if v is True, to 0 if it is False.
If you are turning the bit on, do an OR with a 1 which has been left shifted p times.
To turn a bit off, we can use masking. Left shift a 1 into the position you want to turn off (p places to the left), then invert the pattern with NOT to create the mask you need. Use an AND between n and your mask to get the result you need.
This function is meant to give you a list of the bits of a number, starting with the units bit. That is called the least significant bit (LSB). Make an empty list. Make a for loop repeat as many times as there are bits in n. In the loop, do an AND with 1 and append the value, then right shift the number a place before the next iteration.
You can also do this with a list comprehension for extra cool points. | 1,197 | 4,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.880567 |
https://goprep.co/ex-7.2-q5-integrate-the-functions-sin-ax-b-cos-ax-b-i-1nk1w6 | 1,610,900,693,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00502.warc.gz | 350,542,397 | 48,752 | # Integrate the functions.sin (ax + b) cos (ax + b)
Let I = ∫sin (ax + b) cos (ax + b) dx
We know that,
sin 2A = 2sinA.cosA
Therefore,
sin (ax + b) cos (ax + b) =
Let 2(ax+b) = t
=
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Mathematics - Exemplar
Evaluate the following:
Mathematics - Exemplar | 271 | 1,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-04 | latest | en | 0.797553 |
https://socratic.org/questions/how-do-you-solve-8-y-5-2y-10 | 1,601,583,676,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131986.91/warc/CC-MAIN-20201001174918-20201001204918-00214.warc.gz | 587,071,978 | 5,981 | # How do you solve 8(y-5)+2y=-10?
Jul 2, 2016
$y = 3$
#### Explanation:
Multiply out the bracket and collect like terms.
$8 y - 40 + 2 y = - 10$
$10 y = - 10 + 40 = 30$
Divide both sides by 10.
$y = 3$
(Just to check we have obtained the correct answer, sub back into the original equation)
$8 \left(3 - 5\right) + 2 \left(3\right) = - 16 + 6 = - 10$
So our solution works! | 143 | 384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-40 | latest | en | 0.810501 |
https://betterlesson.com/lesson/resource/1934873/55025/powerpoint-angles-on-a-coordinate-plane | 1,503,277,117,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107065.72/warc/CC-MAIN-20170821003037-20170821023037-00280.warc.gz | 720,206,597 | 22,826 | ## PowerPoint- Angles on a Coordinate Plane - Section 3: Locker Combination Activity
PowerPoint- Angles on a Coordinate Plane
# Angle and Degree Measure
Unit 9: Trigonometric Functions
Lesson 1 of 19
## Big Idea: Students will turn a locker combination into a secret math code (angles).
Print Lesson
11 teachers like this lesson
Standards:
Subject(s):
Math, Degree Measures, Algebra 2, coterminal angles, 11th Grade, master teacher project
50 minutes
### Amelia Jamison
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Phoenix, AZ
Environment: Urban | 296 | 1,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-34 | longest | en | 0.74314 |
https://www.physicsforums.com/threads/ftl-question.24212/ | 1,485,300,937,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00101-ip-10-171-10-70.ec2.internal.warc.gz | 966,273,096 | 15,095 | # FTL Question
1. May 5, 2004
### ccb056
If this equation is true:
V=(V1+V2)/(1+(V1*V2)/c^2)
Then why is it when you plug in c^2 for both V1 and V2 you get the total velocity as 2m/s
2. May 5, 2004
### chroot
Staff Emeritus
You don't.. you get c.
$$V = \frac{2 c}{1 + c^2/c^2} = \frac{2 c}{2} = c$$
- Warren
3. May 6, 2004
### turin
This will give a nonsensical result. You must plug in speed, not squared speed.
4. May 11, 2004
### ccb056
What I meant to say was (3*10^8)^2
5. May 11, 2004
### arildno
It's still meaningless, besides wrong arithmetic:
Putting your value into slots for V1 and V2 yields:
$$\frac{2*(3*10^{8})^{2}}{1+(3*10^{8})^{2}}$$
To clarify, and make a more "accurate" argument:
Let the "velocities" be some big, ugly number on the form: V=kc, k>>1
Then you have by plugging in:
$$\frac{2V}{1+(\frac{V}{c})^{2}}\approx\frac{2c}{k}$$
In your case, k=c; hence the approximate value of 2.
Last edited: May 11, 2004 | 365 | 952 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-04 | longest | en | 0.847576 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/11/lesson/11.4.1/problem/11-115 | 1,721,694,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00172.warc.gz | 245,740,832 | 15,862 | ### Home > APCALC > Chapter 11 > Lesson 11.4.1 > Problem11-115
11-115.
Multiple Choice: The luminous intensity $E$ of a light bulb, measured in lumens/ft2, varies inversely as the square of the distance $s$ from the bulb. $E = 5.2$ lumens/ft2 when $s = 5 \text{ ft}$ for a $100$ watt bulb. If you are moving away from a $100$ watt bulb at a speed of $2$ ft/sec and you are $3$ feet from the bulb, the luminous intensity is changing at the rate of:
1. $- \frac { 520 } { 27 }$ lumens/ft
1. $- \frac { 260 } { 27 }$ lumens/ft
1. $- \frac { 130 } { 27 }$ lumens/ft
1. $- \frac { 130 } { 9 }$ lumens/ft
1. $- \frac { 260 } { 9 }$ lumens/ft
$E=\frac{k}{s^2}$
$5.2=\frac{k}{5^2}\text{ }k=?$
When $s = 3, E =$ ?
$\frac{d}{dt}(Es^2)=\frac{d}{dt}(k)$
$E^\prime s^2+2Ess^\prime=0$
Substitute in the known values and solve for E′. | 335 | 832 | {"found_math": true, "script_math_tex": 18, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.611599 |
https://math.stackexchange.com/questions/1778063/topology-continuity-and-the-induced-topology | 1,720,955,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00615.warc.gz | 328,789,653 | 36,168 | # Topology -- Continuity and the induced topology.
Here is my question.
Let $X=Y=\Bbb R$, with the usual topology. Let $A=[0,1]$ and topologize $A$ with the induced topology from $X$.
Does there exist a continuous function from the topological space $A$ onto $Y$?
Why or why not?
I'd argue No, since we'd be going from a closed to open set. But I also thought that any subset of a topology is an open subset.
Any help would be appreciated.
• Hint: All of $\mathbb{R}$ is a closed set (also an open set; it's clopen). The tan and arctangent functions have the required type of behavior, just rescale them as needed (domain restrict the tan function to make it invertible). Commented May 9, 2016 at 12:55
• @JustinBenfield That works for the open interval $(0,1)$, not the closed interval $[0,1]$. Commented May 9, 2016 at 12:56
No: $[0,1]$ is compact, $\mathbb R$ is not compact, and the image of a compact set under a continuous function is always compact.
The extreme value theorem, from real analysis, states that any continuous function from a compact set to the reals is bounded (and has a maximum). A bounded function cannot be surjective onto $\Bbb R$. | 311 | 1,166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.935622 |
https://math.stackexchange.com/questions/310349/real-analysis-cauchy-sequence-question | 1,725,949,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00156.warc.gz | 373,477,418 | 36,618 | # Real Analysis Cauchy Sequence Question
So I've been been working on the question below, and I have some questions in regards to the validity of my answer.
Let $(a_n)$ be a sequence such that $$\lim_{N \to \infty} \sum_{n=1}^{N} |a_n - a_{n+1}| < \infty .$$ Show that $(a_n)$ is Cauchy.
I make the claim that the distance between the terms of $(a_n)$ must approach zero. As such for every $\epsilon > 0$, there must be be an integer $N$ such that
$$|a_m - a_n| < \epsilon$$
for all $m,n \geq N$. That is, the sequence is Cauchy. To show this, assume that the distances between the terms of $(a_n)$ do not approach zero. Let
$$a=\min \left \{ |a_1 - a_2|,|a_2 - a_3|,...,|a_N - a_{N+1}|,... \right \}.$$
Then we have $a \neq 0$. Observe that
$$\lim_{N \to \infty} Na \leq \lim_{N \to \infty} \sum_{n=1}^{N} |a_{N}-a_{N+1}| < \infty,$$
which, is a contraction, as
$$\lim_{N \to \infty} Na= \infty$$
for any $a \neq 0$. Thus, we must have $a=0$, and the distance between the terms of $(a_n)$ must approach zero and as such the sequence is Cauchy.
I am unsure about setting $a= \min \{...\}$. Any input or comments about my answer would be appreciated.
• What if you change $a_2$ to be the same as $a_1$ keeping all other terms the same. The convergence properties will not change, but $a$ will be 0. Commented Feb 21, 2013 at 17:20
Your $a$ could be $0$, so this does not work, even if the idea is great.
Take $\epsilon>0$.
Since the series converges, there exists $N$ such that $$\sum_{n=N}^{+\infty} |a_{n+1}-a_n|\leq \epsilon.$$
Now for all for $N\leq k<l$, we have $$|a_k-a_l|=|\sum_{n=k}^{l}a_{n+1}-a_n|\leq \sum_{n=k}^{l}|a_{n+1}-a_n|\leq \sum_{n=N}^{+\infty}|a_{n+1}-a_n|\leq \epsilon.$$
So the sequence $a_n$ is indeed Cauchy. | 627 | 1,746 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-38 | latest | en | 0.817667 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-10-exponents-and-radicals-10-8-the-complex-numbers-10-8-exercise-set-page-686/56 | 1,537,943,353,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163704.93/warc/CC-MAIN-20180926061824-20180926082224-00482.warc.gz | 743,023,631 | 15,008 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$5$
Using $(a+b)(a-b)=a^2-b^2$, or the product of the sum and difference of like terms, and using the properties of radicals, the given expression, $(1+2i)(1-2i) ,$ simplifies to \begin{array}{l}\require{cancel} (1)^2-(2i)^2 \\\\= 1-4i^2 \\\\= 1-4(-1) \\\\= 1+4 \\\\= 5 .\end{array} | 129 | 361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-39 | longest | en | 0.683628 |
http://www.carrawaydavie.com/quart-to-gallon-calculator/ | 1,571,191,129,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00306.warc.gz | 232,529,570 | 8,691 | ## What is a gallon quart to gallon calculator?
Gallon is a royal as well as United States normal measurement system volume device. quart to gallon calculator? There is one kind of gallon in the imperial system and also 2 kinds (fluid and completely dry) in the US customary dimension system. 1 US liquid gallon is specified as 231 cubic inches, 1 US dry gallon is 268.8 cubic inches and 1 royal gallon is 277.4 cubic inches. The abbreviation is gall. quart to gallon calculator?
Gallons Conversion:
1 Gallon = 4 Quarts
1 Gallon = 8 Pints
1 Gallon = 16 Cups
1 Gallon = 256 Tablespoons
1 Gallon = 768 Teaspoons
quart to gallon calculator
## What is a liquid ounce (fl oz)quart to gallon calculator?
A liquid ounce is an imperial as well as US customary measurement system volume unit. 1 US fluid ounce equals to 29.5735 mL and 1 royal (UK) fluid ounce equals to 28.4131 mL. The acronym is fl oz. So, quart to gallon calculator?
There are 8 fluid ounces in a United States mug and also 10 imperial liquid ounces in an imperial cup.
1 United States Fluid Gallon = 128 US Fluid Ounces
1 US Dry Gallon = 148.946 United States Fluid Ounces
1 Imperial Gallon (UK) = 160 Imperial Fluid Ounces (UK).
quart to gallon calculator
## Liquid Ounce quart to gallon calculator
Liquid Ounce is utilized for volume, Ounce for mass, and also they are various.
As an example, 1 liquid ounce of honey has a mass of about 1,5 ounces!
But also for water, 1 liquid ounce has a mass of about 1 ounce.
quart to gallon calculator
If you imply an ounce of fluid says fluid ounce (fl oz).
In Summary:.
1 gallon = 4 quarts = 8 pints = 16 mugs = 128 fluid ounces.
quart to gallon calculator | 439 | 1,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.86189 |
http://www.caraudio.com/forums/archive/index.php/t-228261.html?s=02b9178b19d0585ca5a90ca658c14eb9 | 1,500,728,291,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424060.49/warc/CC-MAIN-20170722122816-20170722142816-00404.warc.gz | 403,060,781 | 3,509 | PDA
View Full Version : Is there a way to empirically test the tuning of an enclosure?
jibberjive
04-16-2007, 03:59 PM
As the title says, is there an empirical means of finding the tune of an enclosure other than using the math (or box building program) and the enclosure volume, port area and length?
audioholic
04-16-2007, 04:04 PM
You want a way to find a mathematical answer without using math? I guess you could try praying to God for the correct answer. Im really not following what you think you want here.
jibberjive
04-16-2007, 04:35 PM
You want a way to find a mathematical answer without using math? I guess you could try praying to God for the correct answer. Im really not following what you think you want here.
No, I'm curious if there's another way to test it other than the math (ie. measuring port mach or something).
Twistid
04-16-2007, 04:38 PM
play test tones , where the subwoofer moves the most is the usual tuning
helotaxi
04-16-2007, 04:53 PM
Actually it's where it moves the least. You can test it in the same manner as the T/S parameters are tested, but you need to already know the T/S parameters and then do the math backwards to figure volume. Tuning though is easy, wire a resistor in series and then measure the voltage across the resistor using a series of test tones. Where the voltage drop across the resistor is the least, the impedance of the driver is at its max and that the the resonance point of a ported enclosure. It works exactly the opposite for a sealed box.
Volenti
04-16-2007, 06:43 PM
Actually it's where it moves the least. You can test it in the same manner as the T/S parameters are tested, but you need to already know the T/S parameters and then do the math backwards to figure volume. Tuning though is easy, wire a resistor in series and then measure the voltage across the resistor using a series of test tones. Where the voltage drop across the resistor is the least, the impedance of the driver is at its max and that the the resonance point of a ported enclosure. It works exactly the opposite for a sealed box.
Yes, do this with the enclosure both outside and inside the car, and notice the difference in frequency and impedance value. :graduate:
ngsm13
04-16-2007, 07:52 PM
play test tones , where the subwoofer moves the most is the usual tuning
Completely and utterly incorrect.
:fyi:
nG
Immacomputer
04-16-2007, 11:03 PM
It's very simple to just play some tones about .5hz apart and check for the least amount of cone movement.
I usually do this with a signal generator after I finish an enclosure to see how close the final product turned out.
jibberjive
04-17-2007, 01:12 AM
So the difference in excursion is distinguishable even up to .5 hz I take it then.
play test tones , where the subwoofer moves the most is the usual tuning
No need for me to say it again since it's been said twice;)
Volenti, so the environment outside of the enclosure affects the impedence rise of the enclosure as well? I thought just the enclosure affected it, but idk:scared:
Volenti
04-17-2007, 04:51 AM
Volenti, so the environment outside of the enclosure affects the impedence rise of the enclosure as well? I thought just the enclosure affected it, but idk:scared:
Yes indeed, the last vented enclosure that I did detailed measurements on had a tuning frequency of 47.5hz @ 14.4 ohms impedance (dual 4ohm sub, coils in parallel) in half space (box sitting on the ground) and a tuning frequency of 45hz @ 10.95 ohms rear firing in the boot of my small sedan.
Even bigger changes can be seen in more exotic enclosure designs like transmission lines and horns.
helotaxi
04-17-2007, 09:55 AM
Since the car is a small space, the air in the cabin of the car will affect the movement of the cone since it acts as another part of the enclosure. The cabin will thus affect the tuning of the enclosure and the response of the system.
PV Audio
04-17-2007, 06:38 PM
play test tones , where the subwoofer moves the most is the usual tuning
You're merely boosting your output in that frequency range, not boosting speaker excursion. A frequency much lower than tuned in a large box will move quite more than the Fb.
JohnBlayz142
04-18-2007, 02:09 AM
play test tones , where the subwoofer moves the most is the usual tuning
:suicide: Here for 3 years and you still have this backwards? Just proves post count means nothing around here.
Decipha
04-18-2007, 02:46 AM
post count just determines how much time u spend on the forum
i've seen people here as long as me with over 10k
jibberjive
04-18-2007, 04:03 AM
post count just determines how much time u spend on the forum
i've seen people here as long as me with over 10k
eh, not necessarily. Some people post alot, some don't. I've spent my fair share of time on the forum, yet I don't have an insane high post count.:blush:
PV Audio
04-18-2007, 06:56 AM
He's right, I have 16k posts and haven't learned a **** thing :D | 1,269 | 4,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-30 | latest | en | 0.911207 |
https://doc.cgal.org/latest/Triangulation_2/classCGAL_1_1Triangulation__2.html | 1,723,619,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641104812.87/warc/CC-MAIN-20240814061604-20240814091604-00096.warc.gz | 155,414,215 | 27,921 | CGAL 5.6.1 - 2D Triangulations
CGAL::Triangulation_2< Traits, Tds > Class Template Reference
#include <CGAL/Triangulation_2.h>
## Definition
The class Triangulation_2 is the basic class designed to handle triangulations of set of points $${ A}$$ in the plane.
Such a triangulation has vertices at the points of $${ A}$$ and its domain covers the convex hull of $${ A}$$. It can be viewed as a planar partition of the plane whose bounded faces are triangular and cover the convex hull of $${ A}$$. The single unbounded face of this partition is the complementary of the convex hull of $${ A}$$.
In many applications, it is convenient to deal only with triangular faces. Therefore, we add to the triangulation a fictitious vertex, called the infinite vertex and we make each convex hull edge incident to an infinite face having as third vertex the infinite vertex. In that way, each edge is incident to exactly two faces and special cases at the boundary of the convex hull are simpler to deal with.
The infinite vertex.
The class Triangulation_2 implements this point of view and therefore considers the triangulation of the set of points as a set of triangular, finite and infinite faces. Although it is convenient to draw a triangulation as in figure Triangulation_ref_Fig_infinite_vertex, note that the infinite vertex has no significant coordinates and that no geometric predicate can be applied on it or on an infinite face.
A triangulation is a collection of vertices and faces that are linked together through incidence and adjacency relations. Each face give access to its three incident vertices and to its three adjacent faces. Each vertex give access to one of its incident faces.
The three vertices of a face are indexed with 0, 1 and 2 in counterclockwise order. The neighbor of a face are also indexed with 0,1,2 in such a way that the neighbor indexed by $$i$$ is opposite to the vertex with the same index.
The triangulation class offers two functions int cw(int i) and int ccw(int i) which, given the index of a vertex in a face, compute the index of the next vertex of the same face in clockwise or counterclockwise order. Thus, for example the neighbor neighbor(cw(i)) is the neighbor of f which is next to neighbor(i) turning clockwise around f. The face neighbor(cw(i)) is also the first face encountered after f when turning clockwise around vertex i of f (see Figure Triangulation_ref_Fig_neighbors).
Vertices and neighbors.
Template Parameters
Traits is the geometric traits which must be a model of the concept TriangulationTraits_2. Tds is the triangulation data structure which must be a model of the concept TriangulationDataStructure_2. By default, the triangulation data structure is instantiated by Triangulation_data_structure_2 < Triangulation_vertex_base_2, Triangulation_face_base_2 >.
Traversal of the Triangulation
A triangulation can be seen as a container of faces and vertices. Therefore the triangulation provides several iterators and circulators that allow to traverse it completely or partially.
Traversal of the Convex Hull
Applied on the infinite vertex the above functions allow to visit the vertices on the convex hull and the infinite edges and faces. Note that a counterclockwise traversal of the vertices adjacent to the infinite vertex is a clockwise traversal of the convex hull.
Face_circulator incident_faces(t.infinite_vertex()) const;
Face_circulator incident_faces(t.infinite_vertex(), f) const;
Edge_circulator incident_edges(t.infinite_vertex()) const;
Edge_circulator incident_edges(t.infinite_vertex(), f);
Vertex_circulator incident_vertices(t.infinite_vertex() v) ;
Vertex_circulator incident_vertices(t.infinite_vertex(), f) ;
I/O
The I/O operators are defined for iostream. The format for the iostream is an internal format.
The information output in the iostream is:
• the number of vertices (including the infinite one), the number of faces (including infinite ones), and the dimension.
• for each vertex (except the infinite vertex), the non combinatorial information stored in that vertex (point, etc.).
• for each faces, the indices of its vertices and the non combinatorial information (if any) in this face.
• for each face again the indices of the neighboring faces.
The index of an item (vertex of face) is the rank of this item in the output order. When dimension $$<$$ 2, the same information is output for faces of maximal dimension instead of faces.
Implementation
Locate is implemented by a line walk from a vertex of the face given as optional parameter (or from a finite vertex of infinite_face() if no optional parameter is given). It takes time $$O(n)$$ in the worst case, but only $$O(\sqrt{n})$$ on average if the vertices are distributed uniformly at random.
Insertion of a point is done by locating a face that contains the point, and then splitting this face. If the point falls outside the convex hull, the triangulation is restored by flips. Apart from the location, insertion takes a time time $$O(1)$$. This bound is only an amortized bound for points located outside the convex hull.
Removal of a vertex is done by removing all adjacent triangles, and re-triangulating the hole. Removal takes time $$O(d^2)$$ in the worst case, if $$d$$ is the degree of the removed vertex, which is $$O(1)$$ for a random vertex.
The face, edge, and vertex iterators on finite features are derived from their counterparts visiting all (finite and infinite) features which are themselves derived from the corresponding iterators of the triangulation data structure.
TriangulationTraits_2
TriangulationDataStructure_2
TriangulationDataStructure_2::Face
TriangulationDataStructure_2::Vertex
CGAL::Triangulation_data_structure_2<Vb,Fb>
CGAL::Triangulation_vertex_base_2<Traits>
CGAL::Triangulation_face_base_2<Traits>
Examples:
Triangulation_2/adding_handles.cpp, Triangulation_2/colored_face.cpp, Triangulation_2/draw_triangulation_2.cpp, Triangulation_2/for_loop_2.cpp, Triangulation_2/low_dimensional.cpp, and Triangulation_2/triangulation_prog1.cpp.
## Related Functions
(Note that these are not member functions.)
ostream & operator<< (ostream &os, const Triangulation_2< Traits, Tds > &T)
Inserts the triangulation into the stream os. More...
istream & operator>> (istream &is, const Triangulation_2< Traits, Tds > &T)
Reads a triangulation from stream is and assigns it to the triangulation. More...
## Types
typedef Traits Geom_traits
the traits class.
typedef Tds Triangulation_data_structure
the triangulation data structure type.
typedef Traits::Point_2 Point
the point type.
typedef Traits::Segment_2 Segment
the segment type.
typedef Traits::Triangle_2 Triangle
the triangle type.
typedef Tds::Vertex Vertex
the vertex type.
typedef Tds::Face Face
the face type.
typedef Tds::Edge Edge
the edge type.
typedef Tds::size_type size_type
Size type (an unsigned integral type).
typedef Tds::difference_type difference_type
Difference type (a signed integral type).
## Handles, Iterators, and Circulators
The vertices and faces of the triangulations are accessed through handles, iterators and circulators.
The handles are models of the concept Handle which basically offers the two dereference operators and ->. The handles are also model of the concepts LessThanComparable and Hashable, that is they can be used as keys in containers such as std::map and boost::unordered_map. The iterators and circulators are all bidirectional and non-mutable. The circulators and iterators are convertible to handles with the same value type, so that whenever a handle appear in the parameter list of a function, an appropriate iterator or circulator can be passed as well.
The edges of the triangulation can also be visited through iterators and circulators, the edge circulators and iterators are also bidirectional and non mutable.
In the following, we called infinite any face or edge incident to the infinite vertex and the infinite vertex itself. Any other feature (face, edge or vertex) of the triangulation is said to be finite. Some iterators (the All iterators ) allows to visit finite or infinite feature while others (the Finite iterators) visit only finite features. Circulators visit infinite features as well as finite ones. The triangulation class also defines the following enum type to specify which case occurs when locating a point in the triangulation.
In order to write C++ 11 for-loops we provide range types.
enum Locate_type {
VERTEX =0, EDGE, FACE, OUTSIDE_CONVEX_HULL,
OUTSIDE_AFFINE_HULL
}
specifies which case occurs when locating a point in the triangulation. More...
typedef Tds::Vertex_handle Vertex_handle
handle to a vertex.
typedef Tds::Face_handle Face_handle
handle to a face.
typedef Tds::Face_iterator All_faces_iterator
iterator over all faces.
typedef Tds::Edge_iterator All_edges_iterator
iterator over all edges.
typedef Tds::Vertex_iterator All_vertices_iterator
iterator over all vertices.
typedef unspecified_type Finite_faces_iterator
iterator over finite faces.
typedef unspecified_type Finite_edges_iterator
iterator over finite edges.
typedef unspecified_type Finite_vertices_iterator
iterator over finite vertices.
typedef unspecified_type Point_iterator
iterator over the points corresponding to the finite vertices of the triangulation.
typedef Iterator_range< unspecified_typeAll_face_handles
range type for iterating over all faces (including infinite faces), with a nested type iterator that has as value type Face_handle.
typedef Iterator_range< All_edges_iteratorAll_edges
range type for iterating over all edges (including infinite ones).
typedef Iterator_range< unspecified_typeAll_vertex_handles
range type for iterating over all vertices (including the infinite vertex), with a nested type iterator that has as value type Vertex_handle.
typedef Iterator_range< unspecified_typeFinite_face_handles
range type for iterating over finite faces, with a nested type iterator that has as value type Face_handle.
typedef Iterator_range< Finite_edges_iteratorFinite_edges
range type for iterating over finite edges.
typedef Iterator_range< unspecified_typeFinite_vertex_handles
range type for iterating over finite vertices, with a nested type iterator that has as value type Vertex_handle.
typedef Iterator_range< Point_iteratorPoints
range type for iterating over the points of the finite vertices.
typedef unspecified_type Line_face_circulator
circulator over all faces intersected by a line.
typedef unspecified_type Face_circulator
circulator over all faces incident to a given vertex.
typedef unspecified_type Edge_circulator
circulator over all edges incident to a given vertex.
typedef unspecified_type Vertex_circulator
circulator over all vertices incident to a given vertex.
## Creation
Triangulation_2 (const Traits >=Traits())
Introduces an empty triangulation.
Triangulation_2 (const Triangulation_2 &tr)
Copy constructor. More...
template<class InputIterator >
Triangulation_2 (InputIterator first, InputIterator last, const Traits >=Traits())
Equivalent to constructing an empty triangulation with the optional traits class argument and calling insert(first,last).
Triangulation_2 operator= (const Triangulation_2< Traits, Tds > &tr)
Assignment. More...
void swap (Triangulation_2 &tr)
The triangulations tr and *this are swapped. More...
void clear ()
Deletes all faces and finite vertices resulting in an empty triangulation.
## Access Functions
int dimension () const
Returns the dimension of the convex hull.
size_type number_of_vertices () const
Returns the number of finite vertices.
size_type number_of_faces () const
Returns the number of finite faces.
Face_handle infinite_face () const
a face incident to the infinite vertex.
Vertex_handle infinite_vertex () const
the infinite vertex.
Vertex_handle finite_vertex () const
a vertex distinct from the infinite vertex.
const Geom_traitsgeom_traits () const
Returns a const reference to the triangulation traits object.
const TriangulationDataStructure_2tds () const
Returns a const reference to the triangulation data structure.
## Non const access
Attention
The responsibility of keeping a valid triangulation belongs to the user when using advanced operations allowing a direct manipulation of the tds. This method is mainly a help for users implementing their own triangulation algorithms.
TriangulationDataStructure_2tds ()
Returns a reference to the triangulation data structure.
## Predicates
The class Triangulation_2 provides methods to test the finite or infinite character of any feature, and also methods to test the presence in the triangulation of a particular feature (edge or face).
bool is_infinite (Vertex_handle v) const
true iff v is the infinite vertex.
bool is_infinite (Face_handle f) const
true iff face f is infinite.
bool is_infinite (Face_handle f, int i) const
true iff edge (f,i) is infinite.
bool is_infinite (Edge e) const
true iff edge e is infinite.
bool is_infinite (Edge_circulator ec) const
true iff edge *ec is infinite.
bool is_infinite (All_edges_iterator ei) const
true iff edge *ei is infinite.
bool is_edge (Vertex_handle va, Vertex_handle vb)
true if there is an edge having va and vb as vertices.
bool is_edge (Vertex_handle va, Vertex_handle vb, Face_handle &fr, int &i)
as above. More...
bool includes_edge (Vertex_handle va, Vertex_handle vb, Vertex_handle &vbb, Face_handle &fr, int &i)
true if the line segment from va to vb includes an edge e incident to va. More...
bool is_face (Vertex_handle v1, Vertex_handle v2, Vertex_handle v3)
true if there is a face having v1, v2 and v3 as vertices.
bool is_face (Vertex_handle v1, Vertex_handle v2, Vertex_handle v3, Face_handle &fr)
as above. More...
## Queries
The class Triangulation_2 provides methods to locate a given point with respect to a triangulation.
It also provides methods to locate a point with respect to a given finite face of the triangulation.
Face_handle locate (const Point &query, Face_handle f=Face_handle()) const
If the point query lies inside the convex hull of the points, a face that contains the query in its interior or on its boundary is returned. More...
Face_handle inexact_locate (const Point &query, Face_handle start=Face_handle()) const
Same as locate() but uses inexact predicates. More...
Face_handle locate (const Point &query, Locate_type <, int &li, Face_handle h=Face_handle()) const
Same as above. More...
Oriented_side oriented_side (Face_handle f, const Point &p) const
Returns on which side of the oriented boundary of f lies the point p. More...
Oriented_side side_of_oriented_circle (Face_handle f, const Point &p)
Returns on which side of the circumcircle of face f lies the point p. More...
## Modifiers
The following operations are guaranteed to lead to a valid triangulation when they are applied on a valid triangulation.
Insertion of a point on an edge.
Insertion in a face.
Insertion outside the convex hull.
Removal.
void flip (Face_handle f, int i)
Exchanges the edge incident to f and f->neighbor(i) with the other diagonal of the quadrilateral formed by f and f->neighbor(i). More...
Vertex_handle insert (const Point &p, Face_handle f=Face_handle())
Inserts point p in the triangulation and returns the corresponding vertex. More...
Vertex_handle insert (const Point &p, Locate_type lt, Face_handle loc, int li)
Same as above except that the location of the point p to be inserted is assumed to be given by (lt,loc,i) (see the description of the locate method above.)
Vertex_handle push_back (const Point &p)
Equivalent to insert(p).
template<class PointInputIterator >
std::ptrdiff_t insert (PointInputIterator first, PointInputIterator last)
Inserts the points in the range [first,last) in the given order, and returns the number of inserted points. More...
template<class PointWithInfoInputIterator >
std::ptrdiff_t insert (PointWithInfoInputIterator first, PointWithInfoInputIterator last)
inserts the points in the iterator range [first,last) in the given order, and returns the number of inserted points. More...
void remove (Vertex_handle v)
Removes the vertex from the triangulation. More...
Vertex_handle move_if_no_collision (Vertex_handle v, const Point &p)
If there is not already another vertex placed on p, the triangulation is modified such that the new position of vertex v is p, and v is returned. More...
Vertex_handle move (Vertex_handle v, const Point &p)
If there is no collision during the move, this function is the same as move_if_no_collision . More...
## Specialized Modifiers
The following member functions offer more specialized versions of the insertion or removal operations to be used when one knows to be in the corresponding case.
Vertex_handle insert_first (const Point &p)
Inserts the first finite vertex .
Vertex_handle insert_second (const Point &p)
Inserts the second finite vertex .
Vertex_handle insert_in_face (const Point &p, Face_handle f)
Inserts vertex v in face f. More...
Vertex_handle insert_in_edge (const Point &p, Face_handle f, int i)
Inserts vertex v in edge i of f. More...
Vertex_handle insert_outside_convex_hull (const Point &p, Face_handle f)
Inserts a point which is outside the convex hull but in the affine hull. More...
Vertex_handle insert_outside_affine_hull (const Point &p)
Inserts a point which is outside the affine hull.
void remove_degree_3 (Vertex_handle v)
Removes a vertex of degree three. More...
void remove_second (Vertex_handle v)
Removes the before last finite vertex.
void remove_first (Vertex_handle v)
Removes the last finite vertex.
template<class EdgeIt >
Vertex_handle star_hole (Point p, EdgeIt edge_begin, EdgeIt edge_end)
creates a new vertex v and use it to star the hole whose boundary is described by the sequence of edges [edge_begin, edge_end). More...
template<class EdgeIt , class FaceIt >
Vertex_handle star_hole (Point p, EdgeIt edge_begin, EdgeIt edge_end, FaceIt face_begin, FaceIt face_end)
same as above, except that the algorithm first recycles faces in the sequence [face_begin, face_end) and create new ones only when the sequence is exhausted. More...
## Finite Face, Edge and Vertex Iterators
The following iterators allow respectively to visit finite faces, finite edges and finite vertices of the triangulation.
These iterators are non mutable, bidirectional and their value types are respectively Face, Edge and Vertex. They are all invalidated by any change in the triangulation.
Finite_vertices_iterator finite_vertices_begin () const
Starts at an arbitrary finite vertex.
Finite_vertices_iterator finite_vertices_end () const
Past-the-end iterator.
Finite_edges_iterator finite_edges_begin () const
Starts at an arbitrary finite edge.
Finite_edges_iterator finite_edges_end () const
Past-the-end iterator.
Finite_faces_iterator finite_faces_begin () const
Starts at an arbitrary finite face.
Finite_faces_iterator finite_faces_end () const
Past-the-end iterator.
Point_iterator points_begin () const
Point_iterator points_end () const
Past-the-end iterator.
Finite_vertex_handles finite_vertex_handles () const
returns a range of iterators over finite vertices. More...
Finite_edges finite_edges () const
returns a range of iterators over finite edges.
Finite_face_handles finite_face_handles () const
returns a range of iterators over finite faces. More...
Points points () const
returns a range of iterators over the points of finite vertices.
## All Face, Edge and Vertex Iterators
The following iterators allow respectively to visit all (finite or infinite) faces, edges and vertices of the triangulation.
These iterators are non mutable, bidirectional and their value types are respectively Face, Edge and Vertex. They are all invalidated by any change in the triangulation.
All_vertices_iterator all_vertices_begin () const
Starts at an arbitrary vertex.
All_vertices_iterator all_vertices_end () const
Past-the-end iterator.
All_edges_iterator all_edges_begin () const
Starts at an arbitrary edge.
All_edges_iterator all_edges_end () const
Past-the-end iterator.
All_faces_iterator all_faces_begin () const
Starts at an arbitrary face.
All_faces_iterator all_faces_end () const
Past-the-end iterator.
All_vertex_handles all_vertex_handles () const
returns a range of iterators over all vertices. More...
All_edges all_edges () const
returns a range of iterators over all edges.
All_face_handles all_face_handles () const
returns a range of iterators over all faces. More...
## Line Face Circulator
The triangulation defines a circulator that allows to visit all faces that are intersected by a line.
A face f is considered has being intersected by the oriented line l if either:
• f is a finite face whose interior intersects l, or
• f is a finite face with an edge collinear with l and lies to the left of l, or
• f is an infinite face incident to a convex hull edge whose interior is intersected by l, or
• f is an infinite face incident to a convex hull vertex lying on l and the finite edge of f lies to the left of l.
The circulator has a singular value if the line l intersect no finite face of the triangulation. This circulator is non-mutable and bidirectional. Its value type is Face. Figure Triangulation_ref_Fig_Line_face_circulator illustrates which finite faces are enumerated. Lines $$l_1$$ and $$l_2$$ have no face to their left. Lines $$l_3$$ and $$l_4$$ have faces to their left. Note that the finite faces that are only vertex incident to lines $$l_3$$ and $$l_4$$ are not enumerated.
The line face circulator. A line face circulator is invalidated if the face the circulator refers to is changed.
Line_face_circulator line_walk (const Point &p, const Point &q, Face_handle f=Face_handle()) const
This function returns a circulator that allows to visit the faces intersected by the line pq. More...
## Face, Edge and Vertex Circulators
The triangulation also provides circulators that allows to visit respectively all faces or edges incident to a given vertex or all vertices adjacent to a given vertex.
These circulators are non-mutable and bidirectional. The operator++ moves the circulator counterclockwise around the vertex while the operator- moves clockwise. A face circulator is invalidated by any modification of the face pointed to. An edge or a vertex circulator are invalidated by any modification of one of the two faces incident to the edge pointed to.
Face_circulator incident_faces (Vertex_handle v) const
Starts at an arbitrary face incident to v.
Face_circulator incident_faces (Vertex_handle v, Face_handle f) const
Starts at face f. More...
Edge_circulator incident_edges (Vertex_handle v) const
Starts at an arbitrary edge incident to v.
Edge_circulator incident_edges (Vertex_handle v, Face_handle f) const
Starts at the first edge of f incident to v, in counterclockwise order around v. More...
Vertex_circulator incident_vertices (Vertex_handle v) const
Starts at an arbitrary vertex incident to v.
Vertex_circulator incident_vertices (Vertex_handle v, Face_handle f)
Starts at the first vertex of f adjacent to v in counterclockwise order around v. More...
Vertex_handle mirror_vertex (Face_handle f, int i) const
returns the vertex of the $$i^{th}$$ neighbor of f that is opposite to f. More...
int mirror_index (Face_handle f, int i) const
returns the index of f in its $$i^{th}$$ neighbor. More...
Edge mirror_edge (Edge e) const
returns the same edge seen from the other adjacent face. More...
## Miscellaneous
int ccw (int i) const
Returns $$i+1$$ modulo 3. More...
int cw (int i) const
Returns $$i+2$$ modulo 3. More...
Triangle triangle (Face_handle f) const
Returns the triangle formed by the three vertices of f. More...
Segment segment (Face_handle f, int i) const
Returns the line segment formed by the vertices ccw(i) and cw(i) of face f. More...
Segment segment (const Edge &e) const
Returns the line segment corresponding to edge e. More...
Segment segment (const Edge_circulator &ec) const
Returns the line segment corresponding to edge *ec. More...
Segment segment (const Edge_iterator &ei) const
Returns the line segment corresponding to edge *ei. More...
Point circumcenter (Face_handle f) const
Compute the circumcenter of the face pointed to by f. More...
## Setting
void set_infinite_vertex (const Vertex_handle &v)
This is an advanced function. More...
## Checking
The responsibility of keeping a valid triangulation belongs to the users if advanced operations are used.
Obviously the advanced user, who implements higher levels operations may have to make a triangulation invalid at some times. The following method is provided to help the debugging.
bool is_valid (bool verbose=false, int level=0) const
Checks the combinatorial validity of the triangulation and also the validity of its geometric embedding. More...
Public Member Functions inherited from CGAL::Triangulation_cw_ccw_2
Triangulation_cw_ccw_2 ()
default constructor.
int ccw (const int i) const
returns the index of the neighbor or vertex that is next to the neighbor or vertex with index i in counterclockwise order around a face.
int cw (const int i) const
returns the index of the neighbor or vertex that is next to the neighbor or vertex with index i in counterclockwise order around a face.
## ◆ Locate_type
template<typename Traits , typename Tds >
specifies which case occurs when locating a point in the triangulation.
CGAL::Triangulation_2<Traits,Tds>
Enumerator
VERTEX
when the located point coincides with a vertex of the triangulation
EDGE
when the point is in the relative interior of an edge
FACE
when the point is in the interior of a facet
OUTSIDE_CONVEX_HULL
when the point is outside the convex hull but in the affine hull of the current triangulation
OUTSIDE_AFFINE_HULL
when the point is outside the affine hull of the current triangulation.
## ◆ Triangulation_2()
template<typename Traits , typename Tds >
CGAL::Triangulation_2< Traits, Tds >::Triangulation_2 ( const Triangulation_2< Traits, Tds > & tr )
Copy constructor.
All the vertices and faces are duplicated. After the copy, *this and tr refer to different triangulations: if tr is modified, *this is not.
## ◆ all_face_handles()
template<typename Traits , typename Tds >
All_face_handles CGAL::Triangulation_2< Traits, Tds >::all_face_handles ( ) const
returns a range of iterators over all faces.
Note
While the value type of All_faces_iterator is Face, the value type of All_face_handles::iterator is Face_handle.
## ◆ all_vertex_handles()
template<typename Traits , typename Tds >
All_vertex_handles CGAL::Triangulation_2< Traits, Tds >::all_vertex_handles ( ) const
returns a range of iterators over all vertices.
Note
While the value type of All_vertices_iterator is Vertex, the value type of All_vertex_handles::iterator is Vertex_handle.
## ◆ ccw()
template<typename Traits , typename Tds >
int CGAL::Triangulation_2< Traits, Tds >::ccw ( int i ) const
Returns $$i+1$$ modulo 3.
Precondition
$$0\leq i \leq2$$.
## ◆ circumcenter()
template<typename Traits , typename Tds >
Point CGAL::Triangulation_2< Traits, Tds >::circumcenter ( Face_handle f ) const
Compute the circumcenter of the face pointed to by f.
This function is available only if the corresponding function is provided in the geometric traits.
## ◆ cw()
template<typename Traits , typename Tds >
int CGAL::Triangulation_2< Traits, Tds >::cw ( int i ) const
Returns $$i+2$$ modulo 3.
Precondition
$$0\leq i \leq2$$.
## ◆ finite_face_handles()
template<typename Traits , typename Tds >
Finite_face_handles CGAL::Triangulation_2< Traits, Tds >::finite_face_handles ( ) const
returns a range of iterators over finite faces.
Note
While the value type of Finite_faces_iterator is Face, the value type of Finite_face_handles::iterator is Face_handle.
## ◆ finite_vertex_handles()
template<typename Traits , typename Tds >
Finite_vertex_handles CGAL::Triangulation_2< Traits, Tds >::finite_vertex_handles ( ) const
returns a range of iterators over finite vertices.
Note
While the value type of Finite_vertices_iterator is Vertex, the value type of Finite_vertex_handles::iterator is Vertex_handle.
## ◆ flip()
template<typename Traits , typename Tds >
void CGAL::Triangulation_2< Traits, Tds >::flip ( Face_handle f, int i )
Exchanges the edge incident to f and f->neighbor(i) with the other diagonal of the quadrilateral formed by f and f->neighbor(i).
Precondition
The faces f and f->neighbor(i) are finite faces and their union form a convex quadrilateral.
## ◆ incident_edges()
template<typename Traits , typename Tds >
Edge_circulator CGAL::Triangulation_2< Traits, Tds >::incident_edges ( Vertex_handle v, Face_handle f ) const
Starts at the first edge of f incident to v, in counterclockwise order around v.
Precondition
Face f is incident to vertex v.
## ◆ incident_faces()
template<typename Traits , typename Tds >
Face_circulator CGAL::Triangulation_2< Traits, Tds >::incident_faces ( Vertex_handle v, Face_handle f ) const
Starts at face f.
Precondition
Face f is incident to vertex v.
## ◆ incident_vertices()
template<typename Traits , typename Tds >
Vertex_circulator CGAL::Triangulation_2< Traits, Tds >::incident_vertices ( Vertex_handle v, Face_handle f )
Starts at the first vertex of f adjacent to v in counterclockwise order around v.
Precondition
Face f is incident to vertex v.
## ◆ includes_edge()
template<typename Traits , typename Tds >
bool CGAL::Triangulation_2< Traits, Tds >::includes_edge ( Vertex_handle va, Vertex_handle vb, Vertex_handle & vbb, Face_handle & fr, int & i )
true if the line segment from va to vb includes an edge e incident to va.
If true, vbb becomes the other vertex of e, e is the edge (fr,i) where fr is a handle to the face incident to e and on the right side e oriented from va to vb.
## ◆ inexact_locate()
template<typename Traits , typename Tds >
Face_handle CGAL::Triangulation_2< Traits, Tds >::inexact_locate ( const Point & query, Face_handle start = Face_handle() ) const
Same as locate() but uses inexact predicates.
This function returns a handle on a face that is a good approximation of the exact location of query, while being faster. Note that it may return a handle on a face whose interior does not contain query. When the triangulation has dimension smaller than 2, start is returned.
## ◆ insert() [1/3]
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::insert ( const Point & p, Face_handle f = Face_handle() )
Inserts point p in the triangulation and returns the corresponding vertex.
If point p coincides with an already existing vertex, this vertex is returned and the triangulation remains unchanged.
If point p is on an edge, the two incident faces are split in two.
If point p is strictly inside a face of the triangulation, the face is split in three.
If point p is strictly outside the convex hull, p is linked to all visible points on the convex hull to form the new triangulation.
At last, if p is outside the affine hull (in case of degenerate 1-dimensional or 0-dimensional triangulations), p is linked all the other vertices to form a triangulation whose dimension is increased by one. The last argument f is an indication to the underlying locate algorithm of where to start.
Examples:
## ◆ insert() [2/3]
template<typename Traits , typename Tds >
template<class PointInputIterator >
std::ptrdiff_t CGAL::Triangulation_2< Traits, Tds >::insert ( PointInputIterator first, PointInputIterator last )
Inserts the points in the range [first,last) in the given order, and returns the number of inserted points.
Template Parameters
PointInputIterator must be an input iterator with value type Point.
## ◆ insert() [3/3]
template<typename Traits , typename Tds >
template<class PointWithInfoInputIterator >
std::ptrdiff_t CGAL::Triangulation_2< Traits, Tds >::insert ( PointWithInfoInputIterator first, PointWithInfoInputIterator last )
inserts the points in the iterator range [first,last) in the given order, and returns the number of inserted points.
Given a pair (p,i), the vertex v storing p also stores i, that is v.point() == p and v.info() == i. If several pairs have the same point, only one vertex is created, and one of the objects of type Vertex::Info will be stored in the vertex.
Precondition
Vertex must be model of the concept TriangulationVertexBaseWithInfo_2.
Template Parameters
PointWithInfoInputIterator must be an input iterator with the value type std::pair.
## ◆ insert_in_edge()
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::insert_in_edge ( const Point & p, Face_handle f, int i )
Inserts vertex v in edge i of f.
Precondition
The point in vertex v lies on the edge opposite to the vertex i of face f.
## ◆ insert_in_face()
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::insert_in_face ( const Point & p, Face_handle f )
Inserts vertex v in face f.
Face f is modified, two new faces are created.
Precondition
The point in vertex v lies inside face f.
## ◆ insert_outside_convex_hull()
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::insert_outside_convex_hull ( const Point & p, Face_handle f )
Inserts a point which is outside the convex hull but in the affine hull.
Precondition
The handle f points to a face which is a proof of the location ofp, see the description of the locate method above.
## ◆ is_edge()
template<typename Traits , typename Tds >
bool CGAL::Triangulation_2< Traits, Tds >::is_edge ( Vertex_handle va, Vertex_handle vb, Face_handle & fr, int & i )
as above.
In addition, if true is returned, the edge with vertices va and vb is the edge e=(fr,i) where fr is a handle to the face incident to e and on the right side of e oriented from va to vb.
## ◆ is_face()
template<typename Traits , typename Tds >
bool CGAL::Triangulation_2< Traits, Tds >::is_face ( Vertex_handle v1, Vertex_handle v2, Vertex_handle v3, Face_handle & fr )
as above.
In addition, if true is returned, fr is a handle to the face with v1, v2 and v3 as vertices.
## ◆ is_valid()
template<typename Traits , typename Tds >
bool CGAL::Triangulation_2< Traits, Tds >::is_valid ( bool verbose = false, int level = 0 ) const
Checks the combinatorial validity of the triangulation and also the validity of its geometric embedding.
This method is mainly a debugging help for the users of advanced features.
## ◆ line_walk()
template<typename Traits , typename Tds >
Line_face_circulator CGAL::Triangulation_2< Traits, Tds >::line_walk ( const Point & p, const Point & q, Face_handle f = Face_handle() ) const
This function returns a circulator that allows to visit the faces intersected by the line pq.
If there is no such face the circulator has a singular value.
The starting point of the circulator is the face f, or the first finite face traversed by l , if f is omitted.
The circulator wraps around the infinite vertex: after the last traversed finite face, it steps through the infinite face adjacent to this face then through the infinite face adjacent to the first traversed finite face then through the first finite traversed face again.
Precondition
The triangulation must have dimension 2.
Points p and q must be different points.
If f != nullptr, it must point to a finite face and the point p must be inside or on the boundary of f.
## ◆ locate() [1/2]
template<typename Traits , typename Tds >
Face_handle CGAL::Triangulation_2< Traits, Tds >::locate ( const Point & query, Face_handle f = Face_handle() ) const
If the point query lies inside the convex hull of the points, a face that contains the query in its interior or on its boundary is returned.
If the point query lies outside the convex hull of the triangulation but in the affine hull, the returned face is an infinite face which is a proof of the point's location:
• for a two dimensional triangulation, it is a face $$(\infty, p, q)$$ such that query lies to the left of the oriented line $$pq$$ (the rest of the triangulation lying to the right of this line).
• for a degenerate one dimensional triangulation it is the (degenerate one dimensional) face $$(\infty, p, nullptr)$$ such that query and the triangulation lie on either side of p.
If the point query lies outside the affine hull, the returned Face_handle is nullptr.
The optional Face_handle argument, if provided, is used as a hint of where the locate process has to start its search.
## ◆ locate() [2/2]
template<typename Traits , typename Tds >
Face_handle CGAL::Triangulation_2< Traits, Tds >::locate ( const Point & query, Locate_type & lt, int & li, Face_handle h = Face_handle() ) const
Same as above.
Additionally, the parameters lt and li describe where the query point is located. The variable lt is set to the locate type of the query. If lt==VERTEX the variable li is set to the index of the vertex, and if lt==EDGE li is set to the index of the vertex opposite to the edge. Be careful that li has no meaning when the query type is FACE, OUTSIDE_CONVEX_HULL, or OUTSIDE_AFFINE_HULL or when the triangulation is $$0$$-dimensional.
## ◆ mirror_edge()
template<typename Traits , typename Tds >
Edge CGAL::Triangulation_2< Traits, Tds >::mirror_edge ( Edge e ) const
returns the same edge seen from the other adjacent face.
Precondition
$$0\leq i \leq2$$.
## ◆ mirror_index()
template<typename Traits , typename Tds >
int CGAL::Triangulation_2< Traits, Tds >::mirror_index ( Face_handle f, int i ) const
returns the index of f in its $$i^{th}$$ neighbor.
Precondition
$$0\leq i \leq2$$.
## ◆ mirror_vertex()
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::mirror_vertex ( Face_handle f, int i ) const
returns the vertex of the $$i^{th}$$ neighbor of f that is opposite to f.
Precondition
$$0\leq i \leq2$$.
## ◆ move()
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::move ( Vertex_handle v, const Point & p )
If there is no collision during the move, this function is the same as move_if_no_collision .
Otherwise, v is removed and the vertex at point p is returned.
Precondition
Vertex v must be finite.
## ◆ move_if_no_collision()
template<typename Traits , typename Tds >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::move_if_no_collision ( Vertex_handle v, const Point & p )
If there is not already another vertex placed on p, the triangulation is modified such that the new position of vertex v is p, and v is returned.
Otherwise, the triangulation is not modified and the vertex at point p is returned.
Precondition
Vertex v must be finite.
## ◆ operator=()
template<typename Traits , typename Tds >
Triangulation_2 CGAL::Triangulation_2< Traits, Tds >::operator= ( const Triangulation_2< Traits, Tds > & tr )
Assignment.
All the vertices and faces are duplicated. After the assignment, *this and tr refer to different triangulations: if tr is modified, *this is not.
## ◆ oriented_side()
template<typename Traits , typename Tds >
Oriented_side CGAL::Triangulation_2< Traits, Tds >::oriented_side ( Face_handle f, const Point & p ) const
Returns on which side of the oriented boundary of f lies the point p.
Precondition
f is finite.
## ◆ remove()
template<typename Traits , typename Tds >
void CGAL::Triangulation_2< Traits, Tds >::remove ( Vertex_handle v )
Removes the vertex from the triangulation.
The created hole is re-triangulated.
Precondition
Vertex v must be finite.
## ◆ remove_degree_3()
template<typename Traits , typename Tds >
void CGAL::Triangulation_2< Traits, Tds >::remove_degree_3 ( Vertex_handle v )
Removes a vertex of degree three.
Two of the incident faces are destroyed, the third one is modified.
Precondition
Vertex v is a finite vertex with degree three.
## ◆ segment() [1/4]
template<typename Traits , typename Tds >
Segment CGAL::Triangulation_2< Traits, Tds >::segment ( Face_handle f, int i ) const
Returns the line segment formed by the vertices ccw(i) and cw(i) of face f.
Precondition
$$0\leq i \leq2$$. The vertices ccw(i) and cw(i) of f are finite.
## ◆ segment() [2/4]
template<typename Traits , typename Tds >
Segment CGAL::Triangulation_2< Traits, Tds >::segment ( const Edge & e ) const
Returns the line segment corresponding to edge e.
Precondition
e is a finite edge.
## ◆ segment() [3/4]
template<typename Traits , typename Tds >
Segment CGAL::Triangulation_2< Traits, Tds >::segment ( const Edge_circulator & ec ) const
Returns the line segment corresponding to edge *ec.
Precondition
*ec is a finite edge.
## ◆ segment() [4/4]
template<typename Traits , typename Tds >
Segment CGAL::Triangulation_2< Traits, Tds >::segment ( const Edge_iterator & ei ) const
Returns the line segment corresponding to edge *ei.
Precondition
*ei is a finite edge.
## ◆ set_infinite_vertex()
template<typename Traits , typename Tds >
void CGAL::Triangulation_2< Traits, Tds >::set_infinite_vertex ( const Vertex_handle & v )
This method is meant to be used only if you have done a low-level operation on the underlying tds that invalidated the infinite vertex. Sets the infinite vertex.
## ◆ side_of_oriented_circle()
template<typename Traits , typename Tds >
Oriented_side CGAL::Triangulation_2< Traits, Tds >::side_of_oriented_circle ( Face_handle f, const Point & p )
Returns on which side of the circumcircle of face f lies the point p.
The circle is assumed to be counterclockwise oriented, so its positive side correspond to its bounded side. This predicate is available only if the corresponding predicates on points is provided in the geometric traits class.
## ◆ star_hole() [1/2]
template<typename Traits , typename Tds >
template<class EdgeIt >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::star_hole ( Point p, EdgeIt edge_begin, EdgeIt edge_end )
creates a new vertex v and use it to star the hole whose boundary is described by the sequence of edges [edge_begin, edge_end).
Returns a handle to the new vertex.
This function is intended to be used in conjunction with the find_conflicts() member functions of Delaunay and constrained Delaunay triangulations to perform insertions.
## ◆ star_hole() [2/2]
template<typename Traits , typename Tds >
template<class EdgeIt , class FaceIt >
Vertex_handle CGAL::Triangulation_2< Traits, Tds >::star_hole ( Point p, EdgeIt edge_begin, EdgeIt edge_end, FaceIt face_begin, FaceIt face_end )
same as above, except that the algorithm first recycles faces in the sequence [face_begin, face_end) and create new ones only when the sequence is exhausted.
This function is intended to be used in conjunction with the find_conflicts() member functions of Delaunay and constrained Delaunay triangulations to perform insertions.
## ◆ swap()
template<typename Traits , typename Tds >
void CGAL::Triangulation_2< Traits, Tds >::swap ( Triangulation_2< Traits, Tds > & tr )
The triangulations tr and *this are swapped.
This method should be used instead of assignment of copy constructor. if tr is deleted after that.
## ◆ triangle()
template<typename Traits , typename Tds >
Triangle CGAL::Triangulation_2< Traits, Tds >::triangle ( Face_handle f ) const
Returns the triangle formed by the three vertices of f.
Precondition
The face is finite.
## ◆ operator<<()
template<typename Traits , typename Tds >
ostream & operator<< ( ostream & os, const Triangulation_2< Traits, Tds > & T )
related
Inserts the triangulation into the stream os.
Precondition
The insert operator must be defined for Point.
## ◆ operator>>()
template<typename Traits , typename Tds >
istream & operator>> ( istream & is, const Triangulation_2< Traits, Tds > & T )
related
Reads a triangulation from stream is and assigns it to the triangulation.
Precondition
The extract operator must be defined for Point. | 10,228 | 43,986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.931825 |
https://www.allinterview.com/interview-questions/80-43/civil-engineering.html | 1,537,669,612,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00510.warc.gz | 654,081,983 | 8,473 | Civil Engineering Interview Questions
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# A0_soln - Math 235 Assignment 0 Solutions 1 Determine proj...
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Unformatted text preview: Math 235 Assignment 0 Solutions 1. Determine proj ~v ~x and perp ~v ~x where a) ~v = 2 3- 2 and ~x = 4- 1 3 . Solution: proj ~v ~x = ~x · ~v k ~v k 2 ~v =- 1 17 2 3- 2 = - 2 / 17- 3 / 17 2 / 17 perp ~v ~x = ~x- proj ~v ~x = 4- 1 3 - - 2 / 17- 3 / 17 2 / 17 = 70 / 17- 14 / 17 49 / 17 b) ~v = - 1 2 1- 3 and ~x = 2- 1 2 1 . Solution: proj ~v ~x = ~x · ~v k ~v k 2 ~v =- 1 3 - 1 2 1- 3 = 1 / 3- 2 / 3- 1 / 3 1 . perp ~v ~x = ~x- proj ~v ~x = 2- 1 2 1 - 1 / 3- 2 / 3- 1 / 3 1 = 5 / 3- 1 / 3 7 / 3 2. Prove algebraically that proj ~x ( ~v ) and perp ~x ~v are orthogonal. Solution: We have proj ~x ( ~v ) · perp ~x ~v = ~v · ~x k ~x k 2 ~x · ~v- ~v · ~x k ~x k 2 ~x = ~v · ~x k ~x k 2 ( ~x · ~v )- ~v · ~x k ~x k 2 2 ( ~x · ~x ) = ( ~v · ~x ) 2 k ~x k 2- ( ~v · ~x ) 2 k ~x k 4 k ~x k 2 = ( ~v · ~x ) 2 k ~x k 2- ( ~v · ~x ) 2 k ~x k 2 = 0 . Hence, they are orthogonal. 2 3. Solve the system z 1- z 2 + iz 3 = 2 i (1 + i ) z 1- iz 2 + iz 3 =- 2 + i (1- i ) z 1 + (- 1 + 2 i ) z 2 + (1 + 2 i ) z 3 = 3 + 2 i Solution: Making an augmented matrix and row-reducing we get 1- 1 i 2 i 1 + i- i i- 2 + i 1- i- 1 + 2 i 1 + 2 i 3 + 2 i ∼ 1 0 1 + i i 0 1 1- i 0 0 x 3 does not have a leading one so let x 3 = t ∈ C . Then we have x 1 = i- (1 + i ) t x 2 =- i- t x 3 = t So the general solution is ~x = i- i + t - 1- i- 1 1 . 4. Prove each of the following mappings are linear and find the standard matrix of each. a) proj (2 , 2 ,- 1) . Solution: Let ~n = 2 2- 1 and let ~x,~ y ∈ R 3 and k ∈ R . Then by using properties of the dot product we get proj (2 , 2 ,- 1) ( k~x + ~ y ) = ( k~x + ~ y ) · ~n ) k ~n k 2 ~n = k ~x · ~n k ~n k 2 ~n + ~ y · ~n k ~n 2 k 2 ~n = k proj ~n ~x + proj ~n ~ y Hence, proj (2 , 2 ,- 1) is linear. 3 We have proj (2 , 2 ,- 1) ~e 1 = ~e · ~n ) k ~n k 2 ~n = 2 9 2 2- 1 = 4 / 9 4 / 9- 2 / 9 proj (2 , 2 ,- 1) ~e 2 = ~e · ~n ) k ~n k 2 ~n = 2 9 2 2- 1 = 4 / 9 4 / 9- 2 / 9 proj (2 , 2 ,- 1) ~e 3 = ~e · ~n ) k ~n k 2 ~n =- 1 9 2 2- 1 = - 2 / 9- 2 / 9 1 / 9 Hence [proj (2 , 2 ,- 1) ] = 4 / 9 4 / 9- 2 / 9 4 / 9 4 / 9- 2 / 9- 2 / 9- 2 / 9 1 / 9 ....
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A0_soln - Math 235 Assignment 0 Solutions 1 Determine proj...
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Ask a homework question - tutors are online | 1,345 | 2,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-05 | latest | en | 0.675834 |
https://frieord.no/what-can-a-2000-watt-inverter-run/ | 1,718,862,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00513.warc.gz | 232,731,063 | 58,903 | What can a 2000 watt inverter run?
Most 2000 watt inverter generators can power a range of devices from hot plates, water pumps, deep freezers, light bulbs and electric stoves to small power tools – items common to camping, caravanning, and around the home.
How many batteries do I need for a 2000 watt inverter?
Helpful Expert Reply: Typically two batteries are needed for a 2,000 watt inverter like the part # 34278156 that you referenced.
Innholdsfortegnelse
How much current does a 2000W inverter draw?
The maximum current a 2000 watt inverter can draw is 166 amps… in fact it will be a bit more as the inverter itself needs power, usually to run a couple of cooling fans and its own internal circuits.
How long will a 2000W inverter last?
If you max out the inverter at 2000 watts, you are pulling 2000 watts /12 volts = 166.6 DC amps per hour. If you use a 200-amp 12-volt battery, you would divide the 200-amp battery / 166.6 amps = 1.2 hours of run time.
What can a 2000 watt inverter run? – Related Questions
Is it OK to leave inverter on all the time?
Should the inverter be on all the time? Yes, you should keep your inverter ON all the time. Otherwise, you will lose your battery backup time due to the self-discharge of batteries. You will need to start the inverter manually every time when grid power failed.
Do inverters drain battery?
Inverters will draw power from your batteries when not in use, and the unit is turned on. This can vary from around . 02 amps right up to 2amps depending on the unit and design of their standby systems.
How long will a 12 volt battery last with an inverter?
A 12 volt 100Ah deep-cycle battery with regular depth of discharge 50% would run a fully-loaded 1000 watt inverter for 34 minutes. This calculation takes into account average pure sine wave inverter efficiency of 95%.
How long can I run an inverter on my battery?
Small Inverters: Most automobile and marine batteries will provide an ample power supply for 30 to 60 minutes even when the engine is off. Actual time may vary depending on the age and condition of the battery, and the power demand being placed on it by the equipment being operated by the inverter.
How many hours does an inverter last?
In general, you can expect your inverter battery to last anywhere around 5 to 10 hours when it is fully charged. However, you can easily calculate the accurate battery backup time with a simple formula or use a battery backup calculator.
Will a 2000 watt inverter run a microwave?
A 2,000-watt (running watts) inverter may have a peak (or surge) output of 3000 watts. This inverter could easily handle both the 900 running watt and the 2,700-watt surge (starting draw) requirements of your microwave.
Can I run a fridge off an inverter?
Refrigerator manufacturers have introduced a Smart Inverter technology, which let you power your fridge even with a home inverter. This technology is pretty neat and can help you keep your food fresh even when you experience hours of long power cuts.
What size inverter do I need to run a TV?
Generally speaking, a 300-watt inverter will run a TV with the right solar setup. The average 32″ LED or LCD TV requires between 55 and 70 watts to operate (3). If you’re looking to power something larger, consider a larger power inverter and more battery storage.
How long will an inverter run a refrigerator?
In general, you risk damage if a rechargeable lead-acid battery is drained below 50 percent of capacity, so you safely can draw 105 amps for only 30 minutes. A fridge drawing only 55 DC amps can be run for about 60 minutes of continuous duty off a 105 ampere-hour battery.
Are inverters better than generators?
Although inverter generators can’t produce as much power as conventional generators, they are more efficient due to how the energy how the AC current is produced. The more efficient a generator is the less fuel is used and the smaller the fuel tank can be.
How do I know what size inverter I need?
Computer Load= No x Watt =1×200=200 Watt. Computer Load=(No x Watt)/P.F =(1×200)/0.8= 250VA. Tube Light Load= No x Watt =2×30=60 Watt. Tube Light Load=(No x Watt)/P.F =(2×30)/0.8= 75VA.
Step 2: Size of Inverter:
1. Size of Inverter=Total Load+(1+Af) / Ie VA.
2. Size of Inverter= 475+(1+20%) / 80%
3. Size of Inverter= 712 VA. | 1,069 | 4,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.89407 |
https://readingfeynman.org/tag/poisson-equation/ | 1,606,647,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00347.warc.gz | 454,200,143 | 41,159 | # The method of images
Pre-script (dated 26 June 2020): This post got mutilated by the removal of some illustrations by the dark force. You should be able to follow the main story-line, however. If anything, the lack of illustrations might actually help you to think things through for yourself.
Original post:
In my previous post, I mentioned the so-called method of images, but didn’t elaborate much. Let’s recall the problem. As you know, the whole subject of electrostatics is governed by one equation: the so-called Poisson equation:
2Φ = ∂2Φ/∂x2 + ∂2Φ/∂x2 + ∂2Φ/∂x2 = −ρ/ε0
We get this equation by combining Maxwell’s first law (·Φ = −ρ/ε0) and the E = −Φ formula. Now, if we know the distribution of charges, then we don’t need that Poisson equation: we can calculate the potential at every point – denoted by (1) below – using the following formulas:
And if we have Φ, we have E, because E = –Φ. But, in most actual situations, we don’t know the charge distribution, and then we need to work with that Poisson equation. Of course, you’ll say: if you don’t know the charge distribution, then you don’t know the ρ in the equation, and so what use is it really?
The answer is: most problems will involve conductors, and we do know that their surface is an equipotential surface. We also know that the electric field just outside the surface must be normal to the surface. Let’s take the example of the grounded conducting sheet once again, as depicted below. We know the image charge and the field lines on the left-hand side are not there. In fact, because the sheet is grounded, there is no net charge on it, and the conductor acts as a shield.
We do have a real field on the right-hand side though, and it’s exactly the same as that of a dipole: we only need to cross out the left-hand half of the picture. What charges are responsible for it? It surely cannot be the lone +q charge alone, and it’s isn’t: we also have induced local charges on the sheet. Indeed, the positive charge will attract negative charges to the surface and, hence, while the sheet as a whole is neutral (so it has no net charge), the surface charge density is not zero. We can calculate it. How? It’s quite complicated, but let’s give it a try.
Look at the detail below. Let’s forget about the induced charges for a while, and analyze the field produced by the positive charge in the absence of induced charges, so that’s the E field at point P. The magnitude of its normal component is En+ = E·cosθ, with θ the angle between the two vectors.
θ is an angle of a rectangular triangle, and it’s easy to see that cosθ is equal to a/(a2 + ρ2)1/2. Now, Coulomb’s Law tells us that E = (1/4πε0)·q/[(a2 + ρ2)1/2]= (1/4πε0)·q/(a2 + ρ2). Hence, we can write:
En+ = (1/4πε0a·q/(a2 + ρ2)3/2
[A quick note on the symbols used here: we use ρ (rho) to denote a distance here. That’s somewhat confusing because it usually denotes a volume density. However, we’re interested in a surface density here, for which the σ (sigma) symbol is used. So don’t worry about it. Just note that ρ is some distance here, instead of a charge density.]
Now we know that the induced charges will arrange themselves in such way that the addition of their field makes the field at P look like there was a negative charge of the same magnitude as q at the other side of the sheet. If there was such charge −q, then we could do the same analysis, as shown below. It’s easy to see that the component of the imaginary field along the sheet (i.e. the component that’s perpendicular to the normal) cancels the actual component along the shield of the field created by +q, while its normal component adds to the normal component of the +q field. To make a long story short, the actual field at P is equal to E(ρ) = (1/4πε0)·2a·q/(a2 + ρ2)3/2, and it has two components of strength (1/4πε0a·q/(a2 + ρ2)3/2.
To put it differently, the actual field can be thought as two parts: (1) the (normal) component of the field caused by + q, and (2) the field caused by the surface charge density σ at P, which we denote as σ(ρ). Let’s see what we can do with this.
The analysis of the field of a sheet of charge on a conductor is quite complicated, and not quite like the analysis of just a sheet of charge. The analysis for just a sheet of charge was based on the theoretical situation depicted below. We imagined some box with two Gaussian surfaces of area A, and we then used Gauss’ Law to deduce that, if σ was the charge per unit area (i.e. the surface density), the total flux out of the box should be equal to EA + EA = σA/ε0 and, hence, E = (1/2)·σ/ε0. The illustration below shows we should think of two fields with opposite direction, and with a magnitude of (1/2)·σ/ε0 each.
That’s simple enough. However, a sheet of charge on a conductor produces a different field, as shown below. Because of the shielding effect, we have flux on one side of the box only, and the field strength of this flux is σ/ε0, so that’s two times the (1/2)·σ/ε0 magnitude described above. However, as mentioned, it’s zero on the other side, i.e. the inside of the conductor shown below.
So what happens here? The charges in the neighborhood of a point P on the surface actually do produce a local field (Elocal), both inside and outside of the surface, which respects the Elocal = (1/2)·σ/2ε0 equality, but all the rest of the charges on the conductor “conspire” to produce an additional field at the point P, which also produces two fields, again with opposite direction and with a magnitude of (1/2)·σ/ε0 each. So the net result is that the total field inside goes to zero, and the field outside is equal to E = σ/ε0, so E = 2·Elocal. Note that the example above assumes a positively charged conductor: if the charge on the conductor would be negative, the direction of the field would be inwards, but we’d still have a field on and outside of the surface only.
I know you’ve switched off already but − just in case you didn’t − what equality should we use to find σ in this case, i.e. the grounded sheet with no net charge on it but with some (negative) surface charge density. Well… We’re talking a surface density, and a conductor, and, therefore, I would think it’s the E = σ/ε0, i.e. the formula for a charged sheet on a conductor. So we write:
E = σ(ρ)/ε0 ⇔ σ(ρ) = ε0E
But what E do we take to continue our calculation? The whole field or (1/4πε0a·q/(a2 + ρ2)3/2 only? The analysis above may make you think that we should take (1/4πε0a·q/(a2 + ρ2)3/2 only, so that’s the component that’s related to the imaginary charge only, but… No! We’re talking one actual field here, which is produced by the positive charge as well as by the induced charges. So we should not cut it for the purpose of calculating σ(ρ)! So the grand result is:
σ(ρ) = ε0E = (1/4π)·2a·q/(a2 + ρ2)3/2
The shape of this function should not surprise us: it’s shown below for some different values of q (1 and 2 respectively) and a (1, 2 and 3 respectively).
How do we know our solution is correct? We can check it: if we integrate σ over the whole surface, we should find that the total induced charge is equal to −q. So… Well… I’ll let you do that. Feynman also notes the induced charges should exert a force on our point charge, which we can calculating the force between the surface charges and the charge. It’s again an integral, and it should be equal to
Lo and behold! The force acting on the positive charge is exactly the same as it would be with the negative image charge instead of the plate. Why? Well… Because the fields are the same!
The results we obtained are quite wonderful! Indeed, we said we did not know the charge distribution, and so we used a very different method to find the field: the method of images, which consists of computing the field due to q and some imaginary point charge –q somewhere else. Feynman summarizes the method of images as follows:
“The point charge we “imagine” existing behind the conducting surface is called an image charge. In books you can find long lists of solutions for hyperbolic-shaped conductors and other complicated looking things, and you wonder how anyone ever solved these terrible shapes. They were solved backwards! Someone solved a simple problem with given charges. He then saw that some equipotential surface showed up in a new shape, and he wrote a paper in which he pointed out that the field outside that particular shape can be described in a certain way.”
However, as you can see, the method is actually quite powerful, because we got a substantial bonus here: we calculated the field indeed, but then we could also calculate the charge distribution afterwards, so we got it all! Let’s see if we master the topic by looking at some other applications of the method of images.
Point charges near conducting spheres
For a grounded conducting sphere, we get the result shown below: the point charge q will induce charges on it whose fields are those of an image charge q’ = −aq/b placed at the point below.
You can check the details in Feynman’s Lecture on it, in which you will also find a more general formula for spheres that are not at zero potential. The more general formula involves a third charge q” at the center of the sphere, with charge q” = −q’ = aq/b.
Again, we’ll have a force of attraction between the sphere and the point charge, even if the net charge on the sphere is zero, because it’s grounded. Indeed, the positive charge q attracts negative charges to the side closer to itself and, hence, leaves positive charges on the surface of the far side. As the attraction by the negative charges exceeds the repulsion from the positive charges, we end up with some net attraction. Feynman leaves us with an interesting challenge here:
“Those who were entertained in childhood by the baking powder box which has on its label a picture of a baking powder box which has on its label a picture of a baking powder box which has … may be interested in the following problem. Two equal spheres, one with a total charge of +Q and the other with a total charge of −Q, are placed at some distance from each other. What is the force between them? The problem can be solved with an infinite number of images. One first approximates each sphere by a charge at its center. These charges will have image charges in the other sphere. The image charges will have images, etc., etc., etc. The solution is like the picture on the box of baking powder—and it converges pretty fast.”
Well… I’ll leave it to you to take up that challenge. 🙂
Direct and indirect methods
Let me end this post by noting that I started out with that Poisson equation, but that I actually didn’t use it. Having said that, this method of images did result in some solutions for it. It is what Feynman calls an indirect method of solving some problems, and he writes the following on it:
“If the problem to be solved does not belong to the class of problems for which we can construct solutions by the indirect method, we are forced to solve the problem by a more direct method. The mathematical problem of the direct method is the solution of Laplace’s equation ∇2Φ = 0 subject to the condition that Φ is a suitable constant on certain boundaries—the surfaces of the conductors. [Note that Laplace’s equation is Poisson’s equation with a zero on the right-hand side.] Problems which involve the solution of a differential field equation subject to certain boundary conditions are called boundary-value problems. They have been the object of considerable mathematical study. In the case of conductors having complicated shapes, there are no general analytical methods. Even such a simple problem as that of a charged cylindrical metal can closed at both ends—a beer can—presents formidable mathematical difficulties. It can be solved only approximately, using numerical methods. The only general methods of solution are numerical.”
Well… That says it all, I guess. There are other indirect methods, i.e. other than the method of images, but I won’t present these here. I may write something about it in some other post, perhaps. 🙂
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# Fields and charges (I)
Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂
Original post:
My previous posts focused mainly on photons, so this one should be focused more on matter-particles, things that have a mass and a charge. However, I will use it more as an opportunity to talk about fields and present some results from electrostatics using our new vector differential operators (see my posts on vector analysis).
Before I do so, let me note something that is obvious but… Well… Think about it: photons carry the electromagnetic force, but have no electric charge themselves. Likewise, electromagnetic fields have energy and are caused by charges, but so they also carry no charge. So… Fields act on a charge, and photons interact with electrons, but it’s only matter-particles (notably the electron and the proton, which is made of quarks) that actually carry electric charge. Does that make sense? It should. 🙂
Another thing I want to remind you of, before jumping into it all head first, are the basic units and relations that are valid always, regardless of what we are talking about. They are represented below:
Let me recapitulate the main points:
• The speed of light is always the same, regardless of the reference frame (inertial or moving), and nothing can travel faster than light (except mathematical points, such as the phase velocity of a wavefunction).
• This universal rule is the basis of relativity theory and the mass-energy equivalence relation E = mc2.
• The constant speed of light also allows us to redefine the units of time and/or distance such that c = 1. For example, if we re-define the unit of distance as the distance traveled by light in one second, or the unit of time as the time light needs to travel one meter, then c = 1.
• Newton’s laws of motion define a force as the product of a mass and its acceleration: F = m·a. Hence, mass is a measure of inertia, and the unit of force is 1 newton (N) = 1 kg·m/s2.
• The momentum of an object is the product of its mass and its velocity: p = m·v. Hence, its unit is 1 kg·m/s = 1 N·s. Therefore, the concept of momentum combines force (N) as well as time (s).
• Energy is defined in terms of work: 1 Joule (J) is the work done when applying a force of one newton over a distance of one meter: 1 J = 1 N·m. Hence, the concept of energy combines force (N) and distance (m).
• Relativity theory establishes the relativistic energy-momentum relation pc = Ev/c, which can also be written as E2 = p2c+ m02c4, with mthe rest mass of an object (i.e. its mass when the object would be at rest, relative to the observer, of course). These equations reduce to m = E and E2 = p2 + m0when choosing time and/or distance units such that c = 1. The mass is the total mass of the object, including its inertial mass as well as the equivalent mass of its kinetic energy.
• The relationships above establish (a) energy and time and (b) momentum and position as complementary variables and, hence, the Uncertainty Principle can be expressed in terms of both. The Uncertainty Principle, as well as the Planck-Einstein relation and the de Broglie relation (not shown on the diagram), establish a quantum of action, h, whose dimension combines force, distance and time (h ≈ 6.626×10−34 N·m·s). This quantum of action (Wirkung) can be defined in various ways, as it pops up in more than one fundamental relation, but one of the more obvious approaches is to define h as the proportionality constant between the energy of a photon (i.e. the ‘light particle’) and its frequency: h = E/ν.
Note that we talked about forces and energy above, but we didn’t say anything about the origin of these forces. That’s what we are going to do now, even if we’ll limit ourselves to the electromagnetic force only.
Electrostatics
According to Wikipedia, electrostatics deals with the phenomena and properties of stationary or slow-moving electric charges with no acceleration. Feynman usually uses the term when talking about stationary charges only. If a current is involved (i.e. slow-moving charges with no acceleration), the term magnetostatics is preferred. However, the distinction does not matter all that much because – remarkably! – with stationary charges and steady currents, the electric and magnetic fields (E and B) can be analyzed as separate fields: there is no interconnection whatsoever! That shows, mathematically, as a neat separation between (1) Maxwell’s first and second equation and (2) Maxwell’s third and fourth equation:
1. Electrostatics: (i) ∇•E = ρ/ε0 and (ii) ×E = 0.
2. Magnetostatics: (iii) c2∇×B = j0 and (iv) B = 0.
Electrostatics: The ρ in equation (i) is the so-called charge density, which describes the distribution of electric charges in space: ρ = ρ(x, y, z). To put it simply: ρ is the ‘amount of charge’ (which we’ll denote by Δq) per unit volume at a given point. As for ε0, that’s a constant which ensures all units are ‘compatible’. Equation (i) basically says we have some flux of E, the exact amount of which is determined by the charge density ρ or, more in general, by the charge distribution in space. As for equation (ii), i.e. ×E = 0, we can sort of forget about that. It means the curl of E is zero: everywhere, and always. So there’s no circulation of E. Hence, E is a so-called curl-free field, in this case at least, i.e. when only stationary charges and steady currents are involved.
Magnetostatics: The j in (iii) represents a steady current indeed, causing some circulation of B. The cfactor is related to the fact that magnetism is actually only a relativistic effect of electricity, but I can’t dwell on that here. I’ll just refer you to what Feynman writes about this in his Lectures, and warmly recommend to read it. Oh… Equation (iv), B = 0, means that the divergence of B is zero: everywhere, and always. So there’s no flux of B. None. So B is a divergence-free field.
Because of the neat separation, we’ll just forget about B and talk about E only.
The electric potential
OK. Let’s try to go through the motions as quickly as we can. As mentioned in my introduction, energy is defined in terms of work done. So we should just multiply the force and the distance, right? 1 Joule = 1 newton × 1 meter, right? Well… Yes and no. In discussions like this, we talk potential energy, i.e. energy stored in the system, so to say. That means that we’re looking at work done against the force, like when we carry a bucket of water up to the third floor or, to use a somewhat more scientific description of what’s going on, when we are separating two masses. Because we’re doing work against the force, we put a minus sign in front of our integral:
Now, the electromagnetic force works pretty much like gravity, except that, when discussing gravity, we only have positive ‘charges’ (the mass of some object is always positive). In electromagnetics, we have positive as well as negative charge, and please note that two like charges repel (that’s not the case with gravity). Hence, doing work against the electromagnetic force may involve bringing like charges together or, alternatively, separating opposite charges. We can’t say. Fortunately, when it comes to the math of it, it doesn’t matter: we will have the same minus sign in front of our integral. The point is: we’re doing work against the force, and so that’s what the minus sign stands for. So it has nothing to do with the specifics of the law of attraction and repulsion in this case (electromagnetism as opposed to gravity) and/or the fact that electrons carry negative charge. No.
Let’s get back to the integral. Just in case you forgot, the integral sign ∫ stands for an S: the S of summa, i.e. sum in Latin, and we’re using these integrals because we’re adding an infinite number of infinitesimally small contributions to the total effort here indeed. You should recognize it, because it’s a general formula for energy or work. It is, once again, a so-called line integral, so it’s a bit different than the ∫f(x)dx stuff you learned from high school. Not very different, but different nevertheless. What’s different is that we have a vector dot product F•ds after the integral sign here, so that’s not like f(x)dx. In case you forgot, that f(x)dx product represents the surface of an infinitesimally rectangle, as shown below: we make the base of the rectangle smaller and smaller, so dx becomes an infinitesimal indeed. And then we add them all up and get the area under the curve. If f(x) is negative, then the contributions will be negative.
But so we don’t have little rectangles here. We have two vectors, F and ds, and their vector dot product, F•ds, which will give you… Well… I am tempted to write: the tangential component of the force along the path, but that’s not quite correct: if ds was a unit vector, it would be true—because then it’s just like that h•n product I introduced in our first vector calculus class. However, ds is not a unit vector: it’s an infinitesimal vector, and, hence, if we write the tangential component of the force along the path as Ft, then F•d= |F||ds|cosθ = F·cosθ·ds = Ft·ds. So this F•ds is a tangential component over an infinitesimally small segment of the curve. In short, it’s an infinitesimally small contribution to the total amount of work done indeed. You can make sense of this by looking at the geometrical representation of the situation below.
I am just saying this so you know what that integral stands for. Note that we’re not adding arrows once again, like we did when calculating amplitudes or so. It’s all much more straightforward really: a vector dot product is a scalar, so it’s just some real number—just like any component of a vector (tangential, normal, in the direction of one of the coordinates axes, or in whatever direction) is not a vector but a real number. Hence, W is also just some real number. It can be positive or negative because… Well… When we’d be going down the stairs with our bucket of water, our minus sign doesn’t disappear. Indeed, our convention to put that minus sign there should obviously not depend on what point a and b we’re talking about, so we may actually be going along the direction of the force when going from a to b.
As a matter of fact, you should note that’s actually the situation which is depicted above. So then we get a negative number for W. Does that make sense? Of course it does: we’re obviously not doing any work here as we’re moving along the direction, so we’re surely not adding any (potential) energy to the system. On the contrary, we’re taking energy out of the system. Hence, we are reducing its (potential) energy and, hence, we should have a negative value for W indeed. So, just think of the minus sign being there to ensure we add potential energy to the system when going against the force, and reducing it when going with the force.
OK. You get this. You probably also know we’ll re-define W as a difference in potential between two points, which we’ll write as Φ(b) – Φ(a). Now that should remind you of your high school integral ∫f(x)dx once again. For a definite integral over a line segment [a, b], you’d have to find the antiderivative of f(x), which you’d write as F(x), and then you’d take the difference F(b) – F(a) too. Now, you may or may not remember that this antiderivative was actually a family of functions F(x) + k, and k could be any constant – 5/9, 6π, 3.6×10124, 0.86, whatever! – because such constant vanishes when taking the derivative.
Here we have the same, we can define an infinite number of functions Φ(r) + k, of which the gradient will yield… Stop! I am going too fast here. First, we need to re-write that W function above in order to ensure we’re calculating stuff in terms of the unit charge, so we write:
Huh? Well… Yes. I am using the definition of the field E here really: E is the force (F) when putting a unit charge in the field. Hence, if we want the work done per unit charge, i.e. W(unit), then we have to integrate the vector dot product E·ds over the path from a to b. But so now you see what I want to do. It makes the comparison with our high school integral complete. Instead of taking a derivative in regard to one variable only, i.e. dF(x)/dx) = f(x), we have a function Φ here not in one but in three variables: Φ = Φ(x, y, z) = Φ(r) and, therefore, we have to take the vector derivative (or gradient as it’s called) of Φ to get E:
Φ(x, y, z) = (∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z) = –E(x, y, z)
But so it’s the same principle as what you learned how to use to solve your high school integral. Now, you’ll usually see the expression above written as:
E = –Φ
Why so short? Well… We all just love these mysterious abbreviations, don’t we? 🙂 Jokes aside, it’s true some of those vector equations pack an awful lot of information. Just take Feynman’s advice here: “If it helps to write out the components to be sure you understand what’s going on, just do it. There is nothing inelegant about that. In fact, there is often a certain cleverness in doing just that.” So… Let’s move on.
I should mention that we can only apply this more sophisticated version of the ‘high school trick’ because Φ and E are like temperature (T) and heat flow (h): they are fields. T is a scalar field and h is a vector field, and so that’s why we can and should apply our new trick: if we have the scalar field, we can derive the vector field. In case you want more details, I’ll just refer you to our first vector calculus class. Indeed, our so-called First Theorem in vector calculus was just about the more sophisticated version of the ‘high school trick’: if we have some scalar field ψ (like temperature or potential, for example: just substitute the ψ in the equation below for T or Φ), then we’ll always find that:
The Γ here is the curve between point 1 and 2, so that’s the path along which we’re going, and ψ must represent some vector field.
Let’s go back to our W integral. I should mention that it doesn’t matter what path we take: we’ll always get the same value for W, regardless of what path we take. That’s why the illustration above showed two possible paths: it doesn’t matter which one we take. Again, that’s only because E is a vector field. To be precise, the electrostatic field is a so-called conservative vector field, which means that we can’t get energy out of the field by first carrying some charge along one path, and then carrying it back along another. You’ll probably find that’s obvious, and it is. Just note it somewhere in the back of your mind.
So we’re done. We should just substitute E for Φ, shouldn’t we? Well… Yes. For minus Φ, that is. Another minus sign. Why? Well… It makes that W(unit) integral come out alright. Indeed, we want a formula like W = Φ(b) – Φ(a), not like Φ(a) – Φ(b). Look at it. We could, indeed, define E as the (positive) gradient of some scalar field ψ = –Φ, and so we could write E = ψ, but then we’d find that W = –[ψ(b) – ψ(a)] = ψ(a) – ψ(b).
You’ll say: so what? Well… Nothing much. It’s just that our field vectors would point from lower to higher values of ψ, so they would be flowing uphill, so to say. Now, we don’t want that in physics. Why? It just doesn’t look good. We want our field vectors to be directed from higher potential to lower potential, always. Just think of it: heat (h) flows from higher temperature (T) to lower, and Newton’s apple falls from greater to lower height. Likewise, when putting a unit charge in the field, we want to see it move from higher to lower electric potential. Now, we can’t change the direction of E, because that’s the direction of the force and Nature doesn’t care about our conventions and so we can’t choose the direction of the force. But we can choose our convention. So that’s why we put a minus sign in front of Φ when writing E = –Φ. It makes everything come out alright. 🙂 That’s why we also have a minus sign in the differential heat flow equation: h = –κT.
So now we have the easy W(unit) = Φ(b) – Φ(a) formula that we wanted all along. Now, note that, when we say a unit charge, we mean a plus one charge. Yes: +1. So that’s the charge of the proton (it’s denoted by e) so you should stop thinking about moving electrons around! [I am saying this because I used to confuse myself by doing that. You end up with the same formulas for W and Φ but it just takes you longer to get there, so let me save you some time here. :-)]
But… Yes? In reality, it’s electrons going through a wire, isn’t? Not protons. Yes. But it doesn’t matter. Units are units in physics, and they’re always +1, for whatever (time, distance, charge, mass, spin, etcetera). AlwaysFor whatever. Also note that in laboratory experiments, or particle accelerators, we often use protons instead of electrons, so there’s nothing weird about it. Finally, and most fundamentally, if we have a –e charge moving through a neutral wire in one direction, then that’s exactly the same as a +e charge moving in the other way.
Just to make sure you get the point, let’s look at that illustration once again. We already said that we have F and, hence, E pointing from a to b and we’ll be reducing the potential energy of the system when moving our unit charge from a to b, so W was some negative value. Now, taking into account we want field lines to point from higher to lower potential, Φ(a) should be larger than Φ(b), and so… Well.. Yes. It all makes sense: we have a negative difference Φ(b) – Φ(a) = W(unit), which amounts, of course, to the reduction in potential energy.
The last thing we need to take care of now, is the reference point. Indeed, any Φ(r) + k function will do, so which one do we take? The approach here is to take a reference point Pat infinity. What’s infinity? Well… Hard to say. It’s a place that’s very far away from all of the charges we’ve got lying around here. Very far away indeed. So far away we can say there is nothing there really. No charges whatsoever. 🙂 Something like that. 🙂 In any case. I need to move on. So Φ(P0) is zero and so we can finally jot down the grand result for the electric potential Φ(P) (aka as the electrostatic or electric field potential):
So now we can calculate all potentials, i.e. when we know where the charges are at least. I’ve shown an example below. As you can see, besides having zero potential at infinity, we will usually also have one or more equipotential surfaces with zero potential. One could say these zero potential lines sort of ‘separate’ the positive and negative space. That’s not a very scientifically accurate description but you know what I mean.
Let me make a few final notes about the units. First, let me, once again, note that our unit charge is plus one, and it will flow from positive to negative potential indeed, as shown below, even if we know that, in an actual electric circuit, and so now I am talking about a copper wire or something similar, that means the (free) electrons will move in the other direction.
If you’re smart (and you are), you’ll say: what about the right-hand rule for the magnetic force? Well… We’re not discussing the magnetic force here but, because you insist, rest assured it comes out alright. Look at the illustration below of the magnetic force on a wire with a current, which is a pretty standard one.
So we have a given B, because of the bar magnet, and then v, the velocity vector for the… Electrons? No. You need to be consistent. It’s the velocity vector for the unit charges, which are positive (+e). Now just calculate the force F = qv×B = ev×B using the right-hand rule for the vector cross product, as illustrated below. So v is the thumb and B is the index finger in this case. All you need to do is tilt your hand, and it comes out alright.
But… We know it’s electrons going the other way. Well… If you insist. But then you have to put a minus sign in front of the q, because we’re talking minus e (–e). So now v is in the other direction and so v×B is in the other direction indeed, but our force F = qv×B = –ev×is not. Fortunately not, because physical reality should not depend on our conventions. 🙂 So… What’s the conclusion. Nothing. You may or may not want to remember that, when we say that our current j current flows in this or that direction, we actually might be talking electrons (with charge minus one) flowing in the opposite direction, but then it doesn’t matter. In addition, as mentioned above, in laboratory experiments or accelerators, we may actually be talking protons instead of electrons, so don’t assume electromagnetism is the business of electrons only.
To conclude this disproportionately long introduction (we’re finally ready to talk more difficult stuff), I should just make a note on the units. Electric potential is measured in volts, as you know. However, it’s obvious from all that I wrote above that it’s the difference in potential that matters really. From the definition above, it should be measured in the same unit as our unit for energy, or for work, so that’s the joule. To be precise, it should be measured in joule per unit charge. But here we have one of the very few inconsistencies in physics when it comes to units. The proton is said to be the unit charge (e), but its actual value is measured in coulomb (C). To be precise: +1 e = 1.602176565(35)×10−19 C. So we do not measure voltage – sorry, potential difference 🙂 – in joule but in joule per coulomb (J/C).
Now, we usually use another term for the joule/coulomb unit. You guessed it (because I said it): it’s the volt (V). One volt is one joule/coulomb: 1 V = 1 J/C. That’s not fair, you’ll say. You’re right, but so the proton charge e is not a so-called SI unit. Is the Coulomb an SI unit? Yes. It’s derived from the ampere (A) which, believe it or not, is actually an SI base unit. One ampere is 6.241×1018 electrons (i.e. one coulomb) per second. You may wonder how the ampere (or the coulomb) can be a base unit. Can they be expressed in terms of kilogram, meter and second, like all other base units. The answer is yes but, as you can imagine, it’s a bit of a complex description and so I’ll refer you to the Web for that.
The Poisson equation
I started this post by saying that I’d talk about fields and present some results from electrostatics using our ‘new’ vector differential operators, so it’s about time I do that. The first equation is a simple one. Using our E = –Φ formula, we can re-write the ∇•E = ρ/ε0 equation as:
∇•E = ∇•∇Φ = ∇2Φ = –ρ/ε0
This is a so-called Poisson equation. The ∇2 operator is referred to as the Laplacian and is sometimes also written as Δ, but I don’t like that because it’s also the symbol for the total differential, and that’s definitely not the same thing. The formula for the Laplacian is given below. Note that it acts on a scalar field (i.e. the potential function Φ in this case).
As Feynman notes: “The entire subject of electrostatics is merely the study of the solutions of this one equation.” However, I should note that this doesn’t prevent Feynman from devoting at least a dozen of his Lectures on it, and they’re not the easiest ones to read. [In case you’d doubt this statement, just have a look at his lecture on electric dipoles, for example.] In short: don’t think the ‘study of this one equation’ is easy. All I’ll do is just note some of the most fundamental results of this ‘study’.
Also note that ∇•E is one of our ‘new’ vector differential operators indeed: it’s the vector dot product of our del operator () with E. That’s something very different than, let’s say, Φ. A little dot and some bold-face type make an enormous difference here. 🙂 You may or may remember that we referred to the ∇• operator as the divergence (div) operator (see my post on that).
Gauss’ Law
Gauss’ Law is not to be confused with Gauss’ Theorem, about which I wrote elsewhere. It gives the flux of E through a closed surface S, any closed surface S really, as the sum of all charges inside the surface divided by the electric constant ε(but then you know that constant is just there to make the units come out alright).
The derivation of Gauss’ Law is a bit lengthy, which is why I won’t reproduce it here, but you should note its derivation is based, mainly, on the fact that (a) surface areas are proportional to r2 (so if we double the distance from the source, the surface area will quadruple), and (b) the magnitude of E is given by an inverse-square law, so it decreases as 1/r2. That explains why, if the surface S describes a sphere, the number we get from Gauss’ Law is independent of the radius of the sphere. The diagram below (credit goes to Wikipedia) illustrates the idea.
The diagram can be used to show how a field and its flux can be represented. Indeed, the lines represent the flux of E emanating from a charge. Now, the total number of flux lines depends on the charge but is constant with increasing distance because the force is radial and spherically symmetric. A greater density of flux lines (lines per unit area) means a stronger field, with the density of flux lines (i.e. the magnitude of E) following an inverse-square law indeed, because the surface area of a sphere increases with the square of the radius. Hence, in Gauss’ Law, the two effect cancel out: the two factors vary with distance, but their product is a constant.
Now, if we describe the location of charges in terms of charge densities (ρ), then we can write Qint as:
Now, Gauss’ Law also applies to an infinitesimal cubical surface and, in one of my posts on vector calculus, I showed that the flux of E out of such cube is given by E·dV. At this point, it’s probably a good idea to remind you of what this ‘new’ vector differential operator •, i.e. our ‘divergence’ operator, stands for: the divergence of E (i.e. • applied to E, so that’s E) represents the volume density of the flux of E out of an infinitesimal volume around a given point. Hence, it’s the flux per unit volume, as opposed to the flux out of the infinitesimal cube itself, which is the product of and dV, i.e. E·dV.
So what? Well… Gauss’ Law applied to our infinitesimal volume gives us the following equality:
That, in turn, simplifies to:
So that’s Maxwell’s first equation once again, which is equivalent to our Poisson equation: E = ∇2Φ = –ρ/ε0. So what are we doing here? Just listing equivalent formulas? Yes. I should also note they can be derived from Coulomb’s law of force, which is probably the one you learned in high school. So… Yes. It’s all consistent. But then that’s what we should expect, of course. 🙂
The energy in a field
All these formulas look very abstract. It’s about time we use them for something. A lot of what’s written in Feynman’s Lectures on electrostatics is applied stuff indeed: it focuses, among other things, on calculating the potential in various circumstances and for various distributions of charge. Now, funnily enough, while that E = –ρ/ε0 equation is equivalent to Coulomb’s law and, obviously, much more compact to write down, Coulomb’s law is easier to start with for basic calculations. Let me first write Coulomb’s law. You’ll probably recognize it from your high school days:
Fis the force on charge q1, and Fis the force on charge q2. Now, qand q2. may attract or repel each other but, in both cases, the forces will be equal and opposite. [In case you wonder, yes, that’s basically the law of action and reaction.] The e12 vector is the unit vector from qto q1, not from qto q2, as one might expect. That’s because we’re not talking gravity here: like charges do not attract but repel and, hence, we have to switch the order here. Having said that, that’s basically the only peculiar thing about the equation. All the rest is standard:
1. The force is inversely proportional to the square of the distance and so we have an inverse-square law here indeed.
2. The force is proportional to the charge(s).
3. Finally, we have a proportionality constant, 1/4πε0, which makes the units come out alright. You may wonder why it’s written the way it’s written, i.e. with that 4π factor, but that factor (4π or 2π) actually disappears in a number of calculations, so then we will be left with just a 1/ε0 or a 1/2ε0 factor. So don’t worry about it.
We want to calculate potentials and all that, so the first thing we’ll do is calculate the force on a unit charge. So we’ll divide that equation by q1, to calculate E(1) = F1/q1:
Piece of cake. But… What’s E(1) really? Well… It’s the force on the unit charge (+e), but so it doesn’t matter whether or not that unit charge is actually there, so it’s the field E caused by a charge q2. [If that doesn’t make sense to you, think again.] So we can drop the subscripts and just write:
What a relief, isn’t it? The simplest formula ever: the (magnitude) of the field as a simple function of the charge q and its distance (r) from the point that we’re looking at, which we’ll write as P = (x, y, z). But what origin are we using to measure x, y and z. Don’t be surprised: the origin is q.
Now that’s a formula we can use in the Φ(P) integral. Indeed, the antiderivative is ∫(q/4πε0r2)dr. Now, we can bring q/4πε0 out and so we’re left with ∫(1/r2)dr. Now ∫(1/r2)dr is equal to –1/r + k, and so the whole antiderivative is –q/4πε0r + k. However, the minus sign cancels out with the minus sign in front of the Φ(P) = Φ(x, y, z) integral, and so we get:
You should just do the integral to check this result. It’s the same integral but with P0 (infinity) as point a and P as point b in the integral, so we have ∞ as start value and r as end value. The integral then yields Φ(P) – Φ(P0) = –q/4πε0[1/r – 1/∞). [The k constant falls away when subtracting Φ(P0) from Φ(P).] But 1/∞ = 0, and we had a minus sign in front of the integral, which cancels the sign of –q/4πε0. So, yes, we get the wonderfully simple result above. Also please do quickly check if it makes sense in terms of sign: the unit charge is +e, so that’s a positive charge. Hence, Φ(x, y, z) will be positive if the sign of q is also positive, but negative if q would happen to be negative. So that’s OK.
Also note that the potential – which, remember, represents the amount of work to be done when bringing a unit charge (e) from infinity to some distance r from a charge q – is proportional to the charge of q. We also know that the force and, hence, the work is proportional to the charge that we are bringing in (that’s how we calculated the work per unit in the first place: by dividing the total amount of work by the charge). Hence, if we’d not bring some unit charge but some other charge q2, the work done would also be proportional to q2. Now, we need to make sure we understand what we’re writing and so let’s tidy up and re-label our first charge once again as q1, and the distance r as r12, because that’s what r is: the distance between the two charges. We then have another obvious but nice result: the work done in bringing two charges together from a large distance (infinity) is
Now, one of the many nice properties of fields (scalar or vector fields) and the associated energies (because that’s what we are talking about here) is that we can simply add up contributions. For example, if we’d have many charges and we’d want to calculate the potential Φ at a point which we call 1, we can use the same Φ(r) = q/4πε0r formula which we had derived for one charge only, for all charges, and then we simply add the contributions of each to get the total potential:
Now that we’re here, I should, of course, also give the continuum version of this formula, i.e. the formula used when we’re talking charge densities rather than individual charges. The sum then becomes an infinite sum (i.e. an integral), and qj (note that j goes from 2 to n) becomes a variable which we write as ρ(2). We get:
Going back to the discrete situation, we get the same type of sum when bringing multiple pairs of charges qi and qj together. Hence, the total electrostatic energy U is the sum of the energies of all possible pairs of charges:
It’s been a while since you’ve seen any diagram or so, so let me insert one just to reassure you it’s as simple as that indeed:
Now, we have to be aware of the risk of double-counting, of course. We should not be adding qiqj/4πε0rij twice. That’s why we write ‘all pairs’ under the ∑ summation sign, instead of the usual i, j subscripts. The continuum version of this equation below makes that 1/2 factor explicit:
Hmm… What kind of integral is that? It’s a so-called double integral because we have two variables here. Not easy. However, there’s a lucky break. We can use the continuum version of our formula for Φ(1) to get rid of the ρ(2) and dV2 variables and reduce the whole thing to a more standard ‘single’ integral. Indeed, we can write:
Now, because our point (2) no longer appears, we can actually write that more elegantly as:
That looks nice, doesn’t it? But do we understand it? Just to make sure. Let me explain it. The potential energy of the charge ρdV is the product of this charge and the potential at the same point. The total energy is therefore the integral over ϕρdV, but then we are counting energies twice, so that’s why we need the 1/2 factor. Now, we can write this even more beautifully as:
Isn’t this wonderful? We have an expression for the energy of a field, not in terms of the charges or the charge distribution, but in terms of the field they produce.
I am pretty sure that, by now, you must be suffering from ‘formula overload’, so you probably are just gazing at this without even bothering to try to understand. Too bad, and you should take a break then or just go do something else, like biking or so. 🙂
First, you should note that you know this EE expression already: EE is just the square of the magnitude of the field vector E, so EE = E2. That makes sense because we know, from what we know about waves, that the energy is always proportional to the square of an amplitude, and so we’re just writing the same here but with a little proportionality constant (ε0).
OK, you’ll say. But you probably still wonder what use this formula could possibly have. What is that number we get from some integration over all space? So we associate the Universe with some number and then what? Well… Isn’t that just nice? 🙂 Jokes aside, we’re actually looking at that EE = Eproduct inside of the integral as representing an energy density (i.e. the energy per unit volume). We’ll denote that with a lower-case symbol and so we write:
Just to make sure you ‘get’ what we’re talking about here: u is the energy density in the little cube dV in the rather simplistic (and, therefore, extremely useful) illustration below (which, just like most of what I write above, I got from Feynman).
Now that should make sense to you—I hope. 🙂 In any case, if you’re still with me, and if you’re not all formula-ed out you may wonder how we get that ε0EE = ε0E2 expression from that ρΦ expression. Of course, you know that E = –∇Φ, and we also have the Poisson equation ∇2Φ = –ρ/ε0, but that doesn’t get you very far. It’s one of those examples where an easy-looking formula requires a lot of gymnastics. However, as the objective of this post is to do some of that, let me take you through the derivation.
Let’s do something with that Poisson equation first, so we’ll re-write it as ρ = –ε02Φ, and then we can substitute ρ in the integral with the ρΦ product. So we get:
Now, you should check out those fancy formulas with our new vector differential operators which we listed in our second class on vector calculus, but, unfortunately, none of them apply. So we have to write it all out and see what we get:
Now that looks horrendous and so you’ll surely think we won’t get anywhere with that. Well… Physicists don’t despair as easily as we do, it seems, and so they do substitute it in the integral which, of course, becomes an even more monstrous expression, because we now have two volume integrals instead of one! Indeed, we get:
But if Φ is a vector field (it’s minus E, remember!), then ΦΦ is a vector field too, and we can then apply Gauss’ Theorem, which we mentioned in our first class on vector calculus, and which – mind you! – has nothing to do with Gauss’ Law. Indeed, Gauss produced so much it’s difficult to keep track of it all. 🙂 So let me remind you of this theorem. [I should also show why ΦΦ still yields a field, but I’ll assume you believe me.] Gauss’ Theorem basically shows how we can go from a volume integral to a surface integral:
If we apply this to the second integral in our U expression, we get:
So what? Where are we going with this? Relax. Be patient. What volume and surface are we talking about here? To make sure we have all charges and influences, we should integrate over all space and, hence, the surface goes to infinity. So we’re talking a (spherical) surface of enormous radius R whose center is the origin of our coordinate system. I know that sounds ridiculous but, from a math point of view, it is just the same like bringing a charge in from infinity, which is what we did to calculate the potential. So if we don’t difficulty with infinite line integrals, we should not have difficulty with infinite surface and infinite volumes. That’s all I can, so… Well… Let’s do it.
Let’s look at that product ΦΦ•n in the surface integral. Φ is a scalar and Φ is a vector, and so… Well… Φ•is a scalar too: it’s the normal component of Φ = –E. [Just to make sure, you should note that the way we define the normal unit vector n is such that ∇Φ•n is some positive number indeed! So n will point in the same direction, more or less, as ∇Φ = –E. So the θ angle between ∇Φ = –E and n is surely less than ± 90° and, hence, the cosine factor in the ∇Φ•= |∇Φ||n|cosθ = |∇Φ|cosθ is positive, and so the whole vector dot product is positive.]
So, we have a product of two scalars here. What happens with them if R goes to infinity? Well… The potential varies as 1/r as we’re going to infinity. That’s obvious from that Φ = (q/4πε0)(1/r) formula: just think of q as some kind of average now, which works because we assume all charges are located within some finite distance, while we’re going to infinity. What about Φ•n? Well… Again assuming that we’re reasonably far away from the charges, we’re talking the density of flux lines here (i.e. the magnitude of E) which, as shown above, follows an inverse-square law, because the surface area of a sphere increases with the square of the radius. So Φ•n varies not as 1/r but as 1/r2. To make a long story short, the whole product ΦΦ•n falls of as 1/r goes to infinity. Now, we shouldn’t forget we’re integrating a surface integral here, with r = R, and so it’s R going to infinity. So that surface integral has to go to zero when we include all space. The volume integral still stands however, so our formula for U now consists of one term only, i.e. the volume integral, and so we now have:
Done !
What’s left?
In electrostatics? Lots. Electric dipoles (like polar molecules), electrolytes, plasma oscillations, ionic crystals, electricity in the atmosphere (like lightning!), dielectrics and polarization (including condensers), ferroelectricity,… As soon as we try to apply our theory to matter, things become hugely complicated. But the theory works. Fortunately! 🙂 I have to refer you to textbooks, though, in case you’d want to know more about it. [I am sure you don’t, but then one never knows.]
What I wanted to do is to give you some feel for those vector and field equations in the electrostatic case. We now need to bring magnetic field back into the picture and, most importantly, move to electrodynamics, in which the electric and magnetic field do not appear as completely separate things. No! In electrodynamics, they are fully interconnected through the time derivatives ∂E/∂t and ∂B/∂t. That shows they’re part and parcel of the same thing really: electromagnetism.
But we’ll try to tackle that in future posts. Goodbye for now!
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Some content on this page was disabled on June 17, 2020 as a result of a DMCA takedown notice from Michael A. Gottlieb, Rudolf Pfeiffer, and The California Institute of Technology. You can learn more about the DMCA here: | 14,257 | 57,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-50 | latest | en | 0.916032 |
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Purpose of Assignment
The purpose of this assignment is to allow the student to calculate the project cash flow using net present value (NPV), internal rate of return (IRR), and the payback methods.
Assignment Steps
Resources: Corporate Finance
Create a 350-word memo to management including the following:
Describe the use of internal rate of return (IRR), net present value (NPV), and the payback method in evaluating project cash flows.
Calculate the following time value of money problems:
Calculate the project cash flow generated for Project A and Project B using the NPV method.
Which project would you select, and why?
Which project would you select under the payback method? The discount rate is 10% for both projects.
Sample Template for Project A and Project B:
“Table showing investments and returns for Project A and Project B. Project A has \$10,000 initial investment with \$5,000 returns in each of the first 3 years. Project B has \$55,000 initial investment with \$20,000 in each of the first 3 years.â€
Show all work.
Submit the memo and all calculations.
Resources
### 5.1 Why Use Net Present Value?
capital budgeting for
This chapter, as well as the next two, focuses on capital budgeting, the decision-making process for accepting or rejecting projects. This chapter develops the basic capital budgeting methods, leaving much of the practical application to subsequent chapters. But we donâ€t have to develop these methods from scratch. In Chapter 4, we pointed out that a dollar received in the future is worth less than a dollar received today. The reason, of course, is that todayâ€s dollar can be reinvested, yielding a greater amount in the future. And we showed in Chapter 4 that the exact worth of a dollar to be received in the future is its present value. Furthermore, Section 4.1 suggested calculating the net present value of any project. That is, the section suggested calculating the difference between the sum of the present values of the projectâ€s future cash flows and the initial cost of the project.
The net present value (NPV) method is the first one to be considered in this chapter. We begin by reviewing the approach with a simple example. Then, we ask why the method leads to good decisions.
PAGE 136EXAMPLE
5.1
Net Present Value The Alpha Corporation is considering investing in a riskless project costing \$100. The project receives \$107 in one year and has no other cash flows. The riskless discount rate on comparable riskless investments is 2 percent.
The NPV of the project can easily be calculated as:
(5.1)
From Chapter 4, we know that the project should be accepted because its NPV is positive. This is true because the project generates \$107 of future cash flows from a \$100 investment whereas comparable investments only generate \$102.
The basic investment rule can be generalized to:
Accept a project if the NPV is greater than zero.
Reject a project if the NPV is less than zero.
We refer to this as the NPV rule.
Why does the NPV rule lead to good decisions? Consider the following two strategies available to the managers of Alpha Corporation:
• Use \$100 of corporate cash to invest in the project. The \$107 will be paid as a dividend in one year.
• Forgo the project and pay the \$100 of corporate cash to stockholders as a dividend today.
If Strategy 2 is employed, the stockholder might deposit the cash dividend in a bank for one year. With an interest rate of 2 percent, Strategy 2 would produce cash of \$102 (=\$100 X 1.02) at the end of the year. The stockholder would prefer Strategy 1 because Strategy 2 produces less than \$107 at the end of the year.
Our basic point is:
Accepting positive NPV projects benefits the stockholders.
How do we interpret the exact NPV of \$4.90? This is the increase in the value of the firm from the project. For example, imagine that the firm today has productive assets worth \$V and has \$100 of cash. If the firm forgoes the project, the value of the firm today would simply be:
\$V + \$100
If the firm accepts the project, the firm will receive \$107 in one year but will have no cash today. Thus, the firmâ€s value today would be:
The difference between these equations is just \$4.90, the net present value of Equation 5.1. Thus:
The value of the firm rises by the NPV of the project.
Note that the value of the firm is merely the sum of the values of the different projects, divisions, or other entities within the firm. This property, called value additivity, is quite important. It implies that the contribution of any project to a firmâ€s value is simply the Page 137NPV of the project. As we will see later, alternative methods discussed in this chapter do not generally have this nice property.
The NPV rule uses the correct discount rate.
One detail remains. We assumed that the project was riskless, a rather implausible assumption. Future cash flows of real-world projects are invariably risky. In other words, cash flows can only be estimated, rather than known. Imagine that the managers of Alpha expect the cash flow of the project to be \$107 next year. That is, the cash flow could be higher, say \$117, or lower, say \$97. With this slight change, the project is risky. Suppose the project is about as risky as the stock market as a whole, where the expected return this year is perhaps 10 percent. Then 10 percent becomes the discount rate, implying that the NPV of the project would be:
Because the NPV is negative, the project should be rejected. This makes sense: A stockholder of Alpha receiving a \$100 dividend today could invest it in the stock market, expecting a 10 percent return. Why accept a project with the same risk as the market but with an expected return of only 7 percent?
### Calculating NPVs with a Spreadsheet
Spreadsheets are commonly used to calculate NPVs. Examining the use of spreadsheets in this context also allows us to issue an important warning. Consider the following:
In our spreadsheet example, notice that we have provided two answers. The first answer is wrong even though we used the spreadsheetâ€s NPV formula. What happened is that the “NPV†function in our spreadsheet is actually a PV function; unfortunately, one of the original spreadsheet programs many years ago got the definition wrong, and subsequent spreadsheets have copied it! Our second answer shows how to use the formula properly.
The example here illustrates the danger of blindly using calculators or computers without understanding what is going on; we shudder to think of how many capital budgeting decisions in the real world are based on incorrect use of this particular function. | 1,487 | 6,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.934388 |
https://infospotz.com/converting-30-centimeters-to-inches/ | 1,669,634,974,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00347.warc.gz | 356,617,724 | 34,834 | November 28, 2022
# Converting 30 Centimeters to Inches
If you have a project coming up and need to convert 30 centimeters (CM) to inches, you have probably encountered the problem of how to measure a 30-centimeter object. In this article we will look at a Visual representation of the 30 cm measurement and give you some simple tips to convert 30 cm to inches. We also explain how to calculate the measurements in inches. Using this simple conversion tool, you will be able to get the exact measurements in inches, irrespective of the type of measurement you need to convert.
## Converting 30 centimeters to inches
If you’re a bit confused by the units of length, you can make use of a conversion calculator to convert 30 centimeters to inches. It’s easy to see that a centimeter is a hundredth of an inch, and the same is true of inches. The metric system uses centimeters as their units of measurement, while the imperial system uses inches. To convert 30 cm to inches, enter its length in centimeters and choose the reference value in inches.
The measurement system in the US and Europe is based on the metric system. The inch is the most widely used measurement system in the US. In US customary measurements, a foot is equal to 12 inches, and a yard is equal to 36 inches. The rulers used in the US measure in inches, and one inch is approximately equal to thirty centimeters. If you want to convert 30 cm to inches, you’ll need a ruler that’s twelve inches long.
## Visual representation of 30 centimeters in inches
In a metric system, 30 centimeters equal to 2.5 feet, or 76.2 centimeters. To understand what that means, imagine a meter being made up of four pencils, stacked one on top of the other. In the United States, the unit of measurement is known as an inch. It can be written as “in” or “1 in” or abbreviated as “in.” The inch is also commonly expressed with the ” symbol, also known as a double-quote. In fact, some people will simply write “1 inch” instead of using the double-quote symbol.
## Calculating 30 centimeters in inches
If you’re not sure how to convert a measurement, you can use a conversion calculator. The easiest way to convert 30 centimeters to inches is to multiply the number by a hundredth. The metric system of measurement uses a unit called centimeters (cm).
In addition to meters and feet, the metric system also recognizes the metric system. In the US, inches are widely used. One meter equals 100 centimeters. In Europe, a foot measures 12 inches. A yard is 36 inches. A ruler is typically 12 inches long, and a centimeter is one-hundredth of a metre. The metric system is abbreviated “cm.” In most cases, a centimeter is equivalent to one-hundredth of a metre.
Happy
0 % | 624 | 2,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | longest | en | 0.932006 |
https://insurance-thai.com/s21-vs-oneplus-9/ | 1,675,742,901,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00379.warc.gz | 321,777,943 | 13,417 | A Beginner’s Guide to s21 vs oneplus 9
The numbers we are all familiar with are the ones that show our number of digits that we have reached after we have finished our first number. In our case, the numbers we have reached after we have finished our first s21 are the ones that we have reached after we have finished oneplus 9.
The oneplus is the number of digits that we have reached after we have finished one plus nine. In our case, the number we have reached after we have finished one plus nine is the one that we have reached after we have finished a total of thirteen.
In our case, the number eight is the number that we have reached after we have finished one plus nine.In our case, the number seven is the number that we reached after we finished a total of nine.The oneplus 9 is the number that we reached after we have finished three plus nine.In our case, the number eight is the number that we reached after we finished one plus nine.
So if you’re wondering what the oneplus X is, that is the one we have reached after we have finished thirteen.The oneplus X is the one we have reached after we have finished eight.The oneplus X is the one we reach after we have finished nine.The oneplus X is the one we reached after we have finished thirteen.The oneplus X is the one we reached after we have finished eight.
For our purpose, the oneplus X is the one we have reached after we have finished eight.For our purposes, the oneplus X is the one we have reached after we have finished eight.For our purposes, the oneplus X is the one we reach after we have finished thirteen.The oneplus X is the one we reach after we have finished eight.The oneplus X is the one we reach after we have finished nine.
The oneplus X is the one we reach after we have finished eight.For our purposes, the oneplus X is the one we reach after we have finished eight.For our purposes, the oneplus X is the one we reach after we have finished eight.For our purposes, the oneplus X is the one we reach after we have finished nine.
The oneplus X is the one that reaches after we have finished thirteen.The oneplus X is the one that reaches after we have finished eight.The oneplus X is the one that reaches after we have finished nine.The oneplus X is the one that reaches after we have finished eight.For our purposes, the oneplus X is the one we reach after we have finished eight.For our purposes, the oneplus X is the one we reach after we have finished nine.
It’s not so bad when you think of it this way. The oneplus X is the one we reach after we have finished nine.The oneplus X is the one that reaches after we have finished thirteen.The oneplus X is the one that reaches after we have finished eight.The oneplus X is the one that reaches after we have finished nine.The oneplus X is the one that reaches after we have finished eight.
It’s not an odd name for a phone company, but it’s not the most obvious choice. After all, the name is obviously also the design language of the X itself. But it’s also the name of the company that makes the X, and it’s a nice touch. It’s sort of like the oneplus logo is the entire company’s logo. | 692 | 3,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-06 | latest | en | 0.98244 |
https://math.stackexchange.com/questions/4536086/show-that-a-linear-system-may-have-no-solutions | 1,701,451,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00837.warc.gz | 457,102,566 | 36,984 | # Show that a linear system may have no solutions
I would like your help to show that the system below has at least one solution.
Let $$\mathcal{Y}\equiv \{0,1\}$$. Let $$\mathcal{V}$$ be a finite set containing positive and negative numbers.
Consider the system of equations/inqualities below. The vector of unknowns is $$(x_{y,v}: y\in \mathcal{Y}, v\in \mathcal{V})$$. We know the vectors $$(w_v: v\in \mathcal{V})$$, $$(q_y: y\in \mathcal{Y})$$, and $$(z_{y,v}: y\in \mathcal{Y}, v\in \mathcal{V})$$, and the sets $$\mathcal{V},\mathcal{Y}$$. $$(*) \quad \begin{cases} &(1) \quad \sum_{y\in \mathcal{Y}}x_{y,v} =w_v \quad \forall v \in \mathcal{V},\\ &(2) \quad \sum_{v\in \mathcal{V}} x_{y,v}=q_y\quad \forall y\in \mathcal{Y},\\ & -----------------------\\ &(3) \quad \sum_{v\in \mathcal{V}} x_{1,v} *z_{1,v} \geq \sum_{v\in \mathcal{V}} x_{1,v} *z_{0,v},\\ &(4) \quad\sum_{v\in \mathcal{V}} x_{0,v} *z_{0,v} \geq \sum_{v\in \mathcal{V}} x_{0,v} *z_{1,v} ,\\ &--------------------\\ &(5) \quad \sum_{y\in \mathcal{Y},v\in \mathcal{V}} x_{y,v}=1,\\ &(6) \quad 0\leq x_{y,v}\leq 1 \quad \forall y\in \mathcal{Y}, v\in \mathcal{V},\\ &(7) \quad \sum_{v\in \mathcal{V} } w_v=1,\\ &(8) \quad 0\leq w_v\leq 1 \quad \forall v\in \mathcal{V},\\ &(9) \quad \sum_{y\in \mathcal{Y} } q_y=1,\\ &(10) \quad 0\leq q_y\leq 1 \quad \forall y\in \mathcal{Y}.\\ \end{cases}$$
Question: Show that $$(*)$$ may not have a solution.
Note: I've posed a similar question here for the case where $$z_{y,v}\equiv y*v$$ (hence, constraints (3) and (4) look simpler there). Can the same counterexample be extended to this more general setting?
• Note: Your system of equations consists just of (1), (2), (5). In addition you have (3), (4), (6) as constraints on that system (note that they are not "equations"). (7), (8), (9) (10) are just some facts that the known data happens to satisfy. They may or may not be useful in showing it is possible to not have solutions, but they are not part of the system. I strongly suggest you divide matters up into the equations, the constraints, and the other information, instead of hiding an additional equation and constrains in with the other data. Sep 22, 2022 at 15:23
Let $$\mathcal{Y}\equiv \{0,1\}$$. Let $$\mathcal{V}$$ be a finite set containing positive and negative numbers.
Since the elements of $$\cal Y, V$$ are not used as anything other than indices, it doesn't matter what they are, only that $$|\mathcal Y| = 2$$ and $$n = |\mathcal V|$$ is finite.
We can define vectors in $$\Bbb R^n$$: $$x_0 = (x_{0,v})_{v\in\cal V}\\x_1 = (x_{1,v})_{v\in\cal V}\\z_0 = (z_{0,v})_{v\in\cal V}\\z_1 = (z_{1,v})_{v\in\cal V}\\z = z_1 - z_0\\w = (w_v)_{v\in\cal V}\\c=(1)_{v\in\cal V}$$
Your system of equations is $$x_0 + x_1 = w \tag{1}$$ $$x_0 \cdot c = q_0\tag {2a}$$ $$x_1 \cdot c = q_1\tag {2b}$$ $$x_0 \cdot c + x_1 \cdot c = 1\tag 5$$ with constraints $$x_1 \cdot z \ge 0\tag 3$$ $$x_0 \cdot z \le 0\tag 4$$ $$0 \le x_0, 0 \le x_1\tag 6$$
Where $$(6)$$ is meant to hold coordinate-wise. The upper bound of $$1$$ on the coordinates of $$x_0,x_1$$ follows from $$(6)$$ and $$(5)$$, and thus does not need to be enforced separately. But since we also know that $$q_0 + q_1 = 1$$, equation $$(5)$$ follows from equations $$(2)$$, and thus can also be dropped from the list.
If we simplify the notation by letting $$x = x_0$$, and substitute $$w - x$$ for $$x_1$$, we can reduce the problem further: $$x\cdot c = q_0\\x\cdot c = w\cdot c - q_1\\ x\cdot z \le w \cdot z \\x\cdot z \le 0\\0 \le x, 0 \le w - x$$
But we also know that $$w\cdot c = 1 = q_0 + q_1$$, so the two equations are the same. we can reduce it further to just $$x \cdot c = q_0\\x\cdot z \le \min\{0,w\cdot z\}\\0\le x\le w$$
But when $$0 < z$$ and $$q_0 > 0$$, this is impossible.
• Thanks for your insightful answer, it was very helpful. Now, I'd like to generalise the claim to $\mathcal{V}$ not finite. I've posted my question here math.stackexchange.com/questions/4539423/… It involves a bit of probability notions, thanks in advance if you can help.
– TEX
Sep 26, 2022 at 13:51 | 1,513 | 4,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-50 | longest | en | 0.724485 |
https://number.academy/35051 | 1,653,341,499,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00018.warc.gz | 476,227,799 | 11,901 | # Number 35051
Number 35,051 spell 🔊, write in words: thirty-five thousand and fifty-one . Ordinal number 35051th is said 🔊 and write: thirty-five thousand and fifty-first. The meaning of number 35051 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 35051. What is 35051 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 35051.
## What is 35,051 in other units
The decimal (Arabic) number 35051 converted to a Roman number is (X)(X)(X)(V)LI. Roman and decimal number conversions.
#### Weight conversion
35051 kilograms (kg) = 77273.4 pounds (lbs)
35051 pounds (lbs) = 15899.0 kilograms (kg)
#### Length conversion
35051 kilometers (km) equals to 21780 miles (mi).
35051 miles (mi) equals to 56410 kilometers (km).
35051 meters (m) equals to 114996 feet (ft).
35051 feet (ft) equals 10684 meters (m).
35051 centimeters (cm) equals to 13799.6 inches (in).
35051 inches (in) equals to 89029.5 centimeters (cm).
#### Temperature conversion
35051° Fahrenheit (°F) equals to 19455° Celsius (°C)
35051° Celsius (°C) equals to 63123.8° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
35051 seconds equals to 9 hours, 44 minutes, 11 seconds
35051 minutes equals to 3 weeks, 3 days, 8 hours, 11 minutes
### Zip codes 35051
• Zip code 35051 Columbiana, Alabama, Shelby, USA a map
### Codes and images of the number 35051
Number 35051 morse code: ...-- ..... ----- ..... .----
Sign language for number 35051:
Number 35051 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 35051
### Multiplications
#### Multiplication table of 35051
35051 multiplied by two equals 70102 (35051 x 2 = 70102).
35051 multiplied by three equals 105153 (35051 x 3 = 105153).
35051 multiplied by four equals 140204 (35051 x 4 = 140204).
35051 multiplied by five equals 175255 (35051 x 5 = 175255).
35051 multiplied by six equals 210306 (35051 x 6 = 210306).
35051 multiplied by seven equals 245357 (35051 x 7 = 245357).
35051 multiplied by eight equals 280408 (35051 x 8 = 280408).
35051 multiplied by nine equals 315459 (35051 x 9 = 315459).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 35051
Half of 35051 is 17525,5 (35051 / 2 = 17525,5 = 17525 1/2).
One third of 35051 is 11683,6667 (35051 / 3 = 11683,6667 = 11683 2/3).
One quarter of 35051 is 8762,75 (35051 / 4 = 8762,75 = 8762 3/4).
One fifth of 35051 is 7010,2 (35051 / 5 = 7010,2 = 7010 1/5).
One sixth of 35051 is 5841,8333 (35051 / 6 = 5841,8333 = 5841 5/6).
One seventh of 35051 is 5007,2857 (35051 / 7 = 5007,2857 = 5007 2/7).
One eighth of 35051 is 4381,375 (35051 / 8 = 4381,375 = 4381 3/8).
One ninth of 35051 is 3894,5556 (35051 / 9 = 3894,5556 = 3894 5/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
35051
#### Is Prime?
The number 35051 is a prime number. The closest prime numbers are 35027, 35053.
#### Factorization and factors (dividers)
The prime factors of 35051
Prime numbers have no prime factors less than themselves.
The factors of 35051 are 1 , 35051
Total factors 2.
Sum of factors 35052 (1).
#### Prime factor tree
35051 is a prime number.
#### Powers
The second power of 350512 is 1.228.572.601.
The third power of 350513 is 43.062.698.237.651.
#### Roots
The square root √35051 is 187,219123.
The cube root of 335051 is 32,726543.
#### Logarithms
The natural logarithm of No. ln 35051 = loge 35051 = 10,464559.
The logarithm to base 10 of No. log10 35051 = 4,5447.
The Napierian logarithm of No. log1/e 35051 = -10,464559.
### Trigonometric functions
The cosine of 35051 is -0,968723.
The sine of 35051 is -0,248144.
The tangent of 35051 is 0,256156.
### Properties of the number 35051
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 35051 in Computer Science
Code typeCode value
35051 Number of bytes34.2KB
Unix timeUnix time 35051 is equal to Thursday Jan. 1, 1970, 9:44:11 a.m. GMT
IPv4, IPv6Number 35051 internet address in dotted format v4 0.0.136.235, v6 ::88eb
35051 Decimal = 1000100011101011 Binary
35051 Decimal = 1210002012 Ternary
35051 Decimal = 104353 Octal
35051 Decimal = 88EB Hexadecimal (0x88eb hex)
35051 BASE64MzUwNTE=
35051 SHA1f03c4372c744d917eb44319d863f5634b99a4966
35051 SHA224bb24bb1da1cc0f882657e410d55078defa846d2ed092ab6768430824
35051 SHA256f106bca47fb80745ed020eb6c4b74e2baf510e7379c8661b0a0f7c2df9f12812
35051 SHA38411386f647e3ea4eb75fdd4835e217a1a62ac33ce1828d6a7e2f78b156597aba9c53c2758c38f063f0f41d3e614eb635e
More SHA codes related to the number 35051 ...
If you know something interesting about the 35051 number that you did not find on this page, do not hesitate to write us here.
## Numerology 35051
### Character frequency in number 35051
Character (importance) frequency for numerology.
Character: Frequency: 3 1 5 2 0 1 1 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 35051, the numbers 3+5+0+5+1 = 1+4 = 5 are added and the meaning of the number 5 is sought.
## Interesting facts about the number 35051
### Asteroids
• (35051) 1981 ED47 is asteroid number 35051. It was discovered by S. J. Bus from Siding Spring Observatory on 3/2/1981.
## Number 35,051 in other languages
How to say or write the number thirty-five thousand and fifty-one in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 35.051) treinta y cinco mil cincuenta y uno German: 🔊 (Anzahl 35.051) fünfunddreißigtausendeinundfünfzig French: 🔊 (nombre 35 051) trente-cinq mille cinquante et un Portuguese: 🔊 (número 35 051) trinta e cinco mil e cinquenta e um Chinese: 🔊 (数 35 051) 三万五千零五十一 Arabian: 🔊 (عدد 35,051) خمسة و ثلاثون ألفاً و واحد و خمسون Czech: 🔊 (číslo 35 051) třicet pět tisíc padesát jedna Korean: 🔊 (번호 35,051) 삼만 오천오십일 Danish: 🔊 (nummer 35 051) femogtredivetusinde og enoghalvtreds Dutch: 🔊 (nummer 35 051) vijfendertigduizendeenenvijftig Japanese: 🔊 (数 35,051) 三万五千五十一 Indonesian: 🔊 (jumlah 35.051) tiga puluh lima ribu lima puluh satu Italian: 🔊 (numero 35 051) trentacinquemilacinquantuno Norwegian: 🔊 (nummer 35 051) tretti-fem tusen og femti-en Polish: 🔊 (liczba 35 051) trzydzieści pięć tysięcy pięćdziesiąt jeden Russian: 🔊 (номер 35 051) тридцать пять тысяч пятьдесят один Turkish: 🔊 (numara 35,051) otuzbeşbinellibir Thai: 🔊 (จำนวน 35 051) สามหมื่นห้าพันห้าสิบเอ็ด Ukrainian: 🔊 (номер 35 051) тридцять п'ять тисяч п'ятдесят одна Vietnamese: 🔊 (con số 35.051) ba mươi lăm nghìn lẻ năm mươi mốt Other languages ...
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## Comment
If you know something interesting about the number 35051 or any natural number (positive integer) please write us here or on facebook. | 2,419 | 7,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.628938 |
https://crypto.stackexchange.com/questions/99373/practical-check-the-point-is-on-the-curve?noredirect=1 | 1,718,314,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00702.warc.gz | 168,983,223 | 36,366 | Practical check the point is on the Curve [duplicate]
The curve I am using is secp256r1. Its formulae is
$$y^2 == x^3 + a\cdot x + b$$
$$a$$ = 0xffffffff00000001000000000000000000000000fffffffffffffffffffffffc (115792089210356248762697446949407573530086143415290314195533631308867097853948)
$$b$$ = 0x5ac635d8aa3a93e7b3ebbd55769886bc651d06b0cc53b0f63bce3c3e27d2604b (41058363725152142129326129780047268409114441015993725554835256314039467401291)
And I am checking the base point $$G$$:
$$G_x$$ = 0x6b17d1f2e12c4247f8bce6e563a440f277037d812deb33a0f4a13945d898c296 (48439561293906451759052585252797914202762949526041747995844080717082404635286)
$$G_y$$ = 0x4fe342e2fe1a7f9b8ee7eb4a7c0f9e162bce33576b315ececbb6406837bf51f5 (36134250956749795798585127919587881956611106672985015071877198253568414405109)
Calculating left side $$y^2$$ gives me:
1305684092205373533040221077691077339148521389884908815529498583727542773586739078600732747106020956683600164371063053787771205051084393085089418365301881
Calculating right side $$x^3 + a\cdot x + b$$ gives:
113658155427813365024510503555061841058107074695539734801914243855899581676106121216742031186749037217068373713699401633275460693094202620308271598867055040123401752346577561684789671973397929725392419990583281258891711488349384075
Left and right sides are not equal.
What I am doing wrong in my calculations?
• Does this answer your question? Verify that a point belongs to secp256r1 Exactly the same reason. Commented Mar 29, 2022 at 19:17
• @kelalaka, yes if I performed mod with p on both sides and it goes equal. It works with base point and other constant points on the curve. But I got a problem with points calculated with scalar multiplication. I asked the question in another thread. Now digging in into my implementation of scalar multiplication operation to identify what is wrong. Commented Mar 30, 2022 at 3:27
$$y^2 \equiv x^3 + ax + b \pmod p$$
where $$p$$ is the characteristic of the field that P256 uses. When working in this field, we usually understand that we're in $$GF(p)$$ and not $$\mathbb{Z}$$ (and so we don't need to write out the modulus), however it is important that we realize that it's there.
When don't you reduce each side modulus $$p$$ and see if it then works. | 742 | 2,261 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.630479 |
https://www.esaral.com/q/the-coefficient-of-t-4-in-the-expansion-of-48461 | 1,719,354,524,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00648.warc.gz | 663,509,076 | 11,337 | # The coefficient of t^4 in the expansion of
Question:
The coefficient of $\mathrm{t}^{4}$ in the expansion of $\left(\frac{1-t^{6}}{1-t}\right)^{3}$ is
1. 12
2. 15
3. 10
4. 14
Correct Option: , 2
Solution:
$\left(1-t^{6}\right)^{3}(1-t)^{-3}$
$\left(1-t^{18}-3 t^{6}+3 t^{12}\right)(1-t)^{-3}$
$\Rightarrow$ cofficient of $t^{4}$ in $(1-t)^{-3}$ is
${ }^{3+4-1} C_{4}=6 C_{2}=15$ | 169 | 392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.458936 |
https://www.physicsforums.com/threads/horsepower-engine-is-required.262750/ | 1,516,525,672,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890394.46/warc/CC-MAIN-20180121080507-20180121100507-00265.warc.gz | 973,705,493 | 15,291 | # Horsepower engine is required?
1. Oct 8, 2008
### pbumper1
1. The problem statement, all variables and given/known data
A 80kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 23*degrees hill. The skier is pulled a distance x = 250m along the incline and it takes 2.1 minutes to reach the top of the hill.
If the coefficient of kinetic friction between the snow and skis is uk = 0.10, what horsepower engine is required if 30 such skiers (max) are on the rope at one time?
2. Relevant equations
P= w/t=mgy/t
3. The attempt at a solution
80x9.8xsin23*=306
Fnuk=Fr
80x9.8x.1=78.4
306+78.4=390N
390x9.8x250/126s=773 W X30 =3.1x10^4hp
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Can you offer guidance or do you also need help?
Draft saved Draft deleted | 264 | 869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-05 | longest | en | 0.916179 |
https://numberworld.info/201002121001002 | 1,618,463,416,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038083007.51/warc/CC-MAIN-20210415035637-20210415065637-00294.warc.gz | 531,720,603 | 4,029 | # Number 201002121001002
### Properties of number 201002121001002
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
b6cf7405782a
Base 32:
5mptq0au1a
sin(201002121001002)
-0.99472953499328
cos(201002121001002)
-0.10253366379899
tan(201002121001002)
9.701492155234
ln(201002121001002)
32.934336576176
lg(201002121001002)
14.303200640178
sqrt(201002121001002)
14177521.680498
Square(201002121001002)
4.0401852646901E+28
### Number Look Up
201002121001002 (two hundred one trillion two billion one hundred twenty-one million one thousand two) is a very special figure. The cross sum of 201002121001002 is 12. If you factorisate 201002121001002 you will get these result 2 * 3 * 41 * 817081792687. The figure 201002121001002 has 16 divisors ( 1, 2, 3, 6, 41, 82, 123, 246, 817081792687, 1634163585374, 2451245378061, 4902490756122, 33500353500167, 67000707000334, 100501060500501, 201002121001002 ) whith a sum of 411809223514752. 201002121001002 is not a prime number. The number 201002121001002 is not a fibonacci number. 201002121001002 is not a Bell Number. 201002121001002 is not a Catalan Number. The convertion of 201002121001002 to base 2 (Binary) is 101101101100111101110100000001010111100000101010. The convertion of 201002121001002 to base 3 (Ternary) is 222100200121121102002111110010. The convertion of 201002121001002 to base 4 (Quaternary) is 231230331310001113200222. The convertion of 201002121001002 to base 5 (Quintal) is 202321204320434013002. The convertion of 201002121001002 to base 8 (Octal) is 5554756401274052. The convertion of 201002121001002 to base 16 (Hexadecimal) is b6cf7405782a. The convertion of 201002121001002 to base 32 is 5mptq0au1a. The sine of the figure 201002121001002 is -0.99472953499328. The cosine of 201002121001002 is -0.10253366379899. The tangent of the figure 201002121001002 is 9.701492155234. The root of 201002121001002 is 14177521.680498.
If you square 201002121001002 you will get the following result 4.0401852646901E+28. The natural logarithm of 201002121001002 is 32.934336576176 and the decimal logarithm is 14.303200640178. that 201002121001002 is impressive figure! | 816 | 2,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-17 | latest | en | 0.656522 |
https://programmer.help/blogs/implementation-and-application-of-linear-table.html | 1,632,717,639,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00663.warc.gz | 492,243,602 | 6,318 | # Course practice experiment 5: implementation and application of linear table
## Problem description and requirements
(1) Implement the sequence table class and basic operations and test them.
(2) Realize the single linked list class and basic operation and test it. In addition to the basic operations, add inversion, search by value, statistics by value, deletion and other (recursive and non recursive) algorithms for the single linked list class and test them.
(3) Design and implement the ordered single linked list class, and add algorithms such as value insertion and ordered table merging under the ordered table.
(4) Use the single linked list to complete the simple management of daily expenditure.
Enter n expense items (you can choose to import from file), and output all expense items in turn.
Find the minimum, maximum and average consumption of these n expenditure items.
Find out all the expenses of a day by date.
Find out the cost of the project by date and project.
Find out all the expenses of the project according to the project. For example, ask for a total amount of money spent on shoes.
## Outline design
(1) Understanding and secondary generalization of the experimental content
Use the definition of linear list and linked list to realize the basic operation, and realize the daily expenditure management by defining the data structure of nodes.
(2) Give the function list of the system
Basic operations of sequence table class: increase, decrease, clear, etc
Basic operation of single linked list
Inverse of single linked list, search by value, statistics by value, deletion, etc
Algorithms such as inserting by value and merging ordered tables under ordered tables
Simple management of daily expenses
(3) Interface design of program operation (it can be illustrated, for example: the screen prompt appears first, and the user is asked to select the mode of input configuration, 1 input the live unit coordinates from the keyboard, 2... 3... And then the user)
(4) Determine the overall design idea, which data structure to adopt, which classes to design, the functions of various classes, the introduction of class methods, and the description of the relationship between classes
Using the data structure of sequence table and single linked table, a sequence table class and a single linked table class are designed, and then a single linked table class is designed for the simple management of daily expenditure. Each class has basic operation methods: size, empty, clear, full, reverse, retrieve, replace, remove and insert
## detailed design
Inverse of single linked list: void reverse();
Starting from the head node, make the pointer p point to the head node, then make the next of the first head node point to the third node, and the next of the second node point to the head node, so as to complete the sequential exchange of the first two nodes, and then make the head point to the second node to continue the cycle and exchange in turn.
Search by value of single linked list: error_ code list<list_ entry>::search(const list_ entry&x, int &position)
Start from the head node of the single linked list, and then use position to record the position. If the value of a node is equal to the searched value, the searched position will be recorded.
Attached code:
```#include <iostream>
#include <ctime>
#include <random>
#include <algorithm>
using namespace std;
enum Error_code
{
success, fail, rangeerror, overflow, underflow, range_over
};
int max_list = 10000;
template <class List_entry>
class List
{
public:
List()
{
count = 0;
}//Constructor
int size() const;
bool full() const;
bool empty() const;
void traverse(void(*visit)(List_entry&));
Error_code retrieve(int position, List_entry& x) const;
Error_code remove(int position, List_entry& x);
Error_code insert(int position, const List_entry& x);
void printout()
{
for (int i = 0; i < count; i++)
{
cout << entry[i] << " ";
}
cout << endl;
}
void ListSort()//Sequential table sort
{
for (int i = 0; i < count - 1; ++i)
{
int flag = 0;
for (int j = 0; j < count - i - 1; ++j)
{
if (entry[j]>entry[j + 1])
{
swap(entry[j], entry[j + 1]);
flag = 1;
}
}
if (flag == 0)
break;
}
}
protected: //Private member data
int count;
List_entry entry[10000];//The so-called contiguous means that it is stored in an array
};
template <class List_entry>
int List <List_entry>::size() const
{
return count;
}
template <class List_entry>
bool List <List_entry>::full() const
{
if (count == max_list) return true;
else return false;
}
template <class List_entry>
bool List <List_entry>::empty() const
{
if (count == 0) return true;
else return false;
}
template <class List_entry>
Error_code List<List_entry>::insert(int position, const List_entry& x)
{
if (full()) return overflow;//overflow
if (position<0 || position>count)
return range_over; //Insertion point error
for (int i = count - 1; i >= position; i--)
entry[i + 1] = entry[i]; //Move element (back to front)
entry[position] = x; //Put element
count++; //Increase the number of Statistics
return success;
}
template <class List_entry>
Error_code List<List_entry> ::remove(int position, List_entry &x)
{
if (count == 0) return underflow;
if (position < 0 || position >= count) return rangeerror;
x = entry[position];
count--;
while (position < count - 1) {
entry[position] = entry[position + 1];
position++;
}
return success;
}
template <class List_entry>
Error_code List<List_entry> ::retrieve(int position, List_entry &x) const
//If the position value is illegal, an error message is returned. Otherwise, the position element in the table is obtained according to the position value and returned by x.
{
if (position < 0 || position >= count)
return rangeerror;
x = entry[position];
return success;
}
template <class List_entry>
void List<List_entry>::traverse(void(*visit)(List_entry&))
{
for (int i = 0; i<count; i++)
(*visit)(entry[i]);
//Here, the visit function represents the specific actions performed on each table element during the specified traversal
}
Error_code sequential_search(int x, List<int> a, int &count)
{
for (int i = 0; i < a.size(); i++)
{
int m;
a.retrieve(i, m);
count++;
if (m == x)
{
return success;
}
}
count -= a.size();
return fail;
}
Error_code binary_search_nonrecursive(const int &target, const List<int> &x)
{
int data;
int bottom = 0, top = x.size() - 1;
while (bottom < top) {
int mid = (bottom + top) / 2;
x.retrieve(mid, data);
if (target > data)
bottom = mid + 1;
else
top = mid;
}
if (top < bottom) return fail;
else {
x.retrieve(bottom, data);
if (data == target) return success;
else return fail;
}
}
Error_code binary_search_recursive(const List<int> &x, const int &target, int bottom, int top)
{
{
int data;
if (bottom < top) { // List has more than one entry.
int mid = (bottom + top) / 2;
x.retrieve(mid, data);
if (data < target) // Reduce to top half of list.
return binary_search_recursive(x, target, mid + 1, top);
else // Reduce to bottom half of list.
return binary_search_recursive(x, target, bottom, mid);
}
else if (top < bottom)
return fail; // List is empty.
else { // List has exactly one entry.
x.retrieve(bottom, data);
if (data == target)
return success;
else
return fail;
}
}
}
int main()
{
List<int> unordered_list;
int n;
cout << "Please enter the number of integers stored in the linear table(1~9999):";
cin >> n;
for (int i = 0; i < n; i++)
{
int x=rand()%10000;
unordered_list.insert(i, x);
}
cout << "Unordered table generated:";
unordered_list.printout();
int m;
cout << "Please enter the number of duplicate searches:";
cin >> m;
Error_code result;
int count_success=0, count_fail=0, count_compare=0;
clock_t start_t, end_t;
double suc_t=0, fail_t=0;
for (int i = 0; i < m; i++)
{
start_t = clock();
result = sequential_search(rand()%10000, unordered_list, count_compare);
if (result == 0)
{
end_t = clock();
suc_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_success++;
}
if (result == 1)
{
end_t = clock();
fail_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_fail++;
}
}
cout << endl;
cout << "Find sequentially in an unordered table:" << endl;
cout << "When the search is successful, the absolute execution time of the algorithm is" << suc_t / count_success*m << "s" << endl;
cout << "When the search fails, the absolute execution time of the algorithm is" << fail_t / count_fail*m << "s" << endl;
cout << "When the search is successful, the average number of keyword comparisons is" << count_compare / count_success << endl;
cout << endl;
unordered_list.ListSort();
cout << "After sorting, the generated ordered table is:";
unordered_list.printout();
cout << endl;
count_success = 0, count_fail = 0, count_compare = 0;
suc_t = 0, fail_t = 0;
for (int i = 0; i < m; i++)
{
start_t = clock();
result = sequential_search(rand() % 10000, unordered_list, count_compare);
if (result == 0)
{
end_t = clock();
suc_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_success++;
}
else if (result == 1)
{
end_t = clock();
fail_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_fail++;
}
}
cout << "Find sequentially in an ordered table:" << endl;
cout << "When the search is successful, the absolute execution time of the algorithm is" << suc_t / count_success*m << "s" << endl;
cout << "When the search fails, the absolute execution time of the algorithm is" << fail_t / count_fail*m << "s" << endl;
cout << "When the search is successful, the average number of keyword comparisons is" << count_compare / count_success << endl;
cout << endl;
suc_t = 0, fail_t = 0;
for (int i = 0; i < m; i++)
{
start_t = clock();
result = binary_search_nonrecursive(rand() % 10000, unordered_list);
if (result == 0)
{
end_t = clock();
suc_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_success++;
}
else if (result == 1)
{
end_t = clock();
fail_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_fail++;
}
}
cout << "Using non recursive binary lookup in ordered tables:" << endl;
cout << "When the search is successful, the absolute execution time of the algorithm is" << suc_t / count_success*m << "s" << endl;
cout << "When the search fails, the absolute execution time of the algorithm is" << fail_t / count_fail*m << "s" << endl;
cout << endl;
suc_t = 0, fail_t = 0;
for (int i = 0; i < m; i++)
{
start_t = clock();
result = binary_search_recursive(unordered_list, rand() % 10000, 0, unordered_list.size()-1);
if (result == 0)
{
end_t = clock();
suc_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_success++;
}
else if (result == 1)
{
end_t = clock();
fail_t += (double)(end_t - start_t) / CLOCKS_PER_SEC;
count_fail++;
}
}
cout << "Using recursive binary lookup in ordered tables:" << endl;
cout << "When the search is successful, the absolute execution time of the algorithm is" << suc_t / count_success*m << "s" << endl;
cout << "When the search fails, the absolute execution time of the algorithm is" << fail_t / count_fail*m << "s" << endl;
cout << endl;
system("pause");
}
```
## experimental result
Design and provide corresponding test methods and results according to the requirements of the subject. Specific test cases can be given, and each test case can generally be listed:
Test input: None
Test purpose: check whether the basic function is wrong
Actual output:
Test conclusion: passed
## Experimental analysis and discussion
The main problems encountered in the experiment are how to correctly return the error code in the process of realizing the basic functions, how to improve the program, how to correctly define the data structure of the node and how to respond when applying the basic functions of sequence list and linked list.
Posted on Sun, 05 Sep 2021 16:56:26 -0400 by cap97 | 2,922 | 11,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-39 | latest | en | 0.882163 |
https://www.unitconverters.net/length/micron-to-foot.htm | 1,721,909,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00551.warc.gz | 874,713,011 | 4,118 | Home / Length Conversion / Convert Micron to Foot
# Convert Micron to Foot
Please provide values below to convert micron [µ] to foot [ft], or vice versa.
From: micron To: foot
### Micron to Foot Conversion Table
Micron [µ]Foot [ft]
0.01 µ3.2808398950131E-8 ft
0.1 µ3.2808398950131E-7 ft
1 µ3.2808398950131E-6 ft
2 µ6.5616797900262E-6 ft
3 µ9.8425196850394E-6 ft
5 µ1.64042E-5 ft
10 µ3.28084E-5 ft
20 µ6.56168E-5 ft
50 µ0.000164042 ft
100 µ0.000328084 ft
1000 µ0.0032808399 ft
### How to Convert Micron to Foot
1 µ = 3.2808398950131E-6 ft
1 ft = 304800 µ
Example: convert 15 µ to ft:
15 µ = 15 × 3.2808398950131E-6 ft = 4.92126E-5 ft | 268 | 641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.349482 |
https://www.physicsforums.com/threads/recursive-series-equality.189132/ | 1,477,268,570,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719453.9/warc/CC-MAIN-20161020183839-00394-ip-10-171-6-4.ec2.internal.warc.gz | 966,797,418 | 18,225 | # Recursive Series Equality
1. Oct 4, 2007
### ramsey2879
Let there be two recursive series S and R as follows:
$$S_{0} = 0 \quad S_{1} = s \quad S_{n} = 6S_{n-1} - S_{n-2} + p$$
$$R_{0} = 0 \quad R_{1} = t \quad R_{n} = 6R_{n-1} - R_{n-2} + p$$
Prove $$S_{n}*(R_{n+1} + t - p) = R_{n}*(S_{n+1} + s - p)$$
2. Oct 5, 2007
### dodo
Sn is a linear combination of s and p, as Rn is of t and p. In fact, they can be written as
\begin{align*} S_n &= s A_n + p B_n \\ R_n &= t A_n + p B_n \end{align*}
where
\begin{align*} A_n &= 6 A_{n-1} - A_{n-2} & "(Sloane A001109)" \\ B_n &= 6 B_{n-1} - B_{n-2} + 1 \end{align*}
A few things can be said about An and Bn. It is easy to prove by induction that
$$B_n = \sum_{i=1}^{n-1} A_i$$
Thus $B_{n+1} = A_n + B_n$.
Since $B_{n+1}$ is also $6 B_n - B_{n-1} + 1$,
we have $A_n = 5 B_n - B_{n-1} + 1$.
From the above, it is possible to work by induction to prove
$$B_n (B_n - 1) = B_{n+1} B_{n-1}$$
and from here, to arrive by substitution to
$$A_n (B_{n+1} - 1) = B_n (A_{n+1}+1)$$
At this point it is possible to use these results to prove your claim. It is best to rule out first two simpler cases, (1) s=t and (2) p=0, and finally, the case where both s$\neq$t and p$\neq$0.
More specifically, you'll arrive to something like
$$p (s-t) \left( A_n B_{n+1} - A_n - B_n - A_{n+1}B_n \right) = 0$$
Last edited: Oct 5, 2007
3. Oct 5, 2007
### ramsey2879
I also discovered that "p" can be different for S and R so let p1 correspond to S and p2 correspond to R. My identity becomes
$$S_{n}*(R_{n+1}+t -p2) = R_{n}*(S_{n+1} + s - p1)$$
4. Oct 6, 2007
### ramsey2879
Also any integer can be substituted for the 6 in the recursive formula
5. Oct 6, 2007
### dodo
I was wondering about that. Now what about a second integer coefficient (i.e. other than -1) on the n-2 term?
6. Oct 6, 2007
### MathematicalPhysicist
why not look at:
T(n)=S(n)-R(n)=6(S(n-1)-R(n-1))-(S(n-2)-S(n-1))=6T(n-1)-T(n-2)
T(0)=0
T(1)=s-t
solve for T(n), and now you have a connection between S(n) and R(n), then you can write S(n)=R(n)+(solution of T(n)) and then you write each side of the equation with only one of the expressions which is much easier than both S and R, i think it should be easier, did'nt try it myself though.
7. Oct 6, 2007
### ramsey2879
I tried integers from -9 to 9 for the n-2 term multiplier but only -1 worked. I will work more on the theory now. By the way the LaTex graphic for the A(n) and B(n) definition in your first post doesn't generate on my computer. Is there a problem with it?
8. Oct 7, 2007
### dodo
Yes, thanks. It was supposed to be
\begin{align*} A_0 = 0; \ \ \ A_1 = 1; \ \ \ & A_n = 6 A_{n-1} - A_{n-2} & \mbox{ (Sloane A001109) } \\ B_0 = 0; \ \ \ B_1 = 0; \ \ \ & B_n = 6 B_{n-1} - B_{n-2} + 1 \end{align*}
It showed in the preview, but not in the post. I just thought it was some server load issue. And it was incomplete anyway. And my LaTex sucks.
BTW, did you follow the same route for the proof? Mine was too long. Maybe using the closed form of the sequences is faster, but that's beyond me.
Last edited: Oct 7, 2007
9. Oct 8, 2007
### ramsey2879
It seems that a proof is more involved now that we added more variables to the mix
I like the $$A_{n}$$ and $$B_{n}$$ approach but we need to rephase it using x in place of 6.
There are two tables of coefficients for A_n and B_n each are triangular with T{i,j) = T(i-1,j-1) - T(i-1,j) for j > 0. The left hand column, (j=0) alternates between 1 and -1 for A_n and between 1 and 0 for B_n.
Table of coefficients for $$A_{n}$$
1
-1 1
1 -2 1
-1 3 -3 1
1 -4 6 -4 1
....
The powers of x increase from left to right
You use diagonal lines through this table with a slope of 1 down to 2 across
A(1) = 1
A(2) = x
A(3) = -1 + x^2
A(4) = -2x+x^3
A(5) = 1 -3x^2 + x^4
...
Table of coefficients for $$B_{n}$$
1
0 1
1 -1 1
0 2 -2 1
1 -2 4 -3 1
Here there are two adjacent diagonal lines that make up the polynomial in x
B(2) = 1
B(3) = 1 + x
B(4) = x+ x^2
B(5) = -x + x^2 + x^3
...
I got to close now.
Last edited: Oct 8, 2007
10. Oct 9, 2007
### MathematicalPhysicist
have you even tried my suggestion, cause it seems this appraoch gets you nowhere.
p.s
do you know how to solve recursive equations?
11. Oct 9, 2007
### ramsey2879
I thought of that but the solution I knew was for basic equations with the d variable equal to 0. The known solution would be for $$S_{n} = bS_{n-1} - S_{n-2} + d$$ for d = 0.
Now that you mention it I went back to figure how to add a constant to each term of my series to get to a recursive series that has 0 for the d variable.
Thus I found that if I add $$\frac{d}{b-2}$$ to each term of my series I obtain a new series with the same auxiliary equation $$x^2-bx+1 = 0$$ with roots $$r_{1} , r_{2}$$ but with d = 0. Then A and B are found by solving the following set of equations;
$$S_{n} = Ar_{1}^{n} + Br_{2}^{n} - \frac{d}{b-2}$$
Do you agree with this?
Last edited: Oct 9, 2007
12. Oct 10, 2007
### ramsey2879
$$A_{0} = 0 \quad A_{1} = 1 \quad A_{n} = 6 A_{n-1} - A_{n-2}$$
$$B_{0} = B_{1} = 0 \quad B_{n} = 6 B_{n-1} - B_{n-2} + 1$$
I looked this over again and it indeed does prove my identity even for the added variables. Thanks for your effort. PS I simplified the LaTex that didn't take for some reason. I think your LaTex is much better than mine though.
Again I thank you. | 2,035 | 5,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2016-44 | longest | en | 0.82937 |
https://www.phphelp.com/t/using-arrays-and-foreach-statements-solved/9730 | 1,558,660,180,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257481.39/warc/CC-MAIN-20190524004222-20190524030222-00538.warc.gz | 888,442,054 | 4,534 | Using arrays and foreach statements {SOLVED}
#1
Ok to explain what is happening “\$total” is an equation of a percentage that im mulitplying with the “\$value” if the total is greater than 100% it makes the number bigger than 333, 300, 166. I don’t want the values to be any bigger than the array values. I know the below code works.
[php]
<?php \$arkonor1 = array("\$mega" => 333, "\$trit" => 300, "\$zyd" => 166); ?> <?php foreach(\$arkonor1 as \$key1 => \$value1) { \$get = floor(\$value1*\$total); echo ""; } ?>
You Get Gross Station Fee Net \$key1 \$get
[/php]
so i was thinking of adding something like this but was unsure how to do it any suggestions.
[php]
<?php if (\$get >= 333) echo "333"; else echo \$get; ?>
[/php]
my other problem is I need to make the value a variable for other things that i need to do to it. So if the percentage is equal to 97% it take (333*.97) = 323.01 and then make this a variable. Hope i was able to explain it ok to understand. Thank in advance
#2
If you want to use the value in the variable to be no greater than a predetermined value, you could use something like this:
[php]
if (\$variable >= 333) { \$variable = 333; }
echo \$variable;
[/php]
Your last question tells me you’re not a pro in PHP though Once assigned, a variable keeps its value until either:
• unset() is called on the variable
• the variable gets a new value assigned to it
• the script stops running
#3
Thank you for the code it works great for the first number but when the foreach statement loops back to the second value in the statement i need to make that number no highter than 300 and same thing for the 3rd time where that number can’t be higher than 166. any suggestions. Thanks
#4
[php]
\$maxvalues = array(
333,
300,
166
);
for (\$i = 0; \$i < count(\$maxvalues); \$i++) {
if (\$variable >= \$maxvalues[\$i]) { \$variable = \$maxvalues[\$i]; }
}
[/php]
Seriously, this is basic PHP & array knowledge (dare I say, a no-brainer?). You should easily be able to come up with something like this if you can figure out the foreach() loop (or associated arrays for that matter).
#5
Ok not sure what happened or what i miss typed but the only value that is showing up 3 times in a row is 166. Why would this be happening.
[php]
\$arkvalues = array(333, 300, 166);
<?php foreach(\$arkvalues as \$variable) { \$variable1 = floor(\$variable*\$total); for (\$i = 0; \$i < count(\$arkvalues); \$i++) { if (\$variable1 >= \$arkvalues[\$i]) { \$variable1 = \$arkvalues[\$i]; } } echo " \$variable1"; } ?>
[/php]
#6
I was able to solve it with this code but thanks for the help.
[php]
foreach(\$arkvalues as \$var)
{
if (\$total < 1)
\$var = floor(\$var*\$total);
echo “
\$var”;
}
[/php] | 765 | 2,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-22 | latest | en | 0.836059 |
https://mathematica.stackexchange.com/questions/195271/fit-markov-chain | 1,561,047,681,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999261.43/warc/CC-MAIN-20190620145650-20190620171650-00519.warc.gz | 501,381,520 | 34,334 | # Fit Markov Chain
Assume a simple Markov Chain with the transition matrix
n = 9;
matrix = Table[Piecewise[{{-c[i], And[i == j, i < n]}, {c[i], i + 1 == j}, {0,True}}], {i, n}, {j, n}];
matrix // MatrixForm
and the starting point
ini[start_] := Table[If[i == start, 1, 0], {i, n}];
It looks like this
P[ini_] := ContinuousMarkovProcess[ini, matrix];
Graph[P[ini[1]]]
Now we run this Markov chain repeatedly for random times, beginning at random stages.
values = Table[c[i] -> 10 + Mod[i, 3], {i, n - 1}];
simulation[start_, t_] := Normal[RandomFunction[P[ini[start]] /. values, {0, t}]][[1, -1, 2]];
runs = 100;
dat = Table[ Module[{start = RandomInteger[{1, 3}], t = RandomReal[{.1, .4}]}, {t, {start, simulation[start, t]}}], runs];
For each data point, the first element is the time of the simulation and the second elements are the initial and final stages.
dat[[1 ;; 5]]
{{0.22866, {2, 5}}, {0.252981, {2, 4}}, {0.291232, {2, 6}}, {0.370085, {1, 6}}, {0.103945, {1, 2}}}
Is there an efficient way to fit the parameters $$c[i]$$ to this data?
Edit: If this is difficult, is there an easy way to estimate the mean transition time from stage $$i$$ to $$n$$, $$\sum_{i=1}^{n-1}\frac{1}{c[i]}$$? | 414 | 1,213 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-26 | longest | en | 0.623276 |
https://freevideolectures.com/Course/2508/Engineering-Mathematics/48 | 1,529,348,924,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00207.warc.gz | 618,904,083 | 23,081 | # Engineering Mathematics
## The University of New South Wales , Prof.Chris Tisdell
LECTURE48 :
1438 views
### Lecture Description
A basic introduction on how to integrate over curves (line integrals). Several examples are discussed involving scalar functions and vector fields. Such ideas find important applications in engineering and physics.
### Course Description
Contents:
Vector Revision – Intro to curves and vector functions – Limits of vector functions – Calculus of vector functions – Calculus of vector functions tutorial – Vector functions of one variable tutorial – Vector functions tutorial – Intro to functions of two variables – Partial derivatives-2 variable functions: graphs + limits tutorial – Multivariable chain rule and differentiability – Chain rule: partial derivative of \$arctan (y/x)\$ w.r.t. \$x\$ – Chain rule: identity involving partial derivatives – Chain rule & partial derivatives – Partial derivatives and PDEs tutorial – Multivariable chain rule tutorial – Gradient and directional derivative – Gradient of a function – Tutorial on gradient and tangent plane – Directional derivative of \$f(x,y)\$ – Gradient & directional derivative tutorial – Tangent plane approximation and error estimation – Partial derivatives and error estimation – Multivariable Taylor Polynomials – Taylor polynomials: functions of two variables – Differentiation under integral signs: Leibniz rule – Leibniz’ rule: Integration via differentiation under integral sign
Evaluating challenging integrals via differentiation: Leibniz rule – Critical points of functions. Chris Tisdell UNSW Sydney – Second derivative test: two variables. Chris Tisdell UNSW Sydney – How to find critical points of functions – Critical points + 2nd derivative test: Multivariable calculus – Critical points + 2nd derivative test: Multivariable calculus – How to find and classify critical points of functions – Lagrange multipliers – Lagrange multipliers: Extreme values of a function subject to a constraint – Lagrange multipliers example – Lagrange multiplier example: Minimizing a function subject to a constraint – 2nd derivative test, max / min and Lagrange multipliers tutorial – Lagrange multipliers: 2 constraints-Intro to vector fields – What is the divergence – Divergence + Vector fields – Divergence of a vector field: Vector Calculus – What is the curl? Chris Tisdell UNSW Sydney – Curl of a vector field (ex. no.1): Vector Calculus – Line integrals – Integration over curves – Path integral (scalar line integral) from vector calculus
YOU | 516 | 2,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-26 | latest | en | 0.737216 |
https://www.writework.com/essay/exchange | 1,526,951,330,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864572.13/warc/CC-MAIN-20180521235548-20180522015548-00200.warc.gz | 874,646,225 | 8,607 | Exchange
Essay by sumeetbanker April 2004
All organisms need to exchange substances such as food, waste, gases and heat with their surroundings. These substances must diffuse between the organism and the surroundings. The rate at which a substance can diffuse is given by Fick's law:
So rate of exchange of substances depends on the organism's surface area that's in contact with the surroundings. Requirements for materials depends on the volume of the organism, So the ability to meet the requirements depends on the surface area : volume ratio. As organisms get bigger their volume and surface area both get bigger, but volume increases much more than surface area. This can be seen with some simple calculations for different-sized organisms. Although it's innacurate lets assume the organisms are cube shaped to simplify the maths - the overall picture is still the same. The surface area of a cube with length of side L is LxLx6, while the volume is LxLxL.
ORGANISMLENGTHSA (M")VOL. (M")S/A:VOL
bacterium1 mm6 x 10-1210-186,000,000:1
amoeba100 mm6 x 10-810-1260,000:1
fly10 mm6 x 10-410-6600:1
dog1 m6 x 1001006:1
whale100 m6 x 1041060.06:1
So as organisms get bigger their surface area/volume ratio gets smaller. Bacteria are all surface with not much inside, while whales are all insides without much surface. So as organisms become bigger it is more difficult for them to exchange materials with their surroundings.
Organisms also need to exchange heat with their surroundings, and here large animals have an advantage in having a small surface area/volume ratio: they lose less heat than small animals. Large mammals keep warm quite easily and don't need much insulation or heat generation. Small mammals and birds lose their heat very readily, so need a high metabolic rate in order to keep generating heat, as well as thick insulation. So large mammals can feed once every few days... | 443 | 1,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-22 | longest | en | 0.957702 |
https://www.bartleby.com/solution-answer/chapter-141-problem-1lc-essentials-of-statistics-for-the-behavioral-sciences-8th-edition/9781133956570/for-each-of-the-following-indicate-whether-you-would-expect-a-positive-or-a-negative-correlation/73997454-a41d-11e8-9bb5-0ece094302b6 | 1,571,119,564,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00435.warc.gz | 799,753,019 | 38,762 | Chapter 14.1, Problem 1LC
Essentials of Statistics for the B...
8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Chapter
Section
Essentials of Statistics for the B...
8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Textbook Problem
For each of the following, indicate whether you would expect a positive or a negative correlation. a. Model year and price for a used Honda b. IQ and grade point average for high school students c. Daily high temperature and daily energy consumption for 30 winter days in New York City
a.
To determine
Sign of correlation between model year and price for a used Honda.
Explanation
Given info:
Model year and price for a used Honda.
Justification:
Higher model year is an indicator of decency of its bunch by Honda...
b.
To determine
Sign of correlation between IQ and grade point average for high school boys.
c.
To determine
Sign of correlation between daily high temperature and daily energy consumption for 30 days in New York city.
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Evaluate the integral. x2x2+1dx
Calculus (MindTap Course List)
Through (1, 6), parallel to the line x + 2y = 6
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In Exercises 5-8, find the limit. limx02xx+(x)2x
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True or False: converges.
Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th | 402 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-43 | latest | en | 0.823707 |
https://quant.stackexchange.com/tags/money-management/hot | 1,642,930,190,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00616.warc.gz | 532,646,204 | 22,795 | # Tag Info
20
The short answer is that: Maximizing the expected logarithm leads to more wealth almost surely in the long run. In contrast, maximizing expected return can easily lead to going broke almost surely in the long run! Maximizing expected return results in betting everything on your highest expected return investment. Repeatedly doing that over time typically ...
13
Maximizing $E[\log(G)]$ which corresponds to a concave utility function is a subtle way of incorporating risk aversion in the utility. Maximizing $E[G]$ is basically saying that you have linear utility which corresponds to infinite risk appetite because as soon as you have positive expectation you are willing to bet as much capital as possible no matter the ...
9
This problem can be expressed as the original Merton's portfolio problem. Consider wealth process defined by SDE $$d X _ { t } = \frac { X _ { t } \alpha _ { t } } { S _ { t } } d S _ { t } + \frac { X _ { t } \left( 1 - \alpha _ { t } \right) } { S _ { t } ^ { 0 } } d S _ { t } ^ { 0 }$$ where $\alpha_t$ is proportion of the investment in the risky ...
9
The original paper was concerned with optimizing the long run geometric return. In fact, he does not explicitly optimize either $\mathbb{E}(G)$ or $\mathbb{E}(\log(G))$. He also assumes the probabilities are known. The expectation is implicit in his assumption that $$G=\lim_{N\to\infty}\frac{1}{N}\log\left(\frac{V_N}{V_0}\right).$$ He notes that $$V_N=(... 5 This is indeed a very good question! There were (and still are) very hefty debates, where even academic champions like Paul Samuelson were involved! A very good starting point to get some main arguments is the following chapter 4 from the book "Fortunes Formula" by William Poundstone: https://books.google.de/books?id=xz4y3u-qM04C&lpg=PA179&dq=... 3 "Money management" is the art or business of managing money on behalf of an investor (individual or institutions such as pension fund, college endowment, etc.). Money managers usually organize their work by making decisions in a hierarchical fashion on 3 different levels: Strategic Asset Allocation (highest level, few decisions - infrequently ... 3 Hint: you are looking for weights w1,w2,⋯,w15 such that the linear combination of the 15 stocks daily returns is maximally correlated with the S&P500 index returns published by S&P. There are method(s) in statistics that can find these weights. Some of these techniques restrict the weights to be positive and some do not. There is plenty of historical ... 2 ok I found it 🙂and this works for any distribution, not just the normal distribution f^*=\frac μ {σ^2 + μ^2} \approx \frac μ {σ^2} \space if μ\llσ here the steps: https://www.dropbox.com/s/4nqd5yfk2xcuag5/kelly.pdf 2 When it comes to Kelly, I've always liked the explanation at http://www.financialwisdomforum.org/gummy-stuff/kelly-ratio.htm. At this link there are three versions of Kelly explained, if you don't mind the "chatty" style. The third version brings in the standard deviation of wins and losses, which I think is very useful. 2 According to Skiena (link page 21) the Kelly fraction in the case of wins all equal to W and losses all equal to L is:$$f=\frac{pW-qL}{WL}=p/L-q/W where $q=1-p$ and $p$ is the probability of a win. When the wins and losses are random, with average $W$ and $L$ respectively, I am not sure this formula is completely justified. But it might be a good ...
2
Maginn et al. (2007) suggest one of two approaches: optimization and stratified sampling. Optimization. First, you select some factor model which (you believe) best captures major risk factors in your universe. Then you select portfolio of 15 stocks so that this resulting portfolio would minimize expected tracking risk of original portfolio (S&P500 in ...
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I've recently had to do quite a bit of work on position sizing. Leonard C MacLean, Edward O Thorp, and William T Ziemba have written an incredible amount of literature on this. The following text book encompasses an incredibly deep study of the topic on position sizing, different utility functions and so on. From what I can tell the two broad branches of ...
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The underlying assumption to your mu/simga^2 formula is that the pricing process follows geometric Brownian motion, so your returns are therefore symmetric and normal. The existence of a very high positive return implies the possibility of a very low negative return, even if you haven’t realized it yet in the time series you used to calibrate your sigma. If ...
1
Two thoughts that I'm interested in at the moment... A Deep Hedging-stype approach to risk management (eg. https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3355706) Fundamentally all of derivative pricing quant finance is Model-Driven. Why? Probably mostly because it's easier. However, we now have maybe 30 years of high frequency tick data available ...
1
Kelly maximises the geometric return in each period. I believe this is equivalent to maximising expectation of log wealth in the next period. If you maximise the arithmetic return, your expected wealth will be higher. However the actual return you will experience is -100% i.e. you will go bust with probability 1. If you maximise the geometric return, then ...
1
Book with counterpart SwedishAlphaBank, with whom you margin in SEK: USD RUB SEK SEKPnL 0 0 0 0 Buy 100 SEK worth of USD/RUB, meaning buy USD and sell RUB. 100 -100 0 0 With RUB interest rate at 0 (!), USD/RUB moves to 1.1, USD/SEK stays flat at 1 100 -100 0 9.09 Square up back to SEK on USD and RUB: 0 0 9....
Only top voted, non community-wiki answers of a minimum length are eligible | 1,386 | 5,643 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | latest | en | 0.931173 |
https://community.pointhacks.com/t/when-should-i-pay-credit-card-surcharge-how-do-i-calculate-this/18945 | 1,721,732,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00668.warc.gz | 155,084,115 | 5,880 | # When should I pay credit card surcharge? How do I calculate this?
Hi,
I have read the article on ‘Here’s how and why I value my rewards points’ but I’m afraid I still am getting a bit lost. My main concern is more of the ‘earn cost’ and would appreciate some help.
I have 2 Amex cards, one is the old Amex Plat Reserve card (on the flexible points program) which earns the following :
a. 3 points per \$1 spent at restaurants in Australia
b. 2 points per \$1 when you spend on travel and currency purchases
c. 1 point per \$1 spent elsewhere excluding utilities, insurance (except insurances offered by American Express), telecommunications providers and government bodies in Australia where you will earn 0.5 points
My other card is the NAB Amex Plat card (earns direct to Velocity) which earns 1.5 points per \$1 spent everywhere.
I recently paid for a hotel stay where there was a surcharge of 3% but would have earned me 3 points/\$1. Did I pay too much for that ? So basically want to know if there is a simple formula to calculate how much surcharge I am willing to pay depending on the earn points.
Hi aswang,
Let us use your hotel stay and 3% surcharge as an example.
For your Amex card, you could earn 2 MR points per \$ (because of travel multiplier of 2). The 3% surcharge equates to 3 cents per \$.
2 MR points : \$ spent : 3 cents surcharge
3 cents spent : 2 MR points = 1.5c per point
Using Keith’s max value of AMEX point of 2cpp, you should still pay using this card (because it is cheaper than the dearest price you are willing to pay) if the price you are willing to pay for AMEX point matches Keith’s.
1.5cpp < 2cpp (Keith’s value of AMEX)
For your NAB card, you could earn 1.5 velocity points per \$. The 3% surcharge equates to 3 cents per \$.
1.5 vel points : \$ spent : 3 cents surcharge
3 cents spent : 1.5 vel points = 2c per point
Using Keith’s max value of vel point of 1.2cpp, you should NOT pay using this card (because it is higher than the dearest price you are willing to pay) if the price you are willing to pay for vel point matches Keith’s.
2cpp > 1.2cpp (Keith’s value of vel points)
Hope that clarifies things for you.
When you should pay the surcharge all depends on the use you have in mind for your points. The different ways that you use points (e.g. certain flights vs redeeming for gift cards vs redeeming for hotels etc.) provide different value for your points.
It’s a complicated business, but you just need to look at what the award you’re going for is worth both in points and in dollars. From that you can equate roughly what your points are worth in terms of cents, and you can then from that work out if the extra dollars you’ll pay for the surcharge will equate to points which are worth more or less than the surcharge you’re paying.
Hope that’s not too confusing!
Hi w-hiew, thanks for this, very helpful especially with examples. Now at least I can gauge when to use which card especially when there is a surcharge.
Jimmy, thanks as well. I plan to use my points for a travel redemption in Business class as from what I have been reading, that is where you get the biggest bang for your money (except probably First class).
Hi
There is one additional argument in the equation;
Amex plat are reward points and can be transferred to many airlines and don’t expire.
the Nab Amex points go straight to Velocity and are only to be used in the Velocity group, I haven’t checked their expiry, which could even be an extra argument.
Hank | 856 | 3,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-30 | latest | en | 0.938695 |
https://hackaday.io/project/13122-my-dalek-build/log/46287-sensor-integration | 1,720,953,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00321.warc.gz | 253,004,896 | 11,882 | # Sensor Integration
A project log for My Dalek Build
For next Halloween I want a Dalek for "Trick or Exterminate"
## Sensor Integration
Now that I have something that is basically working, I thought it would be time to think about how I am going to integrate four sensors to the ESP-12E.
The ESP-12E does not have eight free lines after allowance for the stepper control and other planned functions.
Generating in software a 40 kHz drive signal through I2C is not possible (too slow).
It seems like an ideal job for a ATTiny (slave) using I2C to communicate with the EPS-12E (master).
I that case having an LM567 and an ATTiny85 would be a bit of over-kill.
Why not simulate the Phased Locked Loop (PLL) in software on the ATTiny85 (it is something I have previously researched).
While I am thinking of it, here is my DIY PLL (hardware design using LS-TTL, the actual PLL is the 74LS02 at the bottom of the schematic (beginning with the 2_FSK probe) - yes it is that simple):
And the PPL decoding a signal:
The actual code to simulate the above is not difficult (which you may find surprising).
Another option would a Discrete Fourier Transform (DFT).
A search of the Internet suggest Goertzel algorithm would be a very good choice (someone has even written an Arduino library!).
Here is the algorithm:
```double goertzelFilter(int samples[], double freq, int N) {
double s_prev = 0.0;
double s_prev2 = 0.0;
double coeff,normalizedfreq,power,s;
int i;
normalizedfreq = freq / SAMPLEFREQUENCY;
coeff = 2*cos(2*M_PI*normalizedfreq);
for (i=0; i<N; i++) {
s = samples[i] + coeff * s_prev - s_prev2;
s_prev2 = s_prev;
s_prev = s;
}
power = s_prev2*s_prev2+s_prev*s_prev-coeff*s_prev*s_prev2;
return sqrt(power)/N;
}```
Note: I modified the return value for rms of signal voltage.
Going through the calculations for an ATTiny85, 25 samples at 125 kHz (K=8) has a centre frequency of 40 kHz and will take 200 us (i.e. 8 pulses). The bin/band-width would be 5 kHz. The 8 bit (0-5v) ADC has a voltage step of about 20 mv.
## On the multiplexing problem
A CMOS 4052 (2x4 analog multiplexer) or equivalent would work well here.
The pre-amp could even be on the MPU side of the multiplexer but then shielding of the inputs would become necessary - perhaps not!
Only one analog pin and three digital (output) pins would be required for four sets of sonar range finders (easier than using four ATTiny85 and I2C).
So interface directly to the EPS-12E is a good option.
Need to check out the EPS-12E ADC.
No, the conversion speed is about 83 us (12kHz) but OP report less in loop{} (about 8kHz).
Found an external SPI 4/8 chanel ADC (MCP3004/8) which looks interesting and fast enough (75 to 200 ksps depending on the voltage).
I need to see if a digital version would work for this application.
## Simulation of Goertzel Algorithm
Wrote a short C program and plotted the results in Excel.
Here is the analog filter response:
Here is the digital (0 or 1) filter response:
Here is the response to random digital (noise):
Testing with low mark/space ratio:
And with high mark/space ratio:
Although you could build a sonar range finder that just tested for any signal at the pin, the code is pretty neat in that it adds additional filtering.
Especially neat is the digital version version of the code.
Should be pretty easy to read a digital pin quickly (one source says 0.527 us on the ESP8266) on any MPU.
## Not quite sure!
Not quite sure how to set up the analog/digital interface.
Here is one idea:
Cascade two amplifiers and offset the output (the input pin of the MPU):
The gain is just less than 60dB:
At 2 mv pp (at the signal generator) the signal is less than the MPU input trip point (assumed to be 1.65v):
At 10 mv pp the signal will trip the MPU input:
At 20mv pp the signal is not that much different:
Now the question is what should the gain be?
The gain depends on the background (signal) noise.
I found this image on the Internet:
This author has used an adaptive threshold but for his set up, a fixed threshold of 200 mv pp would work just as well (after removing transmit pulse).
The HC-SR04 (and others) uses variable gain (i.e. gain steps) with time.
## A high gain bandwidth amplifier
I spent the day designing a new high gain (60dB) high gain band width (60dB at 1MHz) for 3.3v. Here is the result:
You will notice that the gain is controlled by the "external" resistors (i.e. the 2k2 and the 2M2) just like an OpAmp.
The output voltage (~2 v) is controlled by the diode drop.
The Rolloff capacitor depends on the application (between 100pF and 100nF), higher values are required for lower gains (basically the same as an OpAmp compensation capacitor but not as well behaved).
Here is the output for a 2 mv pp input signal at 40 kHz:
The input impedance varies with frequency but at 40kHz is about 400k:
Not bad for 9 components excluding the power supply decoupling (2) and the external components (3).
I wonder if it is stable outside of the simulator?
## Checking the 3.3v logic trip voltages
In the above images I had assumed that the trip voltage was 1.65v (half the voltage supply). In fact the Vih is 1.55v and Vil is 1.4v (~150 mv hysteresis):
(source: https://www.idt.com/document/apn/124-33v-logic-characteristics)
Other 3.3v logic families are similar (the TI datasheet stated 1.5v for Vih and 1.4 for Vil (~100 mv hysteresis)).
## Data Slicer
After a day or so I research I determined the approach take by others is a "Data Slicer":
Here a comparator compares the signal against its average DC value (just ignore the schmitt trigger in the circuit):
The RC constant should have a corner frequency (=1/(2*Pi*R*C)) approximately equal to the "pulse" frequency (=40 kHz).
But how do you deal with noise (suppression)? After a few more days searching I found this circuit where the average DC voltage is offset:
So here is my design:
Threshold settings (i.e. the 100k variable resistor) with a 13 mv pp input signal:
• 0% (100k) ~2 mV input signal threshold
• 50% (50k) ~4 mV input signal threshold
• 80% (20k) ~9 mv input signal threshold
In the above design the transistor pre-amp has an input impedance ~4k and a gain of x18. The data slicer corner frequency is 40 kHz.
## Conclusion
This is a very interesting approach and deserves further attention but since I have worked out why the HC-SR04 does not work and how to fix it. I will leave it for now.
AlanX | 1,653 | 6,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | longest | en | 0.894692 |
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