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https://cs.stackexchange.com/questions/109945/how-to-find-highest-normal-form-of-a-relation-given-its-functional-dependencies/109949
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# How to find highest normal form of a relation, given its functional dependencies
Let's say that I have a relational model, defined as:
$$X(A,B,C,D,E,F,G)$$ with the functional dependencies: $$f_1: \{A,B,C\} \to \{D,E,F,G\}$$ $$f_2: \{A,B\} \to \{C\}$$
Using this information I need to determine the highest normal form (for us it's BCNF) that this relation will go to. It's clear that the relation is already in it's atomic form, therefore it's already in it's first normal form. The to proceed to the second normal form, I found that the candidate key is $$\{A,B\}$$ and since all non prime attributes are fully dependent on the candidate keys, it's in the second normal form. But from this point I'm not sure where to go because all the non prime attributes (excluding $$C$$), $$D,E,F,G$$ are dependent on $$C$$ which is also a non prime attribute.
How can I separate this relation and proceed to the third normal form?
PS - I'm not sure if this is the right place to ask a database question but since the stack exchange for database is for professionals, I thought I should post it here. If someone knows a better stack exchange for this sort of a question, I'll transfer it over to that one.
• Your schema is already in third normal form as well as in Boyce-Codd normal form if the two dependencies are a cover of the dependencies of X. May 27, 2019 at 21:10
## 1 Answer
In your example $$C$$ is derived attribute from $$A, B$$, but none of: $$D, E, F, G$$ is more dependent on $$C$$ than $$A, B$$, or solely functionaly dependent on $$C$$, so there is no transitive dependency. Since you have 2NF it is in 3NF.
For BCNF it is sufficient to be in 3NF and do not have overlapping candidate keys, which you do not have, hence it is in BCNF.
• That's a more efficient way to do this, I did the whole algorithmic procedure and came to BCNF after 20 minutes. Thank you! May 28, 2019 at 19:45
• And since it's in BCNF, it will also comply to 1NF, 2NF and 3NF. Correct? May 28, 2019 at 20:17
• @SkiMask Yes, if it is BCNF then for sure it is 1NF, 2NF and 3NF, but implications in the answer took 2NF for granted, which is not much of a shortcut, but proving from scratch would be a bit longer.
– Evil
May 28, 2019 at 21:09
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# Resolving Vectors
## Introduction
A quantity with magnitude and direction is defined as a vector quantity. The weight of an object, velocity, and acceleration of a vehicle and the force acting on a bridge are some examples of vector quantity. All vector quantities can be split, or resolved, into two components: Horizontal components and vertical components. The combined effect of the components is the same as the original vector. Resolving vectors into horizontal and vertical components is used in the addition and subtraction of vectors and finding the resultant of multiple vectors.
How to Resolve a Vector
Consider the force illustrated in Figure 1. The force vector at a 400 angle can be split, or resolved, into two vectors or components as illustrated in Figure 2. The following equations shall be used to resolve any vector quantity.
Figure 1: Vector Illustration
Figure 2: Resolution of Vector
A. Example: 1
B. Example: 2
C. Summary
• Any vector quantity can be split into horizontal and vertical components. The combined effect of the horizontal and vertical components of the vector quantity is the same as the original vector.
• Horizontal components of a vector quantity can be calculated by the equation
• Vertical components of a vector quantity can be calculated by the equation,
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# You asked: Do Wheel of Fortune winners keep money?
Contents
After all, the object of the game (aside from solving word puzzles) is to reward winners, not punish them. That’s why “Wheel of Fortune” offers options. Behind door number 1: winners are allowed to forfeit the trip. That’s what contestant Timothy Mark decided to do when he won a \$7,000 trip to Antigua in 2018.
## Do Wheel of Fortune losers keep money?
On Wheel, once you solve a puzzle, that money can’t be wiped out by a Bankrupt. There is thus no incentive to stop playing, because you can’t lose that money, and thus it makes sense to go for as high of a score as possible.
## Can you take cash instead of prizes on Wheel of Fortune?
No, cash can’t be substituted for prizes on Wheel of Fortune. Many winners on the show are awarded expensive prizes such as world-round trips, new cars and much more. However, these can’t be swapped out for the monetary equivalent. Contestants also have to pay taxes on their winnings.
THIS IS IMPORTANT: What should you bet on a slot machine?
## How much do Wheel of Fortune winners pay in taxes?
According to Forbes, all the winnings on both the game shows are considered ordinary income. As a result, the winnings are taxed up to 37 percent by the IRS.
## Do you have to take the trip you win on Wheel of Fortune?
In the case of Wheel of Fortune, when you win trips, the show allows you to find less expensive versions of the trips you win, thereby decreasing your overall tax bill. But, if you’ve won any significant amount of cash and/or prizes, that will still leave you with a hefty tax payment at the end of the show.
## What is Vanna White’s salary?
For her work on the show, Vanna white earns roughly \$10 million annually. White’s net worth is impressive as well, amounting to roughly \$70 million.
## Does Wheel of Fortune pay for contestants travel?
Free trips from Wheel of Fortune? Not exactly. Alas, the rumors are true: prizes won on “Wheel of Fortune” are taxed in the state of California. So, if you win a high-priced trip, in a sense, you’re going to have to pay for it.
## How much does it cost to buy a vowel on Wheel of Fortune?
The price of a vowel hasn’t been adjusted for inflation over the course of 30 years. It’s still only \$250.
## How much does it cost to be on Wheel of Fortune?
You have to be at least 18 to be on the regular shows. How much does it cost to be a contestant on Wheel of Fortune? Applying for the show is free, but you must pay for your travel expenses.
THIS IS IMPORTANT: Quick Answer: What Dice does DnD use?
## How long does Wheel of Fortune Secret Santa last?
At the election of the Show, in its sole discretion, the Sweepstakes will be played up to five (5) days per week (Monday to Friday only) the weeks of December 7, 2020 and December 14, 2020 (for two (2) weeks or the “Show Weeks”, collectively, the “Show Period”).
## What game show gives away the most money on average?
As of January 2020, the top three winners in American game show history all earned the majority of their winnings from the quiz show Jeopardy!, which has aired since 1984 and has had no hard earnings limit since 2003.
## Can you take money instead of prizes on Price is Right?
So why don’t The Price Is Right contestants just take the cash value instead of the prizes? Simple: the game doesn’t offer cash value. “There is no cash value option,” explains Aurora’s Blog. “They make it super clear in all of the paperwork – you take exactly what you won, or you take nothing.”
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# Floor
Rectangular floor of living room has a length 5.4 meters and a circumference 17.2 meters. What is its width?
Result
b = 3.2 m
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
## Next similar examples:
1. Hotel
The hotel has a p floors each floor has i rooms from which the third are single and the others are double. Represents the number of beds in hotel.
2. Playground
Rectangular playground is fenced with 38 m long netting. Its width is 7 m. Calculate its length.
3. Cellar
Cellar for storing fruit has a rectangular base with sides 14 m and 7 meters. You should paint sidewall to 2 m. How m2 surface must be painted?
4. Arm-leg
Calculate the length of the base of an isosceles triangle with a circumference 224 cm if the arm length is 68 cm.
5. A rectangle
A rectangle has an area of 36 cm2. What could the length and width of rectangle be?
6. Rectangle 45
The perimeter of a rectangle is 60cm. If the length of the rectangle is 20cm. a)find the width b)find the area.
7. Alley
Alley measured a meters. At the beginning and end are planted poplar. How many we must plant poplars to get the distance between the poplars 15 meters?
8. Sides of triangle
Triangle circumference with two identical sides is 117cm. The third side measures 44cm. How many cms do you measure one of the same sides?
9. Circle - simple
The circumference of a circle is 930 mm. How long in mm is its diameter?
10. Simplify
Simplify the following problem and express as a decimal: 5.68-[5-(2.69+5.65-3.89) /0.5]
11. Expression
Solve for a specified variable: P=a+4b+3c, for a
12. Rectangle
Calculate perimeter of the rectangle with sides a=2.4 m and b=1.9 m.
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Te Kete Ipurangi
Communities
Schools
### Te Kete Ipurangi user options:
Level Four > Statistics
# AO2: Use simple fractions and percentages to describe probabilities.
Children
Resource
## Greedy Pig
This activity provides students with a fun game context in which to practice their addition skills. It also introduces concepts of probability.
Resource
## Beat It
The purpose of this unit is to use games to gather and present data in a systematic way in order to determine the likely outcomes of some everyday events.
Resource
## The Coloured Cube Question
Two red and two blue multi-link cubes are placed in a bag. The player takes out two cubes. If the player gets two red cubes they win. The unit looks at, analyses, and extends, this game of chance.
Resource
Three coins are tossed. A player wins if two heads and a tail come up. The unit looks at, analyses, and extends, this game of chance.
Resource
## Murphy's Law
In this unit, students will explore the commonly held belief that if anything bad can possibly happen it will and at the most inopportune time. They will be encouraged to look at events involving chance and predict the likelihood of certain outcomes by both trialling the event and analysing it theoretically. There is a lot of work in this unit so you might like to spread it over more than one week or do different parts at different times of the year. Some of the simulations are reasonably complicated so it may take students most of a lesson until they have come to grips with the rules involved.
Resource
## Gambling- Who Really Wins?
In this unit we investigate the link between experimental and theoretical probability and learn about short run variability. We then take part in sports betting simulations designed to illustrate how damaging gambling can be.
Resource
## Penny's Pizza
Make a list of possible outcomes as a method of finding probability
Express the outcome as a fraction
Devise and use problem–solving strategies to explore situations mathematically (make an organised list)
Resource
## Wallowing Whales
This is a level 4 statistics activity from the Figure It Out series.
Resource
## What's the Chance?
This is a level 4 statistics activity from the Figure It Out series.
Resource
## Unlucky Lines
This is a level 4 statistics activity from the Figure It Out series.
Resource
## Paper, Scissors, Rock
This is a level 4 statistics activity from the Figure It Out series.
Resource
## Orders
This is a level 4 probability activity from the Figure It Out theme series.
Resource
## Scratch, Text, and Win
This is a level 4 statistics activity from the Figure It Out series.
Resource
## Gaming Choices
This is a level 4 statistics activity from the Figure It Out series.
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# Horse racing betting tutorial
Published в Btc to cad conversion | Октябрь 2, 2012
I quickly have become hooked on betting the constant stream of races that are always going on. It has a tutorial to get you started and a pretty clean interface. At a live racetrack, bettors wager into a pool which is displayed on a tote board. This tote board displays the odds on each of the horses in the race. Bettors. Pick the top two finishers in a race in any order. The bettor wins if the horses finish first and second in any order. The bettor can include. BEST LIVE BETTING SITES
The bettor wins only if the horses finish first, second, third and fourth in the exact order. Superfecta box Pick the top four finishers in a race in any order of finish. The bettor wins if the horses finish first, second, third and fourth in any order. The bettor can include more than four horses in a superfecta box though the cost can be prohibitive.
Superfecta key A superfecta bet using one horse the "key" to finish first in a race with multiple horses finishing second, third and fourth. A bettor can also key a horse to finish second, third or fourth. Here's how it works: Bettor Bob begins by grouping every horse in each race into four tiers: A top horse or horses , B backup horses , C even deeper backup horses and X throwouts. Many times Bob will have only a single A horse -- or just one A and one B horse -- and designate the other horses as Xs.
Then, Bob constructs his wagers. First, he takes all of his A horses in each of the four legs of the Pick 4 and pairs them onto one ticket. This is his "All-A" ticket. He then takes all of his B selections in the first leg of the Pick 4 and pairs them on one ticket with his A selections in the second, third and fourth legs. This is his first "three-A, one-B" ticket.
He repeats this process with the B selections in the second leg pairing them with the A selections in the first, third and fourth legs , the B selections in the third leg pairing them with the A selections in the first, second and fourth legs and the B selections in the fourth leg pairing them with the A selections in the first, second and third legs. He now has four "three-A, one-B" tickets.
If the budget allows, Bob will use the same process above to construct "two-A, two-B" tickets. Daily double or double A bet selecting the winners of two consecutive races. Pick 3 A bet selecting the winners of three consecutive races. Pick 4 A bet selecting the winners of four consecutive races.
Pick 5 A bet selecting the winners of five consecutive races. Like an OTB, simulcast races were tied directly into the tote board and pools at the actual tracks. This allowed horseplayers to receive track odds at any tracks they wished to play. It also meant that the pools for the races were getting larger, meaning that there was more money to be won on individual races.
You could even make the argument that simulcasting is the reason bets like the Pick Six have grown in popularity and offered life changing payouts that can reach millions of dollars. Because the technology for simulcasting was already in place, it was a relatively simple matter to expand that technology to the Internet. Today, a variety of racebooks allow you to place bets from home. Online horse betting works exactly like simulcasting in most cases.
The online pools are tied directly to the pools at the track. A racebook like Twin Spires will offer you the same betting menu and the same odds that you would receive if you were at the racetrack. You can also watch races in real time through live video streaming. A simple click will allow you to switch between various tracks quickly.
If you would like to go back and watch a race replay from an earlier date, you can also do that. Getting Started Betting Horses Online The first step to betting horses online is to create an account with a racebook. EZ Horse Betting recommends that you try one of our affiliate partners. Once there, you will be given the opportunity to create an account. This will require you to submit your personal information and create a username and password.
Once your account is established, you will be permitted to deposit funds through a variety of methods. You can use a variety of credit cards or link your bank account. Withdrawals are available through direct deposit to your deposit method or you can request a check to be delivered by mail. In some cases, fees will apply when you request a check. After your account is created and you have deposited money, the fun of betting on racehorses begins.
How to Bet Horses Online When you log on to your racebook account, you will see a variety of menus that let you make various choices. The first of these will be to choose which track you would like to play. In most racebooks, there will be a drop down menu on the side of the screen which lists all of the tracks that are currently available for wagering. You will only be allowed to select a track that is currently running a race card.
You will be able to see the same video feed that patrons at the racetrack are watching. Just below the video feed you will see a menu of options. This menu is where you will place your bets. Just beside the menu you will be able to see your account balance and how much money you have to bet with. On the menu each horse is listed by the number they wear in the race. If you hover over the number with your mouse or do a long press, you will usually be able to see the name of the horse as well.
Next you will see boxes for different types of bets. There are boxes for win, place, show, exacta, trifecta, superfecta, and other exotics. You select the type of bet you would like to make by simply placing a check mark in the appropriate box. You will also place a check in a box signifying the amount you would like to bet.
You can bet any amount you like, however. Once you have selected the bet type and the amount of money you would like to bet, the final step is to choose the horse or horses that you will use in your wager. Online Horse Betting is Intuitive You will discover that the horseracing platforms used by most racebooks are very intuitive and take only a few moments to master.
## That horse racing betting type chart pity, that
### SIGNAL FOREX AKURAT PROFIT PASTI
The program is loaded with information to assist them in making that decision. The problem is that while the best horse may win, more often it is another who gets his picture taken in the winner's circle. Were it otherwise Triple Crown winners Kentucky Derby, Preakness, and Belmont Stakes would be common rather than being an almost extinct species. There are many reasons for this phenomena. They have to do with form, tactics, and other matters that those new to racing don't need to understand in order to have a fun day at the races and having fun is the most important thing.
Of course, collecting money is much more fun than tearing up tickets so betting a few winners would be very nice. As long as we don't get to greedy that really isn't very difficult. The first thing that we need to do is admit that we don't know enough about the mysterious "art form" called handicapping to try to decide which horse to bet on our own.
We need help, and fortunately help is available in terms of the "tote board". The "win" odds it displays for each horse are based upon how much money has been bet on them. As such, it represents the weighted hopes and fears of all those playing the race. Since they are more experienced than us, the "tote board" is a logical pace to find the help we need. We could just let the "tote board" do our handicapping and bet the favorite in every race.
In all likelihood that would let us cash some tickets. The favorite is usually the most likely horse to win, it does make mistakes. All of the sudden, bettors were flocking to place bets at even the most obscure of race tracks and within weeks, tracks across the country were breaking records for handle week-by-week and month-by-month. Things have since slowed, but interest in betting the ponies remains good and FoxSports1 or FS2 have racing on the weekends.
Whatever horse or horses you choose, you can pick one or more to finish 1st, 2nd, or 3rd. You should know the favorite wins about a third of the time. Give yourself the best chance to win. How do you go about picking horses? When first learning and not at a track, getting the Daily Racing Form is a great place to start. The amount of information can be overwhelming, but they have guides that show you what to look for.
Primarily, you are looking for how this horse has performed in the past and who the jockey on each horse is. For a horse, you might see — Overall, that would be a pretty good horse to bet on. If you saw this , this would definitely not be a horse to consider to win or place.
Each track will have a jockey standings report. There you will find how many mounts or rides he had this year at that track and how many times he finished in the money An average jockey can take a favorite and prevent him from crossing the finish line first.
### Horse racing betting tutorial gamba osaka vs hiroshima betting experts
How to Bet on Horses - Horse Racing Betting Explained
#### It is filled with the loud voices of people placing bets and cheering for their horses to win.
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## Thank for best forex strategy for beginners opinion
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# CBSE Class 12 Physics Notes Electrostatics – Electric potential
## CBSE Class 12 Physics Notes Electrostatics – What Is Electric Potential?
Physics Notes for Class 12: An electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any kind of acceleration. That reference point may be Earth or a point at Infinity, although any point beyond the influence of the electric field charge can be used as the reference point.
According to classical electrostatics, an electric potential is a scalar quantity denoted by V, equal to the electric potential energy of any charged particle at any location (measured in joules) divided by the charge of that particle (measured in coulombs). By dividing out the charge on the particle a quotient is obtained that is a property of the electric field itself.
The electric potential at the point specified will be the sum of the electric potential from the
left charge (VL) and the electric potential from the right charge (VR).
V = VL + VR
V = k(+2q)/√((3a)2 +(a)2) + k(-q)/√(a)2+(a)2
V = 2kq/√10a – kq/√2a
V = ( 2/√10 -1/√2)kq/a
V =-0.0746 kq/a
Notice that since the electric potential is a scalar, calculating the electric potential is often much easier than calculating the electric field.
The electric field and the electric potential are not two independent fields. They are two independent ways of conceptualizing the effect that an electric charge has on the space surrounding it. Just as problems in mechanics can be analyzed using a force-approach or an energy-approach, problems dealing with electrical phenomenon can be analyzed by focusing on the electric field or on the electric potential.
Additionally, just as it is sometimes necessary in mechanics to transfer between force and energy representations, it is sometimes necessary to transfer between the electric field and electric potential representations. The relationship between the two fields can be understood by examining the expression of work, which relates force to transfer of energy.
For sample papers and NCERT Solution Physics Class 12th visit www.takshilalearning.com
Electric Potential due to point charge:
The electric potential created by a point charge Q, at a distance r from the charge (relative to the potential at infinity), can be shown to be
VE = 1*Q/4Пε0r.
where ε0 is the dielectric constant (permittivity of vacuum). This is known as the Coulomb potential.
The electric potential due to a system of point charges is equal to the sum of the point charges’ individual potentials. This fact simplifies calculations significantly since the addition of potential (scalar) fields is much easier than the addition of the electric (vector) fields.
The equation given above for the electric potential (and all the equations used here) are in the forms required by SI units. In some other (less common) systems of units, such as CGS-Gaussian, many of these equations would be altered.
Watch and understand the concept of Electrical Potential with best animated videos, click Physics Class 12 for details.
UNITS FOR ELECTRIC POTENTIAL:
The SI derived unit of electric potential is the volt (in honor of Alessandro Volta), which is why a difference in electric potential between two points is known as voltage. Older units are rarely used today. Variants of the centimetre gram second system of units included a number of different units for electric potential, including the abvolt and the statvolt.
For best study material, solutions, question bank, keynotes, practice papers of Class 12 Physics, visit our site www.takshilalearning.com. There we offer all kind of offline and online course which will help you for better preparations.
For more information, videos lectures, study material, sample papers, Physics notes for Class 12, register with Takshila Learning.
For more details related to all other topics which come in all kind of bank exams, you can visit our site www.takshilalearning.com. We there provide all kind of online and offline courses as per student convenience for better preparation.
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# Question: How Many Cups Is 200 Grams Of Rice?
## How many cups is 600g of rice?
1 cup rice uncooked = 7 oz / 200 g = 600 g ( 5 cups / 21 oz in weight) cooked [2] (Will serve 5 people)..
## What does 1 serving of rice look like?
The proper portion size for one serving of rice is 1/2-cup cooked, which is about the size of a cupcake wrapper.
## How much rice do you get from 1 cup?
Rice Conversions & Equivalents1 cup uncooked white riceyields3 cups of cooked white rice1 cup brown whole grain riceyields4 cups of cooked rice1 cup long grain riceyields3 cups cooked rice1 cup dry pre-cooked instant riceyields2 cups cooked rice1 cup uncooked wild riceyields3 cups cooked wild rice8 more rows•Nov 8, 2016
## What is 2 cups of rice in grams?
US cup to Gram Conversion Chart – Cooked riceUS cups to grams of Cooked rice1 US cup=250 grams2 US cups=500 grams4 US cups=1000 grams5 US cups=1250 grams19 more rows
## How much is 100g rice when cooked?
100 g rice uncooked = 1/2 cup (3.5 oz in weight) uncooked = 300 g ( 2 1/2 cups / 10.5 oz in weight) cooked [2] (Will serve 2 people) 1 cup rice uncooked = 7 oz / 200 g = 600 g ( 5 cups / 21 oz in weight) cooked [2] (Will serve 5 people).
## How many grams is a cup of rice?
180 gramsA cup of raw, uncooked, rice will weigh about 180 grams. A cup of cooked rice, that’s been boiled or steamed, will weigh about 200 grams. Why this is interesting is when you look at factors such as calorie count for each cup. A cup of raw rice has about 685 calories whereas a cup of cooked rice has 240 calories.
## How many cups is 200 grams of rice flour?
200 grams of rice flour = 1 US cup + 4 tablespoons of rice flour. How many US customary cups in 180 grams of rice flour? 180 grams of rice flour = 1 US cup + 2 tablespoons of rice flour.
## How many cups is 100 grams of rice flour?
0.63One – 100 grams portion of white rice flour converted to US cup equals to 0.63 us cup.
## What is a cup in grams?
How many grams in 1 cup? The number of grams in a cup varies based upon the ingredient because the cup is a unit of volume and the gram is a unit of weight. For flour, 1 cup is equal to around 125g. For sugar, 1 cup is equal to around 200g.
## How many cups is 100g?
White Sugar (Granulated)CupsGramsOunces1/3 cup67 g2.37 oz1/2 cup100 g3.55 oz2/3 cup134 g4.73 oz3/4 cup150 g5.3 oz3 more rows•Nov 19, 2020
## What is 200g of pasta in cups?
2US cup to Gram Conversion Chart – Dry pastaUS cups to grams of Dry pasta1 US cup=100 grams2 US cups=200 grams4 US cups=400 grams5 US cups=500 grams19 more rows
## How do I measure 200 grams of rice?
200 grams of rice = 8 rounded tablespoons of rice + 1 rounded teaspoon of rice.
## Does 1 cup equal 200 grams?
Check out the Cups to Grams conversion chart below….How to convert 200 grams to cups?IngredientGranulated SugarMeasurement in Grams200 gramsEquivalence in Cups (US)1 cupApproximately1 cup7 more columns
## How much is a 100 grams of rice?
The answer is: The change of 1 100g ( – 100 grams portion ) unit in a white medium rice measure equals = into 0.50 cup ( Metric cup ) as in the equivalent measure and for the same white medium rice type.
## How many cups is 700 grams of rice?
3 12Results. 700 grams of rice equals to 3 12 ( ~ 3 1/2 ) US cups.
## How many cups is 250 grams of rice?
Results. 250 grams of rice equals to 1.3 ( ~ 1 1/4 ) US cups. (*) or more precisely 1.2505185910445 US cups.
## How many cups is 250 grams flour?
BAKING MEASUREMENT CONVERSIONSAmount OuncesGrams UnitsCups / Units4-ounces flour125g1 – Cup8 ounces flour250g2 – Cups4 ounces oatmeal124g1-Cup Scant14 more rows
## What is 250 grams to cups?
g to cup conversion table:10 grams = 0.04 cup210 grams = 0.84 cup700 grams = 2.8 cups30 grams = 0.12 cup230 grams = 0.92 cup900 grams = 3.6 cups40 grams = 0.16 cup240 grams = 0.96 cup1000 grams = 4 cups50 grams = 0.2 cup250 grams = 1 cup1100 grams = 4.4 cups60 grams = 0.24 cup260 grams = 1.04 cup1200 grams = 4.8 cups13 more rows
## How many cups is 250 grams of uncooked rice?
250 grams of uncooked rice equals to 1.4 ( ~ 1 1/4 ) US cups. (*) or more precisely 1.3512636949266 US cups.
## How many cups of rice is 100g?
Round short rice weight volume chart:CupGramOunce1/450g1.76 oz1/366.7g2.35 oz1/2100g3.53 oz5 more rows
## How much is 100 grams of meat?
One hundred grams of meat, including beef, pork, poultry or seafood, is equal to 3.5274 ounces or 0.22046 pounds. The physical size of a 100-gram portion of whole meat, such as a steak or chicken breast, is comparable to a deck of playing cards.
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Factor p
Author Message
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Factor p [#permalink] 08 Feb 2004, 07:44
257.
This is DS problem. Could anyone try it and show the workings?
Does the integer k have a factor p such that 1 < p < k?
1. k> 4!
2. 13! + 2 ≤ K ≤ 13! + 13
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A) k > 4! means k > 24 let K be 25 which has factors 1, 5 and 25
1 < 5 < 25
But 29 has factors 1 and 29 and does not satisfy the given condition
So A is insufficient.
B) 13! has all the integer mutiples from 1 and 13
13!+2 is divisible by 2
13!+3 is divisible by 3 and so on
13!+12 is divisible by 12
For all of these 1 < n(from 2 to 12) < 13!+n
B is suffiient.
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anand.. could u pls.. tell me how to crack the sencond sentence.. is any easy or diff way.. what u did is good but a bit confusing..
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shubhangi
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I wont be surprised if the answer is D. The stem of the problem is confusing. But I am pretty sure that B is the answer.
1) k>4! involves infinite numbers. This set holds primes and non primes.
Only primes do not satisfy the condition 1<p<k
2) Now 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 = X
This huge number is divisible by any integer between 1 and 13.
The second condition says that the number K is in between
13!+2 and 13!+13. So what are the possible numbers
13*12*11*10*9*8*7*6*5*4*3*2*1 + 3 divisible by p=3 1<3<(X+3)
13*12*11*10*9*8*7*6*5*4*3*2*1 + 4 divisible by p=4 1<4<(X+4)
----
----
13*12*11*10*9*8*7*6*5*4*3*2*1 + 12 divisible by p=3 1<12<(X+12)
So any number you take it has a factor > 1 and < that number.
Hence B is sufficient
I hope I made it clear.
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# The Excel SKEW Function
Related Function:
SKEW.P Function
Skewness
The Skew of a data set is a measurement of the asymmetry of the distribution about the mean.
• A skew of zero indicates perfect symmetry;
• A positive skew indicates that more values lie below the mean and the distribution has a 'tail' which extends towards the higher values;
• A negative skew indicates that more values lie above the mean and the distribution has a 'tail' which extends towards the lower values.
Examples of positive and negative skewed distributions are shown in the charts below:
Positive Skew Negative Skew
## Function Description
The Excel SKEW function calculates the skewness of the distribution of a supplied set of values.
The syntax of the function is:
SKEW( number1, [number2], ... )
Where the number arguments provide a minimum of 3 numeric values that make up the data set.
In the latest versions of Excel (Excel 2007 and later), you can input up to 255 number arguments to the Skew function, but in Excel 2003, the function can only accept up to 30 arguments. However, each number argument can be an individual value or an array of values.
## Skew Function Example
A B
1 1
2 1
3 2
4 2
5 2
6 2
7 3
8 3
9 3
10 4
11 4
12 5
13 6
14 7
15 8
Column A of the above spreadsheet on the right shows 15 data values. The distribution of these values is shown in the chart below:
The skewness of the values in the spreadsheet can be calculated using the Excel Skew Function as follows:
=SKEW( A1:A15 )
This gives the result 0.863378312, indicating that the data set has a positive skew.
For further details and examples of the Excel Skew function, see the Microsoft Office website.
## Skew Function Errors
If you get an error from the Excel Skew function this is likely to be one of the following:
Common Errors
#DIV/0! - Occurs if either: Fewer than 3 data values have been supplied to the function; The sample standard deviation of the supplied data values is zero. #VALUE! - Occurs if any of the supplied number arguments that are supplied directly to the Skew function are not recognised as numeric values.
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24 terms by kauders
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Electricity
The interaction between electric charges. THE MOVEMENT OF ELECTRONS CAUSED BY UNEQUAL AMOUNTS OF PROTONS AND ELECTRONS. THE MOVEMENT OF ELECTRONS.
Static electricity
*A buildup of charges on an object
*Can be done by running and transferring protons or electrons to another object and creating a charge.
*Build up of charges, discharges, and releases electrons
*Discharges can be small
*Example is static cling--built up charges have a charge and cling to other objects
Static discharge
The loss of static electricity as electric charges transfer from one object to another
Electric circuit
A complete, unbroken path through which electric charges can flow
Voltage
*The difference in electric potential energy between two places in a circuit
*Amount of force to push or move an electric current
*Measured in volt (V)
*Voltage source-a device to push an electric source
Electric current
*The continuous flow of electric charges through a material
*Rate which charges pass through a given point
*Takes path of least resistance, measured in Amps
*Example is an electrical device
Ohm's Law
*The law that states that resistance is equal to voltage divided by current
*Resistance=voltage (v)/current (amps or A)
**Increased voltage does not affect the resistance because if you double voltage, the current will double
Resistance
*The measurement of how difficult it is for charges to flow through a material
*The greater resistance, the less current
*Measured in Ohm
Series circuit
*An electric charge with a single path-only one path for current to travel
*Any break causes bulb to go out
*Resistance increases
Parallel circuit
*An electrical circuit with multiple paths-has more than one path for current to travel
*If breaks doesn't mean won't work
*Resistance decreases
Conductor
*A material where charges can easily flow
*Contain loosely bonded electrons that carry the charge and create an electric charge
Amp
*The unit for the rate of current.
*The number of amps describes the amount of charge flowing past a given point each second
Direct current
*Current consisting of charges that flow in only one direction in a circuit
*Known as DC current
*Examples are battery and motor
Alternating current
*Current that can flow in both directions
*Current consisting of charges that move back and forth in a current
*Known as AC
Balloon and hair
lightning
Examples of conductor
metals, water, acid
Insulator
*Material where charges cannot easily flow
*Electrons bound tightly to atoms and do not move easily
*Used to stop the flow of charges
Examples of insulators
rubber, glass, sand, plastic, wood
equation for voltage
voltage=current*resistance
Ohms
Ways to get an electric charge
1)Friction--the transfer of charge from one object to another by rubbing; electrons move from one object to another when rubbing
2)Conduction-The transfer of charge when electrons move from a charged object to another object by direct contact; objects touch another and electrons transfer; electrons transfer to the positive charged object
3)Induction-The transfer of electrons from one part of an object to another part, caused by the electric field of another object, without the two objects touching; electric field attracts or repels electrons
Example of voltage source
Battery, generator
Factors that determine resistance
1) Composition of Material used
Insulators-don't let electrons flow. Higher reistance so charges don't move easily so less current
Conductors-electrons flow. Lower resistance so charges move easily so more current
2) Length of Materials
Longer-more resistance so less current
Shorter-less resistance so more current
3) Diameter-
Thinner-more resistance so less current
Thicker-less resistance so more current
4)Temperature
Hotter or higher temp-more resistance so less current
Cooler or lower temperature-less resistance so more current
Example:
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# Assignment | computer science | University of north texas at denton
CSCE 5300 Introduction to Big data and Data Science
ICE-3
Lesson Description: Overview of Hadoop and Map Reduce Paradigm. The Lesson focuses on
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In class exercise
1. Matrix Multiplication in Map Reduce
Suppose we have a i x j matrix M, whose element in row i and column j will be denoted and
a j x k matrix N whose element in row j and column k is donated by then the product P = MN
will be i x k matrix P whose element in row i and column k will be donated by ,
where = .
1. Create a Map-Reduce Program to perform the task of matrix multiplication
Reference:
https://lendap.wordpress.com/2015/02/16/matrix-multiplication-with-mapreduce/
2. Breadth First Search using Map Reduce
3. Depth First Search using Map Reduce
4. Apply Map reduce problem using K-Means Clustering Technique. A view
point of the such algorithms are presented in the screenshot.
Convert this into code and use right dataset to implement this scenario.
Marks will be distributed between logic, implementation and UI
Programming elements:
Source Code:
Given in canvas.
ICE Submission Guidelines
1. ICE Submission is individual.
2. ICE code has to be properly commented.
3. The documentation should include the screenshots of your code/results with explanation.
4. Provide the explanation of the dataset/exercise as per your understanding.
5. The similarity score for your document should be less than 15%.
6. All you need to do is submit the source code (properly commented) and documentation
(.pdf/.doc) with explanation and screenshot of source code having input logic and output
results.
7. Submission after the deadline is considered as late submission.
Pages (550 words)
Approximate price: -
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# Difficult integral?
Hello all,
I'm trying to get the integral with respect to x of the following expression:
(a / x + b(y) * x ^ c) ^ d,
where a,c,d are positive constants, and b is a function of some variable y.
I'm not sure what I'm doing wrong. Can you help me?
I got a solution in Mathematica (some hypergeometric function), but Mathematica is a software that I'm trying not to use anymore.
Best,
Alejandro
edit retag close merge delete
Sort by ยป oldest newest most voted
Probably Maxima just doesn't know this integral. They do get some hypergeometric functions as a result of certain integrals, but it is not as strong in HG functions as Mma. I do get some results if I assume c>1, like
sage: integrate((a/x+x^c)^3,x)
3*a^2*x^(c - 1)/(c - 1) - 1/2*a^3/x^2 + 3/2*a*x^(2*c)/c + x^(3*c + 1)/(3*c + 1)
but Maxima asks redundant and other questions on this a bunch.
(%i8) display2d:false;
(%o8) false
(%i9) integrate((a/x+b*x^c)^3,x);
Is c-1 zero or nonzero?
nonzero;
Is 3*c+1 zero or nonzero?
nonzero;
Is c zero or nonzero?
nonzero;
(%o9) b^3*x^(3*c+1)/(3*c+1)+3*a*b^2*x^(2*c)/(2*c)+3*a^2*b*x^(c-1)/(c-1)-a^3
/(2
*x^2)
And once you replace the exponent with a d or some noninteger exponent, it doesn't seem to know what to do.
more
In that case, i should stick to Mathematica, I guess...
( 2012-03-09 11:53:52 +0200 )edit
Perhaps. This is really one of the only areas where Sage is at a significant disadvantage vis-a-vis the competitors in terms of functionality. Since HG functions are so ubiquitous, I'm surprised Maxima can't handle this one, but there you have it. I've asked on the Maxima list, because it *can* do them for specific positive integers d (even largish ones), but not a general such d.
( 2012-03-09 14:33:18 +0200 )edit
This result is similar to that from W-alpha:
sage: integrate((a/x+b*x^c)^d,x,algorithm='mathematica_free')
-(x^c*b + a/x)^d*x*hypergeometric2f1(-(d - 1)/(c + 1), -d, -(d - 1)/(c + 1) + 1,
-b*x^(c + 1)/a)/((d - 1)*(b*x^(c + 1)/a + 1)^d)
more
Oh, I see, i didn't know the mathematica_free algorithm. If i understand correctly, Sage is taking the result from the online integrator, isn't it? The next question is if i can work further with that result. For example, evaluate it, etc.
( 2012-03-09 12:02:24 +0200 )edit
I tried to use the mathematica_free algorithm. There is an additional problem: the constant b is indeed a function some variable y. It is of course a constant, but it is undetermined when integrating the function.
So, Sage returns:
NotImplementedError: Mathematica online integrator can only handle single letter variables.
Alejandro
more
You can make the substitution after integration:
sage: var('a b c d y')
sage: ii=integrate((a/x+b*x^c)^d,x,algorithm='mathematica_free')
sage: b=function('b',y)
sage: ii(b=b(y))
-(x^c*b(y) + a/x)^d*x*hypergeometric2f1(-(d - 1)/(c + 1), -d, -(d - 1)/(c + 1) + 1,
-x^(c + 1)*b(y)/a)/((d - 1)*(x^(c + 1)*b(y)/a + 1)^d)
but I don't think that sage understands hypergeometric2f1
more
Exactly, Sage doesn't recognize it. So, i'm completely stuck :S
( 2012-03-09 14:08:49 +0200 )edit
See http://trac.sagemath.org/sage_trac/ticket/2516. We would *really* welcome any assistance with that; as with many things in Sage, the infrastructure has been there for a long time, the hard part is "wrapping" it so that it plays well with the rest of Sage - and even here, the hard part means "the tedious part".
( 2012-03-09 14:39:39 +0200 )edit
Also, http://ask.sagemath.org/question/1168/how-can-one-use-maxima-kummer-confluent-functions has several ideas which you might find helpful. In particular, you should be able to use the templates referred to from (for instance) the beta function to create a "custom" HG function for your own use; mpmath will certainly evaluate it as well as needed, though it sounds like that is not your use case.
( 2012-03-09 14:41:28 +0200 )edit
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# Monday's maths - 8/6/20
Morning, I hope you had a lovely weekend! For maths today, I would like you to have a go at estimating answers and checking them using the inverse...
You guys are really good at using the column method for both addition and subtraction, however it is important to know if you have made a mistake or not. It is easy to leave it to an adult to check, but to be more independent (or to use one of our school values - Autonomous) it is better if you can check yourself.
Estimating is where you get an idea of what your answer should be close to.
Inverse is where you use the opposite to check. So, if you did an addition calculation, you would use subtraction to check as this is the opposite.
Watch this video clip then try to answer the questions attached:
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Frequent Visitor
## Graph : Lines per category and Average line overall
Hi,
Is there a way to add automatically on average line per month on a graph?
I have in this example a cost per site per month (5 sites ) and i would like to add a line with the average monthly cost
1 ACCEPTED SOLUTION
Community Support
Hi @miguesnet ,
I have created a sample for your reference, please check the following steps as below.
1. Create a calculated table as below and make it related to our fact table.
``Table 2 = UNION(VALUES('Table (2)'[category]),{"avg"})``
2. Then we can achieve that by a measure.
``````Measure =
IF (
MAX ( 'Table 2'[category] ) = "avg",
CALCULATE (
AVERAGE ( 'Table (2)'[value] ),
ALLEXCEPT ( 'Table (2)', 'Table (2)'[date] )
),
SUM ( 'Table (2)'[value] )
)
``````
For more details, please check the pbix as attached.
Community Support Team _ Frank
If this post helps, then please consider Accept it as the solution to help the others find it more quickly.
4 REPLIES 4
Super User
You can have like using Analytics . But in this case you need a measure like
Calculate([Measure],All(Site[Site]))
if you need more help make me @
Frequent Visitor
Hi,
I created a new measure as you describe but i can not combine both measures on the same graph. It must be possible
Community Support
Hi @miguesnet ,
I have created a sample for your reference, please check the following steps as below.
1. Create a calculated table as below and make it related to our fact table.
``Table 2 = UNION(VALUES('Table (2)'[category]),{"avg"})``
2. Then we can achieve that by a measure.
``````Measure =
IF (
MAX ( 'Table 2'[category] ) = "avg",
CALCULATE (
AVERAGE ( 'Table (2)'[value] ),
ALLEXCEPT ( 'Table (2)', 'Table (2)'[date] )
),
SUM ( 'Table (2)'[value] )
)
``````
For more details, please check the pbix as attached.
Community Support Team _ Frank
If this post helps, then please consider Accept it as the solution to help the others find it more quickly.
Super User
March Release has a multi-axis line chart. And I remember that bar -line was only splitting bar. Check and try
https://powerbi.microsoft.com/en-us/blog/power-bi-desktop-march-2020-feature-summary/
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# Determine whether the following functions are even or odd or neither even nor odd: x+x^(2)
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 17-5-2021
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,
Watch 1000+ concepts & tricky questions explained!
Text Solution
neither even nor odd
Solution :
NA
Transcript
TimeTranscript
00:00 - 00:59student question is determine whether the following functions are even or odd on either even not what are the function is expert access by this problem when a function is said to be a lot when value of a pocket is equals to minus x value of a Pop minus plus and when function is event event when a perfect is equals to minus ok no problem now the given function is a perfect which is X + x square ok next I am going to find a 4 - 84 - X is equals to minus x + - x square x square considered
01:00 - 01:59Peppa Pig is not equal to EP of -1 both are not equal and also and also a for pig is not = 2 - EP - ok student and FX is neither had not even ok thank you a student
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Community Trekker
## How can I simulate trial execution model for virtual DoE?
I want to create a computer emulator for a learning purpose. It will emulate a binary outcome and use several Yes/No factors as an input (lets say A B C).
The output will be the % of successes.
The output when all factors are at "-" is 5% of successes.
When
A at "+" 5.5% (0.5% gain)
B at "+" 6.5% (1.5% gain)
C at "+" 5% (0% gain)
A and B at "+" (AB interaction) 7.5% (2.5% gain)
So I want to implement it in a software as a trial execution model with the list of random trials and outputs(depending on levels of factors A-B-C).
Example of software behavior:
Order of trial ABC Output
1st trial + + + Fail
2nd trial + - + Fail
3rd trial + - - Success
.....
10000 trial + - - Fail
I want it to look realistic so that not all the time the combination of A+ B- C- will result to "Success" because the process is not physical, it is a human response.
What is the general steps of constructing such an algorithm?
Can it be done in JMP?
In addition I am considering such tools as Excel, R, simulation software like Anylogic, Promodel is needed.
Big thanks!
Learning DOE
1 ACCEPTED SOLUTION
Accepted Solutions
Staff
## Re: How can I simulate trial execution model for virtual DoE?
This could probably be done more elegantly, but a quick "brute force approach" would be to enter the random formulas into a JMP data table. I made the A, B, and C columns a random integer column between 0 and 1 (I chose 0 instead of -1 for ease). Then for the Result column, I just entered a string of "If" conditions with the various probabilities from a binomial distribution with the appropriate probability of success. Fairly simple in this case since there are only 8 possible combinations of A, B, and C. For a larger problem I would try to be more elegant. The number of rows in your table will be the number of random trials you wish to have. For this simple case I just had 10 rows.
``````New Table( "Untitled",
New Column( "A", Numeric, "Nominal", Format( "Best", 12 ),
Formula( Random Integer( 0, 1 ) ) ),
New Column( "B", Numeric, "Nominal", Format( "Best", 12 ),
Formula( Random Integer( 0, 1 ) ) ),
New Column( "C", Numeric, "Nominal", Format( "Best", 12 ),
Formula( Random Integer( 0, 1 ) ) ),
New Column( "Success?", Numeric, "Nominal", Format( "Best", 12 ),
Formula(
If(
:A == 0 & :B == 0 & :C == 0, Random Binomial( 1, 0.05 ),
:A == 0 & :B == 0 & :C == 1, Random Binomial( 1, 0.05 ),
:A == 0 & :B == 1 & :C == 0, Random Binomial( 1, 0.065 ),
:A == 0 & :B == 1 & :C == 1, Random Binomial( 1, 0.065 ),
:A == 1 & :B == 0 & :C == 0, Random Binomial( 1, 0.055 ),
:A == 1 & :B == 0 & :C == 1, Random Binomial( 1, 0.055 ),
:A == 1 & :B == 1 & :C == 0, Random Binomial( 1, 0.075 ),
:A == 1 & :B == 1 & :C == 1, Random Binomial( 1, 0.075 ),
Random Binomial( 1, 0.05 ) ) ) ) )``````
Dan Obermiller
7 REPLIES 7
Staff
## Re: How can I simulate trial execution model for virtual DoE?
This could probably be done more elegantly, but a quick "brute force approach" would be to enter the random formulas into a JMP data table. I made the A, B, and C columns a random integer column between 0 and 1 (I chose 0 instead of -1 for ease). Then for the Result column, I just entered a string of "If" conditions with the various probabilities from a binomial distribution with the appropriate probability of success. Fairly simple in this case since there are only 8 possible combinations of A, B, and C. For a larger problem I would try to be more elegant. The number of rows in your table will be the number of random trials you wish to have. For this simple case I just had 10 rows.
``````New Table( "Untitled",
New Column( "A", Numeric, "Nominal", Format( "Best", 12 ),
Formula( Random Integer( 0, 1 ) ) ),
New Column( "B", Numeric, "Nominal", Format( "Best", 12 ),
Formula( Random Integer( 0, 1 ) ) ),
New Column( "C", Numeric, "Nominal", Format( "Best", 12 ),
Formula( Random Integer( 0, 1 ) ) ),
New Column( "Success?", Numeric, "Nominal", Format( "Best", 12 ),
Formula(
If(
:A == 0 & :B == 0 & :C == 0, Random Binomial( 1, 0.05 ),
:A == 0 & :B == 0 & :C == 1, Random Binomial( 1, 0.05 ),
:A == 0 & :B == 1 & :C == 0, Random Binomial( 1, 0.065 ),
:A == 0 & :B == 1 & :C == 1, Random Binomial( 1, 0.065 ),
:A == 1 & :B == 0 & :C == 0, Random Binomial( 1, 0.055 ),
:A == 1 & :B == 0 & :C == 1, Random Binomial( 1, 0.055 ),
:A == 1 & :B == 1 & :C == 0, Random Binomial( 1, 0.075 ),
:A == 1 & :B == 1 & :C == 1, Random Binomial( 1, 0.075 ),
Random Binomial( 1, 0.05 ) ) ) ) )``````
Dan Obermiller
Community Trekker
## Re: How can I simulate trial execution model for virtual DoE?
Thank you very very much for this script! That is what I was seaching for.
Could you tell me please the reason why when I generated 10000 rows the Nominal Logistic regression doesnt show that A and AB are significant effects?
Learning DOE
Highlighted
Staff
## Re: How can I simulate trial execution model for virtual DoE?
Of course every simulation is different, but you are looking for very small effect sizes. With 10,000 trials, you will have approximately 5000 trials with A at 1, and 5000 trials with A at 0. Of those 5000 at 0, you expect a 5% "success" rate which is about 250 observations. With 5000 at 1, you expect 5.5% success rate, which is about 275 observations. Only a 25 observation difference out of the 10,000 total. That's a small difference. Add the random noise to this and you can see that the difference might actually be even smaller.
Categorical data typically requires larger sample sizes. That coupled with looking for very small effects leads to non-significance. Stated in more statistical terms, you have very low power to detect your desired effect sizes.
Dan Obermiller
Community Trekker
## Re: How can I simulate trial execution model for virtual DoE?
Many thanks for clarification!
I've done calculation of minimal sample size so it could detect the difference of 0,5 %. However, sometimes the analysis of the simulated results doesn't detect AB interaction as significant with 64000 trials.
Have I done the calculations right?
Learning DOE
Staff
## Re: How can I simulate trial execution model for virtual DoE?
You are fitting models that are more complex than this sample size calculator was intended to do. There are extra degrees of freedom being used to estimate other terms.
I would suggest you go back to the formulas for each of the 8 different scenarios (combinations of A, B, and C) to make sure that you are using the proper probabilities. You had only provided four of the eight situations so I made some assumptions about the other four cases. You may wish to check those to make sure everything is as you expected it to be. I am guessing you are expecting the interaction effect to be a 2.5% gain, but the interaction term is NOT a 2.5% gain. You only get a 2.5% gain when both A and B are positive as you indicated. The interaction also includes what happens for the other 3 combinations of A and B.
As an example: A and B both positive: probability is 7.5%
A and B both zero: probability is 5%. So when they are "the same sign" gives an average probability of 6.25%
A zero, B positive: probability is 6.5%
A positive, B zero: probability is 5.5% for an average probability of 6%.
Therefore, the interaction effect is actually 0.25% which is really small (and pretty close to your A*B interaction estimate in your analysis).
Dan Obermiller
Community Trekker
## Re: How can I simulate trial execution model for virtual DoE?
Big thanks for all your help with simulation!
Could you help me please with modifying this script or suggesting a new approach for this task?
Now I want to simulate a definitive screening design. There are 18 factors.
Effect sizes:
Factor % of Successes A 1,31% B 1,42% D 1,58% E 1,64% F 1,46% H 1,39% K 1,34% M 1,37% DE 1,67% FK 1,54%
When all factors at low "0" level success rate is 1.2%.
Here is the exact design that I want to simulate:
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 0 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 1 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 1 1 0 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 1 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 1 0 0 1 0 0 0 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 1 1 1 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 0 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 0 0 1 1 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 1 0 1 1 1 1 0 1 0 0 1 0 0 1 0 1 1 1 0 0
Learning DOE
Community Trekker
## Re: How can I simulate trial execution model for virtual DoE?
Sorry, maybe I should create a new topic.
However, is it possible to accomplish my goal by using Simulate Response script that is built in JMP?
Learning DOE
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# Re: st: Looking up values in a 2 dimensional table
From Phil Schumm To statalist@hsphsun2.harvard.edu Subject Re: st: Looking up values in a 2 dimensional table Date Thu, 17 Apr 2008 20:30:39 -0500
```On Apr 17, 2008, at 5:26 PM, Mike Lacy wrote:
```
A colleague of mine has a data file with family income and family size for a large sample, with real family income for each of a series of years, in the wide format, e.g., something like:
famid FamSize1985 FamSize1987 FamSize1991 ... Inc1985 inc1987 Inc1991 ...
with family sizes and family incomes recorded for about 20 different year, not at fixed intervals.
He needs to create a poverty status indicator corresponding to each year, based on a 2 dimensional table giving poverty thresholds for each value of year and family size. All the ways I can imagine doing this seem relatively clumsy. (Among other things, I thought about ways of doing this with a matrix but they would require using the value of the family size variable as an index into the matrix, which is beyond my ken.) I also considered something involving reshape and merge, but that seemed awkward as well.
```
On Apr 17, 2008, at 7:58 PM, David Kantor wrote:
```
First, you probably mean Poverty Guideline.
Poverty Guideline is a simple, rough calculation based on family size, income, and year.
Poverty Threshold is a more complex calculation that considers number of children and number of elderly in addition to the other factors. It is the "officially correct" measure, but is more complicated to compute.
Those are U.S. government standards. (USDA, I believe.)
I have ado files to calculate both of these. But for the threshold, it works for 1997 & 1998 only.
For the guideline, it does 1982-2007.
It sounds like David's program might be the easiest solution in this case. However, to address the more general programming question you raised, in terms of both efficiency and simplicity, the reshape -> merge option wouldn't be a bad choice. Specifically, you would
1) reshape the data into long format so that your variables are famid, year, famsize, and income
2) reshape the lookup table (also long) to yield the variables year, famsize, and threshold
3) merge the table onto the data using the combination year and famsize as the merge variables
4) calculate poverty status indicator as
gen byte poverty = (income < threshold)
5) reshape back to wide, if necessary
To implement a general function returning poverty status for a given year and family size, you could load the table of thresholds into a matrix, and then create two additional vectors (one for year and one for family size) that would permit you to lookup the corresponding threshold for a given year and family size. However, this would require a bit of Mata programming, and wouldn't be any faster for the task at hand (I don't believe).
-- Phil
*
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* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
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http://quadibloc.com/maps/mco0302.htm
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[Next] [Up] [Previous]
# The Simple Conic
The simple conic projection was once a very popular projection in atlases, particularly for maps of countries like Canada:
as this map shows, having a standard parallel of 55 degrees North latitude, appropriate at least for the most populated portions of Canada.
As the diagram below illustrates:
the cone contacting the standard parallel, when flattened to a sheet of paper, contains the standard parallel as an arc with radius cot(lat_sp), which is cos(lat_sp)/sin(lat_sp). Meanwhile, the standard parallel itself on the globe has radius cos(lat_sp), and its scale is true on the projection.
Hence, on a conic projection, the meridians project out from the center of curvature of the parallels at angles equal to the longitude multiplied by sin(lat_sp). Each parallel in the simple conic is a circle with radius cot(lat_sp)+(lat-lat_sp), where lat and lat_sp are expressed in radians.
Unlike the conic conformal, when this projection has two standard parallels instead of one, that does change the appearance of the projection, leading to areas between the two parallels being stretched.
So, given lat_sp1 and lat_sp2, where lat_sp1 is higher, the radius of the standard parallel at lat_sp1 is defined by
``` cos(lat_sp1)
----------------------------- * (lat_sp1 - lat_sp2)
cos(lat_sp2) - cos(lat_sp1)
```
and angles from the center of the parallels are reduced by a factor of cos(lat_sp1), the radius of that parallel over the globe, divided by this radius.
Here is another map of Canada, this time drawn with standard parallels at 50 degrees North and 65 degrees North, for comparison:
The dimple at the top of the map is the North Pole; it would be invalid to show places north of the North Pole on the map, as they don't exist.
[Next] [Up] [Previous]
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http://www.geeksforgeeks.org/find-the-maximum-element-in-an-array-which-is-first-increasing-and-then-decreasing/
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# Find the maximum element in an array which is first increasing and then decreasing
Given an array of integers which is initially increasing and then decreasing, find the maximum value in the array.
Input: arr[] = {8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1}
Output: 500
Input: arr[] = {1, 3, 50, 10, 9, 7, 6}
Output: 50
Corner case (No decreasing part)
Input: arr[] = {10, 20, 30, 40, 50}
Output: 50
Corner case (No increasing part)
Input: arr[] = {120, 100, 80, 20, 0}
Output: 120
## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Method 1 (Linear Search)
We can traverse the array and keep track of maximum and element. And finally return the maximum element.
## C
#include <stdio.h>
int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The maximum element is %d", findMaximum(arr, 0, n-1));
getchar();
return 0;
}
## Java
// java program to find maximum
// element
class Main
{
// function to find the
// maximum element
static int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low; i <= high; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
// main function
public static void main (String[] args)
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = arr.length;
System.out.println("The maximum element is "+
findMaximum(arr, 0, n-1));
}
}
Output:
The maximum element is 70
Time Complexity: O(n)
Method 2 (Binary Search)
We can modify the standard Binary Search algorithm for the given type of arrays.
i) If the mid element is greater than both of its adjacent elements, then mid is the maximum.
ii) If mid element is greater than its next element and smaller than the previous element then maximum lies on left side of mid. Example array: {3, 50, 10, 9, 7, 6}
iii) If mid element is smaller than its next element and greater than the previous element then maximum lies on right side of mid. Example array: {2, 4, 6, 8, 10, 3, 1}
## C
#include <stdio.h>
int findMaximum(int arr[], int low, int high)
{
/* Base Case: Only one element is present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and first is greater then
the first element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and second is greater then
the second element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* If we reach a point where arr[mid] is greater than both of
its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next element and smaller than the previous
element then maximum lies on left side of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
else // when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1]
return findMaximum(arr, mid + 1, high);
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 3, 50, 10, 9, 7, 6};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The maximum element is %d", findMaximum(arr, 0, n-1));
getchar();
return 0;
}
## Java
// java program to find maximum
// element
class Main
{
// function to find the
// maximum element
static int findMaximum(int arr[], int low, int high)
{
/* Base Case: Only one element is
present in arr[low..high]*/
if (low == high)
return arr[low];
/* If there are two elements and
first is greater then the first
element is maximum */
if ((high == low + 1) && arr[low] >= arr[high])
return arr[low];
/* If there are two elements and
second is greater then the second
element is maximum */
if ((high == low + 1) && arr[low] < arr[high])
return arr[high];
/*low + (high - low)/2;*/
int mid = (low + high)/2;
/* If we reach a point where arr[mid]
is greater than both of its adjacent
elements arr[mid-1] and arr[mid+1],
then arr[mid] is the maximum element*/
if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
return arr[mid];
/* If arr[mid] is greater than the next
element and smaller than the previous
element then maximum lies on left side
of mid */
if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
return findMaximum(arr, low, mid-1);
else
return findMaximum(arr, mid + 1, high);
}
// main function
public static void main (String[] args)
{
int arr[] = {1, 3, 50, 10, 9, 7, 6};
int n = arr.length;
System.out.println("The maximum element is "+
findMaximum(arr, 0, n-1));
}
}
Output:
The maximum element is 50
Time Complexity: O(Logn)
This method works only for distinct numbers. For example, it will not work for an array like {0, 1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 5, 3, 3, 2, 2, 1, 1}.
# GATE CS Corner Company Wise Coding Practice
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.4 Average Difficulty : 2.4/5.0
Based on 110 vote(s)
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| 23,416,009
| 7,141
|
# ThreeSpaceCategory RΒΆ
The category ThreeSpaceCategory is used for creating three dimensional objects using functions for defining points, curves, polygons, constructs and the subspaces containing them.
=: (%, %) -> Boolean
from BasicType
~=: (%, %) -> Boolean
from BasicType
check: % -> %
check(s) returns lllpt, list of lists of lists of point information about the ThreeSpace s.
closedCurve?: % -> Boolean
closedCurve?(s) returns true if the ThreeSpace s contains a single closed curve component, i.e. the first element of the curve is also the last element, or false otherwise.
closedCurve: % -> List Point R
closedCurve(s) checks to see if the ThreeSpace, s, is composed of a single closed curve component defined by a list of points in which the first point is also the last point, all of which are from the domain PointDomain(m, R) and if so, returns the list of points. An error is signaled otherwise.
closedCurve: (%, List List R) -> %
closedCurve(s, [[lr0], [lr1], ..., [lrn], [lr0]]) adds a closed curve component defined by a list of points lr0 through lrn, which are lists of elements from the domain PointDomain(m, R), where R is the Ring over which the point elements are defined and m is the dimension of the points, in which the last element of the list of points contains a copy of the first element list, lr0. The closed curve is added to the ThreeSpace, s.
closedCurve: (%, List Point R) -> %
closedCurve(s, [p0, p1, ..., pn, p0]) adds a closed curve component which is a list of points defined by the first element p0 through the last element pn and back to the first element p0 again, to the ThreeSpace s.
closedCurve: List Point R -> %
closedCurve(lp) sets a list of points defined by the first element of lp through the last element of lp and back to the first elelment again and returns a ThreeSpace whose component is the closed curve defined by lp.
coerce: % -> OutputForm
coerce(s) returns the ThreeSpace s to Output format.
components: % -> List %
components(s) takes the ThreeSpace s, and creates a list containing a unique ThreeSpace for each single component of s. If s has no components defined, the list returned is empty.
composite: List % -> %
composite([s1, s2, ..., sn]) will create a new ThreeSpace that is a union of all the components from each ThreeSpace in the parameter list, grouped as a composite.
composites: % -> List %
composites(s) takes the ThreeSpace s, and creates a list containing a unique ThreeSpace for each single composite of s. If s has no composites defined (composites need to be explicitly created), the list returned is empty. Note that not all the components need to be part of a composite.
copy: % -> %
copy(s) returns a new ThreeSpace that is an exact copy of s.
create3Space: () -> %
create3Space() creates a ThreeSpace object capable of holding point, curve, mesh components and any combination.
create3Space: SubSpace(3, R) -> %
create3Space(s) creates a ThreeSpace object containing objects pre-defined within some SubSpace s.
curve?: % -> Boolean
curve?(s) queries whether the ThreeSpace, s, is a curve, i.e. has one component, a list of list of points, and returns true if it is, or false otherwise.
curve: % -> List Point R
curve(s) checks to see if the ThreeSpace, s, is composed of a single curve defined by a list of points and if so, returns the curve, i.e. list of points. An error is signaled otherwise.
curve: (%, List List R) -> %
curve(s, [[p0], [p1], ..., [pn]]) adds a space curve which is a list of points p0 through pn defined by lists of elements from the domain PointDomain(m, R), where R is the Ring over which the point elements are defined and m is the dimension of the points, to the ThreeSpace s.
curve: (%, List Point R) -> %
curve(s, [p0, p1, ..., pn]) adds a space curve component defined by a list of points p0 through pn, to the ThreeSpace s.
curve: List Point R -> %
curve([p0, p1, p2, ..., pn]) creates a space curve defined by the list of points p0 through pn, and returns the ThreeSpace whose component is the curve.
enterPointData: (%, List Point R) -> NonNegativeInteger
enterPointData(s, [p0, p1, ..., pn]) adds a list of points from p0 through pn to the ThreeSpace, s, and returns the index, to the starting point of the list.
hash: % -> SingleInteger
from SetCategory
hashUpdate!: (HashState, %) -> HashState
from SetCategory
latex: % -> String
from SetCategory
lllip: % -> List List List NonNegativeInteger
lllip(s) checks to see if the ThreeSpace, s, is composed of a list of components, which are lists of curves, which are lists of indices to points, and if so, returns the list of lists of lists; An error is signaled otherwise.
lllp: % -> List List List Point R
lllp(s) checks to see if the ThreeSpace, s, is composed of a list of components, which are lists of curves, which are lists of points, and if so, returns the list of lists of lists; An error is signaled otherwise.
llprop: % -> List List SubSpaceComponentProperty
llprop(s) checks to see if the ThreeSpace, s, is composed of a list of curves which are lists of the subspace component properties of the curves, and if so, returns the list of lists; An error is signaled otherwise.
lp: % -> List Point R
lp(s) returns the list of points component which the ThreeSpace, s, contains; these points are used by reference, i.e. the component holds indices referring to the points rather than the points themselves. This allows for sharing of the points.
lprop: % -> List SubSpaceComponentProperty
lprop(s) checks to see if the ThreeSpace, s, is composed of a list of subspace component properties, and if so, returns the list; An error is signaled otherwise.
merge: (%, %) -> %
merge(s1, s2) will create a new ThreeSpace that has the components of s1 and s2; Groupings of components into composites are maintained.
merge: List % -> %
merge([s1, s2, ..., sn]) will create a new ThreeSpace that has the components of all the ones in the list; Groupings of components into composites are maintained.
mesh?: % -> Boolean
mesh?(s) returns true if the ThreeSpace s is composed of one component, a mesh comprising a list of curves which are lists of points, or returns false if otherwise
mesh: % -> List List Point R
mesh(s) checks to see if the ThreeSpace, s, is composed of a single surface component defined by a list curves which contain lists of points, and if so, returns the list of lists of points; An error is signaled otherwise.
mesh: (%, List List List R, Boolean, Boolean) -> %
mesh(s, [ [[r10]..., [r1m]], [[r20]..., [r2m]], ..., [[rn0]..., [rnm]] ], close1, close2) adds a surface component to the ThreeSpace s, which is defined over a rectangular domain of size WxH where W is the number of lists of points from the domain PointDomain(R) and H is the number of elements in each of those lists; the booleans close1 and close2 indicate how the surface is to be closed: if close1 is true this means that each individual list (a curve) is to be closed (i.e. the last point of the list is to be connected to the first point); if close2 is true, this means that the boundary at one end of the surface is to be connected to the boundary at the other end (the boundaries are defined as the first list of points (curve) and the last list of points (curve)).
mesh: (%, List List List R, List SubSpaceComponentProperty, SubSpaceComponentProperty) -> %
mesh(s, [ [[r10]..., [r1m]], [[r20]..., [r2m]], ..., [[rn0]..., [rnm]] ], [props], prop) adds a surface component to the ThreeSpace s, which is defined over a rectangular domain of size WxH where W is the number of lists of points from the domain PointDomain(R) and H is the number of elements in each of those lists; lprops is the list of the subspace component properties for each curve list, and prop is the subspace component property by which the points are defined.
mesh: (%, List List Point R, Boolean, Boolean) -> %
mesh(s, [[p0], [p1], ..., [pn]], close1, close2) adds a surface component to the ThreeSpace, which is defined over a list of curves, in which each of these curves is a list of points. The boolean arguments close1 and close2 indicate how the surface is to be closed. Argument close1 equal true means that each individual list (a curve) is to be closed, i.e. the last point of the list is to be connected to the first point. Argument close2 equal true means that the boundary at one end of the surface is to be connected to the boundary at the other end, i.e. the boundaries are defined as the first list of points (curve) and the last list of points (curve).
mesh: (%, List List Point R, List SubSpaceComponentProperty, SubSpaceComponentProperty) -> %
mesh(s, [[p0], [p1], ..., [pn]], [props], prop) adds a surface component, defined over a list curves which contains lists of points, to the ThreeSpace s; props is a list which contains the subspace component properties for each surface parameter, and prop is the subspace component property by which the points are defined.
mesh: (List List Point R, Boolean, Boolean) -> %
mesh([[p0], [p1], ..., [pn]], close1, close2) creates a surface defined over a list of curves, p0 through pn, which are lists of points; the booleans close1 and close2 indicate how the surface is to be closed: close1 set to true means that each individual list (a curve) is to be closed (that is, the last point of the list is to be connected to the first point); close2 set to true means that the boundary at one end of the surface is to be connected to the boundary at the other end (the boundaries are defined as the first list of points (curve) and the last list of points (curve)); the ThreeSpace containing this surface is returned.
mesh: List List Point R -> %
mesh([[p0], [p1], ..., [pn]]) creates a surface defined by a list of curves which are lists, p0 through pn, of points, and returns a ThreeSpace whose component is the surface.
modifyPointData: (%, NonNegativeInteger, Point R) -> %
modifyPointData(s, i, p) changes the point at the indexed location i in the ThreeSpace, s, to that of point p. This is useful for making changes to a point which has been transformed.
numberOfComponents: % -> NonNegativeInteger
numberOfComponents(s) returns the number of distinct object components in the indicated ThreeSpace, s, such as points, curves, polygons, and constructs.
numberOfComposites: % -> NonNegativeInteger
numberOfComposites(s) returns the number of supercomponents, or composites, in the ThreeSpace, s; Composites are arbitrary groupings of otherwise distinct and unrelated components; A ThreeSpace need not have any composites defined at all and, outside of the requirement that no component can belong to more than one composite at a time, the definition and interpretation of composites are unrestricted.
objects: % -> Record(points: NonNegativeInteger, curves: NonNegativeInteger, polygons: NonNegativeInteger, constructs: NonNegativeInteger)
objects(s) returns the ThreeSpace, s, in the form of a 3D object record containing information on the number of points, curves, polygons and constructs comprising the ThreeSpace..
point?: % -> Boolean
point?(s) queries whether the ThreeSpace, s, is composed of a single component which is a point and returns the boolean result.
point: % -> Point R
point(s) checks to see if the ThreeSpace, s, is composed of only a single point and if so, returns the point. An error is signaled otherwise.
point: (%, List R) -> %
point(s, [x, y, z]) adds a point component defined by a list of elements which are from the PointDomain(R) to the ThreeSpace, s, where R is the Ring over which the point elements are defined.
point: (%, NonNegativeInteger) -> %
point(s, i) adds a point component which is placed into a component list of the ThreeSpace, s, at the index given by i.
point: (%, Point R) -> %
point(s, p) adds a point component defined by the point, p, specified as a list from List(R), to the ThreeSpace, s, where R is the Ring over which the point is defined.
point: Point R -> %
point(p) returns a ThreeSpace object which is composed of one component, the point p.
polygon?: % -> Boolean
polygon?(s) returns true if the ThreeSpace s contains a single polygon component, or false otherwise.
polygon: % -> List Point R
polygon(s) checks to see if the ThreeSpace, s, is composed of a single polygon component defined by a list of points, and if so, returns the list of points; An error is signaled otherwise.
polygon: (%, List List R) -> %
polygon(s, [[r0], [r1], ..., [rn]]) adds a polygon component defined by a list of points r0 through rn, which are lists of elements from the domain PointDomain(m, R) to the ThreeSpace s, where m is the dimension of the points and R is the Ring over which the points are defined.
polygon: (%, List Point R) -> %
polygon(s, [p0, p1, ..., pn]) adds a polygon component defined by a list of points, p0 throught pn, to the ThreeSpace s.
polygon: List Point R -> %
polygon([p0, p1, ..., pn]) creates a polygon defined by a list of points, p0 through pn, and returns a ThreeSpace whose component is the polygon.
subspace: % -> SubSpace(3, R)
subspace(s) returns the SubSpace which holds all the point information in the ThreeSpace, s.
BasicType
SetCategory
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# Chapter 2: Diffraction of Light¶
## Example 2.2.1, Page number 2-10¶
In [5]:
import math
#Variable declaration
theta = 30 #angle(degrees)
n = 1
lamda = 6500*10**-8 #wavelength(cm)
#Calculations
a = (n*lamda)/math.sin(theta*math.pi/180)
#Result
print "a =",a/1e-4,"*10^-4 cm"
a = 1.3 *10^-4 cm
## Example 2.2.2, Page number 2-10¶
In [10]:
import math
#Variable declaration
a = 6*10**-4 #width of slit(cm)
n = 1 #first order
lamda = 6000*10**-8 #wavelength(cm)
#Calculations
def deg_to_dms(deg):
d = int(deg)
md = abs(deg - d) * 60
m = int(md)
sd = (md - m) * 60
sd=round(sd,2)
return [d, m, sd]
theta = (math.asin((n*lamda)/a))*180/math.pi
d = 2*theta #angular seperation
#Result
print "Angular seperation between the 1st order minima is",deg_to_dms(d)
Angular seperation between the 1st order minima is [11, 28, 42.03]
## Example 2.2.3, Page number 2-11¶
In [11]:
import math
#Variable declaration
n1 = 2 #for second minimum
n2 = 3 #for third minimum
lamda = 4000 #wavelength(A)
#Calculations
'''For 2nd order,
a sin0 = n1*lamda
For 3rd order,
a sin0 = n2*lamda'''
lamda = (n2*lamda)/n1
#Result
print "Wavelength =",lamda,"A"
Wavelength = 6000 A
## Example 2.2.4, Page number 2-11¶
In [22]:
import math
#Variable declaration
a = 0.16*10**-3 #width of slit(cm)
n = 1 #first order
lamda = 5600*10**-10 #wavelength(cm)
#Calculations
def deg_to_dms(deg):
d = int(deg)
md = abs(deg - d) * 60
m = int(md)
sd = (md - m) * 60
sd=round(sd,2)
return [d, m, sd]
theta = (math.asin((n*lamda)/a))*180/math.pi
#Result
print "Half angular width=",deg_to_dms(theta)
Half angular width= [0, 12, 1.93]
## Example 2.2.5, Page number 2-11¶
In [30]:
import math
#Variable declaration
a = 12*10**-5 #width of slit(cm)
n = 1 #first order
lamda = 6000*10**-8 #wavelength(cm)
#Calculations
theta = (math.asin((n*lamda)/a))*180/math.pi
#Result
print "Half angular width=",(theta),"degrees"
Half angular width= 30.0 degrees
## Example 2.2.6, Page number 2-12¶
In [34]:
import math
#Variable declaration
a = 2*10**-6 #width of slit(cm)
n = 1 #first order
lamda = 6500*10**-10 #wavelength(cm)
#Calculations
theta = (math.asin((n*lamda)/a))*180/math.pi
#Result
print "Angle theta =",round(theta,2),"degrees"
Angle theta = 18.97 degrees
## Example 2.3.1, Page number 2-16¶
In [48]:
#Variable declaration
a = 0.16 #width(mm)
b = 0.8 #distance(mm)
#Calculations & Results
m = ((a+b)/a)
for n in range(1,4):
print "m =",m*n
m = 6.0
m = 12.0
m = 18.0
## Example 2.4.1, Page number 2-24¶
In [2]:
#Variable declaration
lamda1 = 5*10**-5 #cm
lamda2 = 7*10**-5 #cm
a_plus_b = 1./4000 #cm
#Calculations
m_max1 = a_plus_b/lamda1
m_max2 = round((a_plus_b/lamda2),1)
#Results
print m_max2,"orders are visible for 7000 A and",m_max1,"orders for 5000 A and either",m_max2,",4 or",m_max1,"for intermediate wavelengths"
3.6 orders are visible for 7000 A and 5.0 orders for 5000 A and either 3.6 ,4 or 5.0 for intermediate wavelengths
## Example 2.4.2, Page number 2-24¶
In [60]:
#Variable declaration
theta = 30 #angle of diffraction(degrees)
lamda1 = 5400*10**-8 #cm
lamda2 = 4050*10**-8 #cm
#Calculations
m = lamda2/(lamda1-lamda2)
a_plus_b = (m*lamda1)/math.sin(theta*math.pi/180)
N = 1/a_plus_b
#Result
print "Number of lines per cm =",round(N,2)
Number of lines per cm = 3086.42
## Example 2.4.3, Page number 2-25¶
In [62]:
#Variable declaration
lamda = 4992 #A
m1 = 3 #for 3rd order
m2 = 4 #for 4th order
#Calculations
'''For m = 3,
(a+b)sin0 = 3*lamda
For m = 4,
(a+b)sin0 = 4*lamda'''
l = (m2*lamda)/m1
#Results
print "Wavelength =",l,"A"
Wavelength = 6656 A
## Example 2.4.4, Page number 2-25¶
In [69]:
#Variable declaration
lamda = 6328*10**-8 #wavelength(cm)
m1 = 1
m2 = 2
a_plus_b = 1./6000 #cm
#Calculations
theta1 = math.asin((m1*lamda)/a_plus_b)*180/math.pi
theta2 = math.asin((m2*lamda)/a_plus_b)*180/math.pi
#Result
print "01 =",round(theta1,2),"degrees"
print "02 =",round(theta2,2),"degrees"
01 = 22.31 degrees
02 = 49.41 degrees
## Example 2.4.5, Page number 2-26¶
In [2]:
import math
#Variable declaration
m = 2 #2nd order
lamda = 5*10**-5 #wavelength(cm)
theta = 30
#Calculations
a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)
N = 1/a_plus_b
#Result
print "The number of lines/cm of the grating surface is",N
The number of lines/cm of the grating surface is 5000.0
## Example 2.4.6, Page number 2-26¶
In [7]:
import math
#Variable declaration
m = 3 #3rd order
a_plus_b = 1./7000 #no. of lines/cm
sin0 = 1
#Calculation
lamda = (a_plus_b*sin0)/m
#Result
print "Wavelength =",round(lamda/1e-8),"A"
Wavelength = 4762.0 A
## Example 2.4.7, Page number 2-26¶
In [15]:
import math
#Variable declaration
m = 1 #1st order
lamda = 6560*10**-8 #wavelength(cm)
theta = 16+12./60
#Calculations
a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)
N = 1/a_plus_b
w = 2*N
#Result
print "The total number of lines is",round(w)
The total number of lines is 8506.0
## Example 2.4.8, Page number 2-27¶
In [3]:
import math
#Variable declaration
m = 1 #1st order
lamda = 6.56*10**-5 #wavelength(cm)
theta = 18+14./60
#Calculations
a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)
N = 1/a_plus_b
w = 2*N
#Result
print "The total number of lines is",round(w,1)
The total number of lines is 9539.3
## Example 2.4.9, Page number 2-27¶
In [2]:
#Variable declaration
lamda = 6*10**-5 #wavelength(cm)
a_plus_b = 1./5000 #cm
#Calculations
m_max = a_plus_b/lamda
#Result
print "Order =",round(m_max)
Order = 3.0
## Example 2.4.10, Page number 2-28¶
In [4]:
#Variable declaration
lamda = 6000*10**-8 #wavelength(cm)
a_plus_b = 1./5000 #cm
#Calculations
m_max = a_plus_b/lamda
#Result
print "m_max =",round(m_max)
#The rest of the solution is theoretical
m_max = 3.0
## Example 2.4.11, Page number 2-28¶
In [5]:
import math
#Variable declaration
m = 1 #1st order
lamda = 5790*10**-8 #wavelength(cm)
theta = 19.994
#Calculations
a_plus_b = (m*lamda)/math.sin(theta*math.pi/180)
N = 1/a_plus_b
w = 2.54*N
#Result
print "The total number of lines is",round(w)
The total number of lines is 15000.0
## Example 2.6.1, Page number 2-31¶
In [13]:
import math
#Varaible declaration
n = 3000./0.5 #no. of lines per cm
m = 2 #for 2nd order
lamda1 = 5893*10**-8 #wavelength(cm)
lamda2 = 5896*10**-8 #wavelength(cm)
#Calculations
a_plus_b = 1/n
theta1 = math.degrees(math.asin((2*lamda1)/(a_plus_b)))
theta2 = math.degrees(math.asin((2*lamda2)/(a_plus_b)))
d = theta2-theta1 #angular seperation
N = lamda1*10**8/(m*3)
#Result
print "Since N=",round(N),"which is smaller than 3000 lines, the two lines will be resolved in the second order"
Since N= 982.0 which is smaller than 3000 lines, the two lines will be resolved in the second order
## Example 2.6.2, Page number 2-32¶
In [15]:
import math
#Varaible declaration
m = 3 #for 3rd order
lamda = 481 #wavelength(nm)
n = 620 #no. of ruling per mm
w = 5.05 #width(mm)
#Calculations
N = n*w
dl = lamda/(m*N)
#Result
print "Smallest wavelength interval =",round(dl,4),"nm"
Smallest wavelength interval = 0.0512 nm
## Example 2.6.3, Page number 2-33¶
In [17]:
import math
#Varaible declaration
m = 2 #for 2nd order
lamda = 5890 #wavelength(A)
dl = 6 #A
n = 500. #no. of lines per cm
#Calculations
N = lamda/(dl*m)
W = N/n
#Result
print "Least width =",W,"cm"
Least width = 0.98 cm
## Example 2.6.4, Page number 2-33¶
In [24]:
#Varaible declaration
N = 3*5000 #no. of lines per cm
a_plus_b = 1./5000 #cm
lamda = 5890*10**-8 #wavelength(cm)
#Calculations
m_max = round(a_plus_b/lamda)
RP = m_max*N
#Result
print "Maximum value of resolving power =",RP
Maximum value of resolving power = 45000.0
## Example 2.6.5, Page number 2-34¶
In [28]:
#Varaible declaration
m = 2 #for 2nd order
lamda = 5890 #wavelength(A)
dl = 6 #A
w = 3. #width of a line(cm)
#Calculations
lamda2 = lamda+dl
N = lamda/(dl*m)
a_plus_b = w/N
#Results
print "The minimum no. of lines a grating must have is",N
print "The grating element is",round(a_plus_b/1e-3,2),"*10^-3 cm"
The minimum no. of lines a grating must have is 490
The grating element is 6.12 *10^-3 cm
## Example 2.6.6, Page number 2-34¶
In [29]:
#Varaible declaration
N = 40000 #no. of lines per cm
m = 2 #for 2nd order
#Calculations
RP = m*N
#Result
print "Maximum value of resolving power =",RP
Maximum value of resolving power = 80000
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Search a number
112304 = 247019
BaseRepresentation
bin11011011010110000
312201001102
4123122300
512043204
62223532
7645263
oct333260
9181042
10112304
1177415
1254ba8
133c16a
142ccda
152341e
hex1b6b0
112304 has 10 divisors (see below), whose sum is σ = 217620. Its totient is φ = 56144.
The previous prime is 112303. The next prime is 112327. The reversal of 112304 is 403211.
It is a happy number.
It is a super-2 number, since 2×1123042 = 25224376832, which contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (112303) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3494 + ... + 3525.
It is an arithmetic number, because the mean of its divisors is an integer number (21762).
2112304 is an apocalyptic number.
It is an amenable number.
112304 is a deficient number, since it is larger than the sum of its proper divisors (105316).
112304 is an equidigital number, since it uses as much as digits as its factorization.
112304 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 7027 (or 7021 counting only the distinct ones).
The product of its (nonzero) digits is 24, while the sum is 11.
The square root of 112304 is about 335.1178897045. The cubic root of 112304 is about 48.2464179824.
Adding to 112304 its reverse (403211), we get a palindrome (515515).
The spelling of 112304 in words is "one hundred twelve thousand, three hundred four", and thus it is an iban number.
Divisors: 1 2 4 8 16 7019 14038 28076 56152 112304
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https://olimx.net/multiplication-tables/1-from-to-20
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# Multiplication Table from 1 to 20
The multiplication table 1 to 20 is the list of the numbers from 1 to 20. Multiplication table 1 to 20 displays the results of multiplying certain numbers from 1 to 20 by each other. Learning the multiplication table is the bedrock to prepare for other math topics such as algebra, fractions, and division. Learning the times table from 1 to 20 will also help students speed up calculations and save more time. You can save or download the times chart to learn the multiplication table more easily.
1 x 1 = 1
1 x 2 = 2
1 x 3 = 3
1 x 4 = 4
1 x 5 = 5
1 x 6 = 6
1 x 7 = 7
1 x 8 = 8
1 x 9 = 9
1 x 10 = 10
2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10
2 x 6 = 12
2 x 7 = 14
2 x 8 = 16
2 x 9 = 18
2 x 10 = 20
3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12
3 x 5 = 15
3 x 6 = 18
3 x 7 = 21
3 x 8 = 24
3 x 9 = 27
3 x 10 = 30
4 x 1 = 4
4 x 2 = 8
4 x 3 = 12
4 x 4 = 16
4 x 5 = 20
4 x 6 = 24
4 x 7 = 28
4 x 8 = 32
4 x 9 = 36
4 x 10 = 40
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4 = 20
5 x 5 = 25
5 x 6 = 30
5 x 7 = 35
5 x 8 = 40
5 x 9 = 45
5 x 10 = 50
6 x 1 = 6
6 x 2 = 12
6 x 3 = 18
6 x 4 = 24
6 x 5 = 30
6 x 6 = 36
6 x 7 = 42
6 x 8 = 48
6 x 9 = 54
6 x 10 = 60
7 x 1 = 7
7 x 2 = 14
7 x 3 = 21
7 x 4 = 28
7 x 5 = 35
7 x 6 = 42
7 x 7 = 49
7 x 8 = 56
7 x 9 = 63
7 x 10 = 70
8 x 1 = 8
8 x 2 = 16
8 x 3 = 24
8 x 4 = 32
8 x 5 = 40
8 x 6 = 48
8 x 7 = 56
8 x 8 = 64
8 x 9 = 72
8 x 10 = 80
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
10 x 1 = 10
10 x 2 = 20
10 x 3 = 30
10 x 4 = 40
10 x 5 = 50
10 x 6 = 60
10 x 7 = 70
10 x 8 = 80
10 x 9 = 90
10 x 10 = 100
11 x 1 = 11
11 x 2 = 22
11 x 3 = 33
11 x 4 = 44
11 x 5 = 55
11 x 6 = 66
11 x 7 = 77
11 x 8 = 88
11 x 9 = 99
11 x 10 = 110
12 x 1 = 12
12 x 2 = 24
12 x 3 = 36
12 x 4 = 48
12 x 5 = 60
12 x 6 = 72
12 x 7 = 84
12 x 8 = 96
12 x 9 = 108
12 x 10 = 120
13 x 1 = 13
13 x 2 = 26
13 x 3 = 39
13 x 4 = 52
13 x 5 = 65
13 x 6 = 78
13 x 7 = 91
13 x 8 = 104
13 x 9 = 117
13 x 10 = 130
14 x 1 = 14
14 x 2 = 28
14 x 3 = 42
14 x 4 = 56
14 x 5 = 70
14 x 6 = 84
14 x 7 = 98
14 x 8 = 112
14 x 9 = 126
14 x 10 = 140
15 x 1 = 15
15 x 2 = 30
15 x 3 = 45
15 x 4 = 60
15 x 5 = 75
15 x 6 = 90
15 x 7 = 105
15 x 8 = 120
15 x 9 = 135
15 x 10 = 150
16 x 1 = 16
16 x 2 = 32
16 x 3 = 48
16 x 4 = 64
16 x 5 = 80
16 x 6 = 96
16 x 7 = 112
16 x 8 = 128
16 x 9 = 144
16 x 10 = 160
17 x 1 = 17
17 x 2 = 34
17 x 3 = 51
17 x 4 = 68
17 x 5 = 85
17 x 6 = 102
17 x 7 = 119
17 x 8 = 136
17 x 9 = 153
17 x 10 = 170
18 x 1 = 18
18 x 2 = 36
18 x 3 = 54
18 x 4 = 72
18 x 5 = 90
18 x 6 = 108
18 x 7 = 126
18 x 8 = 144
18 x 9 = 162
18 x 10 = 180
19 x 1 = 19
19 x 2 = 38
19 x 3 = 57
19 x 4 = 76
19 x 5 = 95
19 x 6 = 114
19 x 7 = 133
19 x 8 = 152
19 x 9 = 171
19 x 10 = 190
20 x 1 = 20
20 x 2 = 40
20 x 3 = 60
20 x 4 = 80
20 x 5 = 100
20 x 6 = 120
20 x 7 = 140
20 x 8 = 160
20 x 9 = 180
20 x 10 = 200
## Importance of Multiplication Table 1 to 20
Memorizing the times table is necessary for each student so the multiplication table is very important because:
• Helping students solve mathematical problems involving multiplication faster and easier.
• Understanding of the operations involved in multiplication
• A strong foundation to learn the other knowledge
## Multiplication chart 1 to 20
X 1234567891011121314151617181920
11234567891011121314151617181920
2246810121416182022242628303234363840
33691215182124273033363942454851545760
448121620242832364044485256606468727680
55101520253035404550556065707580859095100
66121824303642485460667278849096102108114120
7714212835424956637077849198105112119126133140
881624324048566472808896104112120128136144152160
9918273645546372819099108117126135144153162171180
10102030405060708090100110120130140150160170180190200
## How to Learn The Multiplication Tables 1 - 20
• Learning through the songs: Children learn the times table according to the lyrics with the melody, it will helping them enjoy and remember faster.
• Practicing: A daily practice will helping children remember longer.
• Helping the children if they become stuck: If children have a wrong answer, be patient and help them point out the mistakes.
## Tables 1 to 20 Examples
Example 1: Peter wants to buy 10 cakes, the cost of a cake is 20 cents. What are the cost of 10 cakes?
Solve: The cost of a cake = 20 cents
Peter wants to buy 10 cakes
Then, 10 x 20 = 200 cents
Thus, the cost of 10 cakes are 200 cents.
Example 2: Sam reads 17 books a day. Using tables 1 to 20 find how many books will Sam read in 16 days?
Solve: Sam reads: 5 books a day
Sam will read ? books in 16 days
Then, 16 x 17 = 272 books
Thus, Sam will read 272 books in 16 days.
## Q&A for Tables 1 to 20
Why should we learn multiplication table from 1 to 20?
We should learn multiplication table 1 to 20 because:
• It will help us solve mathematical problems faster and easier.
• It is also the bedrock to prepare for other math topics such as algebra, fractions, and division.
How to practice multiplication tables from 1 to 20?
There are many ways to practice times table however we will recommend several ways to you that we believe it is the most useful and simple. First, you can practice for children with flashcard every day, and ask them any questions on the times table, this will help them build confidence and reinforce knowledge. Furthermore, you can use the times table in your daily life, for example when you want to buy something, you can practice calculations by times tables. With counited practice and patience, you will memorize all multiplication tables.
How can I learn multiplication tables from 1 to 20 quickly?
If you want to memorize the times tables, you should practice them everyday or the traditional rote learning methods. Daily practice will reinforce children’s knowledge and help them remember longer.
What are the advantages of learning tables 1 to 20?
There are several advantages of learning tables 1 to 20 such as:
• Solving mathematical problems simply
• Helping students become more confident
• Understanding of the operations involved in multiplication
• Improving the memory of children.
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# Math.Pow(X2-X1,2) With Code Examples
Math.Pow(X2-X1,2) With Code Examples
We'll attempt to use programming in this lesson to solve the Math.Pow(X2-X1,2) puzzle. This is demonstrated in the code below.
```import java.util.Scanner;
public class CoordinateGeometry {
public static void main(String [] args) {
Scanner scnr = new Scanner(System.in);
double x1;
double y1;
double x2;
double y2;
double pointsDistance;
double xDist;
double yDist;
pointsDistance = 0.0;
xDist = 0.0;
yDist = 0.0;
x1 = scnr.nextDouble();
y1 = scnr.nextDouble();
x2 = scnr.nextDouble();
y2 = scnr.nextDouble();
poinsDistance = Math.sqrt(Math.pow(x2 - x1, 2) + (Math.pow(y2 - y1, 2));
System.out.println(pointsDistance);
}
}
```
As we've seen, a lot of examples were used to address the Math.Pow(X2-X1,2) problem.
## How is Math POW calculated?
pow() is used to calculate a number raise to the power of some other number. This function accepts two parameters and returns the value of first parameter raised to the second parameter. There are some special cases as listed below: If the second parameter is positive or negative zero then the result will be 1.0.17-Sept-2022
## What does Math pow () do?
The Math. pow() method returns the value of x to the power of y (xy).
## Does Math POW return a double?
pow() is used to return the value of first argument raised to the power of the second argument. The return type of pow() method is double.
## Is Math POW a double or int?
pow Example. The method raises a to the power of b and returns the result as double. In other words, a is multiplied by itself b times.27-Jun-2022
8.0e0
## What can I use instead of Math POW?
Introduced in ES2016, the infix exponentiation operator ** is an alternative for the standard Math. pow function. Infix notation is considered to be more readable and thus more preferable than the function notation.
## What is output of print Math POW 3 2 ))?
pow(3, 2))? Explanation: math. pow() returns a floating point number. 9.
## What is a power of 2?
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, (sequence A000079 in the OEIS) Because two is the base of the binary numeral system, powers of two are common in computer science.
343
## How do I return an int from Math POW?
• import java. lang. Math;
• class CalculatePower {
• public static void main( String args[] ) {
• //Calculating 5^4 and casting the returned double result to int.
• int ans1 = (int) Math. pow(5,4);
• System. out. println("5^4 is "+ans1);
• //Calculating 1.5^3 and storing the returned double result to ans2.
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# R/hcf.aov.R In Directional: A Collection of Functions for Directional Data Analysis
#### Documented in hcf.aov
```hcf.aov <- function(x, ina, fc = TRUE) {
## x contains all the data
## ina is an indicator variable of each sample
ina <- as.numeric(ina)
g <- max(ina) ## how many groups are there
p <- dim(x)[2]
n <- dim(x)[1] ## dimensionality and sample size of the data
S <- rowsum(x, ina)
Ri <- sqrt( Rfast::rowsums(S^2) ) ## the resultant length of each group
S <- Rfast::colsums(x)
R <- sqrt( sum(S^2) ) ## the resultant length based on all the data
## Next we stimate the common concentration parameter kappa
kappa <- Directional::vmf.mle(x, fast = TRUE)\$kappa
## kappa is the estimated concentration parameter based on all the data
Ft <- (n - g) * (sum(Ri) - R) /( (g - 1) * (n - sum(Ri)) )
if (fc) { ## correction is used
if (p == 3) {
Ft <- kappa * (1/kappa - 1/(5 * kappa^3)) * Ft
} else if (p > 3) Ft <- kappa * ( 1/kappa - (p - 3)/(4 * kappa^2) - (p - 3)/(4 * kappa^3) ) * Ft
}
pvalue <- pf(Ft, (g - 1) * (p - 1), (n - g) * (p - 1), lower.tail = FALSE)
res <- c(Ft, pvalue, kappa)
names(res) <- c('test', 'p-value', 'kappa')
res
}
```
## Try the Directional package in your browser
Any scripts or data that you put into this service are public.
Directional documentation built on Nov. 8, 2021, 1:07 a.m.
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# SUBTRACTING FRACTIONS WITH UNLIKE DENOMINATORS
Subtracting Fractions with Unlike Denominators :
In this section, you will learn, how to subtract two fractions with different denominators.
We can use one of the following methods to subtract two fractions with unlike denominators.
1. Cross-Multiplication method
2. LCM Method.
## Cross-Multiplication Method
If the denominators of the fractions are co-prime or relatively prime, we have to apply this method.
Fro example, let us consider the subtract of two fractions given below.
1/3 - 1/8
In the above two fractions, denominators are 3 and 8.
For 3 and 8, there is no common divisor other than 1.
So 3 and 8 are co-prime.
Here we have to apply cross-multiplication method to find the subtraction of two fractions 1/3 and 1/8 as shown below.
## LCM Method (Least Common Multiple Method)
If the denominators of the fractions are not co-prime (there is a common divisor other than 1), we have to apply this method.
Fro example, let us consider the subtract of two fractions given below.
5/12 - 1/20
In the above two fractions, denominators are 12 and 20.
For 12 and 20, if there is at least one common divisor other than 1, then 12 and 20 are not co-prime.
For 12 and 20, we have the following common divisors other than 1.
2 and 4
So 12 and 20 are not co-prime.
In the next step, we have to find the LCM (Least common multiple) of 12 and 20.
12 = 22 x 3
20 = 22 x 5
When we decompose 12 and 20 in to prime numbers, we find 2, 3 and 5 as prime factors for 12 and 20.
To get L.C.M of 12 and 20, we have to take 2, 3 and 5 with maximum powers found above.
So, the LCM of 12 and 20 is
= 22 x 3 x 5
= 4 x 3 x 5
= 60
Now, make the denominator of each fraction as 60 using multiplication and then subtract them as shown below.
## Subtracting Fractions with Unlike Denominators - Practice Problems
Problem 1 :
Find the value of :
3/5 - 2/3
Solution :
In the above two fractions, denominators are 5 and 3.
For 5 and 3, there is no common divisor other than 1.
So 5 and 3 are co-prime.
Here we have to apply cross-multiplication method to find the subtraction of two fractions 1/8 and 1/3 as shown below.
3/5 - 2/3 = (9 - 10) / 15
3/5 - 2/3 = - 1/15
Problem 2 :
Find the value of :
7/12 - 5/18
Solution :
In the above two fractions, denominators are 12 and 18.
For 12 and 18, if there is at least one common divisor other than 1, then 12 and 18 are not co-prime.
For 12 and 18, we have the following common divisors other than 1.
2, 3, and 6
So 12 and 20 are not co-prime.
Here, we have to apply LCM method to find the subtraction of the two fractions.
Find the LCM of 12 and 18.
LCM of (12, 18) = 36
Now, make the denominator of each fraction as 36 using multiplication and subtract the fractions with like denominators.
Then, we have
7/12 - 5/18 = 21/36 - 10/36
7/12 - 5/18 = (21 - 10) / 36
7/12 - 5/18 = 11/36
After having gone through the stuff and examples, we hope that the students would have understood, how to subtract two fractions with unlike denominators.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
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Sum of all three digit numbers divisible by 6
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Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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https://www.fractal.garden/fractal-canopy
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• Options
# Fractal Canopy
A fractal canopy is a fractal generated, by sequentially splitting and branching lines from existing lines.
At the end of each line, draw n new lines at certain angles.
There are a lot of parameters to vary here: the number of branches, the angle between them, the original width, as well as the decay of length and width...
When playing around with these you can get all kinds of interesting results. There are a few you can play around with in the presets above, like a shape that resembles 2d-broccoli, an H-Tree, or even a Sierpinski Triangle.
Many things in nature follow this pattern - like trees and plants, but also blood vessels or the patterns generated by an electric arc when it passes through materials (so-called Lichtenberg figures). Even some types of crystal growth resemble this structure.
The code used to generate this pattern looks like this:
function drawTree() {
background(config.color);
resetMatrix();
translate(width / 2, height);
branch(config.rootLength, config.rootWidth, 0);
}
const halfNumBranches = Math.floor(config.branches / 2)
function branch(len: number, weight: number, iteration: number) {
// stop recursion if reached max depth
if (iteration > config.maxIterations) {
return;
}
strokeWeight(weight);
stroke(
// mapping stroke color based on iterations to different colors
map(iteration, 0, 10, 100, 150),
map(iteration, 0, 10, 100, 255),
100
);
line(0, 0, 0, -len);
translate(0, -len);
// figure out angles
const evenBranches = config.branches % 2 === 0
const midpoint = evenBranches
? halfNumBranches - 0.5
: halfNumBranches;
rotate(config.angle * midpoint);
// loop over the amount of new branches
for (let i = 0; i < config.branches; i++) {
push();
rotate(-config.angle * i);
// then recurse, drawing a new branch into each direction
branch(
len * config.lengthDecay,
weight * config.widthDecay,
iteration + 1
);
pop();
}
}
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http://myriverside.sd43.bc.ca/gavinp2016/category/grade-11/
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# Totem response
It has made them much more reliant on Canadian traditions and in some words much more “whitewashed” than they might have been if uninfluenced. Still, some keep with there traditions and old ways but at the same time some also keep up with the times. I believe that even if left to their traditions and techniques they would still have adapted to the rest of the world’s technology. I think that right now everyone is worked up over something that’s not there. Why should I have to pay taxes and give up land that should be shared because some great-great relative was a racist. I think that every person should have equal rights, but I do believe that if they were left to their ways they would end up still adapting to modern day society.
# Persuasive essay Edit
(All edits are in red).
I think I made some very solid points throughout my debate, which reflected well in the essay.
But I think I need to choreograph my thoughts in a much more grammatically correct fashion. even though I know what I mean in the writing doesn’t mean others can as well.
One other thing I think I could improve on would have to be my integration of the quotes. I think I did a somewhat sloppy job with the integration because I explained how the passage related to the novel then added the quote.
I think this image is a great way to reflect my essay and debate.
because I make the argument of a scale just like this to prove my point.
this week in math Precalc I learned about radicals and how to graph them
for example y=$\frac{1}{2x+5}$
so first you want to graph the parent function 2x+5
now you want to look for your invariant points (-1 and +1) then you want to draw two L shapes not touching the x-intercept or the x=x int asymptotes making this
the red line doesn’t actually touch the x int even if it looks like it does because a radical is just flipping the numbers 2 is now 1/2.
# Week 12 – Math 11 – Graphing absolute values
in this week we learned how to graph absolute values
for the first example a linear line y=|x+3|
so first you just graph the parent function y=x+3
when it touches the x-intercept it has to go back up because it always has to stay positive
the right side stays the same because it’s positive and the left side just flips making a v shape
we also learned the absolute value of parabolas, for example, y=|x^2 -3|
so you want to draw the parent function again and just flip the negative parts so they are positive making a w shape
# WEEK 11 – MATH 11 – GRAPHING QUADRATIC inequalitys
This week we learned how to solve quadratic inequalities.
to do this we graph a regular quadratic just as we learned before
then to find out what side is true its if the Y is greater than, shade in the center if its Y is less than, then shade the outside.
# WEEK 10 – MATH 11 – GRAPHING INEQUALITYS
In this week we learned how to graph an inequality.
Say we have 2x + 3y > 6
First, we must make the Y a 0
2x > 6
Then we must divide both sides by 2
x > 3
Then change the > to a =
x = 3
we then repeat the same steps for the Y axis to end up with
y = 2
Once you have these two points plot both on the grid and connect them. after this follow the line to reach each side of your graph.
Next, we must figure out which side is true. to do this we must replace the 2x + 3y with a 0 so we end up with
0 > 6
Now we ask our self, is 0 greater than 6?
since its no, the shaded side will be the side without 0 on it.
# WEEK 9 – MATH 11 – GRAPHING ADVANCED TRINOMIALS
In this week we learned how to graph trinomials quicker
for example in this equation
y = 2X^2 + 6X + 8
first of in this equation you can see the Y-axis (8)
Once you find the Y axis you can then convert it to…
y = 2(X + 3)^2 -1
This one shows the vertex (-3,-1) and tells you where the formula will shape and point (2) it shows by looking at the “a” (2).
# WEEK 8 – MATH 11 – GRAPHING TRINOMIALS
In this week we learned how to graph trinomials.
if the graph is Y=X^2
notice how this graph has its axis at (0,0) and how it is going up 1,3,5
if the graph is Y=X^2 +2
The + 2 makes the parabola just go up to spaces so the axis is now (0,2) and still has the same pattern 1,3,5
if the graph is y= (x-2)^2
the -2 means to go to the right 2 so its the opposite of the symbol to which side you go so if it was +2 you would have gone 2 to the left
if the graph is y=2x^2
the 2 in front of the x makes it stretched out upwards because before there is an imaginary 1 in front of the x and now its 2 so it goes from 1,3,5 to 2,6,10 which makes the graph look skinner
# WEEK 7 – MATH 11 – FACTORING TRINOMIALS
In this week we learned how to factor Trinomials
For example 16X^2 + 36X + 8
First, multiply 16 * 8 and find the factors of numbers that multiply towards it until you find the factors that add together to create 36.
So 16 * 8 = 128
128 = 4 * 32
4 + 32 = 36
so once you have done this you will end up with 4X and 32X in your empty squares.
and finally, you will end with (4 + 32X) and (4X + 8).
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http://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Index/problemdetail/pid/3146.html
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### Integer division 2
Time Limit: 1000 ms Memory Limit: 65536 KiB
#### Problem Description
This is a very simple problem, just like previous one.
You are given a postive integer n, and you need to divide this integer into m pieces. Then multiply the m pieces together. There are many way to do this. But shadow95 want to know the maximum result you can get.
#### Input
First line is a postive integer t, means there are T test cases.
Following T lines, each line there are two integer n, m. (0<=n<=10^18, 0 < m < log10(n))
#### Output
Output the maximum result you can get.
#### Sample Input
1
123 2
#### Sample Output
36
#### Hint
You can divide "123" to "12" and "3".
Then the maximum result is 12*3=36.
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http://mathoverflow.net/feeds/question/55685
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| 130,857,580
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faithful unipotent representations of (finite) $p$-groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:57:23Z http://mathoverflow.net/feeds/question/55685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups faithful unipotent representations of (finite) $p$-groups Igor Rivin 2011-02-17T02:15:45Z 2012-09-06T17:38:56Z <p>The title pretty much summarizes the question: does every $p$-group have a <em>faithful</em> unipotent representation (with coefficients in $\mathbb{F}_p$ or some finite extension thereof)? </p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55686#55686 Answer by Mark Sapir for faithful unipotent representations of (finite) $p$-groups Mark Sapir 2011-02-17T03:12:11Z 2011-02-18T02:17:07Z <p>Yes, of course, it is how Bogopolsky proves Sylow theorems in his book (Bogopolski, Oleg, Introduction to group theory. Translated, revised and expanded from the 2002 Russian original. EMS Textbooks in Mathematics. European Mathematical Society (EMS), Zürich, 2008. x+177 pp.). </p> <p>More precisely, a finite $p$-group $G$ of order $n$ embeds into $GL=GL_n({\mathbb F}_p)$ (Cayley theorem). The subgroup $UT$ of upper triangular unipotent matrices of $GL$ is a Sylow p-subgroup of $GL$ (proof by computing the order of $GL$), so $G$ is conjugate to a subgroup of $UT$.</p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55687#55687 Answer by Yuri Zarhin for faithful unipotent representations of (finite) $p$-groups Yuri Zarhin 2011-02-17T03:13:33Z 2011-02-17T03:25:32Z <p>I guess you mean finite $p$-groups. Then every finite-dimensional representation of a finite $p$-group $G$ is unipotent in characteristic $p$. (Indeed, all eigenvalues of every element of $G$ are $p$-power roots of unity and therefore must be equal to $1$. This means that every element of $G$ acts as a unipotent operator.) An example of a faithful unipotent representation of $G$ is provided by its regular representation over any field of characteristic $p$. In particular, the regular representation of $G$ over the prime field $F_p$ is faithful and unipotent.</p> <p>(The same construction works for infinite $p$-groups $G$ as well. Of course, in this case the regular" representation space of functions on $G$ with finite support will be infinite-dimensional.)</p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55814#55814 Answer by Yuri Zarhin for faithful unipotent representations of (finite) $p$-groups Yuri Zarhin 2011-02-18T02:44:44Z 2011-02-18T02:44:44Z <p>Here is a slightly different approach. If $G$ is a $p$-group (infinite or finite) and $k$ is a field of characteristic $p$ then in the group algebra $k[G]$ we have $(x-1)^{p^r}=x^{p^r}-1$ for all $x \in k[G]$. (Actually, if elements $a$ and $b$ of any $k$-algebra do commute then $(a-b)^{p^r}=a^{p^r} - b^{p^r}$, thanks to divisibility properties of binomial coefficients.) Applying it to $x=g$ where $g$ is an element of $G$ of order $p^r$, we conclude that $(g-1)^{p^r}=g^{p^r}-1=1-1=0$ in $k[G]$. Since every representation space $V$ of $G$ over $k$ is a module over $k[G]$, we conclude that $g-1$ acts on $V$ as a nilpotent operator.</p> <p>An example of a faithful representation of $G$ is provided by the regular representation where $V$ is the space of all $k$-valued functions on $G$. Another example is provided by its $G$-invariant space $V_0$ of all functions $f: G \to k$ with finite support (i.e., vanishing at all but finitely many points of $G$). Notice that for each $g \in G$ the space $V_0$ splits into a direct sum of $g$-invariant finite-dimensional subspaces (that correspond to finite left cosets of the cyclic group generated by $g$).</p>
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| 3.171875
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https://www.cazoommaths.com/maths-worksheets/gcse-maths-revision-worksheets/
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# GCSE Maths Revision Worksheets
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## Statistics and probability GCSE REVISION WORKSHEETS
This is Tooltip!
### List of Topics
Our printable maths worksheets cover the full range of topics. See below the list of topics covered. All our maths worksheets can be accessed here.
#### Algebra
• BODMAS
• Expanding Brackets
• Factorising
• Indices
• Inequalities
• Linear Functions
• Real Life Graphs
• Rearranging Equations
• Sequences
• Simplification
• Solving Equations
• Substitution
#### Number
• Calculator Methods
• Decimals
• Fractions
• Fractions Decimals Percentages
• Mental Methods
• Negative Numbers
• Percentages
• Place Value
• Powers
• Proportion
• Ratio
• Rounding
• Time
• Types of Number
• Written Methods
#### Geometry
• 2D Shapes
• 3D Shapes
• Area and Perimeter
• Bearings Scale and Loci
• Circles
• Compound Measures
• Constructions
• Coordinates
• Lines and Angles
• Polygons
• Pythagoras
• Similarity and Congruence
• Transformations
• Volume and Surface Area
#### Statistics
• Histograms and Frequency Polygons
• Mean Median Mode
• Pie Charts and Bar Charts
• Probability
• Scatter Graphs
• Stem-and-Leaf Diagrams
• Two-Way Tables and Pictograms
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# What are the chances of a woman having 4 consecutive male children?
Hey there. Me again. Mathematically, the odds of predicting correctly four male children is 1 in 16, or 6.25%
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# SHM Question
We have been given this as a sample exam question.
For a car store a small model car (75kg) is hung from the ceiling. The model is hanging on an elastic spring with force constant 1500 N/m. The model is oscillating 10 times per second with an amplitude of 0.02 m. (Assume a zero-phase shift).
a) What is the initial extension of the spring?
b)Determine the velocity when the model passes through the equilibrium point.
c) What is the velocity when the model is -0.01 m from equilibrium and what direction has the velocity?
mass = 75kg
force/weight = mg = 735.75N
k = 1500N
Amplitude = 0.02m
frequency = 10Hz
For part a) I have calculated, using the expression F = kx, that the extension of the spring (x) is 0.4905m.
For part b) I know that the maximum velocity will occur at equilibrium as this is where potential energy is 0 and there is maximum kinetic energy. Using the equation v=ωAcos(ωt) and the fact that the max. value of cos(ωt) = I have found that maximum velocity is v=ωA.
Should this be calculated using ω=2πf so that the solution is:
Vmax = ωA = 2πfA = 2π(10)(0.02) = 1.26 m/s
or should ω be calculated as ω=sqrt(k/m) so the solution is:
Vmax = ωA = sqrt(k/m)A = sqrt(1500/75)(0.02) = 0.089 m/s
Should these not give equivalent answers?
Also for part c) I have calculated the following:
x(t) = -0.01m v(t) = ?
-0.01 = Asin(ωt)
solving this for t gives t = 1/ωcos(sin^-1(-0.01/A))
Substituting this into the equation for velocity v=ωAcos(ωt) gives v = ωAcos(sin^-1(-0.01/A))
However to finish this I have the same problem as above in b).
There is probably just something simple that I am overlooking and any help with this would be very much appreciated.
Thank You,
Sarah. :)
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# Electromagnetic Induction
by physgrl
Tags: electromagnetic, induction
P: 138 1. The problem statement, all variables and given/known data An electric generator consists of n = 500 turns of wire formed into a rectangular loop with a length of 5 cm and width of 3 cm placed in a uniform magnetic field of 2.50 T. What is the maximum value of the emf produced when the loop is spun at f = 100 rpm about an axis perpendicular to B? a. 19.6 V b. 144 V c. 95.3 V d. 79.2 V e. 60.3 V 2. Relevant equations ε=-NΔ∅B/Δt ∅B=BAcosβ 3. The attempt at a solution For a maximum emf there must be a maximum change in either B or A, but because A is constant B must be the variable changing. As the loop rotates B varies from 0T to 2.5T and it takes half a loop to do that. ε=NAB/Δt ε=500*(.03m*.05m)*(2.5T)/Δt Δthalf a revolution: =>100rpm*1min/60s=1.67rev/sec => 1 revolution takes .6s => .5 revolutions take .3s ε=500*(.03m*.05m)*(2.5T)/Δt ε=500*(.03m*.05m)*(2.5T)/.3s ε=6.25V The answer I get is not in the options, and I dont see what I am doing wrong? Thanks!
P: 581 The emf, ε, is related to the instantaneous rate of change in the flux, d∅B/dt. You used the value of the constant magnetic field times the area for this, which is incorrect. The value you used, (.03m*.05m)*(2.5T), is the maximum flux though the coil. Instead this should be a maximum magnitude* rate of change of this flux (a value of d∅B/dt at some time). You could start by expressing ∅B as a function of time. *I would pick the most negative value of d∅B/dt, that way ε takes on the highest positive value.
P: 138 I dont understand. The greatest change is from 0T to 2.5T so woudnt that cause the greatest change in ∅B hence the greater emf?
P: 614 Electromagnetic Induction disregard
HW Helper P: 6,187 Hi physgrl! The correct form is ##\mathcal{E}=-N{d\Phi \over dt}##. where ##{d\Phi \over dt}## is the derivative of ##\Phi## with respect to t. You already know that ##\Phi=B A \cos \beta##. And actually B and A are both constant, but ##\beta## is not. ##\beta## is the angle of the loop with the magnetic field. Its derivative is the angular velocity. Do you know how to find the angular velocity from the frequency with which the loop is spun? Furthermore, do you know how to take the derivative of ##\Phi = BA\cos \beta## with respect to t?
P: 138 The angular velocity is given by 100rpm.
HW Helper
P: 6,187
Quote by physgrl The angular velocity is given by 100rpm.
Can you turn that into radians per second?
P: 138 its 10.5rad/s
HW Helper
P: 6,187
Quote by physgrl its 10.5rad/s
Good!
Now you need the derivative of ##BA\cos(\beta(t))##.
For that you will need to apply the chain rule of differentiation.
Do you know how that works, or is that outside of the scope of your class material?
If it is, we may need to find another method to find the result.
P: 138 its BA*-sin(β(t))*β(t)' but what is β(t)? is it 10.5*t?
HW Helper
P: 6,187
Quote by physgrl its BA*-sin(β(t))*β(t)' but what is β(t)? is it 10.5*t?
Yes.
An angle is equal to the (constant) angular velocity multiplied with the time.
P: 138 So the derivative is 2.5T*(.03*.05)*sin(10.5t)*10.5t but what is the time I should plug in? how do i know what makes the max emf?
HW Helper
P: 6,187
Quote by physgrl So the derivative is 2.5T*(.03*.05)*sin(10.5t)*10.5t but what is the time I should plug in? how do i know what makes the max emf?
You should have a derivative of 2.5T*(.03*.05)*-sin(10.5t)*10.5.
What are the maximum and minimum values the sine can take?
P: 138 oh yes my bad! sin(theta)=1 is the max so is the max emf supposed to be=2.5T*(.03*.05)*10.5??
HW Helper
P: 6,187
Quote by physgrl oh yes my bad! sin(theta)=1 is the max so is the max emf supposed to be=2.5T*(.03*.05)*10.5??
Yep. :)
P: 138 ohh wait *500
HW Helper
P: 6,187
Quote by physgrl ohh wait *500
Oh yeah, right, very good.
I'm sorry. I totally forgot about that.
P: 138 ok! thanks for the help!
Related Discussions Introductory Physics Homework 3 General Physics 8 Classical Physics 17 Introductory Physics Homework 1 Introductory Physics Homework 0
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# Cheat sheet r statistics tutorial
Sheet statistics
## Cheat sheet r statistics tutorial
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Tip: check out this survminer cheat sheet After this tutorial, you will be able to take advantage of these data to answer questions such as the following: do.
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Cody
# Problem 1797. 03 - Matrix Variables 1
Solution 1985738
Submitted on 22 Oct 2019 by Ngo Minh Ngoc
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
ref = ones(9,9)*2; user = MatrixFunc(); assert(isequal(user,ref))
aMat = 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
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# Reversed
Reversed Age
#### Riddle:
The ages of a father and son add up to 66. The father's age is the son's age reversed. How old could they be? * There are THREE possible solutions
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# How to animate an arch forming itself?
I have for a student job to animate and explain through animations how arches works. This one is quite complicated and I can't seem to find a solution to animate it correctly.
I'm using freestyle because that is what they want for the type of render. I'll try to explain what I have to do to explain my problem : An 8 part arch is form by the translation of 4 arches in a concentric direction. The part obtained by this translation is a concentric arch ( above ) and a faceted vault ( below ). I have to animate, as shown below, 4 arches translating, showing their intersection in blue, and separate them apart.
The animation itself is not really the problem, the problem is that I don't know how to find a "clean" way to do it. Right now I'm using moving cubes as booleans to make the arches appear. But as you can see, Blender booleans are really quickly limited and junky, they tend to appear where they should not, and even with the fix proposed by blender ( self-interaction etc.. ), it does not work and they appear multiple times in the animation.
How would you do this? I'm taking any workflow, because I'm basically new to animation.
The task is quite tricky, I think. Of course, you could solve it very mathematically, but I took the more logical way using Node Noodles.
Since all boolean operators in Blender give such strange results, the main thing is to apply these operations in a clever order.
Here is a possible solution:
1. First, get the dimensions of the arc and calculate the length of the extrusion. This is done by the formula $$e = a * ( 1 + √2 )$$. $$e$$ is now the length of the curve along which you extrude your profile.
2. The current frame controls the factor of the node Trim Curve, which you can use to animate the extrusion.
3. Then you extrude your profile along this curve with the node Curve to Mesh.
4. Extrude again with Curve to Mesh (activate Fill Caps) and duplicate the mesh once by $$45°$$ offset by once by $$-45°$$, so that you can feed the node Mesh Boolean with it.
5. Here the order is crucial: first combine with Mesh Boolean the two offset meshes, and then subtract them from your first mesh.
However, to get a clean result, remove the faces that are not needed here.
6. To finally get the complete top part, duplicate this once with an offset of $$90°$$ and this again by $$45°$$ and combine the resulting meshes. In the end the node Merge by Distance helps to make it all a bit cleaner.
7. For the bottom part, first create two meshes with your factor-controlled curve and profile at $$45°$$ to each other. With the node Mesh Boolean you get the intersection of these two meshes.
8. You repeat the same again, but here you create a union of both meshes, applying the whole length instead of the shortened curve.
9. The difference of these two meshes is then the inner arc for your vault, and if you remove the superfluous points, you get a clean arc.
10. But to get the part you are looking for, you mirror part of the mesh so that it fits exactly under the vault.
11. Then, as you did with the roof before, multiply this mesh with an offset of $$90°$$ and again by $$45°$$.
12. Then, when you combine these two results, you have a frame-controlled vault.
Here is the node group at a glance:
And here is the blend file:
Note: If you use these nodes, your arc must be symmetrical or it won't work!
Model out a shape that is completely solid in the shape of the arch Volume:
Give it a shape key for the elongation part of the animation.
You animate this value to extend it back and forth as needed.
I inset the interior just a bit to separate the two vertex groups I want to make.
Add those interior faces to a vertex group
Add a mask modifier to hide those verts, that way we can still keep the solid mesh for Boolean operations but it can look hollow. You might have to invert the Mask modifier with that arrow button like I did.
Duplicate this mesh 2x and rotate on the z-axis 45 and -45 degrees. Use two booleans to subtract them from the main one. Make sure you take the vertex group off of the bool objects Mask modifiers, otherwise it will un mask the "carved out" areas:
With that animated you have something that looks like this.
Add that to their own collection and hide it. Make sure the center of your objects is at the origin of the world.
Then just add 4 instances of that collection and rotate them around eachother. Parent all of those to another object to control the whole build.
End result:
• Hi, thanks for the response, but I must say I didnt quite understood. Can you please add more pics ? Commented Apr 27, 2022 at 13:27
• I updated my answer. Commented Apr 27, 2022 at 16:25
• @Jakemoyo Great solution! But you get only the roof and unfortunately not the underlying structure which is also visible in the picture. Commented Apr 27, 2022 at 17:53
• @quellenform You could probably do the exact same thing except set the boolean to intersect instead of difference and get that result His problem was regarding boolean errors during animation. I'm not trying to do the entire project for him. Commented Apr 27, 2022 at 18:57
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FutureStarr
4 1 3 As an Improper Fraction
## 4 1 3 As an Improper Fraction
The 4 1 3 As a Improper Fraction works out in decimal form as 3/16. In binary form, it is 0. 1111111010101101.
### Great
via GIPHY
Fractions that are greater than 0 but less than 1 are called proper fractions. In proper fractions, the numerator is less than the denominator. When a fraction has a numerator that is greater than or equal to the denominator, the fraction is an improper fraction. An improper fraction is always 1 or greater than 1. And, finally, a mixed number is a combination of a whole number and a proper fraction.
A fraction can be identified as proper or improper by comparing the numerator and the denominator. Fractions that are less than one are known as proper fractions, and the numerator (the top number) is less than the denominator (the bottom number). A fraction with a numerator that is greater than or equal to the denominator is known as an improper fraction. It represents a number greater than or equal to one. Numbers that are not whole numbers, but are greater than one, can be written as improper fractions or mixed numbers. A mixed number has a whole number part and a fraction part. (Source: www.montereyinstitute.org)
### Improper
via GIPHY
Improper fractions can also be represented as a mixed number. To convert an improper fraction into a mixed number, divide the numerator by the denominator. The resultant becomes the whole number, and the remainder becomes the numerator of the new fraction. The denominator of the new fraction is the same as the original denominator. If there is no remainder, then there is no fraction--the result is simply a whole number.
This is called a mixed fraction. Thus, an improper fraction can be expressed as a mixed fraction, where quotient represents the whole number, remainder becomes the numerator and divisor is the denominator. A fraction, where the numerator is less than the denominator is called the proper fraction for example, $$\frac{2}{3}$$, $$\frac{5}{7}$$, $$\frac{3}{5}$$ are proper fractions. A fraction with numerator 1 is called a unit fraction.Each whole circle has 3 pieces. You can multiply the number of whole circles, 2, by 3 to find how many one-third pieces are in the two whole circles. Then you add 1 for the one-third piece in the final, incomplete circle. As you can see from the diagram, there are 7 individual one-third pieces. The improper fraction for (Source: www.montereyinstitute.org)
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# Course in Probability, A, 1st edition
• Neil A. Weiss
###### Course in Probability, A
ISBN-13: 9780201774719
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### What's included
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## Overview
This text is intended primarily for readers interested in mathematical probability as applied to mathematics, statistics, operations research, engineering, and computer science. It is also appropriate for mathematically oriented readers in the physical and social sciences. Prerequisite material consists of basic set theory and a firm foundation in elementary calculus, including infinite series, partial differentiation, and multiple integration. Some exposure to rudimentary linear algebra (e.g., matrices and determinants) is also desirable. This text includes pedagogical techniques not often found in books at this level, in order to make the learning process smooth, efficient, and enjoyable.
KEY TOPICS: Fundamentals of Probability: Probability Basics. Mathematical Probability. Combinatorial Probability. Conditional Probability and Independence. Discrete Random Variables: Discrete Random Variables and Their Distributions. Jointly Discrete Random Variables. Expected Value of Discrete Random Variables. Continuous Random Variables: Continuous Random Variables and Their Distributions. Jointly Continuous Random Variables. Expected Value of Continuous Random Variables. Limit Theorems and Advanced Topics: Generating Functions and Limit Theorems. Additional Topics.
MARKET: For all readers interested in probability.
(Chapter Opener and Review appear in each chapter).
I. FUNDAMENTALS OF PROBABILITY.
1. Probability Basics.
Biography: Girolamo Cardano.
From Percentages to Probabilities.
Set Theory.
2. Mathematical Probability.
Biography: Andrei Kolmogorov.
Sample Space and Events.
Axioms of Probability.
Specifying Probabilities.
Basic Properties of Probability.
3. Combinatorial Probability.
Biography: James Bernoulli.
The Basic Counting Rule.
Permutations and Combinations.
Applications of Counting Rules to Probability.
4. Conditional Probability and Independence.
Biography: Thomas Bayes.
Conditional Probability.
The General Multiplication Rule.
Independent Events.
Bayes' Rule.
II. DISCRETE RANDOM VARIABLES.
5. Discrete Random Variables and Their Distributions.
Biography: Siméon-Dennis Poisson.
From Variables to Random Variables.
Probability Mass Functions.
Binomial Random Variables.
Hypergeometric Random Variables.
Poisson Random Variables.
Geometric Random Variables.
Other Important Discrete Random Variables.
Functions of a Discrete Random Variable.
6. Jointly Discrete Random Variables.
Biography: Blaise Pascal.
Joint and Marginal Probability Mass Functions: Bivariate Case.
Joint and Marginal Probability Mass Functions: Multivariate Case.
Conditional Probability Mass Functions.
Independent Random Variables.
Functions of Two or More Discrete Random Variables.
Sums of Discrete Random Variables.
7. Expected Value of Discrete Random Variables.
Biography: Christiaan Huygens.
From Averages to Expected Values.
Basic Properties of Expected Value.
Variance of Discrete Random Variables.
Variance, Covariance, and Correlation.
Conditional Expectation.
III. CONTINUOUS RANDOM VARIABLES.
8. Continuous Random Variables and Their Distributions.
Biography: Carl Friedrich Gauss.
Introducing Continuous Random Variables.
Cumulative Distribution Functions.
Probability Density Functions.
Uniform and Exponential Random Variables.
Normal Random Variables.
Other Important Continuous Random Variables.
Functions of a Continuous Random Variable.
9. Jointly Continuous Random Variables.
Biography: Pierre de Fermat.
Joint Cumulative Distribution Functions.
Introducing Joint Probability Density Functions.
Basic Properties of Joint Probability Density Functions.
Marginal and Conditional Probability Density Functions.
Independent Continuous Random Variables.
Functions of Two or More Continuous Random Variables.
Sums and Quotients of Continuous Random Variables.
Multidimensional Transformation Theorem.
10. Expected Value of Continuous Random Variables.
Biography: Pafnuty Chebyshev.
Expected Value of a Continuous Random Variable.
Basic Properties of Expected Value.
Variance, Covariance, and Correlation.
Conditional Expectation.
The Bivariate Normal Distribution.
IV. LIMIT THEOREMS AND ADVANCED TOPICS.
11. Generating Functions and Limit Theorems.
Biography: William Feller.
Moment Generating Functions.
Joint Moment Generating Functions.
Laws of Large Numbers.
The Central Limit Theorem.
Biography: Sir Ronald Fisher.
The Poisson Process.
Basic Queueing Theory.
The Multivariate Normal Distribution.
Sampling Distributions.
Appendices.
Index.
## For teachers
All the material you need to teach your courses.
Discover teaching material
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Jumping on the Earth ```Name: N/A Status: N/A Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: WOULD A PERSON JUMPING ON THE GROUND HAVE ANY EFFECT WHAT SO EVER ON THE MOVEMENT OF THE EARTH? MY FREIND HAS STATED THAT IT WOULD MOVE THE EARTH BECAUSE OF NEWTONS LAW ( ACTIONS HAVING EQUAL ) MY ARGUMENT IS THAT THE EARTH IS NOT A STATIONARY DEAD OBJECT, IT IS LIVING WITH MILLIONS OF EVENTS HAPPENING ALL THE TIME I.E. VOLCANOES ETC AND IT IS ALSO MOVING THROUGH SPACE. ALL THIS WOULD MEAN AN ACTION BY ONE TINY OBJECT SUCH AS A PERSON JUMPING OULD HAVE ALMOST 0 EFFECT PERHAPS 000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000001 Replies: Kevin, It sounds to me like you are both right. Newton's law certainly holds despite the large differences in mass. The fact that the earth has millions of such events occurring all the time is irrelevant, each one has an effect. Because of the large differences in masses, a person jumping on the earth would have a very tiny effect on the earth's motion. Remember also that there are two forces acting when a person jumps. There is the force from the person's legs propelling them upward and there is the force of gravity. The center of mass of the earth/person combination will stay the same. When the person jumps the earth will be pushed downward while the person is pushed upward. Gravity will act to bring the two back together with the same center of mass that they had at the beginning of the jump. Essentially there will be no lasting effect on either of them. Bradburn Click here to return to the Physics Archives
NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.
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Update: June 2012
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# What Is Contribution Margin? How To Calculate It?
Wajiha Danish | July 13 2023
In the dynamic business world, understanding financial metrics is essential for making informed decisions and assessing a company’s performance. Two crucial measures that provide insights into profitability are the contribution margin and gross profit margins. While both these metrics analyze revenue and costs, they serve distinct purposes and offer unique perspectives on a company’s financial health.
In this blog post, we delve into the comparison between contribution and gross profit margins, highlighting their definitions, calculations, and applications. By grasping the key differences between these metrics, you’ll better understand how they can guide your strategic decision-making and contribute to your business’s success.
## What is Contribution Margin
Contribution margin is a financial metric that measures a product’s or service’s profitability by determining the amount of revenue remaining after deducting variable costs. It helps businesses understand the portion of each sale, contributing to covering fixed costs and generating a profit.
In simple words, contribution margin measures the profitability of each unit sold or each dollar of revenue generated after deducting the variable costs directly associated with producing the goods or providing the services. It helps businesses understand how much revenue is available to cover fixed expenses and generate a profit.
### Contribution Margin Formula
The contribution margin formula is as follows:
Contribution Margin = Revenue – Variable Costs
Alternatively, it can be calculated on a per-unit basis using the following formula:
Contribution Margin per Unit = Selling Price per Unit – Variable Cost per Unit
By analyzing the contribution margin, businesses can evaluate the profitability of individual products, services, or business segments. It provides insights into the company’s financial health and helps in decision-making related to pricing, cost management, and product mix.
## How can one calculate contribution margin?
To calculate the contribution margin, follow these steps:
#### 1. Determine the Selling Price
Identify the selling price per unit of the product or service. It is the price at which it is sold to customers.
#### 2. Calculate the Variable Costs
Identify all the costs directly associated with producing or delivering the product or service that vary with the level of production or sales. These costs may include raw materials, direct labor, and other variable expenses. Sum up all the variable costs.
#### 3. Subtract Variable Costs from Selling Price
Subtract the total variable costs calculated in the previous step from the selling price. The result will be the contribution margin per unit.
#### 4. Calculate the Contribution Margin Ratio
To express the contribution margin as a percentage, divide the contribution margin per unit by the selling price per unit and multiply by 100.
## Here’s an example to illustrate the calculation
Let’s say a company sells widgets for \$50 each. The variable costs associated with each widget amount to \$20. We can calculate the contribution margin as follows:
Selling Price per Unit: \$50
Variable Costs per Unit: \$20
Contribution Margin per Unit = Selling Price per Unit – Variable Costs per Unit
Contribution Margin per Unit = \$50 – \$20 = \$30
Contribution Margin Ratio = (Contribution Margin per Unit / Selling Price per Unit) * 100
Contribution Margin Ratio = (\$30 / \$50) * 100 = 60%
In this example, the contribution margin per unit is \$30, which means that for each widget sold, \$30 is available to cover fixed costs and contribute to the company’s profit. The contribution margin ratio of 60% indicates that the contribution margin represents 60% of the selling price.
## What is the importance of contribution margin?
The contribution margin is an important financial metric business used to analyze the profitability of individual products or services. It represents the amount of revenue remaining after subtracting the variable costs associated with producing or delivering a product or service. The contribution margin helps assess the financial viability of specific offerings and supports decision-making processes in several ways:
#### 1. Profitability Analysis
The contribution margin provides insight into the profitability of individual products or services. By comparing the contribution margin of different offerings, businesses can identify which products or services generate higher profits and contribute more significantly to overall financial performance.
#### 2. Cost-Volume-Profit Analysis
The contribution margin is a key component in cost-volume-profit (CVP) analysis, which helps determine the breakeven point and assess the impact of sales volume changes on profitability. It helps businesses understand how sales, costs, or price changes will affect their bottom line.
#### 3. Pricing Decisions
Understanding the contribution margin is essential for setting appropriate product or service prices. By considering the variable costs and desired profit margins, businesses can establish pricing strategies that ensure profitability while remaining competitive.
#### 4. Product Mix Decisions
The contribution margin assists in evaluating the profitability of different product lines or variations. Businesses can determine which products or variations should be emphasized or discontinued by comparing the contribution margins of various offerings to optimize overall profitability.
#### 5. Decision-Making
A contribution margin is a valuable tool in making decisions related to resource allocation, such as determining which products to promote, which marketing initiatives to pursue, or which cost-reduction strategies to implement. It helps prioritize investments and allocate resources to maximize profitability.
## Endnote
Overall, the contribution margin is crucial for businesses as it provides insights into the profitability of products or services, aids in pricing decisions, supports cost-volume-profit analysis, and facilitates informed decision-making. By understanding the contribution margin, companies can optimize their operations, enhance profitability, and make strategic choices that align with their financial goals.
## Author Bio
Wajiha is a Brampton-based CPA, CGA, and Controller with 17+ years of experience in the financial services industry. She holds a Bachelor of Science Degree in Applied Accounting from Oxford Brookes University and is a Chartered Certified Accountant. Wajiha spearheads Monily as its Director and is a leader who excels in helping teams achieve excellence. She talks about business financial health, innovative accounting, and all things finances.
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# Dirt roads may evoke the bucolic simplicity of another
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Dirt roads may evoke the bucolic simplicity of another century, but financially strained townships point out that dirt roads cost twice as much as maintaining paved roads.
(B) dirt roads cost twice as much to maintain as paved roads do
(C) maintaining dirt roads costs twice as much as paved roads do
(D) maintaining dirt roads costs twice as much as it does for paved roads
(E) to maintain dirt roads costs twice as much as for paved roads
[Reveal] Spoiler: OA
Last edited by Narenn on 25 Sep 2013, 10:00, edited 1 time in total.
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16 Jun 2008, 00:36
Dirt roads may evoke the bucolic simplicity of another century, but financially strained townships point out that dirt roads cost twice as much as maintaining paved roads.
(B) dirt roads cost twice as much to maintain as paved roads do
(C) maintaining dirt roads costs twice as much as paved roads do
(D) maintaining dirt roads costs twice as much as it does for paved roads
(E) to maintain dirt roads costs twice as much as for paved roads
B Is the right answer, Righ usage of Sub+ Verb agreements (Cost) and Idiom As Much As.
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16 Jun 2008, 06:48
I will go with C.
1. Subject verb agreement..maintaining is the subject..maintaining costs.....
3. correct idiom used. X costs twice as much as Y
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16 Jun 2008, 06:57
B.
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ritula wrote:
Dirt roads may evoke the bucolic simplicity of another century, but financially strained townships point out that dirt roads cost twice as much as maintaining paved roads.
(B) dirt roads cost twice as much to maintain as paved roads do
(C) maintaining dirt roads costs twice as much as paved roads do
(D) maintaining dirt roads costs twice as much as it does for paved roads
(E) to maintain dirt roads costs twice as much as for paved roads
dirt roads cost twice as much to maintain as paved roads do
= dirt roads to maintain cost twice as much as paved roads do [to main]
other choices compare incorrectly.
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16 Jun 2008, 07:30
B is the correct answer here
costs (subject-verb agreement error here) so C D E are eliminated
B uses the correct idiom, and hence is the correct choice
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16 Jun 2008, 07:34
vdhawan1 wrote:
B is the correct answer here
costs (subject-verb agreement error here) so C D E are eliminated
B uses the correct idiom, and hence is the correct choice
C,D, E are not out because of (Sub-Verb agreement)
here subject is "maintaing dirt roads" not "dirt roads". they are out because of incorrect comparision.
"maintaining dirt roads" costs --- no issue.
"to maintain dirt roads" costs --- no issue
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16 Jun 2008, 07:46
x2suresh wrote:
vdhawan1 wrote:
B is the correct answer here
costs (subject-verb agreement error here) so C D E are eliminated
B uses the correct idiom, and hence is the correct choice
C,D, E are not out because of (Sub-Verb agreement)
here subject is "maintaing dirt roads" not "dirt roads". they are out because of incorrect comparision.
"maintaining dirt roads" costs --- no issue.
"to maintain dirt roads" costs --- no issue
Hi suresh
thanks for pointing this out
i think i really need to work hard on verbal
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16 Jun 2008, 08:19
Thanks 2 all for the discussion. OA is B
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16 Jun 2008, 09:46
i didn't get why "D" is won , can someone pease explain it to me?
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### Show Tags
16 Jun 2008, 10:19
Dirt roads may evoke the bucolic simplicity of another century, but financially strained townships point out that dirt roads cost twice as much as maintaining paved roads.
(B) dirt roads cost twice as much to maintain as paved roads do
(C) maintaining dirt roads costs twice as much as paved roads do
(D) maintaining dirt roads costs twice as much as it does for paved roads
An incorrect sentence can use correct grammar, but use too many words. Plus, here the use of "it" is kind odd. It is not entirely clear that "it" refers to "maintaining". B says the same thing in a much clearer manner. Sentence Correction questions say to pick the "best answer" not the only correct one.
gmat2ndtime wrote:
i didn't get why "D" is won , can someone pease explain it to me?
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21 Apr 2010, 15:15
Can anybody explain the wrong comparision in C??
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08 May 2014, 16:06
Can we kindly get an analysis of answer choices? In particular C and D please?
Would be happy to throw some Kudos out there!
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Can we kindly get an analysis of answer choices? In particular C and D please?
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Okay, let me go about explaining this.
C. There is an incorrect comparison between "maintaining dirt roads" and "paved roads." (X twice as much as Y structure.) Then there is a subject-verb disagreement as well. Something like this would have been awesome :- "maintaining dirt roads costs twice as much as maintaining paved roads."
D. Even though I chose this, at the hindsight this option looks awkward. "It" might refer to the proper antecedent "maintaining" but the construction "does for paved roads" doesn't look good to me (it did then )
B. Comparison between dirt roads and paved roads in terms of their action "cost" seems okay and there's no subject-verb disagreement.
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Dirt roads may evoke the bucolic simplicity of another century, but financially strained townships point out that dirt roads cost twice as much as maintaining paved roads.
(A) dirt roads cost twice as much as maintaining paved roads not comparing similar elements
(B) dirt roads cost twice as much to maintain as paved roads do
(C) maintaining dirt roads costs twice as much as paved roads do the subject who is performing these two actions is different
(D) maintaining dirt roads costs twice as much as it does for paved roads "it" is not clear and the rest is also not parallel
(E) to maintain dirt roads costs twice as much as for paved roads not comparing similar elements
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09 Oct 2016, 09:42
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KUDOS
Expert's post
Responding to a PM: Why B is better than D?
D has two problems: the pronoun "it" does not have an antecedent.
The compared elements are not parallel:
maintaining dirt roads costs twice as much as it does for paved roads.
The correct construction would be :
maintaining dirt roads costs twice as much as maintaining paved roads does.
In option B, the compared elements are parallel:
dirt roads cost twice as much to maintain as paved roads cost.
It is allowed to replaced the verb with "do" in the second element. Hence:
dirt roads cost twice as much to maintain as paved roads do.
Re: Dirt roads may evoke the bucolic simplicity of another [#permalink] 09 Oct 2016, 09:42
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# Business Analytics: Correlation is Not Causation
The New York Times reports that a respected psychology journal is due to publish a paper purporting to show “strong evidence” for extra-sensory perception:
The New York Times reports that a respected psychology journal is due to publish a paper purporting to show “strong evidence” for extra-sensory perception:
“A software program randomly posted a picture behind one curtain or the other — but only after the participant made a choice. Still, the participants beat chance, by 53 percent to 50 percent, at least when the photos being posted were erotic ones. They did not do better than chance on negative or neutral photos.”
Crucially, no “topflight statisticians” were part of the peer review. When I was at university, struggling to use a sophisticated statistics package on a mainframe as part of my econometrics degree, I dreamed of having a program that would just cruise through all the possible combinations of variables, and tell me which ones were correlated. That ability now exists, but the danger is that few people realize how much higher the bar must be set for a result to be deemed significant in such circumstances.
Given a large enough set of random numbers, you will always be able to find a “significant” relationship – especially if that’s exactly what you’re looking and hoping for.
To me, the experiment above sounds like it may have this problem – for example, if there were lots of different categories of photos, and the “significant” relationship was cherry-picked from the available results. And even if the level of significance has indeed been increased to take account of this, the result could still be random (if there’s a choice between changing everything we know about science and it being a fluke result, I’m going with the latter).
In science, thankfully, it’s easy for somebody else to repeat the experiment and validate the correlation, ideally before a respected journal makes a fool of itself (although I suspect they’re simply making a calculated bid for more publicity, and it’s working very successfully).
In business, it’s much harder to know if your “results” are valid. The same problem exists – people are looking for a certain type of result, and keep running the numbers until they find something that looks like a relationship: “Look! Customer satisfaction is correlated with their age!” . But it’s much harder to “rerun the experiment”, and businesses don’t always have/take the time to check their results.
Despite having worked in BI for over twenty years (or maybe because of it), I’m deeply distrustful of most corporate analytics. I believe business analytics is essential, but that it’s also essential to assume that any relationship you find is a working hypothesis, to be validated through further analysis (e.g. correlation is not causation), and expert discussion (as with peer-reviewed science papers, the best way to deal with potential analysis problems is greater transparency — social BI technologies like Streamwork are becoming increasingly important).
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# Using Padé approximants for the quaternion exponential
On a lark, I wanted to know if one can use Padé approximants to compute the exponential $\exp(z)$ of a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$. Since Mathematica has a package meant for manipulating quaternions, as well as computing the exponential of a quaternion, I decided to try things out.
The numerator $N_{pq}(z)$ and denominator $D_{pq}(z)$ of the $(p,q)$ Padé approximant for the exponential function can be computed using the following formulae:
$\displaystyle N_{pq}(z)=\sum_{j=0}^p \frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}z^j$
$\displaystyle D_{pq}(z)=\sum_{j=0}^q \frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}(-z)^j$
after which the Padé approximant is $R_{pq}(z)=\frac{N_{pq}(z)}{D_{pq}(z)}$.
For simplicity (and since they are the ones most used in practice), I will only consider here the "diagonal" case, $p=q$.
The case where $z$ is a matrix is well-studied, and it is known that $R_{pq}(z)=N_{pq}(z)\cdot D_{pq}(z)^{-1}=D_{pq}(z)^{-1}\cdot N_{pq}(z)$. Banking on the isomorphism of quaternions to specially constructed matrices, I expected that the order of multiplication won't matter in the quaternion case, and experimentation seems to show that this holds.
What was mystifying in my experiments is that, although in general $|R_{pp}(z)-\exp(z)|$ can be made small enough by taking $p$ large enough, one pattern I saw seems to indicate that
$R_{pp}(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})\approx \overline{\exp(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})}$
for $b>0$, where the bar denotes conjugation.
I can't see an immediate equivalent of this phenomenon in the theory of the matrix exponential, so I'm wondering why the Padé approximants to the quaternion exponential behave this way.
For those who want to retry my Mathematica experiments:
<<AlgebraQuaternions
z=Quaternion[1.,-1.,1.,1.](*Quaternion[2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1]*)
p=q=8;
num=Sum[(((p+q-j)!p!)/((p+q)!j!(p-j)!)) z^j, {j, 0, p}];
den=Sum[(((p+q-j)!q!)/((p+q)!j!(q-j)!)) (-z)^j, {j, 0, q}];
{num**(den^-1),(den^-1)**num,Exp[z],Conjugate[Exp[z]]}
-
Hmm. Can this be explained starting from the observation that if $z=a+bi+cj+dk$, then $izi^{-1}=a+bi-cj-dk$? IOW different kind of conjugations working somehow together? Anyway, you seem to be mostly working inside the ring $\mathbf{R}[z]$, and that is just a copy of $\mathbf{C}$: write $z=a+ru$, with $r=\sqrt{b^2+c^2+d^2}$ and $u$ a suitable unit vector. Therefore $u^2=-1$, and you may as well equate $z$ with the complex number $a+ri$. Looks like your Padé approximants never leave this copy of $\mathbf{C}$, so the results from the complex numbers should carry over to quaternions. – Jyrki Lahtonen Oct 30 '11 at 20:16
Wow, I haven't looked at this in a while! Interesting thoughts, @Jyrki. – Ｊ. Ｍ. Oct 30 '11 at 22:46
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# Levers are the subject of this part of the unit
Subjects:
Art, Language Arts, Math, Science
3, 4
Title – Simple Machines Unit Lesson 6
By – Debbie Haren
Subject – Math, Science, Art, Language Arts
Grade Level – 3/4
Simple Machines Unit Lesson 6
Question:
Can anyone tell me what a lever is and what it is used for?
I will pass out a paper towel roll to each of you I want you to cut it in half, length wise. To make a stable place to balance items on.
On top of the paper towel holder I want you to put a ruler. What do you think we might be building? Does it remind you of seesaw? Place the paper tube flat side down
Balance a ruler across the tube. Put a counter on one end. Have students chart in their journal what happens. Put a counter on the other end and then see what happens. Some things to use as counters are small bear counters, dice, marbles or chain.
Discussion:
What happened when you put a counter on one end of the seesaw?
What happened when you put a counter on each end?
Where is the turning point of the seesaw?
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# What is the logarithm or root operation in type-space?
I was recently reading The Two Dualities of Computation: Negative and Fractional Types. The paper expands on sum-types and product-types, giving semantics to the types a - b and a/b.
Unlike addition and multiplication, there are not one but two inverses of exponentiation, logarithms and rooting. If function types (a → b) are type-theoretic exponentiation, given the type a → b (or b^a) what does it mean to have the type logb(c) or the type a√c?
Does it make sense to extend logarithms and roots to types at all?
If so, has there been any work in this area, and what are some good directions on how to comprehend the repercussions?
I tried looking up information on this via logic, hoping the Curry-Howard correspondence could help me, but to no avail.
A type $C$ has a logarithm to base $X$ of $P$ exactly when $C \cong P\to X$. That is, $C$ can be seen as a container of $X$ elements in positions given by $P$. Indeed, it's a matter of asking to what power $P$ we must raise $X$ to obtain $C$.
It makes sense to work with $\mathop{log}F$ where $F$ is a functor, whenever the logarithm exists, meaning $\mathop{log}\!_X(F\:X)$. Note that if $F\:X\cong \mathop{log}F\to X$, then we certainly have $F\:1\cong 1$, so the container tells us nothing interesting other than its elements: containers with a choice of shapes do not have logarithms.
Familiar laws of logarithms make sense when you think in terms of position sets
$$\begin{array}{rcl@{\qquad}l} \mathop{log} (\mathop{K}1) &=& 0 & \mbox{no positions in empty container}\\ \mathop{log} I &=& 1 & \mbox{container for one, one position}\\ \mathop{log} (F\times G) &=& \mathop{log}F+\mathop{log}G & \mbox{pair of containers, choice of positions} \\ \mathop{log} (F\cdot G) &=& \mathop{log}F\times\mathop{log}G & \mbox{container of containers, pair of positions} \end{array}$$
We also gain $\mathop{log}\!_X(\nu Y. T) = \mu Z. \mathop{log}\!_X T$ where $Z=\mathop{log}\!_XY$ under the binder. That is, the path to each element in some codata is defined inductively by iterating the logarithm. E.g.,
$$\mathop{log}\mathop{Stream} = \mathop{log}\!_X(\nu Y. X\times Y) = \mu Z. 1 + Z = \mathop{Nat}$$
Given that the derivative tells us the type in one-hole contexts and the logarithm tells us positions, we should expect a connection, and indeed
$$F\:1\cong 1 \;\Rightarrow\; \mathop{log}F\cong \partial F\:1$$
Where there is no choice of shape, a position is just the same as a one-hole context with the elements rubbed out. More generally, $\partial F\:1$ always represents the choice of an $F$ shape together with an element position within that shape.
I'm afraid I have less to say about roots, but one could start from a similar definition and follow one's nose. For more uses of logarithms of types, check Ralf Hinze's "Memo functions, polytypically!". Gotta run...
• The answer from Da Man himself. Welcome Conor! Commented Mar 25, 2013 at 15:17
• Hmm, I am interested to see what root types are, as they would require types having imaginary numbers of inhabitants. Unless I am wrong. I will accept your answer, but if you have the time to elaborate on roots that would be much appreciated. Commented Mar 26, 2013 at 1:56
• Can this be related to the Taylor series of ln(1+x) somehow? Commented Mar 26, 2013 at 15:22
• With logarithms and exponentials, I wonder... what do we need to construct a Napier object? (e.g. the supposedly unique object e such that ∂e = e) Commented Mar 29, 2013 at 12:50
I don't know of any work that pursues this line, but a few moments thought about it led me to this hypothesis: wouldn't the "root" of the exponential type just be the codomain, and the "logarithm" of the exponential just the domain?
• Right, so I think your intuition is good but your conclusion is off. The root operation and the logarithm operation are what you get when you "invert" the codomain or the domain respectively, not the (co)domain's themselves. The question is, what do we mean by invert and what is the binary type operation it produces? Commented Mar 24, 2013 at 22:11
• Not sure that's right. If I have $x^y$, the $y$th root is $x$ and the base $x$ logarithm is $y$. The inversion is of the operation, not the components. Commented Mar 24, 2013 at 22:35
• Sorry, I have not been totally clear in my terminology. I do not mean to ask "what is the root, what is the result of applying the logarithm function". I am wondering what the operation of rooting is. What the operation of finding the logarithm is. If → is expenentiation, what is a two types under the root operation. What is two types under the logarithm operation. What I mean by "invert the argument" is something there is not time to explain here. I will clarify my question, thanks. Commented Mar 25, 2013 at 5:33
• The paper I linked provides a semantics for the type a - b and the type a / b. I am not concerned with the result of reducing the operations logarithm and root, but in understanding their semantics as binary type operators. Commented Mar 25, 2013 at 5:50
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Select Page
How do I construct the contingency table of the relative frequencies?2:20 AM Mon Jun 17 fl blackboardcppedu ‘1" 3%E} QUESTION 08 (10 points) – Contingency Table 8: Marginal DistributionsA hypothetical survey of car owners generates the following contingency table. SUM(a)[2] Construct the contingency table of relative frequencies. (5 decimals; not percentages)(b)[1] Find the marginal relative frequency distribution for Choice of Automotive Magazines. (c)[1] Find the marginal relative frequency distribution for Car Ownership. ?? (d)[4] Draw a 2-D column chart and a 3-D column chart in Excel to visualize the table in part (a).(e)[2] Based on part (cl), is there any possible relation between the two variables? Describe!4 QUESTION 09 (5 points) – A Past Exam QuestionRandom variable 2 follows the standard normal distribution. In each of the following, sketch the
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Open Source RDBMS - Seamless, Scalable, Stable and Free
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Versions available for this page: CUBRID 8.4.0 | CUBRID 8.4.1 | CUBRID 8.4.3 | CUBRID 9.0.0 |
STDDEV/STDDEV_POP Functions
Description
The functions STDDEV and STDDEV_POP are used interchangeably and they return a standard deviation of the values calculated for all rows. The STDDEV_POP function is a standard of the SQL:1999. Only one expression is specified as a parameter. If the DISTINCT or UNIQUE keyword is inserted before the expression, they calculate the standard deviation after deleting duplicates; if keyword is omitted or ALL, they it calculate the standard deviation for all values.
The return value is the same with the square root of it's variance (the return value of VAR_POP Function) and it is a DOUBLE type. If there are no rows that can be used for calculating a result, NULL is returned.
The following is a formula that is applied to the function.
STDDEV_POP = [ (1/N) * SUM( { xI - AVG(x) }2 ) ]1/2
• SUM: Sum
• AVG: Average
Warning In CUBRID 2008 R3.1 or earlier, the STDDEV function worked the same as the STDDEV_SAMP Function.
Syntaxs
STDDEV_POP( [ { DISTINCT | DISTINCTROW } | UNIQUE | ALL] expression )
• expression: Specifies an expression that returns a numeric value.
• ALL: Calculates the standard deviation for all data (default).
• DISTINCT or UNIQUE: Calculates the standard deviation without duplicates.
Example
CREATE TABLE test_table (d DOUBLE);
INSERT INTO test_table VALUES(78), (63.65), (230.54), (32), (17.2), (195.7689), (57.57);
SELECT STDDEV_POP(d) FROM test_table;
stddev_pop(d)
==========================
7.672456168942171e+01
SELECT STDDEV_POP(POWER(d,2)+d*2+1) FROM test_table;
stddev_pop( power(d, 2)+d*2+1)
================================
1.995964904967644e+04
TRUNCATE TABLE test_table;
SELECT STDDEV_POP(d) FROM test_table;
stddev_pop(d)
==========================
NULL
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0
# Is -12 smaller than-13
Wiki User
2017-03-24 10:57:47
1
Harry Rogahn
Lvl 10
2021-02-25 22:17:54
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Wiki User
2017-03-24 10:57:47
No because minus 12 is greater than minus 13
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Author Message
Maton
Registered: 03.12.2003
From: Baltimore
Posted: Monday 20th of Aug 19:06 Hello folks ! I have a serious problem about math and I was hoping that someone might have the ability to help me out a little . I have a algebra test pretty soon and even though I have been studying math seriously, there are still a few parts that cause a lot of problems , such as equations and graphs and algebraic signs especially. Last week I had a class with a math teacher, but many things still remain unclear to me. Can you suggest a good way of studying or a good private teacher that you know already?
ameich
Registered: 21.03.2005
From: Prague, Czech Republic
Posted: Tuesday 21st of Aug 10:52 Aaah May Jesus save us students from the wrath of equations and graphs. I used to face same problems that you do when I was there. I always used to be confused in Remedial Algebra, Intermediate algebra and Basic Math. I was worst in equations and graphs until I came to know of Algebrator. It is really effective and I would surely recommend it. The best feature of the software is that it will also help you learn algebra and not just give your answers. I found Algebrator effective and am sure it will help you too. Cheers.
sxAoc
Registered: 16.01.2002
From: Australia
Posted: Tuesday 21st of Aug 15:19 I remember I faced similar difficulties with graphing lines, x-intercept and difference of cubes. This Algebrator is truly a great piece of algebra software program. This would simply give step by step solution to any math problem that I copied from workbook on clicking on Solve. I have been able to use the program through several Algebra 1, Remedial Algebra and Algebra 2. I seriously recommend the program.
MichMoxon
Registered: 21.08.2001
From:
Posted: Wednesday 22nd of Aug 12:28 I remember having often faced problems with midpoint of a line, sum of cubes and adding numerators. A really great piece of math program is Algebrator software. By simply typing in a problem homework a step by step solution would appear by a click on Solve. I have used it through many math classes – College Algebra, Intermediate algebra and Remedial Algebra. I greatly recommend the program.
slockerbay
Registered: 15.07.2007
From:
Posted: Thursday 23rd of Aug 10:10 Friends , Thanks a lot for the responses that you have offered. I just had a look at the Algebrator available at: http://www.mathpoint.net/algebra-1.html. The best part that I liked was the pay back guarantee that they are extending there. I went ahead and purchased Algebrator. It is really easy to handle and turns up to be a noteworthy tool for Remedial Algebra.
Ashe
Registered: 08.07.2001
From:
Posted: Friday 24th of Aug 21:56 It is available at http://www.mathpoint.net/parabolas.html. You can take a look at the demos, compare its features with others available in the market and see for yourself. I am sure you will be impressed.
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Drawing gamma function in LaTeX
Is there an elegant way to plot gamma function (or some similar non-elementary function),
http://en.wikipedia.org/wiki/Gamma_function
via TikZ or pgfplots? The graph I have in mind is something similar to the following one,
http://intmath.com/blog/wp-content/images/2010/05/gamma-4.gif
The only solution that falls to my mind is the calculation of a list of coordinates (x,\Gamma(x)) with some other (non-TeX) software and then plotting those, but I would like to know if it is possible to avoid this path.
• Hello. Have you tried searching this forum (or anywhere else on the web) for ways how to plot a function in LaTeX?
– yo'
Commented Jan 20, 2015 at 21:41
• TikZ has a set of built-in (predefined) elementary functions (such as exp, ln, sin, etc.) so there is no problem with those. I'm note sure if there is some other way to draw \Gamma(x), apart from e.g. generating the coordinates (x,\Gamma(x)) with some other software, and then plotting those in TikZ or pgf? Commented Jan 20, 2015 at 21:48
• Well, the it would be nice to show your effort, or to be more specific. This way it looks like you didn't search at all. And no, I don't think that the gamma function exists...
– yo'
Commented Jan 20, 2015 at 21:53
• I'd try gnuplot. It has an implementation of the gamma function. Commented Jan 20, 2015 at 22:36
• Using essentially the same code I posted for plotting the zeta function, you can plot gamma. You need to import that function, though, as is done here.
– DJP
Commented Jan 20, 2015 at 22:42
There is an elegant way… with the pst-func package:
\documentclass[x11names]{article}
\usepackage{pst-func}
\usepackage{auto-pst-pdf}
\begin{document}
\psset{unit=1.25cm}
\begin{pspicture*}(-4.8,-4.8)(4.8,4.8)
\psaxes[ticksize=2pt -2pt]{->}(0,0)(-4.8,-4.8)(4.8,4.8)[$x$, -120][$y$, -140]
\psset{linecolor=Tomato3,linewidth=1.2pt,plotpoints=100,algebraic}
\psplot{-4.995}{-4.005}{GAMMA(x)}
\psplot{-3.995}{-3.01}{GAMMA(x)}
\psplot{-2.99}{-2.05}{GAMMA(x)}
\psplot{-1.95}{-1.05}{GAMMA(x)}
\psplot{-0.9}{-0.1}{GAMMA(x)}
\psplot{.1}{5.8}{GAMMA(x)}
\psset{linewidth = 0.6pt, linecolor = LightSteelBlue3}
\multido{\i=-4+1}{4}{\psline(\i, -5.8)(\i, 5.8)}
\end{pspicture*}
\end{document}
• I suppose the implication is that GAMMA is a built-in function? neat! Commented Jan 21, 2015 at 2:41
• It's defined in the pst-math module, that called by pst-func. Also defined: GAMMALN. Commented Jan 21, 2015 at 9:39
• @Bernard: Absolutely great! I couldn't ask for more elegant solution. Thank you! Commented Jan 21, 2015 at 9:41
1. With pgfplots and gnuplot:
% arara: pdflatex: {shell: yes}
\documentclass[margin=5mm, tikz]{standalone}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}[]
\begin{axis}[
xmin = -4.9, xmax = 5.1,
%ymin = -3.5, ymax = 3.5,
restrict y to domain=-6:6,
axis lines = middle,
axis line style={-latex},
xlabel={$x$},
ylabel={$y$},
%enlarge x limits={upper={val=0.2}},
enlarge y limits=0.05,
x label style={at={(ticklabel* cs:1.00)}, inner sep=5pt, anchor=north},
y label style={at={(ticklabel* cs:1.00)}, inner sep=2pt, anchor=south east},
]
domain = 0:5] gnuplot{gamma(x)};
\foreach[evaluate={\N=\n-1}] \n in {0,...,-5}{%
domain = \n:\N] gnuplot{gamma(x)};
%
\addplot [domain=-6:6, samples=2, densely dashed, thin] (\N, x);
}%
\end{axis}
\end{tikzpicture}
\end{document}
2. Only pgfplots as an approximation:
We can see a good accordance with the exact gnuplot-curve:
\documentclass[margin=5mm, tikz]{standalone}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}[]
\begin{axis}[
xmin =0, xmax = 5.1,
%ymin = -3.5, ymax = 3.5,
restrict y to domain=-6:6,
axis lines = middle,
xlabel={$x$},
ylabel={$y$},
enlarge x limits=0.05,
enlarge y limits=0.1,
x label style={at={(ticklabel* cs:1.00)}, inner sep=5pt, anchor=north},
y label style={at={(ticklabel* cs:1.00)}, inner sep=2pt, anchor=south east},
]
domain = 0:5] {sqrt(2*pi)*x^(x-0.5)*exp(-x)*exp(1/(12*x))};
\end{axis}
\end{tikzpicture}
\end{document}
The Gamma function is also built in Asymptote. Here is a reproduction of Bernard's graph with this program.
import graph;
unitsize(1.5cm);
real Xmin=-5, Xmax=5, Ymin=-5, Ymax=5;
// Graphs
for (real x = Xmin; x < 0; x=x+1) {draw(graph(gamma, x+0.001, x+0.999), red);};
draw(graph(gamma, 0.1, Xmax), red);
// Clipping (cut the parts beyond the borders)
clip((Xmin, Ymin) -- (Xmax, Ymin) -- (Xmax, Ymax) -- (Xmin, Ymax) -- cycle);
// Asymptotes (no pun here)
for (real x=Xmin; x<0; x=x+1) {draw((x, Ymin)--(x, Ymax), 0.8*white);};
// Axes
xaxis(xmin=Xmin, xmax=Xmax, Ticks(Step=1, OmitTick(0, Xmax)), arrow=Arrow);
yaxis(ymin=Ymin, ymax=Ymax, Ticks(Step=1, OmitTick(0, Ymin, Ymax)), arrow=Arrow);
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3.2 Linear Regression
Regression is the analytical technique for determining the relationship between a dependent variable and an independent variable. When the two variables have a linear correlation, you can develop a simple mathematical formula of the relationship between the two variables by finding the line of best fit.
Interpolation (estimating between data points) and Extrapolation (estimating beyond the range of data) are used for predicting the equation for this line.
Generally, you can "eyeball" a good estimate of the line of best fit on a scatter plot when the linear correlation is strong. But an analytical method of using a least square fit gives more accurate results, especially for the weaker correlations.
For the line of best fit in the least-squares method,
• the sum of the residuals is zero (the positive and negative residuals cancel out)
• the sum of the squares of the residuals has the least possible value
It can be shown that this line has the equation
y=ax+b, where a= n(∑xy) – (∑x)( ∑y)/n(∑x2) – (∑x)2
Example 1: Applying the Least-Squares Formula
This table shows data for the full- time employees of a small packaging company.
Age (years) Annual Income (\$000)
33 33
25 31
19 18
44 52
50 56
54 60
38 44
29 35
a) Use a scatter plot to classify the correlation between age and income.
The scatter plat shows a strong linear correlation between age and income level.
b) Find the equation of the line of best fit analytically.
Age ,x Income,y x2 xy
33 33 1089 1089
25 31 625 775
19 18 361 342
44 52 1936 2288
50 56 2500 2800
54 60 2916 3240
38 44 1444 1672
29 35 841 1015
∑x=292 ∑y=329 ∑x2=11712 ∑xy=13221
Substitute these totals into the formula for a.
a= n(∑xy) – (∑x)( ∑y)
n(∑x2) – (∑x)2
=8(13221) – (292)(329)
8(11712) – (292)2
=1.15
Now, you need the means of x and y.
Mean of x= ∑x/n mean of y=∑y/n b=41.125 – 1.15(36.5)
=292/8 =329/8 = -0.85
=36.5 =41.125
Now substitude the values of a and b into the equation for the line of best fit.
y = ax + b
1.15x – 0.85
Therefore, the equation of the line of best fit is y = 1.15x – 0.85.
c) Predict the income for a new employee who is 21 and for an employee retiring at 65.
Substitute:
y = ax + b
For the 21 year old: y = 1.15(21) – 0.85
= 23.3
For the 65 year old: y = 1.15(65) – 0.85
=73.9
Therefore, if predicted properly the new employee’s income would be \$23300 and for the retiring employee would have an income of about \$73900.
Comment:
By: Osman Osman
Nicely done…. Good job
I think u have the best explanation among all..!!
Peace *__*
page revision: 11, last edited: 09 May 2007 17:09
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-Share Alike 2.5 License.
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# 03 kinetics _energetics
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### 03 kinetics _energetics
1. 1. Medical Chemistry Lecture 3 2007 (J.S.) Kinetics of chemical reactions Chemical equilibrium Energy in chemical reactions Free Gibbs energy – the driving force of chemical reactions
2. 2. The fundamental terms in reaction kinetics Kinetics studies the rates (and mechanisms) of chemical reactions. The term velocity (symbol v ) is the reaction rate expressed in terms of change in the concentrations of reactants: For the simple reaction S P, the velocity is defined as S – substrate, P – product, ν – reaction stoichiometric coefficients (if there are any) Factors affecting velocities of reactions: temperature, concentrations of reactants, catalysts or inhibitors. ] S [ t – v 1 ν ] P [ t 1 ν v = c – t Because , velocity is expressed in mol × l –1 × s –1
3. 3. where k is the kinetic constant that includes the specific reaction features as well as the temperature term ( k = A × e –Ea I R T ) . Due to decreasing concentrations of reactants, there must be always a gradual decrease of reaction velocity in closed systems till the reaction reaches the equilibrium . The sum of all exponents in velocity equations (m + n + ….) indicates the reaction order . The equation mentioned above is a (m+n) th -order reaction. Velocity depends on the concentrations of reactants This dependence is described in the velocity equation : For the reaction mA + nB xC, v = k [ A ] m [ B ] n
4. 4. Progress curves ( kinetic curves ) The progress - the time course of a reaction - is shown by a plot of the concentration of any of the substrates or products against time. The instantaneous velocity v x at any particular time t x is then given by the slope of the tangent to the curve at that time. For the first-order reactions [S] t = [S] 0 e – k t or v t = v 0 e – k t [S] 0 – initial concentration of S, v 0 – initial velocity, in the first moments of the reaction Example : Both curves hold for the reaction S P. It is a first-order reaction according to the velocity equation v = k [S] . At equilibrium the net reaction velocity is zero. [ S ] t x tg α = d [S] / d t = v x α t x [ P ] t [ S ] t (equilibrium)
5. 5. First-order reactions are most commonly decomposition reactions, radioactive decay, isomerization and rearrangements, and simple enzyme-catalyzed reactions involving a single substrate at low concentration. The kinetic constant k of a first-order reaction can be found from the slope of the straight line in the plot of log [S] versus time. The half-life t ½ of a first-order reaction is the time it takes for one-half of a reactant to undergo a reaction. The half-life of a first-order reaction is independent of the concentration of the reactant and equals t ½ = 0,69 / k . [S] 0 t [S] 0 /2 [S] 0 /4 t ½ 2 t ½
6. 6. Second- and higher-order reactions will not be discussed in this course. Zero-order reactions The velocity of a zero-order reaction does not depend on the reactant concentration, it is constant ( v = k [S] 0 = k ). The velocity of such reactions is controlled by other factors than by collisions of the reactants involved. In enzyme-catalyzed reactions , the velocity depends on the concentration of the enzyme-substrate complex. If the substrate concentration is higher than required for the full enzyme saturation, the reaction is zero order for some time . However, after the concentration decreases and the enzyme is not fully saturated, the reaction becomes first- or higher-order. [ S ] t Zero-order kinetics First- or higher-order kinetics
7. 7. If v 1 = v 2 in equilibrium state, then there exists a constant ratio for the particular reaction – the equilibrium constant K : Chemical equilibrium In closed systems, the reactions proceed to certain point and then apparently stop and leave considerable amounts of unaffected reactants. A dynamic, chemical equilibrium is reached because the velocities of the forward and reverse reactions are equal. Reversible reactions do not go to completion. Basically, every reaction (even a " irreversible“) can be viewed as the formation of the equilibrium between the starting reactants and the reaction products. The forward reaction rate v 1 = k 1 [A] m [B] n The reverse reaction rate v 2 = k 2 [C] p [D] q Reversible reaction m A + n B p C + q D v 1 v 2 k 1 k 2 K c =
8. 8. The law of chemical equilibrium (Guldberg and Waage, 1867) When a closed system is at equilibrium, it will remain in this state indefinitely unless the equilibrium is affected in some manner by external factors . "Position“ of the equilibrium: K » 1 – the equilibrium is "on the right“, it favours the reaction products K « 1 – the equilibrium is established "on the left“ favouring the reactants K ≈ 1 – the reaction is "perfectly reversible“ Catalysts do not change the value of K , they cause a system to reach equilibrium more quickly. For any reversible reaction at chemical equilibrium and particular temperature, there exist a fixed ratio among concentrations of products and reactants expressed as the equilibrium constant K .
9. 9. External factors upsetting equilibria Changes in equilibrium concentrations by adding more of the reactants or by removing of some of the products change the reaction rates: the rates of the forward and reverse reactions will continue to change until equilibrium that corresponds to the value of equilibrium constant K is reached again. The value of K does not change ; at restored equilibrium, the concentrations of the reactants and products will be different from those before the change . Changes in temperature change the value of the constant K : An increase in temperature favours the endothermic reaction (which may be either the forward or the reverse reaction), a decrease favours the exothermic reaction. Changes in pressure cause significant changes in equilibria only where the number of moles of gaseous products and reactants differ. An increase in pressure shifts an equilibrium in the direction that produces the smaller number of molecules in the gas phase. The value of the constant K remains unchanged .
10. 10. The general principle that underlies all changes in equilibria is Le Chatelier´s principle: If a system in equilibrium is subjected to a change in concentration, temperature, and pressure, the system will react in a way that tends to relieve the change.
11. 11. Energy in chemical reactions The driving force of reactions
12. 12. The fundamental terms in chemical thermodynamics Energy can be defined as ability of producing heat or doing work . Energy takes any of several forms, such as mechanical, thermal, chemical, osmotic, electrical. Energy release or consumption accompanying chemical reactions is a consequence of bonds cleavage and formation. System is a portion of universe under study, surroundings is everything that is not part of a system under study. Insulated systems – without any communication with their surroundings (no exchange of matter nor energy) Closed systems – can release or absorb energy from the surroundings but no exchange of matter is possible. Open systems exchange energy, matter, even information) with the surroundings.
13. 13. The internal energy U is the sum of all the energy of all atoms, molecules, or ions that comprise a chemical system. The total internal energy U of a system cannot be determined. However, the internal energy change of a system Δ U can be both measured and calculated. It is the amount of energy exchanged with the surroundings during a chemical or physical change of the system. U = U final – U initial or U = U 2 – U 1 U > 0 an increase in the internal energy of the system U < 0 a decrease of the internal energy of the system An increase in the internal energy of a chemical system can result – in the temperature increase, – in melting, vaporization or a change in crystalline form, resp., – in a endothermic chemical reaction.
14. 14. The first law of thermodynamics U = q + w work heat Heat lost by a system or work done by a system on the surroundings are given negative values. Although work can be transformed completely into heat, it does not follow that heat can be transformed completely to work. Then heat is taken as a less utilizable form of energy . The energy of the insulated system is constant. Energy can be converted to one form to another, but cannot be destroyed. In the interaction between a closed system and its surroundings, the internal energy change of the system equals the heat exchanged by the system plus the work done on or by the system.
15. 15. In this course of chemistry, the only two kinds of work will be discussed – the work when a gas expands or contracts (pressure-volume work) and work associated with electrochemical changes in galvanic cells. For a gas expanding against a constant external pressure, the work equals w = – p Δ V . Chemical reactions realized at constant external pressure , most often atmospheric pressure, are very common. In those reactions, the internal energy change equals U = q – p V . This reaction heat q is defined as a quantity called enthalpy H and then the enthalpy change H = U + p V Most reactions in living systems (in aqueous environment) are realized without any pressure-volume work or the contribution of p Δ V is negligable; then H = U
16. 16. The enthalpy change Δ H is equal to the heat of reaction . It expresses the difference in bond energy of reaction products and reactants. H < 0 exothermic reaction , the enthalpy of the reaction products is lower (the bonds are more stable) than that of the reactants H > 0 endothermic reaction , the enthalpy of the products is higher than that of reactants. Standard enthalpy changes are expressed as the quantity of heat per one mole of the substance or substances in question in the standard state and at specified temperature (usually at 298 K, i.e. 25 °C). Then the symbol is Δ H ° or simply Δ H ° . Example: H 2 ( g ) + ½O 2 ( g ) H 2 O (l ) H ° = 286 kJ mol 1 the reaction is strongly exothermic
17. 17. The standard state of any substance is the physical state at which it is most stable at atmospheric pressure (101.3 kPa) and a specified temperature. The usual specified temperature is 298 K (25 °C , roughly room temperature). For some specific processes, the standard enthalpy change Δ H ° is named specifically, e.g. the standard enthalpy change of (or heat of) formation Δ H ° f (of a substance from elements), combustion Δ H ° c (substance + oxygen( g ) -> combustion products), neutralization Δ H ° neutralization (acid + base -> salt( aq ) + H 2 O( l ) ), solution Δ H ° soln (1 mol solute + n mol solvent -> 1 mol solute in solvent). Δ H of a reaction is the same whether the reaction takes place in one step or several steps (Hess´s law for combining Δ H values). Δ H for a reaction in one direction is equal in magnitude to Δ H for the reaction in the reverse direction, but is opposite in sign .
18. 18. Animals use the energy released by the oxidative breakdown of nutrients to H 2 O and CO 2 (proteins give also nitrogenous catabolites). Chemical energy of nutrients corresponds with the heat of combustion. Energetic yield of nutrients ( heat of combustion ) in per one gram of a pure nutrient: Saccharides 17 kJ / g (4.1 kcal / g) Triacylglycerols (fat) 38 kJ / g (9.1 kcal / g) Proteins 24 kJ / g (5.7 kcal / g) when combusted to H 2 O, CO 2 , and N 2 in a calorimeter, 17 kJ / g (4.1 kcal / g) in human bodies (catabolism to H 2 O, CO 2 , and urea CO(NH 2 ) 2 Remember that the heat evolved in going from the initial state to the final state is the same no matter by what route the reaction takes place, whether in calorimeter or within living cells.
19. 19. Spontaneous chemical reactions or physical changes, defined better as " thermodynamically favourable ", are those that can happen without any continuing outside influence. In insulated systems only such processes can proceed which tend to less organization, to more simple compounds, which result in total entropy increase . Entropy is a thermodynamic property, a measure of disorder . It is defined as the amount of energy (heat) in the system that cannot be transformed to work: S = Q rev / T . A change in entropy is Δ S = S 2 – S 1 . The opposite to entropy is information (a negative entropy), a measure of order or organization. The second law of thermodynamics Every spontaneous chemical and physical change increases the entropy of the universe as a whole.
20. 20. If Δ S system decreases ( Δ S of a negative value), this change must be accompanied by a simultaneous and larger increase in Δ S surroundings , because the condition for spontaneous process expressed by the second law of thermodynamics is Δ S universe = Δ S system + Δ S surroundings > 0 In contrast to insulated systems, in spontaneous processes in closed and open systems the entropy can either increase or decrease . S surroundings is proportional to the heat that the system under study releases into the surroundings or absorbs from it. Because also the enthalpy change Δ H of the system (reaction heat) must be taken into account , the mere entropy change Δ S system is not the best criterion of the spontaneity of a given chemical reaction in closed and open systems.. Increase in entropy (Δ S of a positive value) is the driving force of processes in the universe as a whole and also in insulated systems under study.
21. 21. The entropy change of a system Δ S system Entropy increases Increasing temperature Melting a solid Evaporating a liquid Dissolving a solid in a liquid Mixing two substances in the same phase Increasing the number of particles during a reaction (e.g., decompositions of molecules) Entropy decreases Decreasing temperature Freezing a liquid Condensing a gas Precipitation of a product in a solution Separating two substances in the same phase Decreasing the number of particles during a reaction (e.g., syntheses)
22. 22. If Δ G is negative , the reaction can proceed spontaneously and the value of Δ G represents the maximal amount of useful energy (work) that the system can perform in the reaction at constant temperature and pressure. The Gibbs free energy G of a system is the energy that is available to do useful work as the result of chemical or physical change at constant temperature and pressure. The free energy change Δ G (= G 2 – G 1 ) is defined as In closed and open systems , the driving force of chemical reactions or physical changes is the free energy change Δ G . G = H – T S useful work heat lost or absorbed due to entropy change enthalpy change (reaction heat)
23. 23. G the criterion of process feasibility G < 0 exergonic reaction prone to proceed spontaneously G > 0 endergonic reaction that cannot proceed spontaneously under given circumstances (the reverse reaction is spontaneous) G = 0 the system is at the equilibrium state There is no relation between Δ G and the velocity of a reaction !! H the heat of reaction H < 0 exothermic reaction H > 0 endothermic reaction S the entropy change S > 0 the final state is very probable S < 0 a very low probability of the final state The meaning of thermodynamic functions T S the product (the entropic member) is critically dependent on the temperature T
24. 24. – the entropy of the system increases and heat is released, – the chemical change is endothermic but accompanied with a marked entropy increase, and/or – the change is highly exothermic in spite of it is accompanied with an entropy decrease. Spontaneous chemical changes take place In closed systems when Reactions can occur spontaneously (i.e. without any continuing outside influence) only if they are exergonic – only if the free energy change G is of negative value . positive at all temperatures; the forward reaction cannot be spontaneous (the reverse re a ction is always spontaneous) positive (endothermic) as far as T Δ S < Δ H (at lower temperatures), the reaction becomes more favourable negative (exothermic) positive negative (syntheses) as far as T Δ S > Δ H (at higher temperatures), the reaction becomes spontaneous (more favourable) positive (endothermic) always negative; the reaction is spontaneous, practically irreversible negative (exothermic) negative positive (decompositions) Δ G = Δ H – T Δ S Δ H – Δ S Δ S
25. 25. This state is reached in spontaneous chemical reactions by means of – the evolution or absorption of heat ( Δ H ) , – the changes in entropy (more simple products), resulting in the changes in concentrations of the substances in the system so as to comply with the value of the equilibrium constant K . The general tendency of any spontaneous process is to reach an equilibrium state – a state of the most thermodynamic stability. The more far-apart are the concentrations of participants from the equilibrium concentrations, the higher is the Δ G. At constant temperature and pressure, a closed system at the equilibrium state has its minimum of Gibbs free energy , at equilibrium the free energy change Δ G = 0 .
26. 26. Δ G is a measure of disagreement between the initial concentrations of reactants and products of the reaction and their equilibrium concentrations. The expression takes the same form as the equilibrium constant but is used for a initial state, not for a reaction at equilibrium. The initial non - equilibrium concentrations of the substances taking part in reaction a A + b B c C + d D are used to calculate the reaction quotient Q : The equilibrium state is defined as The value of Q indicates what changes will occur in reaching equilibrium: When Q < K , the reaction has a chance to proceed in the forward direction. When Q > K , the reaction has a chance to proceed in the reverse direction. [A] a i [B] b i [C ] c i [D] d i Q = K = [A] a eq [B] b eq [C ] c eq [D] d eq
27. 27. between the system that exists at the beginning of the process in its standard state (the reaction quotient Q = 1), i.e. all reactants , both reactants and products are of unit activity, in aqueous solution their concentration c = 1 mol l –1 (if H + is a reactant, then also [H + ] = 1, pH = 0), at specified temperature (usually 25 °C equal to 298 K), and atmospheric pressure 101.3 kPa, and the reaching the state with a minimal G value, that is the equilibrium state of the system in which the reactants and products has reached the concentrations corresponding with the equilibrium constant K . Standard Gibbs free energy change Δ G ° for a reversible process represents the free energy change (In biological systems , the standard state is defined by pH = 7,0 ; then the free energy changes are marked as Δ G °´ .)
28. 28. The Δ G of a reaction depends on the particular kind of reaction (expressed by the Δ G º term) and the initial concentrations of reactants and products (expressed by the second term equal to Q ). If the equilibrium concentrations are put in as initial ones, the system is in its equilibrium state, G = 0 : 0 = G + RT ln K and G = – RT ln K The relationship is sometimes written in the form G = RT ln K + RT ln Q Δ G = Δ G º + RT ln [A] a i [B] b i [C ] c i [D] d i a A + b B c C + d D The relation between free energy and equilibrium for any reaction (here the type is used) is given by the mathematical expression
29. 29. If the reaction starts in the standard state (all concentrations [A], [B], [C], and [D] equal 1 mol l –1 ), the second term equals zero so that the definition of Δ G ° is obtained: G = G + RT ln 1 = G Values of G are very important characteristics of chemical reactions, but they should be never overemphasized. Even though the value of G is negative, the spontaneous reaction could proceed in the reverse direction (positive Δ G value) due to the high reaction quotient Q at high initial concentration of the reaction products.
30. 30. Transformation of energy in living organisms Living organisms are open systems that have to receive permanently nutrients – compounds of high enthalpy (energy) and low entropy (due to their complex structure). Nutrients are transformed into waste metabolites of low enthalpy and high entropy (simplified structures). The part of free energy gained by exergonic breakdown of nutrients drives endergonic reactions and processes (synthesis of complex molecules, performance of mechanical or osmotic work, etc.). The remaining part of acquired energy is released as heat into the surroundings.
31. 31. Endergonic reaction cannot proceed spontaneously, but these thermodynamically unfavourable reactions are driven by exergonic reactions to which they are coupled. Coupling occurs because the two reactions share a common reactant or intermediate. Example: The overall net free energy change is negative ( Δ G º´ = – 13.4 kJ mol –1 ), the conversion of malate to aspartate is exergonic. Energetic coupling in open systems Malate Fumarate H 2 O NH 3 Aspartate Δ G º´ 1 = + 2 kJ mol –1 Δ G º´ 2 = – 15.4 kJ mol –1
32. 32. ATP + H 2 O ADP + Pi G °´ (at pH 7) = – 30.5 kJ mol –1 O O H O H 2 O O C H O P O P O O O ~ P O O O ~ + H + H N N N N N H 2 ATP + P O O O H O 2 O O O H O H N N N N N H 2 2 ADP P O O O C H O P O O O ~ Adenosine triphosphate (ATP) is a high-energy compound that serves as the "universal currency" of free energy in biological systems. ATP hydrolysis drives metabolism by shifting the equilibrium of coupled reactions .
33. 33. The reaction which is used to drive endergonic ones is very oft the hydrolysis of ATP. Example: Glucose Glucose 6-phosphate ATP ADP G o ´ = + 13.8 kJ mol –1 G o ´ = – 30.5 kJ mol –1 = – 16.7 kJ mol –1
34. 34. Examples of high-energy phosphates and Δ G ´ of the hydrolysis – 30.5 – 62 – 50 – 52 – 43 acid anhydride ester mixed acid anhydride* mixed acid anhydride* amid ATP Phosphoenolpyruvate 1,3-Bisphosphoglycerate Carbamoyl phosphate Creatine phosphate Δ G ´ kJ mol –1 Phosphate derivative High-energy phosphate * acyl phosphate
35. 35. Photosynthetic autotrophs Heterotrophs CO 2 H 2 O O 2 Nutrients rich in H hν 4 H 4 H + O 2 2 O 2– CO 2 Biological oxidations (dehydrogenations) Decarboxylations Reducing equivalents (reducing power) 4 e – 2 H 2 O Nutrients rich in H Heterotrophs:
36. 36. Most of the Gibbs´ free energy in the body originates in the exergonic synthesis of water (2H 2 + O 2 2H 2 O, 25 °C ) : Δ G ° = – 474.3 kJ mol –1 Fatty acids of fats are a more efficient fuel source than saccharides such as glucose because the carbon in fatty acids is more reduced
37. 37. A steady state of an open system is a dynamic state of an open system which receives and gives off, within a given interval, the same amount of substances and energy so that the concentration of intermediates in the system remains unchanged . Nutrients rich in hydrogen Oxygen Water Heat and work Low-energy metabolites
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Joined: 12 Aug 2011
Posts: 2
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### Show Tags
16 Sep 2011, 04:39
do warwick and lse look at the verbal scores for the admissions for masters in finance???
Last edited by karanjhanwer on 21 Sep 2011, 13:11, edited 2 times in total.
Senior Manager
Status: 750+ or Burst !
Joined: 01 May 2011
Posts: 386
Location: India
Concentration: General Management, Strategy
GMAT 1: 670 Q48 V35
GPA: 3.5
Followers: 24
Kudos [?]: 115 [0], given: 26
### Show Tags
21 Sep 2011, 12:52
karanjhanwer wrote:
hey guys,i appeared for the gmat and scored 660(q 50,v 28).I am in the final year of my engineering.Done good quality mathematics(probability and stochastic processes,calculus 1,calculus 2,numerical methods with mat-lab,basics of economics,complex theory and graph theory)and also secured A grade in each of them.I have an intern at Arcelor Mittal.Current cpi of 8.3.Is there any chance for LSE,CASS or WARWICK??
LSE would be a stretch since the minimum requirements for LSE is a CPGA of 8.5.
Nevertheless, I strongly recommend that you check out the Financial Engineering programs in US. Check out http://www.quantnet.com/
_________________
GMAT done - a mediocre score but I still have a lot of grit in me
The last 20 days of my GMAT journey
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Joined: 12 Aug 2011
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### Show Tags
24 Sep 2011, 13:04
akhileshgupta05 wrote:
karanjhanwer wrote:
hey guys,i appeared for the gmat and scored 680(q 50,v 33).I am in the final year of my engineering.Done good quality mathematics(probability and stochastic processes,calculus 1,calculus 2,numerical methods with mat-lab,basics of economics,complex theory and graph theory)and also secured A grade in each of them.I have an intern at Arcelor Mittal.Current cpi of 8.3.Is there any chance for LSE,CASS or WARWICK??
Senior Manager
Status: 750+ or Burst !
Joined: 01 May 2011
Posts: 386
Location: India
Concentration: General Management, Strategy
GMAT 1: 670 Q48 V35
GPA: 3.5
Followers: 24
Kudos [?]: 115 [0], given: 26
### Show Tags
26 Sep 2011, 04:55
karanjhanwer wrote:
do warwick and lse look at the verbal scores for the admissions for masters in finance???
Depending on your score, they do consider the fact the generally non-natives are at a disadvantage on the test. But I don't think the completely ignore it.
Nevertheless, if I were you, I would send the adcom an email to address this issue.
_________________
GMAT done - a mediocre score but I still have a lot of grit in me
The last 20 days of my GMAT journey
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# Navier-Stokes treating convection with IMEX scheme
I’m looking to solve navier-Stokes equation with an IMEX method which gives a convective term on the form
c(u_{n+1}, u_n, v) = \int_{\Omega} \nabla u_{n+1} u_n \cdot v d\Omega
as this gives unconditional stability. I have tried to do this with a bilinear form which includes a gridfunction un. Below is is the setup of the spaces and convective part of the bilinear form:
V = VectorH1(mesh,order=k, dirichlet=“wall|cyl|inlet”)
Q = H1(mesh,order=k-1)
X = VQ
(u,p), (v,q) = X.TnT()
gfu = GridFunction(X)
velocity, pressure = gfu.components
uin = CoefficientFunction((1.5
4y(0.41-y)/(0.410.41),0))
velocity.Set(uin, definedon=mesh.Boundaries(“inlet”))
un = gfu.components[0]
conv = BilinearForm(X)
c = InnerProduct(Grad(u) * un, v)
dx
conv += c
Assembly of conv and updating of un is done while time stepping. This method seems to work for quite a few time steps, until much of the data suddenly become nan values. Any help with figuring out how to do implement this scheme properly would be much appreciated!
Hi,
if you want to trat the convection semi-implicitly, you need to at the c form to the BilinearForm with all the other implicit terms. As a consequence, you will need to reassemble the matrix and inverse in every time-step.
Best, Henry
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6.4 C
New York
Friday, March 1, 2024
# C Program to Check Leap Year
### What is a Leap Year?
A year exactly divisible by 4 is known as leap year except for centurial years. A centurial year will be a leap year if it is exactly divisible by 400, for example, the year 1700 is not a leap year whereas 2000 is a leap year.
As 1700 is a centurial year which is not divisible by 400 whereas 2000 is a centurial year divisible by 400.
In this tutorial, you will learn how to write C program to check if the input year is leap year on not.
Approach of Program:
1) Take input year from the user
2) Use if-else statement to check the following conditions:
• Check if the year is divisible by 4 it is a leap year or else year is not leap year
• Check if the year is divisible by 100 it is not a leap year unless the year is also divisible by 400 then it’s a leap year
• Check if the year is divisible by 400 if yes the year is a leap year
3) Print the Output
## C Program to Check Leap Year
``````//Program to find the leap year
//Including the standered input output header library
#include <stdio.h>
//Main Function
void main(){
int x;
//Giving a hint to user what to enter
printf("Enter A Year : ");
// Take year input from user
scanf("%d",&x);
//Checking the entered number modulation is 0 or not
if(x % 4 == 0){
//If Yes
//Checking the entered Year modulation is 0 or not with 100
if( x % 100 == 0)
{
//If Yes
//Checking the entered Year modulation is 0 or not with 400
if ( x % 400 == 0)
{
//If Yes
printf("%d is a leap year.", x);
}
else
{
//If No
printf("%d is not a leap year.", x);
}
}
else
{
//If No
printf("%d is a leap year.", x );
}
}else{
//If No
printf("%d is not a leap year.", x);
}
}
``````
Output 1 (Leap Year)
``````Enter A Year: 2312
2312 is a leap year``````
Output 2 (Not A Leap Year)
``````Enter A Year: 2019
2312 is not a leap year``````
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# Queries for maximum difference between prime numbers in given ranges
Given n queries of the form range [L, R]. The task is to find the maximum difference between two prime numbers in the range for each query. If there are no prime in the range then print 0. All ranges are below 100005.
Examples:
```Input : Q = 3
query1 = [2, 5]
query2 = [2, 2]
query3 = [24, 28]
Output : 3
0
0
In first query, 2 and 5 are prime number
in the range with maximum difference which
is 3. In second and third queries, there
is only 1 prime number in range, so output
is 0.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to compute Prime numbers using Sieve of Eratosthenes along with some precomputing.
Below are the step to solve the question:
Step 1: Find the prime numbers using Sieve of Eratosthenes algorithm.
Step 2: Make an array, let say prefix[], where prefix[i] represents largest prime number smaller or equal to i.
Step 3: Make an array, let say suffix[], where suffix[i] represents smallest prime number greater or equal to i.
Step 4: Now for each query having [L, R], do the following:
``` if (prefix[R] R)
return 0;
else
return prefix[R] - suffix[L];```
Below is the implementation of this approach:
## C++
`// CPP program to find maximum differences between ` `// two prime numbers in given ranges ` `#include ` `using` `namespace` `std; ` `#define MAX 100005 ` ` ` `// Precompute Sieve, Prefix array, Suffix array ` `void` `precompute(``int` `prefix[], ``int` `suffix[]) ` `{ ` ` ``bool` `prime[MAX]; ` ` ``memset``(prime, ``true``, ``sizeof``(prime)); ` ` ` ` ``// Sieve of Eratosthenes ` ` ``for` `(``int` `i = 2; i * i < MAX; i++) { ` ` ``if` `(prime[i]) { ` ` ``for` `(``int` `j = i + i; j < MAX; j += i) ` ` ``prime[j] = ``false``; ` ` ``} ` ` ``} ` ` ` ` ``prefix[1] = 1; ` ` ``suffix[MAX - 1] = 1e9 + 7; ` ` ` ` ``// Precomputing Prefix array. ` ` ``for` `(``int` `i = 2; i < MAX; i++) { ` ` ``if` `(prime[i]) ` ` ``prefix[i] = i; ` ` ``else` ` ``prefix[i] = prefix[i - 1]; ` ` ``} ` ` ` ` ``// Precompute Suffix array. ` ` ``for` `(``int` `i = MAX - 1; i > 1; i--) { ` ` ``if` `(prime[i]) ` ` ``suffix[i] = i; ` ` ``else` ` ``suffix[i] = suffix[i + 1]; ` ` ``} ` `} ` ` ` `// Function to solve each query ` `int` `query(``int` `prefix[], ``int` `suffix[], ``int` `L, ` ` ``int` `R) ` `{ ` ` ``if` `(prefix[R] < L || suffix[L] > R) ` ` ``return` `0; ` ` ``else` ` ``return` `prefix[R] - suffix[L]; ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `q = 3; ` ` ``int` `L[] = { 2, 2, 24 }; ` ` ``int` `R[] = { 5, 2, 28 }; ` ` ` ` ``int` `prefix[MAX], suffix[MAX]; ` ` ``precompute(prefix, suffix); ` ` ` ` ``for` `(``int` `i = 0; i < q; i++) ` ` ``cout << query(prefix, suffix, L[i], R[i]) << endl; ` ` ` ` ``return` `0; ` `} `
## Java
`// Java program to find maximum differences between ` `// two prime numbers in given ranges ` ` ` `public` `class` `GFG { ` ` ` ` ``final` `static` `int` `MAX = ``100005``; ` ` ` `// Precompute Sieve, Prefix array, Suffix array ` ` ``static` `void` `precompute(``int` `prefix[], ``int` `suffix[]) { ` ` ``boolean` `prime[] = ``new` `boolean``[MAX]; ` ` ``for` `(``int` `i = ``0``; i < MAX; i++) { ` ` ``prime[i] = ``true``; ` ` ``} ` ` ` ` ``// Sieve of Eratosthenes ` ` ``for` `(``int` `i = ``2``; i * i < MAX; i++) { ` ` ``if` `(prime[i]) { ` ` ``for` `(``int` `j = i + i; j < MAX; j += i) { ` ` ``prime[j] = ``false``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``prefix[``1``] = ``1``; ` ` ``suffix[MAX - ``1``] = (``int``) 1e9 + ``7``; ` ` ` ` ``// Precomputing Prefix array. ` ` ``for` `(``int` `i = ``2``; i < MAX; i++) { ` ` ``if` `(prime[i]) { ` ` ``prefix[i] = i; ` ` ``} ``else` `{ ` ` ``prefix[i] = prefix[i - ``1``]; ` ` ``} ` ` ``} ` ` ` ` ``// Precompute Suffix array. ` ` ``for` `(``int` `i = MAX - ``2``; i > ``1``; i--) { ` ` ``if` `(prime[i]) { ` ` ``suffix[i] = i; ` ` ``} ``else` `{ ` ` ``suffix[i] = suffix[i + ``1``]; ` ` ``} ` ` ``} ` ` ``} ` ` ` `// Function to solve each query ` ` ``static` `int` `query(``int` `prefix[], ``int` `suffix[], ``int` `L, ` ` ``int` `R) { ` ` ``if` `(prefix[R] < L || suffix[L] > R) { ` ` ``return` `0``; ` ` ``} ``else` `{ ` ` ``return` `prefix[R] - suffix[L]; ` ` ``} ` ` ``} ` ` ` `// Driven Program ` ` ``public` `static` `void` `main(String[] args) { ` ` ``int` `q = ``3``; ` ` ``int` `L[] = {``2``, ``2``, ``24``}; ` ` ``int` `R[] = {``5``, ``2``, ``28``}; ` ` ` ` ``int` `prefix[] = ``new` `int``[MAX], suffix[] = ``new` `int``[MAX]; ` ` ``precompute(prefix, suffix); ` ` ` ` ``for` `(``int` `i = ``0``; i < q; i++) { ` ` ``System.out.println(query(prefix, suffix, L[i], R[i])); ` ` ``} ` ` ` ` ``} ` `} ` `/*This code is contributed by Rajput-Ji*/`
## Python3
`# Python 3 program to find maximum ` `# differences between two prime numbers ` `# in given ranges ` `from` `math ``import` `sqrt ` ` ` `MAX` `=` `100005` ` ` `# Precompute Sieve, Prefix array, Suffix array ` `def` `precompute(prefix, suffix): ` ` ``prime ``=` `[``True` `for` `i ``in` `range``(``MAX``)] ` ` ` ` ``# Sieve of Eratosthenes ` ` ``k ``=` `int``(sqrt(``MAX``)) ` ` ``for` `i ``in` `range``(``2``, k, ``1``): ` ` ``if` `(prime[i]): ` ` ``for` `j ``in` `range``(i ``+` `i, ``MAX``, i): ` ` ``prime[j] ``=` `False` ` ` ` ``prefix[``1``] ``=` `1` ` ``suffix[``MAX` `-` `1``] ``=` `int``(``1e9` `+` `7``) ` ` ` ` ``# Precomputing Prefix array. ` ` ``for` `i ``in` `range``(``2``, ``MAX``, ``1``): ` ` ``if` `(prime[i]): ` ` ``prefix[i] ``=` `i ` ` ``else``: ` ` ``prefix[i] ``=` `prefix[i ``-` `1``] ` ` ` ` ``# Precompute Suffix array. ` ` ``i ``=` `MAX` `-` `2` ` ``while``(i > ``1``): ` ` ``if` `(prime[i]): ` ` ``suffix[i] ``=` `i ` ` ``else``: ` ` ``suffix[i] ``=` `suffix[i ``+` `1``] ` ` ``i ``-``=` `1` ` ` `# Function to solve each query ` `def` `query(prefix, suffix, L, R): ` ` ``if` `(prefix[R] < L ``or` `suffix[L] > R): ` ` ``return` `0` ` ``else``: ` ` ``return` `prefix[R] ``-` `suffix[L] ` ` ` `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ``q ``=` `3` ` ``L ``=` `[``2``, ``2``, ``24``] ` ` ``R ``=` `[``5``, ``2``, ``28``] ` ` ` ` ``prefix ``=` `[ ``0` `for` `i ``in` `range``(``MAX``)] ` ` ``suffix ``=` `[ ``0` `for` `i ``in` `range``(``MAX``)] ` ` ``precompute(prefix, suffix) ` ` ` ` ``for` `i ``in` `range``(``0``, q, ``1``): ` ` ``print``(query(prefix, suffix, ` ` ``L[i], R[i])) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar `
## C#
`// C# program to find maximum differences between ` `// two prime numbers in given ranges ` `using` `System; ` ` ` `public` `class` `GFG { ` ` ` ` ``static` `readonly` `int` `MAX = 100005; ` ` ` `// Precompute Sieve, Prefix array, Suffix array ` ` ``static` `void` `precompute(``int` `[]prefix, ``int` `[]suffix) { ` ` ``bool` `[]prime = ``new` `bool``[MAX]; ` ` ``for` `(``int` `i = 0; i < MAX; i++) { ` ` ``prime[i] = ``true``; ` ` ``} ` ` ` ` ``// Sieve of Eratosthenes ` ` ``for` `(``int` `i = 2; i * i < MAX; i++) { ` ` ``if` `(prime[i]) { ` ` ``for` `(``int` `j = i + i; j < MAX; j += i) { ` ` ``prime[j] = ``false``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``prefix[1] = 1; ` ` ``suffix[MAX - 1] = (``int``) 1e9 + 7; ` ` ` ` ``// Precomputing Prefix array. ` ` ``for` `(``int` `i = 2; i < MAX; i++) { ` ` ``if` `(prime[i]) { ` ` ``prefix[i] = i; ` ` ``} ``else` `{ ` ` ``prefix[i] = prefix[i - 1]; ` ` ``} ` ` ``} ` ` ` ` ``// Precompute Suffix array. ` ` ``for` `(``int` `i = MAX - 2; i > 1; i--) { ` ` ``if` `(prime[i]) { ` ` ``suffix[i] = i; ` ` ``} ``else` `{ ` ` ``suffix[i] = suffix[i + 1]; ` ` ``} ` ` ``} ` ` ``} ` ` ` `// Function to solve each query ` ` ``static` `int` `query(``int` `[]prefix, ``int` `[]suffix, ``int` `L, ` ` ``int` `R) { ` ` ``if` `(prefix[R] < L || suffix[L] > R) { ` ` ``return` `0; ` ` ``} ``else` `{ ` ` ``return` `prefix[R] - suffix[L]; ` ` ``} ` ` ``} ` ` ` `// Driven Program ` ` ``public` `static` `void` `Main() { ` ` ``int` `q = 3; ` ` ``int` `[]L = {2, 2, 24}; ` ` ``int` `[]R = {5, 2, 28}; ` ` ` ` ``int` `[]prefix = ``new` `int``[MAX]; ` ` ``int` `[]suffix = ``new` `int``[MAX]; ` ` ``precompute(prefix, suffix); ` ` ` ` ``for` `(``int` `i = 0; i < q; i++) { ` ` ``Console.WriteLine(query(prefix, suffix, L[i], R[i])); ` ` ``} ` ` ` ` ``} ` `} ` ` ` `/*This code is contributed by 29AjayKumar*/`
## PHP
` 1; ``\$i``--) ` ` ``{ ` ` ``if` `(``\$prime``[``\$i``]) ` ` ``\$suffix``[``\$i``] = ``\$i``; ` ` ``else` ` ``\$suffix``[``\$i``] = ``\$suffix``[``\$i` `+ 1]; ` ` ``} ` `} ` ` ` `// Function to solve each query ` `function` `query(``\$prefix``, ``\$suffix``, ``\$L``, ``\$R``) ` `{ ` ` ``if` `(``\$prefix``[``\$R``] < ``\$L` `|| ``\$suffix``[``\$L``] > ``\$R``) ` ` ``return` `0; ` ` ``else` ` ``return` `\$prefix``[``\$R``] - ``\$suffix``[``\$L``]; ` `} ` ` ` `// Driver Code ` `\$q` `= 3; ` `\$L` `= ``array``( 2, 2, 24 ); ` `\$R` `= ``array``( 5, 2, 28 ); ` ` ` `\$prefix` `= ``array_fill``(0, ``\$MAX` `+ 1, 0); ` `\$suffix` `= ``array_fill``(0, ``\$MAX` `+ 1, 0); ` `precompute(``\$prefix``, ``\$suffix``); ` ` ` `for` `(``\$i` `= 0; ``\$i` `< ``\$q``; ``\$i``++) ` ` ``echo` `query(``\$prefix``, ``\$suffix``, ` ` ``\$L``[``\$i``], ``\$R``[``\$i``]) . ``"\n"``; ` ` ` `// This code is contributed by mits ` `?> `
Output:
```3
0
0
```
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Given the function f(x) = 3x – 7, determine which of the following values will result in f(x) = -1.
Question
Given the function f(x) = 3x – 7, determine which of the following values will result in f(x) = -1.
in progress 0
1 week 2021-09-10T12:23:09+00:00 1 Answer 0
x = 2
Step-by-step explanation:
We are trying to find what value x equals to get -1.
-1 = 3x – 7
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# I Klein-Gordon propagator
1. Jun 28, 2017
### Silviu
Hello! I am reading Peskin's book on QFT and in the first chapter (pg. 30) he introduces this: $<0|[\phi(x),\phi(y)]|0> = \int\frac{d^3p}{(2\pi)^3}\int\frac{dp^0}{2\pi i}\frac{-1}{p^2-m^2}e^{-ip(x-y)}$ and then he spends 2 pages explaining the importance of choosing the right contour integral (i.e. the right prescription of going around the poles at $\pm E_0$). However I asked a question here just to make sure and I have been told that the way you go around the poles has no influence on the final value of the integral. So now I am confused. If the way you do the integral give the same result in the end, why is Peskin talking about, exactly? Am I missing something? Thank you!
2. Jun 28, 2017
### Orodruin
Staff Emeritus
It is not a question of how you go around the poles, but of if you go around the poles (depending on how you. Lose the contour).
3. Jun 28, 2017
### Silviu
I am not sure I understand what you mean. You have to go around the poles otherwise you encounter a singularity point.
4. Jun 28, 2017
### Orodruin
Staff Emeritus
You are reading it wrong. You do not have to go around the poles if your contour closes in the other direction. Only the enclosed poles affect the integral.
Reminds me of the riddle "What is the contour integral around western Europe?"
Zero. The Poles are in eastern Europe!
5. Jun 28, 2017
### Silviu
Ok the joke is funny but still. I understand that you don't need to enclose the poles necessary, but my point is, whether you enclose them or not, you get the same result in the end (the same numerical value for the integral).
6. Jun 28, 2017
### Orodruin
Staff Emeritus
No you don't. That is the point.
7. Jun 28, 2017
### Silviu
OK, sorry I am confused. From what I know from Complex Analysis, for a holomorphic function, the path you choose shouldn't matter, just the end points. Why is it now different?
8. Jun 28, 2017
### Orodruin
Staff Emeritus
The integrand is not holomorphic. It has poles.
9. Jun 28, 2017
### Silviu
So the way you go around the poles gives you different results? But in the end you have a real integal, which should have a definite, fixed value. Using this complex integration, and going around the poles is just an easier way to solve an integral hard to solve just in real numbers. But in the end, as any definite integral on the real line, it must have only one value. So I am still a bit confused about why you get multiple values, based on the path.
10. Jun 28, 2017
### Orodruin
Staff Emeritus
A priori, your integral is not well defined. It has poles on the real line. That is why you need the pole prescription. Different choices will give you different Green's functions.
11. Jul 2, 2017
### vanhees71
Now you are talking! Contrary to your previous claims, indeed how you avoid the poles is crucial to get the correct Green's function needed for your problem in question. Since the confusion is thanks to Peskin&Schroeder, I suppose what you want is the time-ordered Green's function, defined (for a neutral scalar field) as
$$\Delta(x)=-\mathrm{i} \langle \Omega|\mathcal{T}_c \hat{\phi}(x) \hat{\phi}(0)|\Omega \rangle,$$
where
$$\mathcal{T}_c \hat{\phi}(x) \hat{\phi}(0)=\Theta(x^0) \hat{\phi}(x) \hat{\phi}(0) + \Theta(-x^0) \hat{\phi}(0) \hat{\phi}(x).$$
In vacuum it's identical with the Feynman propagator.
If, however, you have to calculate something in linear-response theory, you need the retarded propagator, defined as
$$\Delta_{\text{ret}}(x) =-\mathrm{i} \langle \Omega|\Theta(x^0) [\hat{\phi}(x),\hat{\phi}(0)] |\Omega \rangle.$$
Thanks to the $\Theta$ unit-step functions, the way to circumvent the poles are uniquely determined.
A pedagogically much better way to introduce the free propagators in QFT is to use the plane-wave mode decomposition with creation and annihilation operators and first evaluate the Mills representation, i.e., the Green's functions Fourier transformed with respect to $\vec{x}$ only. Then you first keep the time and can do the final Fourier transform wrt. to time at the end!
12. Jul 2, 2017
### Orodruin
Staff Emeritus
When did I claim that it does not matter how you avoid the poles? What I am saying in #2 is that it does not matter what the contour looks like, only if the pole prescription places the poles inside the contour or not. If you look at the thread linked in the OP, the confusion is whether or not it matters what the contour that encloses the pole looks like - it does not as long as it encloses the pole. You can deform the contour as you like as long as you do not change what poles are inside it.
Edit: I am never suggesting integrating through the pole. By "going around" I mean that the pole is enclosed by the contour and by "not going around" I mean that the pole is outside of the contour.
13. Jul 2, 2017
### vanhees71
Ok, then I misunderstood your statement in #2.
14. Jul 2, 2017
### Orodruin
Staff Emeritus
It seems the OP did too, so perhaps I could have formulated it better. I guess this is a problem of me using "going around" in the sense of enclosing and a possible meaning of "not going around" could be perceived as "going through". The misspelling in the parenthesis probably indicates I was on my phone at the time and perhaps not paying too much attention to formulation...
I think this was also the source of the OPs confusion in the first place with the other thread using "going around" in the same meaning as I did (which may be why I chose that nomenclature) whereas "way of going around" as used in Peskin might refer to the pole prescription (no access to Peskin now).
15. Jul 3, 2017
### vanhees71
Peskin&Schroeder do the usual thing in going right away to the 4D Fourier transform, i.e., in momentum space, where you have the naive formula
$$(p^2-m^2) \Delta(p)=1,$$
which of course doesn't make sense without deforming the original path of the Fourier integral along the real $p^0$ axis. That it is the temporal component comes from the fact that the different propagators are distinguished by the $\Theta$ functions of time.
Let's do the calculation, so that we have it as a record here in PF. Everything can be derived from the fixed-ordered correlation function (two-point Wightman function),
$$\mathrm{i} \Delta^{>}(x)=\langle \Omega|\hat{\phi}(x) \hat{\phi}(0)|\Omega \rangle.$$
To do so, we use the mode decomposition of the field operator
$$\hat{\phi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi)^3 2 E}}[\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x) + \hat{a}^{\dagger}(\vec{p}) \exp(\mathrm{i} p \cdot x)],$$
where the normalization is such that
$$[\hat{a}(\vec{p}_1) ,\hat{a}(\vec{p}_2)]=\delta^{(3)}(\vec{p}_1-\vec{p}_2)$$
and $p^0=E=\sqrt{\vec{p}^2+m^2}$.
For the Wightman function we thus have
$$\mathrm{i} \Delta^{>}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_1 \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_2 \frac{1}{\sqrt{(2 \pi)^3 2 E_1}} \frac{1}{\sqrt{(2 \pi)^3 2 E_2}} \exp(-\mathrm{i} x \cdot p_1) \langle \Omega|\hat{a}(\vec{p}_1) \hat{a}^{\dagger}(\vec{p}_2)|\Omega \rangle.$$
Now using
$$\Omega|\hat{a}(\vec{p}_1) \hat{a}^{\dagger}(\vec{p}_2)|\Omega \rangle=\Omega|[\hat{a}(\vec{p}_1) ,\hat{a}^{\dagger}(\vec{p}_2)]|\Omega \rangle = \delta^{(3)}(\vec{p}_1-\vec{p}_2),$$
we get
$$\mathrm{i} \Delta^{>}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_1 \frac{1}{(2 \pi)^3 2 E_1} \exp(-\mathrm{i} x \cdot p_1)|_{p_1^0=E_1}.$$
From translation invariance of the vacuum state we get for the other Wightman function
$$\mathrm{i} \Delta^{<}(x)=\langle \Omega |\phi(0) \phi(x) |\Omega \rangle = \langle \Omega |\phi(-x) \phi(0)|\Omega = \mathrm{i} \Delta^{>}(-x) \\ = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_1 \frac{1}{(2 \pi)^3 2 E_1} \exp(\mathrm{i} x \cdot p_1)|_{p_1^0=E_1}.$$
Now it's easy to do the Fourier transform for these two functions
$$\mathrm{i} \tilde{\Delta}^{>}(p)=\int_{\mathbb{R}^4} \mathrm{d}^4 x \exp(\mathrm{i} x \dot p) \mathrm{i} \Delta^{>}(x)=\frac{1}{2E} (2 \pi) \delta(p^0-E)$$
and
$$\mathrm{i} \tilde{\Delta}^{<}(p)=\frac{1}{2E} (2 \pi) \delta(p^0+E),$$
where $E=\sqrt{\vec{p}^2+m^2}$.
To get the time-ordered propagator, we only need the Fourier transformation of the $\Theta$ unit-step functions
$$\tilde{\Theta}(p^0)=\int_{\mathbb{R}} \mathrm{d} t \exp(\mathrm{i} p^0 t) \Theta(t).$$
Of course, this integral doesn't exist, and we have to regularize it. The only way is to make $p^0$ complex. Obviously the integral converges if $\mathrm{Im} p^0>0$. So for real $p^0$ we write $p^0 \rightarrow p^0+\mathrm{i} \epsilon$. This gives
$$\tilde{\Theta}(p^0)=\frac{\mathrm{i}}{p^0+\mathrm{i} \epsilon}.$$
In the same way we find
$$\tilde{\Theta}^{-}(p^0)=\int_{\mathbb{R}} \mathrm{d} t \exp(\mathrm{i} t p^0) \Theta(-t)=-\frac{\mathrm{i}}{p^0-\mathrm{i} \epsilon}.$$
Now we can use the convolution theorem for Fourier transformations to get the time-ordered Green's function
$$\Delta(x)=\Theta(x^0) \Delta^{>}(x) + \Theta(-x^0) \Delta^{<}(x),$$
i.e.
$$\mathrm{i} \tilde{\Delta}(p) = \int_{\mathbb{R}} \frac{\mathrm{d} k^0}{2 \pi} [\tilde{\Theta}(k^0) \mathrm{i} \tilde{\Delta}^{>}(p^0-k^0,\vec{p}) + \tilde{\Theta}^{-}(k^0) \mathrm{i} \Delta^{<}(p^0-k^0,\vec{p})].$$
Plugging in the above expressions one finds by evaluating the $\delta$ distributions in the Wightman functions,
$$\mathrm{i} \tilde{\Delta}(p)=\frac{\mathrm{i}}{(p^0)^2-(E-\mathrm{i} \epsilon)^2} = \frac{\mathrm{i}}{p^2-m^2+\mathrm{i} \epsilon},$$
where one has to read $\epsilon=0^+$, i.e., an "infinitesimal small positive real number".
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# Chapter 1
LINEAR PROGRAMMING
1.1. Using optimal limited resources
One practical optimization problem, which appears frequently in the management activity in every organization is the following: we have several limited resources such as materials, labor, nancial etc and using these resources we can perform several activities. The optimization problem must comply an optimal (maximize or minimize) criteria and consist in determination of the level of these activities, constrained by the available limited resources. Let us denote by i, 1 i m, the resource type and with bi the level of available resource of type i. Denote by j , 1 j n, the type of activity and by xj the level (unknown) at which will be performed this activity. Finally, denote by aij the resource of type i, 1 i m, required for the production of one unit of type j , 1 j n, (in general, activity of type j ). We suppose here that aij depends only of the resource type and process type and does not depend of the level at which these activity will be performed. Using the introduced notations we can express the total quantity of resource of i-th type which is used in the production phase: ai1 x1 + ai2 x2 + . . . + ain xn . Because is not allowed to use from i-th resource more than the available quantity bi , it result that the following constraints must be accomplished:
n
aij xj bi , 1 i m.
j =1
(1.1)
LINEAR PROGRAMMING
Because xj represents the level at which is performed the j -th activity will result that the following constraints must be accomplished: xj 0, 1 j n. (1.2)
Inequalities 1.1 are called the constraints of the problem and 1.2 are called the non negativity conditions of the problem. The system of inequalities can have an innite of solutions, a unique solution or no solution (incompatible system). In many cases when the practical problems are well dened we are in the situation in which the system 1.1 and 1.2 has an innite number of solutions. Thus is possible to organize the production process for the products of type j , 1 j n, in several ways which respects the constraints 1.1 of using the limited resources. Adoption of plan is made using some economic criteria such as: the revenue, the labor time, production costs etc. Let us assume that we can integrate all these economic criteria in one criteria which allows us to adopt the decision. Moreover we shall assume that criteria is a linear one. An example of such criteria we obtain in the following way. If we denote by cj the revenue per unit obtained by the activity of type j , 1 j n, then the total revenue is:
n
cj xj .
j =1
(1.3)
The problem which must be solved, is to nd the solutions of the systems of linear inequalities 1.1, 1.2 which gives the maximum value for the total revenue 1.3. In others words, from mathematical point of view we need to solve the problem: n sup cj xj j =1
n
j =1
aij xj bi , 1 i m,
xj 0, 1 j n,
which is called linear programming problem (or linear program).The linear function which must be maximized is called objective function, criteria function or eciency function. The above problem can be solved with simplex algorithm or simplex dual algorithm. These algorithms are presented in every basic optimization book. Some software products, such as MAPLE have specialized routines for solving linear optimizations problems.
1.2.
## Forms of linear programming problems
Standard form of a linear programming problem is: min(max)c x Ax = b x0 Let us remark that a maximum problem can be transformed in a minimum problem using the formulamax(f ) = min(f ). Canonical form of a linear programming problem is: min c x Ax b x0 or max c x Ax b x0 MAPLE code for transforming in canonical form a maximum problem is: > standardize(set of constraints); or > convert(set of constraints, stdle); A constraint of a linear programming problem is called concordant if it is an inequality for the minimum problem and an for the maximum problem. The mixed form of a linear programming problem contains constraints (some times we use the term of restrictions) which are equations. Because, using mathematical operations, every linear programming problem can be transformed in canonical form, for the minimum optimization problem, we shall work only with such of these problems, thus with problems in canonical form. Denition 1.2.1. Let be the linear programming problem in standard form. Then, we dene the set of programs by: P = {x Rn |Ax = b, x 0}. A minimum point of the objective function z = c x over the set of programs P is called optimal solution and the set of optimal solutions will be denoted with: P = {x P | min c x = c x}.
xP
LINEAR PROGRAMMING
Dene the base solution of the system Ax = b a solution Rn for which to nonzero components corresponds linear independent columns. If B is a basis formed with the columns aj1 , . . . , ajm of the matrix A then the system of equations Ax = b can be written in the explicit form: xB = B1 b B1 RxR where R is the matrix obtained from A by elimination the columns j1 , . . . , jm . If we denote:
B , 1 j n, B1 b =x , B1 aj = yj B
results:
B xB =x j R
B yj xj
equivalently xB i =xi
j R B B yij xj , i B .
where B = {j1 , . . . , jm } si R = {1, . . . , n} B. B si xR = 0. It The base solution which corresponds to the basis B is xB =x results that these base solution is a program if we have: B1 b 0. A basis B which meet the above inequality is called primal feasible basis. In practice these kind of basis are determined using the articial basis method. In many situations this basis is available directly, the feasible basis being the identity (unit) matrix. We have the following theorem called the fundamental theorem of the linear programming. Theorem 1.2.1. i) If the linear programming problem: min(max)c x Ax = b x0 has an optimum, then the problem has a basis program. ii) If the above problem has a optimum program, then it has a optimum basic program. The fundamental algorithm for solving the linear programming programs is called simplex primal algorithm and it proposed by George Dantzig in 1951. This algorithm is described in every fundamental book of Operational Research and is implemented in every dedicated software package.
1.3.
## Simplex algorithm (Dantzig)
For solving linear programming problems we can use simplex algorithm developed by Dantzig (1951). This algorithm allows systematically searching, in the set of base programs of the linear programming problem in standard form, the transition form a base program to another program as least as good as the previous. The algorithm gives us criteria for situations in which the optimum is innity or for the case in which the set of programs is the empty set. STEP 0. Transform the optimization problem in standard form: inf cT x Ax = b, x 0. (If a problem has inequality constraints rs we transform this inequalities; subtracting the slack variables y, the maximum problem will be transformed in a minimum problem etc.). A is a matrix with m rows and n columns for which we have rang(A) = m < n. Denote by z the objective function, thus z = cT x. STEP 1. We nd a primal feasible basis B (available directly or using articial base method) and compute: xB = B1 b, B B z = cT B x , B 1 yj = B aj , 1 j n, z B c , 1 j n. j j These values are written in the simplex table (table 1.1) and we go to the next step. Let us denote by B the set of j indexes which determines the matrix B and by R = {1, . . . , n} B . The initial simplex table has the form:
TABLE 1.1
cB
B.V. xB z
V.B.V xB zB
c1 x1 B y1 Bc z1 1
## ... ... ... ...
cj xj B yj Bc zj j
## ... ... ... ...
cn xn B yn Bc zn n
where we denoted by B.V. the set of base variables and by V.B.V. the values of the B base variables (x ). B c 0 j R, we stop (STOP): xB is an optimum program. STEP 2. If zj j
## 6 Otherwise we nd the set (non empty):
LINEAR PROGRAMMING
B R+ = {j R|zj cj > 0}
and go to the next step. B 0 we stop (STOP): optimum STEP 3. If there exists j R+ for which yj of the problem is innity. Otherwise, we nd k R+ with entering variable choice rules: B B max(zj cj ) = zk ck
j B > 0} with leaving variable choice rules: si r B+ = {i B|yik
iB+
min (
xB xB i r ) = B B yik yrk
## B is called pivot. Go to the next step. The element yrk
STEP 4. Let us consider the basis B obtained by B replacing the ar column with the column ak , and compute the values (using the rules for changing variables)
B , z B c and go to the Step 2 replacing the basis B with the basis xB , z B , yj B. j j The computations can by simplied using the rules for simplex table transformation: i) the elements from the pivot lined are divided at the pivot value; ii) the elements from the pivot column became zero, with the exception of the pivot which became 1; iii) the others values are transformed using the rectangle rule: if we imagine the B rectangle with the diagonal determined by the element under transformation yij B , then the new values y and the pivot yrk ij is obtained by dividing to the pivot the B y B placed on the above dened dierences between the product of the elements yij rk B y B placed on the other diagonal of the rectangle. diagonal and the product yrj ik B c < 0 j R then Remarks. i) If at the ending of the algorithm we have zj j the solution of the problem is unique. ii) During algorithm execution in is possible to have the cycling phenomena (passing to another basis which was already processed). There are several techniques to avoid this. iii) We have also, in bidimensional case, a geometrical interpretation of the solutions of a linear programming problem. The feasible domain can be: -a convex polyhedron, in this case one of its vertex is the solution of the optimization problem. Also we may have the solution being one side of the polyhedron, in this case we have many solutions; B
THE DUAL OF THE LINEAR PROGRAMMING PROBLEM -unbounded domain, in this case the optimum is innity; -empty set, in this case the optimization problem has no solution.
1.4.
## The dual of the linear programming problem
The primal problem deals with physical quantities. With all inputs available in limited quantities, and assuming the unit prices of all outputs is known, what quantities of outputs to produce so as to maximize total revenue? The dual problem deals with economic values. With oor guarantees on all output unit prices, and assuming the available quantity of all inputs is known, what input unit pricing scheme to set so as to minimize total expenditure? To each variable in the primal space corresponds an inequality to satisfy in the dual space, both indexed by output type. To each inequality to satisfy in the primal space corresponds a variable in the dual space, both indexed by input type. The coecients that bound the inequalities in the primal space are used to compute the objective in the dual space, input quantities in this example. The coecients used to compute the objective in the primal space bound the inequalities in the dual space, output unit prices in this example. Both the primal and the dual problems make use of the same matrix. In the primal space, this matrix expresses the consumption of physical quantities of inputs necessary to produce set quantities of outputs. In the dual space, it expresses the creation of the economic values associated with the outputs from set input unit prices. Since each inequality can be replaced by an equality and a slack variable, this means each primal variable corresponds to a dual slack variable, and each dual variable corresponds to a primal slack variable. This relation allows us to complementary slackness. Thus if we have a linear programming problem we can construct the dual problem using the following rules: a) the free terms from the primal problem became coecients of the objective function from the dual problem; b) coecients of the objective function from the dual problem became free terms in the dual problem; c) the dual of a maximization (minimization) primal problem is a minimization (maximization) problem; d) the matrix of the constraints from the dual problem is the transposed of the matrix of the primal problem; e) dual (primal) variables associated with concordant primal (dual) constraints are supposed to the non negative constraints;
LINEAR PROGRAMMING
f) primal (dual) variables associated with non-concordant dual (primal) constraints are supposed to the non positivity g) dual (primal) variables associated with primal (dual) constraints which are equations have no constraints regarding the sign. Let us remark that the dual of a problem in the canonical form is also in canonical form. Here is the fundamental the fundamental theorem of duality: Theorem 1.4.1. If we have the couple of dual problems: min c x Ax b x0 and max b y Ayc y0
only one of the following problems is true: a) both problems have programs. In this case, both problems have optimal programs and the optimal values of the objective functions are the same; b) one of the problems have programs and the other one has no programs. In this case, the problem which has programs has optimum innity; c) none of the problems have programs. The simplex dual algorithm solve the primal problem by the dual problem. Similarly, after solving the primal algorithm we can construct the solution of the dual problem. MAPLE software has a procedure for the construction of the dual problem. The syntax is the following: > dual(objective function, set of constraints, dual variable set);
1.5.
Transportation problems
Assume that we have m warehouses and n shops. A homogenus product is stored in the quantities ai , 1 i m, in the warehouses and it is requested by the shops in the quantities bj , 1 j n,. We assume that the following conditions are satised: ai 0, 1 i m, bj 0, 1 j n, a1 + . . . + am = b1 + . . . + bn . (1.4)
TRANSPORTATION PROBLEMS
In others words the available quantities and the requested quantities are non negative and the request it is equal with the availability. Such a problem is called a balanced transportation problem. The problem consists in planning the transportation from the warehouses to the shops with the minimum transportation cost. Denote by xij the quantity (unknown) which will be supplied from the warehouse i to the shop j . The total quantity transported from the warehouse i to the all shops must be equal with the availability of the warehouse i, thus:
n
xij = ai , 1 i m.
j =1
(1.5)
Similarly, all the total requested quantity of the shop j must be equal with the total quantities transported from all the warehouses to this shop, thus:
m
xij = bj , 1 j n.
i=1
(1.6)
## The transported quantities are non negative: xij 0, 1 i m, 1 j n. (1.7)
The system of linear inequalities 1.5-1.7 in the constraints 1.4 has an innite number of solutions. For adopting a transportation plan we need a economic criteria given by the transportation cost. Denote by cij the transportation cost per unit from the warehouse i to the shop j ; we shall assume that the cost per unit does not depends of the total quantity transported from i to j . The transportation cost is:
m n
cij xij .
i=1 j =1
The transportation problem consist in determination the solution of the system 1.5-1.7 for which the total cost is minimum, thus: m n cij xij , inf i=1 j =1 n x = a , 1 i m, ij i j =1 m xij = bj , 1 j n, i=1 xij 0, 1 i m, 1 j n.
10
LINEAR PROGRAMMING
1.6.
Applications
Exercise 1.6.1. Find the dual of the linear programming problem: min(2x1 + 3x2 + x3 ), x1 + x2 + 3x4 3, 2x2 + 5x3 + 4x4 = 5, x + x3 2, 1 x1 , x2 0, x3 arbitrary, x4 0. Find the optimal solution of the primal problem. Solution. Using the rules for construction the dual problem we get: max(3u1 + 5u2 2u3 ), u1 + u3 2, u1 + 2u2 3, 5u2 + u3 = 1, 3u1 + 4u2 0, u1 0, u2 arbitrary, u3 0. The solution of the primal problem can be found with simple dual or primal algorithm. MAPLE code for solving this problem is: > with(simplex) : > objective := 2 x1 + 3 x2 + x3 ; > constraints := {x1 + x2 +3 x4 >= 3, 2 x2 +5 x3 +4 x4 = 5, x1 + x3 <= 2}; > minimize(objective, constraints union {x1 >= 0, x2 >= 0, x4 <= 0}); Solution given by the program will be: x1 = 0, x2 = 15/2, x3 = 2, x4 = 0, the value (optimum) for the objective function will be in this case 41/2. Exercise 1.6.2. Solve the following linear programming problem. Also solve its dual. min(2x1 + 3x2 + x3 ), x1 + x2 + 3x4 3, 2x2 + 5x3 + 4x4 = 5, x + x3 2, 1 x1 , x2 0, x3 arbitrary, x4 0. where is a real parameter.
APPLICATIONS Exercise 1.6.3. Solve the linear programming problem: min(2x1 + 3x2 + x3 ), x1 + x2 + 3x4 3 + , 2x2 + 5x3 + 4x4 = 5 , x + x3 2 + 2, 1 x1 , x2 0, x3 arbitrary, x4 0. where is a real parameter. Also solve the dual problem.
11
Solution. Exercises 1.6.2-1.6.3 can be solved using simplex primal algorithm or using post optimization technique in the nal simplex table, for the problem 1.6.1, with = 0. Exercise 1.6.4. Using simplex algorithm solve the following linear programming problem: max(2x1 + x2 ), x1 x2 4, 3x1 x2 18, x1 + 2x2 6, x1 , x2 0. Exercise 1.6.5. Minimize 2x1 + 3x2 under the constraints: x1 x2 + x3 = 1, x1 + x2 x4 = 1, x1 2x2 + x5 = 1, 2x1 + x3 x4 = 2, xi 0, i = 1, . . . , 5.
Solution. Applying simplex algorithm we nd the solution x 1 = 1, x2 = 0, x3 = 0, x4 = 0, x5 = 0, and the minimum for the objective function z = 2.
Exercise 1.6.6. Solve the following problem: min(x1 + 6x2 ), 2x1 + x2 3, x + 3x2 4, 1 x1 , x2 0.
12
LINEAR PROGRAMMING
Fig. 1.1: Feasibility domain and the objective function. Solution. We shall solve the problem graphically. We represent in bi dimensional space the feasibility domain. On the same graphic 1.1 we represent the function 1 x1 + 6x2 = 0. Now we draw parallel lines with x2 = x1 until we intersect the 6 feasibility domain. The rst line which intersect this domain gives us the minimum of the objective function zmin = 4. Optimal solution is x 1 = 4, x2 = 0. Exercise 1.6.7. Find the minimum of the function 2x1 + 3x2 + x3 with the constraints: x1 + x2 + 3x4 3, 2x2 + 5x3 + 4x4 = 5, x + x3 2, 1 x1 , x2 , x3 , x4 {0, 1}. Exercise 1.6.8. Solve the following linear programming problem: min(2x1 + 3x2 ), x1 x2 + x3 = 1, x1 + x2 x4 = 1, x1 2x2 + x5 = 1, 2x1 + x3 x4 = 2, xi 0, 1 i 5.
APPLICATIONS
13
Solution. For solving the linear programming problem by simplex algorithm rst we solve the following problem: min(x6 + x7 + x8 ), x 1 x2 + x3 + x6 = 1, x1 + x2 x4 + x7 = 1, x1 2x2 + x5 = 1, 2 x1 + x3 x4 + x8 = 2, xi 0, 1 i 8. (we have introduced the slack variables x6 , x7 , x8 in the constraints of the initial problem, with an exception for the third constraint where the variable x5 is associated with a unit vector of the matrix of the coecients). The primal admissible basis is constructed from the columns of matrix A corresponding to the variables x6 , x7 , x5 , x8 . The simplex associated table 1.2 is:
TABLE 1.2
B.V. x6 x7 x5 x8
V.B.V. 1 1 1 2 4
x1 1 1 1 2 4
x2 1 1 2 0 0
x3 1 0 0 1 2
x4 0 1 0 1 2
x5 0 0 1 0 0
x6 1 0 0 0 0
x7 0 1 0 0 0
x8 0 0 0 1 0
Using entering variable choice rules the vector which entry in the basis will be a1 . Leaving variable choice rules indicates that we can eliminate any of the basis variables: we choose for example the variable that corresponds to x6 , thus a6 . We obtain the simplex table 1.3:
TABLE 1.3
B.V. x1 x7 x5 x8
V.B.V. 1 0 0 0 0
x1 1 0 0 0 0
x2 1 2 1 2 4
x3 1 1 1 1 2
x4 0 1 0 1 2
x5 0 0 1 0 0
x6 1 1 1 2 4
x7 0 1 0 0 0
x8 0 0 0 1 0
From table 1.3 we see that the value of the objective function is 0. All the slack variables are 0 but x7 and x8 are still basis variables. It is easy to see that the variable x7 can be eliminated from the basis and replaced with x2 , because the pivot B = 2 is dierent from 0. After the computations we obtain the simplex table: y72
14
TABLE 1.4
LINEAR PROGRAMMING
B.V. x1 x2 x5 x8
V.B.V. 1 0 0 0 0
x1 1 0 0 0 0
x2 0 1 0 0 0
x5 0 0 1 0 0
## x7 1/2 1/2 1/2 1 2
x8 0 0 0 1 0
The slack variable x8 can not be eliminated from the basis variables because all the values y8j , 1 j 5, are equal to 0. This shows us that the four-th equation, from the initial linear programming problem, is a consequence of the others equations (the four-th equation is the sum of the rst two equations, situation that can be observed from the beginning of computations). In this case, the four-th equation can be eliminated and also the corresponding line from the simplex table. The remainder table does not contain any slack variables in the basis, thus the rst phase is nished. In the second phase, we solve the initial linear programming problem, using initial basis the basis obtained in table 1.4, after the elimination the last row of the table. The corresponding simplex table is the following:
TABLE 1.5
B.V. x1 x2 x5 z
V.B.V. 1 0 0 2
x1 1 0 0 0
x2 0 1 0 0
## x4 1/2 1/2 1/2 5/2
x5 0 0 1 0
Because zj cj 0 for all j, 1 j 5, results that we obtained the optimal solution (degenerated): x 1 = 1, x2 = 0, x3 = 0, x4 = 0, x5 = 0, and the optimal value of the objective function is z = 2. Exercise 1.6.9. (diet problem) In a bakery the are two kinds of cakes: brownies, which cost 50 cents each, and minicheesecakes, which cost 80 cents. The bakery is service-oriented and can sell a fraction of any item. The bakery requires three ounces of chocolate to make each brownie (no chocolate is needed in the cheesecakes). Two ounces of sugar are needed for each brownie and four ounces of sugar for each cheesecake. Finally, two ounces of cream cheese are needed for each brownie and ve ounces for each cheesecake. A snack consumer health-conscious, has decided that he needs at least six total ounces
APPLICATIONS
15
of chocolate in his snack, along with ten ounces of sugar and eight ounces of cream cheese. The consumer wishes to optimize his purchase by nding the least expensive combination of brownies and cheesecakes that meet these requirements. The data is sumarized in the following table:
TABLE 1.6
## Brownie Cheesecake Requirements
Chocolate 3 0 6
Sugar 2 4 10
Cream Cheese 2 5 8
Cost 50 80
Find the optimal solution. Solution. The problem that the consumer must to solve is: min(50x + 80y ), 3x 6, 2x + 4y 10, 2x + 5y 8, x, y 0, where x and y represent the number of brownies and cheesecakes purchased, respectively. By applying the simplex method of the previous selection, we nd that the unique solution is (2, 3/2) , thus the value of the objective function, computed for the optimal solution is 220. We now adopt the perspective of the wholesaler who supplies the baker with the chocolate, sugar, and cream cheese needed to make the goodies. The baker informs the supplier that he intends to purchase at least six ounces of chocolate, ten ounces of sugar, and eight ounces of cream cheese, to meet the consumer minimum nutritional requirements.He also shows the supplier the other data from the table. The supplier now solves the following optimization problem: How can I set the prices per ounce of chocolate, sugar, and cream cheese so that the baker will buy from me, and so that I will maximize my revenue? The baker will buy only if the total cost of raw materials for brownies is below 50 cents; otherwise he runs the risk of making a loss if the student opts to buy brownies. This restriction imposes the following constraint on the prices: 3u1 + 2u2 + 2u3 50. Similarly, he requires the cost of the raw materials for each cheesecake to be below 80 cents, leading to a second constraint:
16
LINEAR PROGRAMMING
4u2 + 5u3 80. Clearly, all the prices must be nonnegative. Moreover, the revenue from the guaranteed sales is 6u1 + 10u2 + 8u3 . In summary, the problem that the supplier solves to maximize his guaranteed revenue from the consumer snack is as follows (dual problem): max(6u1 + 10u2 + 8u3 ) 3u1 + 2u2 + 2u3 50 4u2 + 5u3 80 u1 , u2 , u3 0. The solution of this problem is u = (10/3, 20, 0) , thus the value of the dual objective function is 220 (which is the same with the value of the objective function computed for optimal solution of the primal problem). Exercise 1.6.10. A gardener produce two types of mixtures for planting: gardening mixture and potting mixture. A package of gardening mixture requires 2 kg of soil, 1 kg of peat moss and 1 kg of fertilizer. A package of potting mixture requires 1 kg of soil, 2 kg of peat moss and 3 kg of fertilizer. The gardener has at most 16 kg of soil, 11 kg of peat moss and 15 kg of fertilizer. A package of garden mixture sells for 3 EURO and a package of potting mixture sells for 5 EURO. How many packages of each type of mixture must the gardener to produce to maximize the revenue? Solution. The problem that the gardener must to solve is: max(3x + 5y ), 2x + y 16, x + 2y 11, x + 3y 15, x, y 0, where x and y are the packeges of gardening mixture and potting mixture. The gardener must produce 7 packages of gardening mixture and 2 packages of potting mixture, the revenue is 31 EURO. Let see the problem from the point of view of the supplier which is informed by the gardener that he want to sell with at least 3 EURO a package of gardening mixture and at least 5 EURO a package of potting mixture. Also, the gardener
APPLICATIONS
17
will inform the supplier that he intend to buy 16 kg of soil, 11 kg of peat moss and 15 kg of fertilizer. The supplier will setup the prices u,v and w for 1 kg of soil, 1 kg of peat moss respectivelly 1 kg of fertilizer such that to minimize the function 16u + 11v + 15w (otherwise the gardener will buy from another supplier). Also the restriction regarding the cost of gardening mixture and potting mixture must be imposed: 2u + v + w 3 respectivelly u + 2v + 3w 5. Thus, the problem which must be solved by the supplier is the dual problem of the gardener: min(16u + 11v + 15w), 2u + v + w 3, u + 2v + 3w 15, u, u, w 0.
18
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CC-MAIN-2019-43
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-7-section-7-1-radical-expressions-and-functions-exercise-set-page-512/66
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## Intermediate Algebra for College Students (7th Edition)
RECALL: When the index, $n$ is even, the value of $a$ in $\sqrt[n]{a}$ cannot be negative since the nth root will not be a real number. The given radical expression has an even index. Since the number inside the radical sign is negative, then its nth root is not a real number.
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CC-MAIN-2018-47
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http://www.cs.cmu.edu/~jcl/classnotes/physics/quantum_mechanics/schrodinger.html
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The schrodinger equation in 3-d is:
H Psi = i hb d/dt (Psi)
with H=1/2m*p^2+V(r_x,r_y,r_z)=1/2m * (p_x^2+p_y^2+p_z^2) + V(r_x,r_y,r_z)
p_i=hb/i * d/dx
More concisely, H=-hb^2/2m del^2 + V(r_x,r_y,r_z)
Typically, the differential equation for Psi will quantize as:
H Psi_n = E_n Psi_n
with E_n= quantized energy states indexed by n and Psi_n = an eigenstate.
If E<0 => discrete spectrm while E>0 => continuous spectrum.
If V-> 0 as r -> infinity => Psi -> 0 as r -> infinity, something necessary for normalization.
An analysis of the schrodinger equation in spherical coordinates indicates that it can be written as:
H=p_r^2/2m + L^2/2mr^2 +V(r)
This is solved by the method of seperation of variables, first seperation into Psi(r,theta,phi)=R(r)Y(theta,phi)
Then into Y(theta,phi)=P(theta) Phi(phi)
The BC on phi is that Phi(phi)=Phi(phi+2 PI) => Phi(phi)=1/(2 PI)^.5 e^(i m phi) for m in Z
P(theta) is transformed into P(cos(theta)). Demanding non-infinite solutions => associated Legendre polynomials are a solution.
The solution of the radial equation is Hermite polynomials.
source jl@crush.caltech.edu index
variational_method
WKB_bound_state
TDPT
WKB
TDSE
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https://pupman.com/listarchives/2005/Apr/msg01188.html
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# Re: Current Limiting and Impedence
• To: tesla@xxxxxxxxxx
• Subject: Re: Current Limiting and Impedence
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Wed, 27 Apr 2005 11:00:46 -0600
• Delivered-to: testla@pupman.com
• Delivered-to: tesla@pupman.com
• Old-return-path: <teslalist@twfpowerelectronics.com>
• Resent-date: Wed, 27 Apr 2005 11:02:12 -0600 (MDT)
• Resent-from: tesla@xxxxxxxxxx
• Resent-message-id: <5Rg8KC.A.pzD.JW8bCB@poodle>
• Resent-sender: tesla-request@xxxxxxxxxx
`Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>`
`Gerry:`
```Thanks for all your help. Now if you can bear with me and help me solve
the original problem with the math(BTW <> is "not equal to")....```
```I have (2) inductors in series. My current limiter and my transformer.
Mains is 120V/60 Hz/1 ph.```
```The current limiter is 15 mH, X = 5.65 Ohms, R = 0.3 Ohms.
The transformer is Z = 1.2 Ohms, R = 0.3 Ohms, so X = 1.16 Ohms```
`So System Impedence Z = 6.82 Ohms`
```This gives I = 120/6.82 = 17.6 Amps and is near exactly what I am
getting.```
```The measured voltage between the inductor and transformer is about 90 to
95 volts. Have gotten as high a 100 volts. This is the voltage drop
across the transformer. This is inconsistent with the above values and
in fact seems reversed from what the math suggests. I have double
checked it.```
`For more confusion, lets compute power:`
```True power = R*I^2 = 0.6*17.6^2 = 186 watts
Reactive power = X*I^2 = 6.81*17.6^2 = 2109 var
Apparent power = 2117 VA
Power Factor: 186/2117 = .08 (makes no sense)```
```I know I have 120*17.6 = 2112 VA input and approx 7500*.2=1500 watts
output from the transformer secondary. Thus I have a power factor of
approximately .71. I get 48 to 60" sparks from the coil/toroid so I've
got to have near that kind of power.```
```I believe that I am improperly accounting for the power that is
transferred from transformer primary to secondary in the above
equations. It's as if some of the reactive power "created" by the X of
the inductor is actually true power for the transformer.```
```I tested the current limiter and don't think it is saturating.
(BTW - I'm a "math guy" so I can handle the vector addition and complex
variables no problem. But struggling with setting up equations and how
to apply them.)```
```Thanks again for your help.
Mark```
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https://www.askiitians.com/forums/9-grade-science/the-ratio-4-16-is-in-its-lowest-form-the-ratio-4_269079.htm
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# The ratio 4 :16 is in its lowest form. The ratio 4 :16 is in its lowest form.
Harshit Singh
3 years ago
Dear Student
The Given statement is False
4 : 16
= 4/16
Divide both numerator and denominator by 4, = 1/4
Therefore, lowest form of 4: 16 is 1/4
Thanks
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https://www.gradesaver.com/textbooks/science/physics/college-physics-4th-edition/chapter-8-problems-page-311/17
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## College Physics (4th Edition)
Published by McGraw-Hill Education
# Chapter 8 - Problems - Page 311: 17
#### Answer
We can rank the situations in order of the magnitude of the torque applied to the handle, from smallest to largest: $e \lt a = b = d \lt c$
#### Work Step by Step
The magnitude of the torque can be expressed as $\tau = r~F~sin~\theta$, where $r$ is the displacement between the rotation axis and the point where the force is applied, $F$ is the force, and $\theta$ is the angle between the force vector and the displacement vector $r$. We can find the magnitude of the torque for each situation: (a) $\tau = r~ F~sin~\theta$ $\tau = (0.50~m)(20~N)~sin~90^{\circ}$ $\tau = 10~N\cdot m$ (b) $\tau = r~ F~sin~\theta$ $\tau = (0.25~m)(40~N)~sin~90^{\circ}$ $\tau = 10~N\cdot m$ (c) $\tau = r~ F~sin~\theta$ $\tau = (0.25~m)(80~N)~sin~60^{\circ}$ $\tau = 17.3~N\cdot m$ (d) $\tau = r~ F~sin~\theta$ $\tau = (0.25~m)(80~N)~sin~30^{\circ}$ $\tau = 10~N\cdot m$ (e) $\tau = r~ F~sin~\theta$ $\tau = (0.50~m)(40~N)~sin~0^{\circ}$ $\tau = 0$ We can rank the situations in order of the magnitude of the torque applied to the handle, from smallest to largest: $e \lt a = b = d \lt c$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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http://www.gamefaqs.com/gba/468548-golden-sun/faqs/16530
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## Enemy List by MathU
Version: 2.0 | Updated: 04/06/02 | Search Guide | Bookmark Guide
```Enemy Roster for Golden Sun
Version 2.0
2. Enemy Roster
3. Boss Roster
4. Rare Item Drop Statistics
6. Credits
Version 2.0, 2/27/02-Don't worry, there will be many updates to come!
Version 2.1, 4/2/02-Fixed a large amount of things, including the spaces
between lines. I e-mailed this to my PC and sent it in from there, fixing
the conversion problem CJayC had been encountering with my Macintosh.
2. Enemy Roster
Notes
-Enemies are in order of appearance.
-See the location chart to see what the numbers for location mean.
-Any items that have (rare) at the end of them mean that there is less of
a possibility to get them at the end of a battle (see Rare Item Drop
-When you do damage to the enemy with the element that they are weak to,
you will see three exclamation marks instead of the normal one. If you do
damage with the element that they are resistant to, you will just see a
period.
-Last, but not least, one of the most overlooked things in the game is the
fact that you can increase the amount of experience points and coins you
get from an enemy and increase the chances of getting it to drop an item
by killing it with a Djinni of the element that they it is weak to. You
will know when this has happened because the enemy will change colors and
add an extra cry before it dies. This is one of the most effective ways
to get a certain item from an enemy. Where I have listed what experience
points and coins you get, the second amount is the one you get when you
kill the enemy with the Djinni of the element they're weak to. I have
neglected to put in this amount for enemies that have a base experience or
coins that is below seven because if you try to use this on them, the
amount they give up adds a random number for extra experience or extra
coins instead of the normal number created by a thirty percent increase
then taking off the decimal.
1-Vale
2-Sol Sanctum
3-World Map from Vale to Goma Cave
4-Goma Cave
5-World Map from Goma Cave to Bilibin Cave to Kolima Bridge
6-Kolima Forest
7-Tret Tree
8-Bilibin Cave
9-World Map from Bilibin Cave to Mercury Lighthouse
10-Mercury Lighthouse
11-World Map from Kolima Bridge to Mogall Forest
12-Fuchin Falls Cave
13-Mogall Forest
14-World Map from Mogall Forest to Lamakan Desert
15-Altin Peak
16-Lamakan Desert
17-World Map from Lamakan Desert to Kalay Docks (not including anywhere past
fixed bridge [bridge is fixed when you get past Lamakan Desert])
18-Vale Cave
19-Vault Cave
20-World Map from Tolbi Docks to Altmiller Cave to Kalay Docks to Gondowan
Cave to Gondowan Passage (this is a very large area)
21-Altmiller Cave
22-Gondowan Cave
23-Lunpa Fortress
24-World Map from Gondowan Passage to Road to Babi Lighthouse
25-Suhalla Desert
26-Suhalla Gate
27-Venus Lighthouse (first half)
28-Babi Lighthouse Entry, Babi Lighthouse, and Tunnel Ruins
29-Venus Lighthouse (second half)
Vermin (a)
Location-1, 2
HP-20
Weak-Fire
Resist-Wind
Experience Points-2
Coins-2
Drop-Herb
Bat (a)
Location-1, 2
HP-17
Weak-Wind
Resist-Earth
Experience Points-1
Coins-1
Drop-Herb
Wild Mushroom (a)
Location-1, 2
HP-18
Weak-Fire
Resist-Water
Experience Points-1
Coins-2
Drop-Smoke Bomb
Amaze (a)
Location-2
HP-28
Weak-Wind
Resist-Water
Experience Points-3
Coins-3
Drop-Oil Drop
Slime (a)
Location-2
HP-22
Weak-Fire
Resist-Water
Experience Points-2
Coins-2
Drop-Herb
Amaze (b)
Location-3
HP-39
Weak-Wind
Resist-Water
Experience Points-5
Coins-6
Drop-Oil Drop
Bat (b)
Location-3
HP-30
Weak-Wind
Resist-Earth
Experience Points-2
Coins-3
Drop-Herb
Slime (b)
Location-3
HP-32
Weak-Fire
Resist-Water
Experience Points-4
Coins-4
Drop-Herb
Wild Mushroom (b)
Location-3
HP-34
Weak-Fire
Resist-Water
Experience Points-3
Coins-3
Drop-Smoke Bomb
Vermin (b)
Location-3, 4
HP-36
Weak-Fire
Resist-Wind
Experience Points-4
Coins-4
Drop-Herb
Zombie
Location-3, 4, 5
HP-55
Weak-Fire
Resist-Water
Experience Points-7, 9
Coins-8, 10
Drop-Herb
Ghost
Location-4, 5
HP-56
Weak-Wind
Resist-Water
Experience Points-9, 13
Coins-9, 14
Drop-Oil Drop
Location-4, 5, 6
HP-54
Weak-Earth
Resist-Water
Experience Points-9, 13
Coins-10, 13
Drop-Smoke Bomb
Skeleton
Location-4, 5, 6, 12
HP-60
Weak-Earth
Resist-Water
Experience Points-10, 13
Coins-11, 14
Drop-Herb
Rat
Location-5, 6, 7
Weak-Fire
Resist-Wind
Drop-Herb
Rat Soldier
Location-5, 6, 7, 8
HP-58-
Weak-Fire
Resist-Water
Experience Points-
Coins-
Drop-Smoke Bomb
Drone Bee
Location-6, 7, 8
Weak-Wind
Resist-Earth
Drop-Elixir
Troll
Location-6, 7, 8
Weak-Fire
Resist-Wind
Drop-Nut
Creeper
Location-7
Weak-Wind
Resist-Water
Drop-Weasel's Claw
Spider
Location-7
Weak-Fire
Resist-Earth, Wind, and Water
Drop-Antidote
Gnome
Location-7, 8, 9, 10
HP-84-85
Weak-Water
Resist-Fire
Experience Points-29, 37
Coins-40, 52
Drop-Oil Drop
Ghoul
Location-8, 10
HP-99
Weak-Fire
Resist-Water
Experience Points-34, 44
Coins-42, 54
Drop-Antidote
Ooze
Location-8, 9, 10, 11, 12
HP-72
Weak-Fire
Resist-Water
Experience Points-26, 33
Coins-40, 52
Drop-Herb
Mauler
Location-9, 10, 11, 12
HP-107-108
Weak-Fire
Resist-Wind
Experience Points-37, 48
Coins-45, 58
Drop-Nut
Cuttle
Location-10
HP-78
Weak-Fire
Resist-Water
Experience Points-27, 35
Coins-49, 63
Drop-Sleep Bomb
Siren
Location-10
HP-112-121
Weak-Fire
Resist-Water
Experience Points-45, 58
Coins-56, 72
Drop-Crystal Powder
Lizard Man
Location-10, 11, 12
HP-129-133
Weak-Fire
Resist-Water
Experience Points-42, 54
Coins-52, 67
Drop-Nut
Mole
Location-11, 12, 13, 14
Weak-Fire
Resist-Wind
Drop-Bramble Seed
Dirge
Location-12, 13, 14
Weak-Wind
Resist-Earth
Drop-Nut
Spirit
Location-12, 13, 14
Weak-Wind
Resist-Water
Drop-Crystal Powder
Bone Fighter
Location-12, 13, 14, 15
Weak-Earth
Resist-Water
Drop-Sleep Bomb
Location-13, 14
Weak-Earth
Resist-Water
Drop-Nut
Ape
Location-13, 14, 15
Weak-Fire
Resist-Wind
Drop-Vial
Calamar
Location-15
Weak-Fire
Resist-Water
Drop-Elixir
Slime Beast
Location-15
Weak-Fire
Resist-Water
Drop-Herb
Tarantula
Location-15
Weak-Fire
Resist-Earth, Wind, and Water
Drop-Antidote
Rat Fighter
Location-15, 16
Weak-Fire
Resist-Water
Drop-Smoke Bomb
Ant Lion
Location-16
Weak-Water
Resist-Fire
Drop-Vial
Grub
Location-16
Weak-Fire
Resist-Earth, Wind, and Water
Drop-Antidote
Orc
Location-16
Weak-Wind
Resist-Earth
Drop-Nut
Salamander
Location-16
Weak-Water
Resist-Fire
Drop-Oil Drop
Fighter Bee
Location-16, 17, 18, 19
Weak-Wind
Resist-Earth
Drop-Elixir
Location-17, 18
Weak-Fire
Resist-Wind
Drops-Herb
Death Cap
Location-17, 18, 19
Weak-Fire
Resist-Water
Drop-Sleep Bomb
Location-17, 18, 19
Weak-Fire
Resist-Water
Drop-Antidote
Ghost Mage
Location-18, 19
Weak-Wind
Resist-Water
Drops-Bramble Seed
Ravager
Location-18, 19
Weak-Fire
Resist-Wind
Drop-Elixir
Cave Troll
Location-19
Weak-Fire
Resist-Wind
Drop-Nut
Gargoyle
Location-19
Weak-Earth
Resist-Wind
Drop-Sleep Bomb
Rabid Bat
Location-20
Weak-Wind
Resist-Earth
Drop-Sleep Bomb
Armored Rat
Location-20, 21
Weak-Fire
Resist-Wind
Drop-Herb
Foul Dirge
Location-20, 21
Weak-Wind
Resist-Earth
Drop-Smoke Bomb
Gnome Mage
Location-20, 21
Weak-Earth
Resist-Wind
Drop-Crystal Powder
Location-20, 21
Weak-Fire
Resist-Wind
Drop-Bramble Seed
Clay Gargoyle
Location-21
Weak-Earth
Resist-Wind
Drop-Nut
Location-21
Weak-Water
Resist-Fire
Drop-Prophet's Hat (Rare)
Golem
Location-21
Weak-Wind
Resist-Earth
Drop-Vial
Wight
Location-21
Weak-Fire
Resist-Water
Drop-Elixir
Worm
Location-21
Weak-Fire
Resist-Earth, Wind, and Water
Drop-Herb
Rat Warrior
Location-22, 23, 24
Weak-Fire
Resist-Water
Drop-Bramble Seed
Kobold
Location-22, 23, 24, 25
Weak-Wind
Resist-Earth
Drop-Sleep Bomb
Dirty Ape
Location-22, 24
Weak-Fire
Resist-Earth
Drop-Antidote
Warrior Bee
Location-22, 24
Weak-Wind
Resist-Earth
Drop-Elixir
Brigand
Location-23
Weak-Everything
Resist-None, because they are weak to everything
Drop-Crystal Powder
Vile Dirge
Location-23, 24, 25
Weak-Wind
Resist-Earth
Drop-Weasel's Claw
Brutal Troll
Location-24, 25
Weak-Fire
Resist-Wind
Drop-Nut
Orc Captain
Location-24, 25
Weak-Wind
Resist-Earth
Drop-Smoke Bomb
Acid Maggot
Location-25
Weak-Fire
Resist-Earth, Wind, and Water
Drop-Sleep Bomb
Harridan
Location-25
Weak-Wind
Resist-Earth
Drop-Nut
Magicore
Location-25
Weak-Water
Resist-Fire
Drop-Aura Gloves (Rare)
Roach
Location-25
Weak-Water
Resist-Fire
Drop-Bramble Seed
Stone Soldier
Location-25
Weak-Water
Resist-Fire
Drop-Elixir
Location-25
Weak-Earth
Resist-Wind
Drop-Weasel's Claw
Tempest Lizard
Location-25, Crossbone Isle World Map
Weak-Earth
Resist-Wind
Drop-Potion
Cannibal Ghoul
Location-26
Weak-Fire
Resist-Water
Drop-Antidote
Mole Mage
Location-26
Weak-Fire
Resist-Wind
Drop-Antidote
Earth Golem
Location-26, 27
Weak-Wind
Resist-Earth
Drop-Giant Axe (Rare)
Goblin
Location-26, 27
Weak-Wind
Resist-Earth
Drop-Vial
Horned Ghost
Location-26, 27
Weak-Wind
Resist-Water
Drop-Elixir
Gnome Wizard
Location-26, 27, 28
Weak-Wind
Resist-Earth
Drop-Crystal Powder
Nightmare
Location-26, 27, 28
Weak-Earth
Resist-Wind
Drop-Healing Ring (Rare)
Ice Gargoyle
Location-28
Weak-Earth
Resist-Wind
Drop-Potion
Orc Lord
Location-28
Weak-Wind
Resist-Earth
Drop-Lucky Medal
Plated Rat
Location-28
Weak-Fire
Resist-Wind
Drop-Elixir
Chimera Mage
Location-28, 29
Weak-Water
Resist-Fire
Drop-Spiked Armor (Rare)
Manticore King
Location-28, 29
Weak-Water
Resist-Fire
Drop-Psy Crystal
Skull Warrior
Location-28, 29
Weak-Earth
Resist-Water
Drop-Vial
Wild Gryphon
Location-28, 29
Weak-Wind
Resist-Earth
Drop-Feathered Robe (Rare)
Willowisp
Location-28, 29
Weak-Earth
Resist-Water
Drop-Nut
Boulder Beast
Location-29Weak-Water
Resist-Fire
Drop-Water of Life
Fenrir
Location-29
Weak-Fire
Resist-Water
Drop-Kikuichimonji (Rare)
Grand Golem
Location-29
Weak-Wind
Resist-Earth
Drop-Zodiac Wand (Rare)
Recluse
Location-29
Weak-Fire
Resist-Earth, Wind, and Water
Drop-Unicorn Ring (Rare)
Thunder Lizard
Location-29
Weak-Earth
Resist-Wind
Drop-Blessed Mace (Rare)
3. Boss Roster
The same rules apply for this roster as in the Enemy Roster.
Mars Djinni (Forge)
Location-4
HP-172
Weak-Water
Resist-Fire
Experience Points-28, 36
Coins-85, 110
Drop-nothing
4. Rare Item Drop Statistics
I have put this in my FAQ because these have been asked for many times on the
message board and because they have to do with enemy item drops.
All rare items take a random amount of time to obtain. It may take you only
the time it takes to get through a single area, or it could take you a month
to get a rare item. Keep this in mind when trying to obtain any. I have
taken out the Healing Ring and Unicorn Ring because you can already get them
earlier in the game and because I don't think they're that hard to obtain.
Prophet's Hat
Equip-anyone
Defense +30
Predicts being downed-has the same effect of Curse Psynergy
Notes-Very useful if gotten at first chance, but eventually replaced at
Lalivero.
Aura Gloves
Equip-anyone
Defense +36
Use to resist elements-raises one person's Resistance by 40
Notes-Try to get two, as towards the end of the game they will be your second
best shield/armlet.
Giant Axe
Equip-Isaac and Garet
Attack +114
Unleashes Meltdown-lowers Defense
Element-Mars
Notes-Quite powerful and useful, but eventually replaced.
Feathered Robe
Equip-Ivan and Mia
Defense +45
Jupiter Power +20
Jupiter Resist +30
Agility +30
Notes-THE strongest armor for Mia and Ivan, especially Ivan, him being a
Spiked Armor
Equip-Isaac and Garet
Attack +10
Defense +34
Critical Hits Increase
Notes-Good enough to equip if gotten at first chance, but quickly replaced.
This item is not very hard to get. I have 6 as of 1/10/02.
Zodiac Wand
Equip-Ivan and Mia
Attack +102
Unleashes Shining Star-looks cool and causes Delusion
Element-Jupiter
Notes-By the time you get this, it is already outclassed by all other
weapons. Not worth getting unless you want everything.
Kikuichimonji
Equip-Isaac, Garet, and Ivan
Attack +128
Unleashes Asura-looks very cool and does decent damage
Element-Jupiter
Notes-Almost definitely Jupiter element, THE best weapon for Ivan.
Definitely worth getting.
Blessed Mace
Attack +126
HP recovery +2
Replenishes 200 HP-not an Unleash, a use
Element-None, because it has no Unleash
Notes-Considering this weapon is a mace and it recovers 200 HP like Ply Well,
it should probably be equipped on Mia. It is worth getting and is the
strongest weapon for Mia, but may not be the best, considering her weapon
before it, the Righteous Mace, actually had an Unleash.
This FAQ is Copyright (c) 2001-2002 MathU
Copying this FAQ in part is forbidden; if you wish to copy it you must copy
it in full, including the Copyright information. Feel free to use this FAQ's
information, but you must give me credit. Also if you want to use this
information on other site you must ask me first, then e-mail me the URL at
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# Policies for Simultaneous Estimation and Optimization
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3 uncrtainty in th stimat of b As Π, th optimal input gos to zro h xpctd cost, for ρ =, simplifis to φ = σ (y ds ) + E = +ˆb Π ˆb As Π, th minimum xpctd cost approachs an uppr bound, which is th cost of slcting a zro input Again, small Π,,Π yild a larg xpctd cost h policy is suboptimal bcaus th slction of u dos not ta into account its ffct on Π,,Π (which ar quadratic in u ) his is, in ffct, a grdy policy: At ach tim indx, u is slctd to minimiz th immdiat xpctd cost, E((y y ds ) ), without rgard for futur costs As in, thr is no dsign for stimation, in th sns that th bnfits to b gaind from slcting inputs that ma th Π larg ar not considrd Drivativ of th xpctd cost with rspct to information An invaluabl and gnrally ovrlood fact is that, for many rgularizd or robust policis, th drivativ of th cost with rspct to th information matrix is asily computd For th rgularizd passiv larning policy with ρ =,w hav that d σ + d Π (yds ) +ˆb Πˆb (y ds ) = ˆbˆb +ˆb Πˆb 4 Prsistncy of xcitation, dithring, and maximally informativ inputs W hav sn that th cost incurrd at tim may b larg if λ min(π ) is small In othr words, if th covarianc Σ =Π is larg, ˆb is an unrliabl stimat of th systm paramtrs b, and this lads to poor prformanc h most immdiat solution is to nsur that som masur of information, such as λ min(π )=λ min(π + σ U U ), is larg his can b translatd into a rquirmnt that U b wllconditiond, which is what is usually mant by prsistncy of xcitation Informally, w want th u to span th whol input spac 4 Dithring Svral solutions hav bn proposd to satisfy th prsistncy of xcitation rquirmnt Dithring is a randomizd fasibl policy that consists of adding to th inputs som whit nois (i, normal, zro man and indpndnt random trms) his can wor wll sinc, givn a high nough nois lvl, Π will b wll-conditiond with high probability Although it has th advantag of vry simpl implmntation, dithring is obviously a sub-optimal policy, and th slction of a good nois lvl can b problmatic Exampl Considr th problm dscribd by 6 n =, ˆb =, Π =, σ =, =, Y ds = [ ] W implmnt a dithrd policy basd on rgularizd passiv larning ( ) Figur plots th xpctd cost as th varianc of th random trms (th input nois lvl) rangs from 9 to hs valus wr obtaind by Mont Carlo simulation, with runs at ach input nois lvl (th corrsponding rror bars ar also plottd) Not that this may sm countrintuitiv: h prformanc of th policy is improvd by adding indpndnt nois to th inputs As th nois lvl gos to zro, w approach th non-dithrd rgularizd passiv larning policy, which has an avrag cost of 9 (±4) At th optimal input nois lvl (slctd a postriori), th avrag cost is 4 (±9) Of cours, any practical dithring policy must slct th input nois lvl a priori, which may b difficult <φ> Figur : log σdith Expctd cost as a function of dithring lvl 4 Masuring and valuing information A mor thoughtful approach is to slct a prturbation that, for a givn lvl of control disturbanc, maximizs th information gathrd For this purpos, w nd a masur of information A list of possibl masurs, using th naming convntion from xprimnt dsign, is E-optimal: λ max (Σ ) = λ min(π ) D-optimal: log dt Π A-optimal: r (Σ ) = r Π h E-optimal and A-optimal masurs may b scald to account for th drivativ of th xpctd cost with rspct to information ( ), whil th D-optimal masur is invariant with scaling For a numrically ffctiv huristic, w would li a masur that is concav in th inputs hs masurs ar concav-quadratic, and linarizing Π in U mas thm concav his linarization can b xpctd to wor wll if th prturbations introducd for th purpos of idntification ar small W rturn to this point in 55 4 Maximally informativ inputs On approach consists of xprssing xplicitly th trad-off btwn control and information, by adding to th objctiv function an xtra trm valuing information A vry simpl xampl of such a policy is to slct th u that minimizs u Π u +(ˆb u y ds ) + ρu u γλ min(π ), and liwis for u,,u, with th appropriat updating of Π and ˆb h xtra trm mas this policy, in part, an xprimnt dsign problm As with dithring, a prturbation will b introducd in th input, what w might now call
4 an intllignt nois h factor γ> wighs th trad-off btwn idntification and control Slcting γ prsnts th sam difficultis as for slcting th dithring lvl An altrnativ approach is what has bn trmd plantfrindly idntification Although it is ssntially a solution to a diffrnt problm, plant-frindly idntification can b usd as a huristic for simultanous stimation and control h ida is to slct th maximally informativ input from within th st of inputs that p som masur of th tracing rror within a bound For a simpl xampl, w us a constraint on th absolut tracing rror (spcifid by M R) h policy is dfind by th program maximiz λ min(π ) subjct to ˆb u y ds M h bound M can b sn as th trad-off factor, with a rol similar to γ in th prvious problm Howvr, M is a mor physically maningful numbr, and should b asir to slct in practical applications h constraint on prformanc usd hr disrgards th uncrtainty in b A robust constraint can b usd in its plac (such a constraint is convx in th inputs in fact, it is a scond-ordr con constraint, s S Boyd t al []) Both ths problms ar convx if w linariz Π in u, in which cas thy ar radily solvd If this procdur linarization followd by th solution of a convx program is itratd, a (local) minimum of th non-convx problm will b rachd Not that ths huristics do not us th futur dsird outputs y ds, which w assumd nown his is part of th suboptimal natur of th huristics, and has th ffct of rducing th snsitivity of thir prformanc with rspct to th futur trajctory his rducd snsitivity may b a positiv fatur in applications whr th futur trajctory is not fully crtain 5 Optimal policy, dynamic program and approximation hs huristic approachs still lav us with som qustions, in particular about ) what masur of information to us, and ) how to dcid on th invitabl trad-off btwn th informativnss of u and th output rror xpctd to rsult from its application Roughly spaing, th answr to th scond qustion is that th trad-off should b such that th currnt loss in tracing prformanc (incurrd for th sa of informativnss) quals th total xpctd futur gains in tracing prformanc (du to improvd information about th systm) his, in turn, lads to an answr for th first qustion: h information masur should b such that it capturs th xpctd futur gain in tracing prformanc h tru solution to th problm is givn by a dynamic program, of which w will outlin th drivation his dynamic program is, howvr, hard to solv W propos an approximation which rsults in a smidfinit program 5 Optimal policy for = W assum, from hr on, ρ = Considr th simplst cas, whr = An input u is to b slctd so as to minimiz th xpctd cost φ = = E b, y y ds = E b, b u + y ds = E b, b u ßÞÐ = u Π u ßÞ Ð +(ˆb u y ds ßÞ Ð +(ˆb u y ds ßÞ Ð ) ) + σ ßÞÐ + ßÞÐ h thr trms mard can b intrprtd as ) th cost du to inaccuracy in th stimat of b, ) th cost du to dviation from th crtainty quivalnc policy, and ) th cost du to output nois h input that minimizs this function, obtaind by diffrntiating and quating to zro, is u = ψ (ˆb, Π,σ,y ds )= Π +ˆb ˆb ˆby ds Not that Π can b sn as a rgularization trm As Π bcoms small, th optimal input gos to zro h minimum xpctd cost is ) φ =(ˆb, Π,σ,y ds )=σ + (yds +ˆb Πˆb, whr w usd th matrix invrsion lmma for a ran on updat Not that φ = is convx in Σ and concav in Π For small Π, th minimum xpctd cost approachs an uppr bound, which is th cost of slcting a zro input 5 Optimal policy for = For =, th xpctd cost is φ = = E b,, y y ds + y y ds = E b,, ( b u +(ˆb u y ds )+ ) +( b u +(ˆb u y ds )+ ) = E b, ( b u +(ˆb u y ds )+ ) +E y E b,, ( b u +(ˆb u y ds )+ ) y = u Π u +(ˆb u y ds ) + σ +E y u Π u +(ˆb u y ds ) + σ whr (from ) Π =Π +σ u u, ˆb =Π Π ˆb+σ, (4) u y W usd th towr proprty of conditional xpctation, and th fact that, if y is givn, thn ˆb and u ar constants and b has zro man and covarianc Π Also, it is trivial to s that b and ar indpndnt, and b and ar indpndnt φ = is to b minimizd ovr u = ψ and u = ψ,withψ a function of y and u (both ψ and ψ ar also functions of ˆb,Π,σ,y ds and y ds, but for clarity ths paramtrs
5 will b omittd) h minimum of φ = can b found by minimizing first ovr ψ (i, finding th minimizing scond input u as a function of th first input u and output y ) o find th minimum of (4) w will nd to comput inf E y ψ (u,y ) Π ψ (u,y ) ψ (, ) +ˆb ψ (u,y ) y ds + σ = = E y inf ψ (u,y ) Π ψ (u,y ) ψ (, ) +ˆb ψ (ψ,y ) y ds + σ = E y φ =(ˆb, Π,σ,y ds ) = σ +(y ds ) E y +ˆb Πˆb (5) W conclud that th minimum xpctd cost is φ =(ˆb, Π,σ,y ds ) = inf u u Π u +(u ˆb y ds ) + σ + E y φ =(ˆb, Π,σ,y ds ) Not that w hav just drivd Bllman s principl of optimality from first principls for this particular problm h solution rquirs computing an intgral of th form E X, X N(,σ ) a X + a X + a whr a = σ 4 u Π u, a =ˆb u σ, a =+ˆb Π ˆb, X= b u + N(,σ ), σ = u Π u +σ h dnominator polynomial can b shown to b positiv for all X If an itrativ optimization procdur is to b usd, this xpctation must b valuatd numrically at ach itration Altrnativly, w will propos using a simpl approximation Exampl Considr th prvious xampl (in 4), but with a shortr horizon In particular, =, Y ds =[ ], and n, ˆb, Π,σ as bfor For a givn u, th xpctd cost φ is computd assuming that u is slctd optimally at = h xpctation in (5) is valuatd by numrical intgration Ranging ovr valus for th two ntris of u, this producs Figur h optimum is achivd at u = 998, for which th xpctd cost is φ =568 his is to b compard with th standard procdur of minimizing th xpctd squar rror at ach tim stp (i, th rgularizd passiv larning policy), which yilds u =,andφ=9 5 Approximat solution for = Considr th approximation E X a X + a X + a a With this approximation, th problm bcoms that of minimizing u Π u +(u ˆb y ds ) +σ + (yds ) +ˆb Πˆb φ u, u, Figur : Expctd cost as a function of first and scond ntry of u (with u optimal) ovr u R n Undoing th minimization ovr u,ws that this is quivalnt to minimizing f = = u Π u + u (Π + σ u u ) u +(u ˆb y ds ) +(u ˆb y ds ) +σ ovr u,u R n his approximation is quivalnt to maing th approximation ˆb ˆb in (4), which will b th motivation for an xtnsion of th approximation for any > W will ta ˆb ˆb, =,,, in th quivalnt xprssion for th xpctd cost An intuitiv dscription of this approximation is as follows First, not that th aprioridistribution of b can b dscribd by th llipsoid (x ˆb ) Π (x ˆb ) (th maximum volum st with a givn probability) Liwis, th conditional distribution of b givn y can b dscribd by th llipsoid (x ˆb ) Π (x ˆb ) h total cost will dpnd on both th cntrs (ˆb, ˆb ) and th volums (dfind by Π, Π ) of th two llipsoids From on tim indx to th nxt, with th addd nowldg of y, th cntr and volum of th llipsoid chang (s Figur ) h cntr changs randomly, and this is th trm that introducs incrasd complxity in th dynamic program (as a sid not, this random chang has a zro man normal distribution that dpnds on th inputs, and is asily computd) On th othr hand, th volum changs in a dtrministic fashion Givn th inputs, this chang in volum can b prcisly prdictd With th approximation dscribd, w ar assuming that th chang in volum is mor important in dtrmining th cost than th chang in cntr, i, w assum that th cost is much lss snsitiv to th man of th distribution than to its covarianc his is rasonabl for systms that ar not ovrdtrmind, which includs our problm ˆb Figur : Changs in th conditional distribution of b ˆb Σ Σ
6 Exampl With th sam xampl as in 5, for a givn u w comput f = Again w assum that u is slctd optimally Ranging ovr valus for th two ntris of u, this producs Figur 4 h minimum of th approximat objctiv function f = is achivd at u = h ap- proximat xpctd cost at this point is f = =997, and th tru xpctd cost is φ = =59 h prformanc dgradation rlativ to th optimal policy is 9% with th approximation, as compard to 55% with th rgularizd passiv larning policy Figur 5 plots th approximation rror as a function of u Not th small rror in th rgion whr th optimum is locatd, which sms to b a gnral fatur of this approximation With this approximation, w can rmov th nstd conditional xpctation, and group th inf oprators, so that φ inf f, u,,u with f = = u Π u + = (ˆb u y ds ) + σ Finding this minimum is not a convx program, which gratly limits our ability to solv larg scal problms in practic 55 Convx approximation (linarization of Π ) A convx approximation of th objctiv function abov can b obtaind by linarizing th information matrix in th inputs Writing U = U + U,andfor U small, f 4 Figur 4: f φ u, u, Approximation of th xpctd cost u ßÞ Ð u, ßÞ Ð ) + σ u, Figur 5: Approximation rror 54 Optimal policy and approximation for > Following th prvious analysis for = and =,andby induction on, th problm of minimizing φ can b writtn as a dynamic program h optimum is givn by φ = ϕ, with ϕ = inf u E((y y ds ) )+E(ϕ +)) =inf u u Π +(ˆb u y ds +E(ϕ +), ßÞÐ for =,,, and ϕ + = All infimums ar ovr th spac of fasibl policis, i, ovr all u masurabl σ(y,,y ) h thr mard trms can b intrprtd as ) th cost du to inaccuracy in th stimat of b, )th cost du to th prturbation introducd to improv stimation of b, and ) th cost du to output nois o ma th dynamic program tractabl w ta th sam approach as bfor, and us th approximation ˆb ˆb, ˆb ˆb, ˆb ˆb Liwis, Π U U (U) U +(U) U + U U = (U) U +U U (U) U Π + σ (U ) U + U U (U ) U = P h trm omittd is O(σ U ) It is positiv smidfinit, hnc th approximation undrvalus information W can xpct that a solution basd on this approximation will b consrvativ in th introduction of prturbations for th purpos of idntification h problm now involvs a sum of matrix fractional and quadratic trms, all of which ar convx, minimiz = u P u + = (ˆb u y ds ) whr P is as abov, and th variabls ar u,,u R n his is a matrix-fractional and scond-ordr con program, which is quivalnt to th smidfinit program minimiz = (α + β ) subjct to α (ˆb u y ds ) (ˆb u y ds, ) β u, =,, u P P =Π +σ j= =,, u j(u j) + u ju j u j(u j), =,,, whr th variabls ar α,,α,β,,β R, and u,,u R n Algorithms for solving smidfinit programs ar of polynomial complxity h complxity of solving this particular problm with an intrior-point mthod is boundd by O 7 9 n For mor on smidfinit programming s, g, Vandnbrgh and Boyd [4] 56 Algorithm A possibl practical algorithm is as follows Find a nominal input squnc u,,u È according to a simpl policy, such as minimizing = u Π u + È = (ˆb u y ds ) his amounts to solving without accounting for th bnfits of xtra information
### Non-Homogeneous Systems, Euler s Method, and Exponential Matrix
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Th Matrix Exponntial (with xrciss) 92.222 - Linar Algbra II - Spring 2006 by D. Klain prliminary vrsion Corrctions and commnts ar wlcom! Th Matrix Exponntial For ach n n complx matrix A, dfin th xponntial
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Job shop schduling with unit procssing tims Nikhil Bansal Tracy Kimbrl Maxim Sviridnko Abstract W considr randomizd algorithms for th prmptiv job shop problm, or quivalntly, th cas in which all oprations
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# Metric system
The metric system is a decimal system of measurement that was developed in France in the late 18th century. The metric system was refined over the course of history into the International System of Units (SI), the most widely used measurement system in the world today.
## What is the metric system
The metric system is a measurement system that was developed with the goal of standardizing units of measurement in such a way that the system would be adopted universally.
### Who uses the metric system
Most of the world has fully adopted the International System of Units (SI), the modern form of the metric system. Three notable exceptions include the United States, Myanmar, and Liberia. Although these countries have not officially adopted SI, they still use the systems in many areas such as science, industry, and other contexts in which standardization is important.
### Properties of the metric system
The metric system is characterized by the following:
• Base units - The modern form of the metric system, SI, has 7 base units. Each base unit represents some physical quantity (e.g. length, time, mass, temperature) such that no quantity can be expressed in terms of another. For example, we cannot express time in terms of length; there is no way to express a meter in terms of time. Thus, a base unit represents a base quantity, and other quantities within the metric system are expressed in terms of multiples or submultiples of the corresponding base unit.
• Decimal ratios - The metric system is based on multiples of 10. For example, the meter is the base unit of length in the metric system. Other metric units of length are related to the meter by powers of 10. The centimeter is a unit of length smaller than the meter and it is related to the meter by a factor of 1/100. A kilometer is a larger unit of length than the meter and is related by a factor of 1000. In other words, 1 centimeter = 0.01 meters and 1 kilometer = 1,000 meters.
• Derived units - A derived unit is a unit that expresses any other physical quantity that is not a base unit. Derived units are expressed in terms of a product of powers of base units. For example, speed is a derived unit that is expressed in terms of length and time. Specifically, speed has the derived unit meters per second.
• Prefixes for multiples and submultiples - In the metric system, decimal-based prefixes are used to indicate multiplication or division of a base unit by the corresponding power of 10. For example, the prefix "kilo-" indicates multiplication by 1,000 while the prefix "centi-" indicates division by 100. A kilometer is a multiple of a meter while a centigram is a submultiple of a kilogram.
• Coherence - Derived units are directly related to base units without a need for intermediate conversion factors.
These properties make the metric system easy to use and widely applicable, which is why the International System of Units has been adopted by the vast majority of the world.
## Base unit
There are 7 base units in the International System of Units, the modern form of the metric system: meter, kilogram, kelvin, second, ampere, candela, and mole. The following table shows the base units, the physical quantities they measure, and the symbol used to denote each base unit.
Base unit Physical quantity Symbol
meter length m
kilogram mass kg
kelvin temperature K
second time s
ampere current A
candela luminous intensity cd
mole amount of substance mol
### SI base unit definitions
The modern definitions of the base units are shown below. Note that the definitions are complex and are shown here just for reference.
• meter - the length traveled by light in a vacuum in of a second.
• kilogram - the fixed numerical value of the Planck constant h (6.62607015 × 10-34) when expressed in units of J·s, which is equal to kg·m2·s-1.
• kelvin - the fixed numerical value of the Boltzmann constant k (1.380649 × 10-23) when expressed in units of J·K-1, which is equal to kg·m2·s-2K-1.
• second - the fixed numerical value of the caesium frequency ΔVcs, 9,192,631,770 when expressed in units of Hz, which is equal to s-1.
• ampere - the fixed numerical value of the elementary charge (1.602176634 × 10-19) when expressed in coulombs, which is equal to A·s.
• candela - the candela is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation 540 × 1012 Hz, Kcd, to be 683 when expressed in units of lm·W-1, which is equal to cd·sr·W-1 or cd·sr·kg-1·m-2·s3.
• mole - the fixed numerical value of the Avogadro Constant (6.02214076 × 1023) when expressed in units of mol-1.
## Metric prefix
A metric prefix is a prefix (e.g. milli-, centi-) that denotes a decimal multiple or submultiple of the base unit it precedes. The prefix indicates the power of 10 by which the base unit is multiplied or divided. For example, the prefix centi- indicates division by 102, or 100. Its counterpart, "hecto-", indicates multiplication by 100. Thus, a centimeter is 1/100th of a meter, while a hectometer is 100 times larger than 1 meter. When the metric system was first developed in 1795, there were 8 prefixes. More prefixes were added over the years up through 2022, when 4 more were added making 24 the total number of SI prefixes 24. These prefixes are shown in the table below.
SI Prefixes table
Prefix Symbol Factor Factor name
quetta Q 1030 nonillion
ronna R 1027 octillion
yotta Y 1024 septillion
zetta Z 1021 sextillion
exa E 1018 quintillion
tera T 1012 trillion
giga G 109 billion
mega M 106 million
kilo k 103 thousand
hecto h 102 hundred
deka da 101 ten
100 one
deci d 10-1 tenth
centi c 10-2 hundredth
milli m 10-3 thousandth
micro μ 10-6 millionth
nano n 10-9 billionth
pico p 10-12 trillionth
atto a 10-18 quintillionth
zepto z 10-21 sextillionth
yocto y 10-24 septillionth
ronto r 10-27 octillionth
quecto q 10-30 nonillionth
## Derived units and non-SI units
The table below shows some commonly used SI derived units and non-SI units that are accepted for use within SI. There are many, not all of which are included here.
Name Symbol Quantity
hertz Hz frequency
newton N force, weight
pascal Pa pressure, stress
joule J energy, work, heat
coulomb C electric charge, quantity of electricity
volt V voltage, electrical potential difference, electromotive force
ohm Ω electrical resistance
degree Celsius °C temperature
lumen lm luminous flux
## History of the metric system
The development of the metric system is widely credited to the French. In 1670, Gabriel Mouton proposed a decimal system of measurement that the French further developed over the course of over a century. In 1790, the national assembly of France called for a system that used a unit of length based on the circumference of the Earth as its basis. This unit became known as the meter, and the standard that it represented was designed to be equal to a fraction of the distance from the North Pole to the equator. The system was also a decimal-based system in which larger and smaller units were arrived at by multiplying or dividing by powers of 10. This was the earliest form of the metric system, and many different versions were developed over the course of history before arriving at the International System of Units (SI), the current global standard. Below is a timeline of some of the milestones in the development of the metric system in use today.
• 1832 - Gauss uses the astronomical second as the base unit in defining the gravitation of the earth. Along with the millimeter and milligram, this was the first system of mechanical units.
• 1860s - The centimetre-gram-second system of units (CGS) was developed and formally promoted by the British Association for Advancement of Science. This was the first coherent metric system. It was characterized by the expression of density in g/cm3, force in dynes, mechanical energy in ergs, and thermal energy in calories.
• 1893 - The 1893 International Electrical Congress defines the international ampere and ohm with definitions based on the metre, kilogram, and second.
• 1901 - Giovanni Giorgi shows that the addition of a fourth base unit, one for electrical charge, resolves the anomalies in various electromagnetic systems. Metric systems such as the metre-kilogram-second-coulomb and metre-kilogram-second-ampere systems were then developed.
• 1960 - The General Conference on Weights and Measures (CGPM) promulgates the International System of Units (SI). SI is based on the metre-kilogram-second-ampere system with the addition of numerous coherent derived units such as watts, lumens, and joules and the addition of three more base units, the kelvin, candela, and mole.
## Metric system vs imperial
The imperial system is a system that was first defined in 1824 in the British Weights and Measures Act 1824. It was further developed over time and is the system from which the US customary system of units is derived. It was predominantly used in the United Kingdom and other Commonwealth countries, but has mostly been replaced by SI. The US customary system is the system still primarily used in day-to-day life in the US, though SI is also used in many contexts.
### Main differences between the metric and imperial systems
Below are some of the key differences between the metric and imperial systems of measurement.
• The metric system is based around the meter. The imperial system (and US customary system) are not based around a specific unit, though its units are exactly defined in relation to metric units.
• Metric units can be easily converted with an understanding of SI prefixes since units are simply multiples or submultiples of a base unit and can be arrived at by multiplying or dividing by the appropriate power of 10. On the other hand, imperial units have no pattern of conversion, and it is necessary to know the appropriate conversion factors to convert between units, even if they measure the same quantity.
• The imperial and US customary systems are not coherent systems of measurement, while SI is since the derived units of the metric system can be directly related to its base units without the need for intermediate conversion factors; this is not true of either the US customary or imperial systems.
### Why doesn't the US use the metric system
The reason that the US has not fully adopted the metric system is because at the time of the development of the US customary system, the metric system was not as widespread as it is today. Since all industries in the US were set up using the US customary system, overhauling its entire infrastructure would be extremely costly and time consuming. Since the US already uses the metric system in areas such as science, the military, and most any context in which standardization is important, at this point it is unlikely that it will overhaul its current infrastructure without a major impetus.
### Metric vs imperial units
The table below provides some comparisons between units used in the metric and US customary systems of measurement.
Measurement Metric units US customary units
Length centimeter, meter, kilometer inch, foot, yard, mile
Mass/weight gram, kilogram ounce, pound, ton
Volume liter, cubic centimeter cup, pint, quart, gallon
## Metric conversions
Since most countries around the world use the metric system, it is useful to be able to convert between common metric units and US customary or imperial units.
### Metric conversions of length
The following are some conversions from meters to measurements of length in the US customary system of measurement.
1 meter = 39.3701 inches = 3.28084 feet = 1.09361 yards = 0.00062136931818182 miles
### Metric conversions of mass
The following are some conversions from kilograms to measurements of mass in the US customary system of measurement.
1 kilogram = 35.274 ounces = 2.20462 pounds = 0.001 metric tons
### Metric conversions of volume
The following are some conversions from cubic meters to other commonly used measurements of volume. Some measures are units accepted for use within SI while others are US customary units of volume.
1 cubic meter = 264.172 gallons = 1056.69 quarts = 33,814 fluid ounces = 1,000 liters (metric unit) = 10,000 milliliters (metric unit)
### Metric conversions of area
The following are some conversions from square meters to US customary units of area.
1 square meter = 1550 square inches = 10.7639 square feet = 1.19599 square yards = 0.0000003861 square miles = 0.000247105 acres
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380_checklist_1_Sp10[1]
# 380_checklist_1_Sp10[1] - & su cient conditions for f (...
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Math 380 Spring 2010 Midterm Exam Checklist Date of Exam: Friday, March 12, 2010 What to bring: Pencil, eraser, calculuator (not really needed). The exam will be closed book and closed notes. Complex Arithmetic mulitplication, addition, division, complex conjugation real and imaginary parts, modulus, and polar form of a complex number geometric view of complex addition, subtraction, multiplication, and division ±nding nth roots of a complex number De±nitions for planar sets: interior point, boundary point, path-connected set, open set, closed set, domain the complex derivative f 0 ( z ) f is analytic at z 0 f is di/erentiable in the complex sense at z 0 harmonic function harmonic conjugate of a harmonic function a zero or pole of order m of a rational function e z ; sin z , cos z , arg ( z ) ; Arg ( z ) , log ( z ) , Log ( z ) Be able to state precisely: general necessary conditions for f 0 ( z 0 ) to exist
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Unformatted text preview: & su cient conditions for f ( z ) to exist & su cient conditions for f to be constant & the Fundamental Theorem of Algebra & For a given function f = u + iv , be able to & determine where f ( z ) exists and where f is analytic & compute f ( z ) in terms of the partial derivatives of u and v & For a real valued function u on a domain be able to & determine if u is harmonic & nd a harmonic conjugate for u in the case that u is harmonic & For the basic analytic functions built from e z , sin z , cos z , Log ( z ) ; z n (where n is an integer) & evaluate at specic values of z & di/erentiate & nd the image of a simple set under the specied function & determine if the function is one-to-one on a given domain & Find all values of arg z; log z , z a for a given value of z: 1...
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## This note was uploaded on 09/16/2011 for the course MATH 380 taught by Professor Staff during the Spring '11 term at S.F. State.
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# Fridericia Formula
##### Description
A formula for calculating the QT interval on an electrocardiogram (ECG) according to heart rate. Using this formula, Fridericia postulated that QT values >0.43 seconds were considered prolonged
##### History of the Fridericia Formula
1891Augustus Desiré Waller provided a series of values for the duration of mechanical systole with different heart rates. He demonstrated that the period of mechanical contraction (systole) was shortened at faster heart rates
1920 – The Danish physician Louis Sigurd Fridericia (1881 – 1947) hypothesised a relationship between the duration of the ventricular electrocardiogram (s) and the duration of the pulse period (p). He studied 50 healthy individuals (age 3-81 years; 28 males and 32 females) at rest to determine the accuracy and the error rate in the measurement of both the pulse period (p) and the duration of the ventricular electrocardiogram (s).
He found that QT could be predicted accurately from the following formula: s = K (p0.3558) which he then simplified (p0.3558) to ³√p without any significant error; then calculated k (=8.22) for normal healthy individuals.
He concluded that in a normal, resting subject, the duration of electrical systole on an ECG is proportional to the cube root of the duration of the pulse period.
By applying this equation to his data set, Fridericia defined an average (Gauss) error of 0.015 seconds for a single duration. Variations of the observed duration of electrical systole >3 times the average error were considered to be pathological. Therefore the upper value for a ‘normal QT interval’ at a heart rate of 60 bpm was defined as 0.427 seconds (0.382 sec+0.045 s), and so QT values>0.43 seconds were deemed prolonged.
##### Controversies
Charbit B et al in a study of 108 patients found that automatic QT correction using Bazett formula had a sensitivity for detection of QT prolongation of 54% while automatic QT correction using Fridericia formula had 100% sensitivity
##### References
Original articles
Review articles
## eponymictionary
the names behind the name
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# MANOVA SPSS Statistics - 18498
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Description
Main Task: Activity #8 You will submit one Word document for this activity. You will create this Word document by cutting and pasting SPSS output into Word.
Part A. SPSS Activity In this exercise, you are playing the role of a researcher that is testing new medication designed to improve cholesterol levels. When examining cholesterol in clinical settings, we look at two numbers: low-density lipoprotein (LDL) and high-density lipoprotein (HDL). You may have heard these called “good” (HDL) and “bad” (LDL) cholesterol. For LDL, lower numbers are better (below 100 is considered optimal). For HDL, 60 or higher is optimal.
In this experiment, you will be testing three different versions of the new medication. In data file “Activity 8.sav” you will find the following variables: group (0=control, 1=Drug A, 2=Drug B, 3=Drug C), LDL, and HDL (cholesterol numbers of participants after 12 weeks).
Using a MANOVA, try to ascertain which version of the drug (A, B or C) shows the most promise. Perform the following analyses and paste the SPSS output into your Word document. 1. Exploratory Data Analysis. a. Perform exploratory data analysis on the relevant variables in the dataset. When possible, include appropriate graphs to help illustrate the dataset. b. Compose a one to two paragraph write up of the data. c. Create an APA style table that presents descriptive statistics for the sample.
2. Perform a MANOVA. Using the “Activity 8.sav” data set, perform a MANOVA. “Group” is your fixed factor and LDL and HDL are your dependent variables. Be sure to include simple contrasts to distinguish between the drugs (group variable). In the same analysis, include descriptive statistics and parameter estimates. Finally, be certain to inform SPSS that you want a post-hoc test to help you determine which drug works best. a. Is there any statistically significant difference in how the drugs perform? If so, explain the effect. Use the post hoc tests as needed. b. Write up the results using APA style and interpret them.
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MANOVA SPSS Statistics Solution
\$40.00 no category Cirara 0 time(s)
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August 2011, Volume 17, Issue 1
FEATURES Canine Conjectures: Using Data for Proportional ReasoningArla Westenskow and Patricia S. Moyer-PackenhamA family dog supplies the measurements for scatter plots and variables so that students can explore relationships among data sets—not to mention paws and tails. Lining Up Arithmetic SequencesCarol J. BellAsk your students to analyze geometric representations of arithmetic sequences for a meaningful problem-solving experience. Representations and RaftsKimberly Sipes HartwegBuilding a rod raft allows students to make mathematical connections among a model, a table, a formula, and a graph. A Geometric Path to the Concept of FunctionScott Steketee and Daniel ScherTransformations using dynamic software can provide a unique perspective on a common topic. Additional InformationExamples from the Geometer's Sketchpad (ZIP file) Second Look:Transformations
Departments Readers WriteReaders Write - August 2011 Informing PracticeInforming Practice: The Process of Abstracting in Students’ Representations Quick ReadsQuick Reads: A Handy Multiplication Mystery Solve It!Solve It! Student Thinking: Change for a Dollar Cartoon CornerCartoon Corner: That's Perfect! Palette of Problems/Menu of ProblemsPalette of Problems - August 2011 Mathematical ExplorationMath Explorations: Hanging in the Balance Window on ResourcesWindow on Resources: Books - August 2011 Math for RealMath for Real: Hurricane Force Math
Second Look - Transformations Transformations and Technology: What Path to Follow?A technology-based approach for dealing with a misconception that some students hold concerning the equivalency of geometric transformations. Students can begin to conceptualize transformations in terms of the relationship between preimage and image figures rather than focus on the motion of a figure. Teachers will learn how to use technology to implement this activity in the mathematics classroom. It's Friezing in Here: Tessellations through Art, Architecture, and Cultural ArtifactsThe integration of transformational geometry, mathematics, history, art, and cultural studies in a middle-grades mathematics classroom. From Tessellations to Polyhedra: Big PolyhedraStudents explore relationships among polygons to discover which combinations tessellate; which combinations form polyhedra. Activity sheets, and answers, included. Illuminations Activity: Frieze PatternsStudents can experiment with the seven classes of Frieze patterns with this activity. They can explore the transformations that constitute each of the seven categories.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Change of State
## Explore the energy required to convert a substance between gas, liquid, and solid
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Change of State
TEKS
6E: Describe how the macroscopic properties of a thermodynamic system such as temperature, specific heat, and pressure are related to the molecular level of matter, including kinetic or potential energy of atoms.
6G: Analyze and explain everyday examples that illustrate the laws of thermodynamics, including the law of conservation of energy and the law of entropy.
### Objective
• Students will understand how the heat of fusion and heat of vaporization energy is used to create a phase change and how a phase change can be used to do work.
### Equations
Q=mcΔt\begin{align*}Q = mc \Delta t\end{align*}
Q=mHf\begin{align*}Q = mH_f\end{align*}
Q=mHv\begin{align*}Q = mH_v\end{align*}
Steam traction engine depended on water changing from a liquid to gaseous state [Figure1]
Before the internal combustion engine was invented, steam engines were the power source for ships, locomotives, tractors, lumber saws, and most industrial machines. Coal or wood was burned to boil water into steam, which ran the engine. As the water is boiled it creates steam under pressure. The steam is allowed to expand to do positive work on the pistons. The work done on the piston to turn the wheels is at the expense of the steams internal energy. The steam loses its internal energy by being able to expand under lower pressure of the outside atmosphere. The largest problem with steam engines was their size! A large oven had to support fire which was used to heat the water to a boil. The work done by a steam engine does not come free.
### Change of State
Most substances may exist in any of the three common states of matter. In the gaseous state, the molecular motion has completely overcome any attraction between the particles and the particles are totally separate from each other. There are large spaces between the particles and they move large distances between collisions. In the liquid state, the molecular motion and the molecular attractions are more balanced. While the particles stay more or less in contact with each other, they are still free to move and can slide past one another easily. In the solid state, the attractive forces dominate. The particles are pulled together into a tightly packed pattern which does not allow the particles to pass each other. The molecular motion in this form is essentially reduced to vibration in place. Increasing the temperature of a substance means increasing the molecular motion (kinetic energy) of the molecules in the substance. The phase in which a substance exists is the result of a competition between attractive forces and molecular motion.
Credit: Christopher Auyeung
Source: CK-12 Foundation
For most substances, when the temperature of the solid is raised high enough, the substance changes to a liquid, and when the temperature of the liquid is raised high enough, the substance changes to a gas. We typically visualize a solid as tiny particles in constant motion held together by attractive forces. As we add heat to the solid, the motion, or the kinetic energy, of the particles increases. At some temperature, the motion of the particles becomes great enough to overcome the attractive forces. The thermal energy that was added to the solid up to this point was absorbed by the solid as kinetic energy, increasing the speed of the molecules. The lowest temperature at which the particles are able to exist in the liquid form is called the freezing point as it turns from a liquid to a solid. The temperature at which a solid begins to turn into a liquid is called the melting point
In order for the molecules to actually separate from each other, more energy must be added. This energy, called latent heat of fusion or heat of melting, is absorbed by the particles as potential energy as the solid changes to a liquid. Recognize that, once the temperature of a solid has been raised to the melting point, it is still necessary for the solid to absorb additional thermal energy in the form of potential energy as the molecules separate.
The boiling point of a liquid is the temperature at which the particles have sufficient molecular motion to exist in the form of a gas and will rapidly break away from the liquid state. This process can happen slowly in a process called evaporation. The temperature that gasses condense into liquids is called the condensation point. Once again, however, in order for the particles to separate to the gaseous form, they must absorb a sufficient amount of potential energy. The amount of potential energy necessary for a phase change to gaseous form is called the latent heat of vaporization. Consider the heating curve shown below.
Credit: Christopher Auyeung
Source: CK-12 Foundation
Heating curve of water [Figure3]
The heating curve shown is for water but other substances have similarly shaped heating curves. Suppose you begin with solid water (ice) at -30°C and add heat at a constant rate. The heat you add in the beginning will be absorbed as kinetic energy and the temperature of the solid will increase. When you reach a temperature of 0°C (the melting point for water), the heat you add is no longer absorbed as kinetic energy. Instead, the added heat is absorbed as potential energy and the particles separate from each other. During the flat part of the curve labeled “melting”, heat is being added constantly but the temperature does not increase. At the left edge of this flat line, the water is solid; by the time enough heat has been added to get to the right edge, the water is liquid, but maintains the same temperature. Once all the water is in the liquid form, the added heat will once again be absorbed as kinetic energy and the temperature will increase again. During the time labeled “water being heated as a liquid”, all the added heat is absorbed as kinetic energy.
When a temperature of 100°C (the boiling point of water) is reached, the added heat is once again absorbed as potential energy and the molecules separate from liquid form into gaseous form. When all the substance has been converted into gas, the temperature will again begin to rise.
Here is a simulation where you can interactively change the state of matter of molecules.
Here is a follow-up activity to complete while using this simulation.
Substance Heat of Fusion, Hf (J/kg)−−−−−−−−\begin{align*}\underline{H_f \ (J/kg)}\end{align*} Heat of Vaporization, Hv (J/kg)−−−−−−−−\begin{align*}\underline{H_v \ (J/kg)}\end{align*} Copper 2.05×105\begin{align*}2.05 \times 10^5\end{align*} 5.07×106\begin{align*}5.07 \times 10^6\end{align*} Gold 6.30×104\begin{align*}6.30 \times 10^4\end{align*} 1.64×106\begin{align*}1.64 \times 10^6\end{align*} Iron 2.66×105\begin{align*}2.66 \times 10^5\end{align*} 6.29×106\begin{align*}6.29 \times 10^6\end{align*} Methanol 1.09×105\begin{align*}1.09 \times 10^5\end{align*} 8.78×105\begin{align*}8.78 \times 10^5\end{align*} Water 3.34×105\begin{align*}3.34 \times 10^5\end{align*} 2.26×106\begin{align*}2.26 \times 10^6\end{align*}
When the temperature of a substance is changing, we can use the specific heat to determine the amount of heat that is being gained or lost. When a substance is changing phase, we can use the heat of fusion or heat of vaporization to determine the amount of heat being gained or lost. When a substance freezes from liquid to solid, the amount of heat given off is exactly the same as the amount of heat absorbed when the substance melts from solid to liquid. The equations for heat gained or lost are given here:
The heat gained or lost during a temperature change: Q=mcΔt\begin{align*}Q = mc \Delta t\end{align*}.
The heat gained or lost during a phase change of solid to liquid: Q=mHf\begin{align*}Q = mH_f\end{align*}.
The heat gained or lost during a phase change of liquid to gas: Q=mHv\begin{align*}Q = mH_v\end{align*}.
#### Change in Volume Can Do Work:
The First Law of Thermodynamics describes how heat added or taken away from a system can change the internal energy of the system and if it is a gas substance then the increase in heat added will also change the ability of the system to do work due to the volume change of a gas due to changing temperatures. According to the ideal gas law, gasses will have a direct change in pressure or volume based on any changes in the gases temperature. As long as the number of gas molecules do not change in the gas sample, the pressure will increase if volume is kept constant and temperature is increased. On the other hand if pressure is held constant, the volume of a gas will increase as temperature increases.
[Figure4]
The change in volume during a phase change from a solid to a liquid is not very large so its uses for work are very limited. However when a liquid goes through a phase change in to the vapor or gas state, the gas state takes up considerable more volume than the liquid state. This change in volume can be used to do work like in the steam engine. Once the water in the steam engine begins to boil, any additional heat added to the steam by the oven will increase the internal energy of the steam by increasing the average kinetic energy of the molecules of steam. This in turn increase the pressure in the steam being contained in the plumbing of the steam engine. This stored steam now has stored potential energy that can be used to do some work.
The diagram below shows the equations used to quantify the values of heat, internal energy and work for the system. Simply put, since pressure is force per unit area and that force is moved the distance of a piston stroke, the steam does work on the piston as it expands to push the piston. The energy for the work comes from the loss of internal energy of the steam by letting it expand to decrease its internal energy. Therefore the initial phase change from liquid to vapor was required to super heat the steam and get some work out of all the energy put into the system from the oven fire.
[Figure5]
Example Problem: 5000. Joules of heat is added to ice at 273 K. All the heat goes into changing solid ice into liquid water. How much ice is melted?
Solution: m=QHf=5000 J3.34×105 J/kg=0.0150 kg\begin{align*}m=\frac{Q}{H_f}=\frac{5000 \ J}{3.34 \times 10^5 \ J/kg}=0.0150 \ kg\end{align*}
Example Problem: Beginning with 1.00 kg of ice at -20.0°C, heat is added until the substance becomes water vapor at 130.0°C. How much heat was added? The specific heat of ice is 2108 J/kgC\begin{align*}2108 \ J/kg^\circ C\end{align*}, the specific heat of liquid water is 4187 J/kgC\begin{align*}4187 \ J/kg^\circ C\end{align*}, and the specific heat of water vapor is 1996 J/kgC\begin{align*}1996 \ J/kg^\circ C\end{align*}.
Solution: 5 steps.
1. Calculate the heat required to raise the sample from -20.0°C to 0°C.
2. Calculate the heat required to melt the sample.
3. Calculate the heat required to raise the sample from 0°C to 100°C.
4. Calculate the heat required to vaporize the sample.
5. Calculate the heat required to raise the sample from 100°C to 130°C.
The solution is the sum of these steps.
1.QHS=mciceΔt=(1.00 kg)(2108 J/kg.C)(20.0C)=42160 J\begin{align*}1. Q_{HS} = mc_{\text{ice}} \Delta t = (1.00 \ kg)(2108 \ J/kg.^\circ C)(20.0^\circ C) = 42160 \ J\end{align*}
2.QMelt=mHf=(1.00 kg)(334000 J/kg)=334000 J\begin{align*}2. Q_{\text{Melt}} = mH_f = (1.00 \ kg)(334000 \ J/kg) = 334000 \ J\end{align*}
3.QHL=mcwaterΔt=(1.00 kg)(4187 J/kg.C)(100.0C)=418700 J\begin{align*}3. Q_{HL} = mc_{\text{water}} \Delta t = (1.00 \ kg)(4187 \ J/kg.^\circ C)(100.0^\circ C) = 418700 \ J\end{align*}
4.QVap=mHv=(1.00 kg)(2260000 J/kg)=2260000 J\begin{align*}4. Q_{\text{Vap}} = mH_v = (1.00 \ kg)(2260000 \ J/kg) = 2260000 \ J\end{align*}
5.QHV=mcvaporΔt=(1.00 kg)(1996 J/kg.C)(30.0C)=59880 J\begin{align*}5. Q_{\text{HV}} = mc_{\text{vapor}} \Delta t = (1.00 \ kg)(1996 \ J/kg.^\circ C)(30.0^\circ C) = 59880 \ J\end{align*}
Total Heat=3.11×106 J\begin{align*}\text{Total Heat} = 3.11 \times 10^6 \ J\end{align*}
### Summary
• Most substances may exist in any of the three common states of matter, solid, liquid, or gas.
• The phase in which a substance exists is the result of a competition between attractive forces and molecular motion.
• The potential energy absorbed by a solid as it changes to a liquid is called the heat of fusion or the heat of melting.
• The amount of potential energy necessary for a phase change to gaseous form is called the heat of vaporization.
• The heat gained or lost during a temperature change is given by, Q=mcΔt\begin{align*}Q = mc \Delta t\end{align*}.
• The heat gained or lost during a phase change of solid to liquid is given by, Q=mHf\begin{align*}Q = mH_f\end{align*}.
• The heat gained or lost during a phase change of liquid to gas is given by, Q=mHv\begin{align*}Q = mH_v\end{align*}.
### Vocabulary
• boiling point: The temperature at which a liquid rapidly begin to change phase into a vapor or gas depending on the substance.
• first law of thermodynamics: Describes how work and heat are related to a system's internal energy. Using energy conservation, the change in internal energy of a system and a systems ability to do work are related to any heat added or taken away from the system.
• latent heat of fusion: The heat input required for a phase change from solid to liquid or the heat released for a phase change from liquid to solid
• heat of vaporization: The heat input required for a phase change from liquid to vapor or gas or the heat released
• ideal gas law: Describes the relationships between pressure, volume, temperature and numbers of gas molecules in a system as one or more of these conditions change. This law is ideal and is limited to discrete temperature and pressure changes near STP.
• kinetic energy: The energy of motion of a particle. In a gas substance the moving gas particles have kinetic energy.
• melting point: The temperature at which the phase change from liquid to solid occurs for a particular substance.
• phase change: A change in the state of a substance from solid to liquid, or from a liquid to a gas. Phase change also goes in the reverse direction from gas to liquid and from liquid to solid.
• potential energy: Energy stored in a system due to a change in its position or a change in the systems internal energy.
### Review
Questions
1. A 200. g sample of water at 60.0°C is heated to water vapor at 140.0°C. How much heat was absorbed?
2. A 175 g lump of molten lead at its melting point (327°C) is placed into 55.0 g of water at 20.0°C. The specific heat of lead is 130 J/kgC\begin{align*}130 \ J/kg \cdot ^\circ C\end{align*} and the Hf\begin{align*}H_f\end{align*} of lead is 20,400 J/kg.
1. When the lead has become solid but is still at the melting point, what is the temperature of the water?
2. When the lead and the water have reached equilibrium, what is the temperature of the mixture?
After reviewing the questions above, here's a quiz on changing the states of matter.
### Practice Problems
Questions
The following video explains heat of fusion and vaporization. Use this video to answer the questions that follow.
1. For water, which takes more energy, melting or evaporating?
2. When are there two phases present at the same time in the pot?
### Hands on Lab Activity
the following link will send you to a pdf file involving the latent heats of fusion and vaporization of ice to water to steam and the calculations based on the data obtained in the lab.
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# College Chemistry
posted by .
Calculate the volume of carbon dioxide produced from .025 g of aluminum carbonate and excess HCl at 25 degrees c and 745 toor according to the following reaction:
Al(2)(CO(3)(3)) (aq) + HCl (aq) ----- AlCl(3) (aq) + H(2)O + CO(2) (g)
im using the numbers in () to show a number that is really little beside the numbers.
• College Chemistry -
PV=nRT
P=745 Torr/760 Torr = 0.98atm
V=?
n= 0.025g of Al/ 26.98g/mol of Al
R=0.0821
T=298K
V= (0.0009266*0.0821*298)/0.98
V=0.0231L
## Similar Questions
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Calculate the volume of Carbon dioxide gas generated when 40.0 g of sodium carbonate reacts with hydrocholric acid according to the following reaction. The gas is to be collected at STP Na2CO3 + HCL-----> NaCL +H2O + CO2(g)
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OK I just need help with two final problems that I am stuck on: The problem states the calcium carbonate in limestone reacts with HCl to produce a calcium chloride solution and carbon dioxide gas CaCO3(s)+2HCl(aq)-->CaCl2(aq)+H2O(l)+CO2(g) …
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You weigh out 1.70 g of sodium carbonate monohydrate, dissolve it in water and react it with 15.00 mL of 2.00 M HCl. Write an equation for the reaction and calculate the expected mass of carbon dioxide that will be produced.
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# Why Watt Density Matters When Choosing Heating Elements
Watt density refers to the measure of power from a heating element divided by the heated surface area of the element itself. Of course, that’s just the surface (excuse the pun) of this concept. It has major implications when choosing a heater and its components (i.e., flanged elements). This post will explain how watt density works, how to calculate it, and how it factors when purchasing certain heating elements.
## Clarifying watt density
Industrial operators use watt density to help determine the most effective heating elements (and heaters) for their application. As mentioned at the article’s outset, watt density is the heating element’s power divided by its surface area that’s actively heated. The calculation equation looks like this:
Watt Density = Power / Surface Area = Heat Flux
Watt density is typically expressed in watts per square inch, although in some countries, the unit is watts per square millimetre.
## Importance of calculating watt density for industrial applications
Watt density essentially predicts a heating element’s efficiency compared to other options. Calculating watt density helps you identify whether a heating element will produce heat adequately and its potential lifespan. Essentially, it helps answer the question of whether it will be a suitable choice or not.
Of course, greater heat efficiency and a longer lifespan are ideal for most (if not all) industrial applications. But the right watt density for your operation depends on the process and the heat required to sustain an ideal temperature at a cost-efficient rate.
Understanding the watt density of a heating element can help you determine whether a specific heating method is superior (or not). For example, industrial staff may determine whether conduction, irradiation or convection is better for a particular application. The equation can also help determine whether a specific element is beneficial, such as an open coil, tubular or finned-tubular flanged element.
## The balance between low vs high-density flanged elements
Although ideal watt densities vary considerably depending on the application, extremes can also be problematic. When the watt density is too low, the heater’s price will be high. Ultimately, a heater of this nature is not cost-effective since its price may be too high for its limited heating capability.
Contrarily, a watt density that’s too high can lead to equipment failure. In such instances, what happens is that the heat overwhelms the element, leading to damage or complete burnout. That’s why it is common to choose the one with a lower watt density for safety when deciding between one of two heating elements.
Remember that watt density can vary significantly based on the size of the heating element and the material used in its sheath. Take, for example, a 4” low-density steel sheath and a 4” high-density Incoloy sheath.
Both have the same diameter and can both weigh 150 lbs, yet the Incoloy sheath has a watt density of 65.8 w/in2 compared to the steel sheath, which has a density of just 10 w/in2. Size and material should be deciding factors when choosing a heating element for your industrial application.
## Wattco recommendations for watt densities and flanged elements
Watt density is a crucial factor that helps you choose the most efficient flanged elements (and, by extension, flanged heaters). That ultimately enables you to save time, cut costs, and optimize your heating process while safeguarding your equipment.
Here at Wattco, we offer a range of options for flanged heaters and elements so that you can find a suitable unit for your application. Our representatives can also provide insights on the ideal watt density for your application based on its requirements. Get a quote or request more information about a flanged heating element today.
Could a Wattco staff provide a sample calculation of watt density.
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# A report consisting of 2,600 words is divided into 23 paragr
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A report consisting of 2,600 words is divided into 23 paragraphs. A 2-paragraph preface is then added to the report. Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120 ?
(1) Each paragraph of the preface has more than 100 words.
(2) Each paragraph of the preface has fewer than 150 words.
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A report consisting of 2,600 words is divided into 23 paragraphs. A 2-paragraph preface is then added to the report. Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120 ?
The question asks whether the average number of words per paragraph for all 25 paragraphs is less than 120, or whether the total number of words for all 25 paragraphs is less than 25*120=3,000. Thus, the question basically asks whether the total number of words for 2-paragraph preface is less than 3,000-2,600=400.
(1) Each paragraph of the preface has more than 100 words. This implies that the total number of words for 2-paragraph preface is more than 200. Not sufficient.
(2) Each paragraph of the preface has fewer than 150 words. This implies that the total number of words for 2-paragraph preface is less than 300. Sufficient.
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the report already has 2600 words and 23 paragraphs .. if 2 paragraphs are added that would take the total no. of paragraphs to 25.
for the average no .of of words per paragraph to be less than 120, the total no. of words need to be less than 3000.
statement I says , the paragraphs have more than 100 words .. but do not extly how much .. could be 500 each in which case the toal number of words in the report may go well beyond 3000.
statement II says, the new paragraphs have less than 150 words.. so the two new paragraphs together have less than 300 words.. so the total no .of words in the paragraph will be less than 3000.
so II alone is sufficient to answer the question.
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If the average is 120 words per paragraph, then for 25 paragraphs, there will be a total of 25*120=3000 words. So any total less than 3000 words will give an average less than 120.
(1) Each paragraph of the preface has more than 100 words.
The total number of words may or may not exceed 3000. Insufficient
(2) Each paragraph of the preface has fewer than 150 words.
The total number of words cannot exceed 2750 so average is less than 120. SUFFICIENT
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Question. I tried 2 ways. First way worked, second way didn't. Could someone have a look and let me know where I went wrong with the 2nd method?
FIRST WAY (CORRECT):
Background info:
Y/N question
23 paragraphs, 2,600 words
Added 2 preface paragraphs. Let x be total of new words added.
Thus question is: (2,600+x)/25<120? OR, simply, is x<400?
Stem (1): Each preface paragraph has 100+ words.
If each preface paragraph has 101 words then x= 2*(101) = 202 words. This would make x<400
Alternatively if each preface paragraph has 250 words then x=2*(202) = 404 words. This would mean x>400. Stem 1 is Insufficient.
Stem (2) Each preface paragraph has less than 150 words. This means x is AT MOST 2*(149) = 298 words. I.e. x MUST BE <400. Stem 1 is Sufficient.
WAY TWO (INCORRECT)
I worked out the average word count for the current 23 paragraphs: 2,600/23= approx 113 words
I then thought, if the new average (inclusive of 2 preface paragraphs) is 120 then the difference b/w the 2 averages is 7 words.
The total number of words added due to preface paragraphs would then = 7*25 = 175 words
The answer stem has to tell me if the 2 new paragraphs has less or more than 175 words.
Stem (1): Each preface paragraph has 100+ words. Total must be more than 175 words. Sufficient
Stem (2): Each preface paragraph has less than 150 words. So could be 100 words, 200 words, 298 words. NOT sufficient
Thanks guys!
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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03 Apr 2015, 16:38
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Hi ColdSushi,
There's nothing wrong with the second approach that you took (although the "rounding" that did could be a problem in certain questions that require specifics/accuracy).
The "7 word difference" you calculated IS correct, but you also have to account for the 120 words that would need to appear in EACH of the 2 preface paragraphs. Your calculation didn't do that.
Using your approximation, your calculation should have been (7)(25) + 2(120) = 415 words
Using this value, you would have gotten the correct answer.
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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09 Apr 2015, 18:56
EMPOWERgmatRichC wrote:
Hi ColdSushi,
There's nothing wrong with the second approach that you took (although the "rounding" that did could be a problem in certain questions that require specifics/accuracy).
The "7 word difference" you calculated IS correct, but you also have to account for the 120 words that would need to appear in EACH of the 2 preface paragraphs. Your calculation didn't do that.
Using your approximation, your calculation should have been (7)(25) + 2(120) = 415 words
Using this value, you would have gotten the correct answer.
GMAT assassins aren't born, they're made,
Rich
Thanks very much Rich. Appreciate your help and it makes sense now!
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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20 May 2016, 08:45
ColdSushi wrote:
EMPOWERgmatRichC wrote:
Hi ColdSushi,
There's nothing wrong with the second approach that you took (although the "rounding" that did could be a problem in certain questions that require specifics/accuracy).
The "7 word difference" you calculated IS correct, but you also have to account for the 120 words that would need to appear in EACH of the 2 preface paragraphs. Your calculation didn't do that.
Using your approximation, your calculation should have been (7)(25) + 2(120) = 415 words
Using this value, you would have gotten the correct answer.
GMAT assassins aren't born, they're made,
Rich
Thanks very much Rich. Appreciate your help and it makes sense now!
Hey,
I too did the same way as you did in the second method .
however I still didnt get the point - "account for the 120 words that would need to appear in EACH of the 2 preface paragraphs."
what I thought was :
consider avg to be 120 , so difference in avg is 7 , that means 7 words per para is extra . to find out what extra was divided equally -->> 7*25= 175 extra words was divided equally among 25 to increase the avg to 120.
this 125 extra words came from 2 paras combined so each para approx 87.5 words
so basically q is asking does each para less than 87.5 words?
OMG! while typing this and reviewing it, I found out what was wrong in my analysis, I will still continue to write since it will help those who thought this same thing as mentioned before..
125 words are EXTRA words...again " EXTRA" words.. that means those 2 paras to fit into the avg of 113 it should first satisfy its requirement i.e. 120 words/para
so 2 paras have 120*2 words and then extra 25*7 to increase the avg to 120
therefore those 2 paras have -->> 2*120+25*7 = 415 words ... that means 207.5 each para
tht means Q is asking " if each para < 207.5 then avg is <120 "
I hope this is correct.
thanks
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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20 May 2016, 09:31
A report consisting of 2,600 words is divided into 23 paragraphs. A 2-paragraph preface is then added to the report. Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120 ?
(1) Each paragraph of the preface has more than 100 words.
(2) Each paragraph of the preface has fewer than 150 words.
Information Given:-
1) 2 paragraphs are added to existing 23 paragraphs, making total paragraphs to be 25
2) Number of words in existing 23 paragraphs= 2600
Average number of words <120
Inference:-
Total number of words <120*25
T<3000
Initially words were 2600, so if newly added words are less then 400 then we can answer the question.
(1) Each paragraph of the preface has more than 100 words.
If the words added are 101 with each paragraph then 'yes'. But if 300 words are added with each paragraph then 'No'
(2) Each paragraph of the preface has fewer than 150 words.
Basically, this statement tells that fewer than 300 words are added. And this was the information we were looking for
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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26 Jul 2016, 11:08
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A report consisting of 2,600 words is divided into 23 paragraphs. A 2-paragraph preface is then added to the report. Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120 ?
(1) Each paragraph of the preface has more than 100 words.
(2) Each paragraph of the preface has fewer than 150 words.
Target question: Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120?
When it comes to averages, we know that average value = (sum of n values)/n
We can rewrite this into a useful formula: sum of n values = (average value)(n)
So, if 25 paragraphs were to have an average EQUAL TO 120 words, then the TOTAL number of words = (120)(25) = 3000
We want to determine whether or not the average is LESS THAN 120. This will occur when the total number of words is LESS THAN 3000
So, we could rephrase the target question as follows...
REPHRASED target question: Is the total number of words less than 3000?
IMPORTANT: When I scan ahead, I see that the two statements tell us about the number of words in the PREFACE. So, let's rephrase the target question so that it pertains to the preface only.
Since the report (without the preface) contains 2600 words, the preface must have FEWER THAN 400 words in order to keep the entire word count under 3000 words. So, we can RE-REPHRASE the target question as follows:
RE-REPHRASED target question: Does the preface contain fewer than 400 words?
Now that we've rewritten the target question to suit the statements, the rest of the question is a breeze.
Statement 1: Each paragraph of the preface has more than 100 words
In other words, the preface contains more than 200 words. There are several possibilities. Here are two:
Case a: the preface contains 220 words, in which case the preface contains fewer than 400 words
Case b: the preface contains 450 words, in which case the preface DOES NOT contain fewer than 400 words
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: Each paragraph of the preface has fewer than 150 words
In other words, the preface contains fewer than 300 words.
If the preface contains FEWER than 300 words, then we can be certain that the preface contains fewer than 400 words
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
[Reveal] Spoiler:
B
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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27 Jul 2016, 04:31
A report consisting of 2,600 words is divided into 23 paragraphs. A 2-paragraph preface is then added to the report. Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120 ?
(1) Each paragraph of the preface has more than 100 words.
(2) Each paragraph of the preface has fewer than 150 words.
(1) Each paragraph of the preface has more than 100 words.
each para can hypothetically have 20k words as no upper limit is given
insufficient
(2) Each paragraph of the preface has fewer than 150 words
let each para has 149 words
both combined = 398
total words 2600+398 = 2898
max avg = 2898/25 = ~117 words
sufficient.
B
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink]
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02 Aug 2016, 00:41
A report consisting of 2,600 words is divided into 23 paragraphs. A 2-paragraph preface is then added to the report. Is the average (arithmetic mean) number of words per paragraph for all 25 paragraphs less than 120 ?
(1) Each paragraph of the preface has more than 100 words.
(2) Each paragraph of the preface has fewer than 150 words.
THIS IS A YES /NO TYPE DS
Original scenario
23 paragraphs 2600 words
average words per paragraph = 2600/23 = 113
Revised scenario
AIM :- IS average word per paragraph <120
(1) Each paragraph of the preface has more than 100 words.
The new preface paragraphs may have 30 words or 300 words.
YES <120 or NO >120
INSUFFICIENT
(2) Each paragraph of the preface has fewer than 150 words.
Now either we will get a either only a YES or NO
SUFFICIENT
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Re: A report consisting of 2,600 words is divided into 23 paragr [#permalink] 09 Sep 2017, 11:30
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# Beginner's Information: May 2007 Archives
## Real Estate Numbers: Calculators and Spreadsheets Are Your Friend But They Have Limits
On a fairly regular basis I get email asking what I think of this or that loan calculator on the web, this or that predictive model for real estate prices or loan rates, etcetera.
For spreadsheets, what you can get is usually an analysis of one variable per spreadsheet. I've programmed a loan comparison spreadsheet, but it only compares two alternatives at a time and it's not really suitable for use with the public, because you have to understand the limitations and GIGO factor. Just like I've got spreadsheets that answer the "rent or buy" question, among others, but you have to understand the limitations on the results imposed by your model.
As a computer programmer, I make a pretty decent loan officer. In order to compare financial information via spreadsheets, you have to understand what points of comparison the calculations are meant to compare. If your data is out of whack, if your assumptions are away from reality, or if you're trying to apply the comparison outside its design limits, what you get is useless.
I have several spreadsheets I have programmed and use. All of them have limits that need to be understood in order to get useful information out of them.
The second limitation upon this sheet is that it's assuming smooth increases. This is not what happens, as anyone over the age of ten ought to know. Over longer periods of time, the data may tend towards an aggregate average, but that says nothing about any given year. In reality, some years are plus thirty percent while other years are minus twenty. Even if my assumptions for averages are good, the spreadsheet that predicts the next thirty years is useful mainly to predict overall level of the market many years out. The numbers for any particular year are so much garbage, as far as the real world goes, where a 5% differential between estimate and actual is often enough to render something worse than useless. Even if my assumptions for average return are right on the money (and if I didn't think they were pretty close, I'd use others), any particular year could be at the top of a peak or the bottom of a market trough. If you know what state the market will be in in a particular year three decades out, why the heck aren't you richer than the ten richest billionaires in the world combined? Knowing what the market was going to do these past few years is a lot easier than knowing what it'll be like thirty years from now! I have what I think are good predictions based upon good models, but I don't have any god-level knowledge of where any part of the economy will be thirty years from now, and neither does anyone else. We see the future dimly, reflected through the present and the past.
Speaking of which, let's drag one of the standard disclaimers out and air the dirty laundry. "Past performance is not indicative of future results." Averages of past results may be the only way we have of predicting the future, but those results depend upon unknowable factors. Somebody could invent something tomorrow that utterly changes the face of housing thirty years out. You think the urban planners of the 1920s foresaw urban sprawl? I know for a fact that they didn't. What no model of the future can predict is unforeseen factors. I can't tell you what they will be or what effects they will have, but I can promise you there will be some. In 1894, Michaelson (who first measured the speed of light) said, "Our future discoveries must be looked for in the sixth decimal place." This just a few years after the formulation of Maxwell's equations, and within a year Rutherford had changed the atomic model forever, while the basis of quantum mechanics was being laid, and less than ten years later were Einstein and relativity. Michaelson was right in a technical sense that precise measurements were the key to unlocking future discoveries, but wrong in the sense he meant it, that all the major discoveries had already been made. My predictive model is more detailed than most, and I do my best to include all of the factors I see, but I have no way of including factors that I can't see, and one thing I can promise you is that there are some. It may work out that I guess right anyway, but that doesn't mean there weren't any unforeseen factors, just that I got lucky despite them. The further out the model goes, the more it is dependent upon subsequent events no one can predict. Someone could announce man-portable fusion power tomorrow, or "Star Trek" transporters, or any of dozens of new potential technologies that could alter the world, and that's just the technological possibilities. Politics and demographics will utterly change in the next thirty years (In 1977, more people were predicting the world conquest of communism than the collapse of the communist system. Mr. Carter's presidency was not the United States' shining hour).
Just because we know that the precise numbers are wrong, however, doesn't mean that those numbers have no value in predicting the future. The way the numbers will move relative to each other is much more important information. Population is increasing and will continue to increase. Demand in major urban areas and desirable areas will continue to rise faster than supply, and since such areas are where most of us live or want to live, the price of real estate will quite likely continue to increase faster than inflation. Particularly types of housing which are universally desired, such as detached single family residences sitting on a certain amount of land owned basically fee simple. PUDs and townhomes are less desirable for most folks, true condominiums less desirable yet, and below that are apartments. Offer most people the chance to move up on the ladder of desirability, and they'll take it. Since the only thing preventing most people from doing so is price, price is what's going to make it ever harder to make that transition to more desirable housing. Living space in a desirable location is a scarce good. Living space, desirable location or not, is a limited good. The only way to change this is to somehow manufacture more space or arrange to have fewer people to share it. I'm not aware of any plans to manufacture enough space to make a difference to the billions of people on earth, so I'm guessing that barring worldwide nuclear or biological warfare, population density is going to increase, demand for housing is going to increase, and supply is going to stay pretty much right where it is. Nonetheless, this is only a guess. My guess is that housing will be about four to five times as expensive as it is today thirty years out, If it's only twice, we'll all still live in million dollar houses. If it's eight times, we'll be in four million dollar houses. The wider the net, the more probability I have of being right - and the less useful the information is. Unless the price right now is something like two cents, nobody sane is going to invest money for that long without a better idea of what the payoff will be.
Whether I'm right or not is something nobody knows right now, or even how close. Actually, not being quite that much of an egotist, the question in my mind is more akin to "how far off will I be?" But the data is still useful, because it tells me that as long as my assumptions are anything like real, we're all looking at living in million dollar real estate - the only question is exactly when. It tells me what people will be need to be able to pay every month, at least in a general sense, and it tells me that more and more people are going to get priced out of real estate, or down into less desirable housing, and that real estate is therefore going to be a quite satisfactory vehicle for creating personal wealth.
On the other hand, no system of projecting the future is better than the limitations imposed upon it by limited foresight. If the population of the United States drops to 1789 levels all of a sudden - or 1607 levels - all bets are off. Of course if that happens, most of us won't be here to worry about it, and the ones that are will have bigger problems than the price of real estate. It's pointless to waste time worrying about the price of real estate in such possible circumstances, where the price of real estate will be the least of our worries.
Caveat Emptor
## There's No Such Thing as Full Service Agency for a Discount Price
The local dog target gave discounters several hundred thousand dollars of free puffery recently.
I'm not against discounters. I'm perfectly willing to do a discounters work for a discounter's price. Fifty percent for the pay for less than ten percent of the work and almost none of the liability is a real win as far as I'm concerned. The difference is that I'm not willing to pretend that you're getting the same value from me. In fact, the amount of value the buyer receives from their alleged "agent" is pretty much negligible.
Let's illustrate with a real example from last week. Some full service clients of mine had gotten interested in a property. They wanted a fixer property with potential and a view, and they asked me to check this one out. Yes, it had a view, but the view was of a high school stadium, making peaceful enjoyment of the property rather hit and miss, subject to the local sports schedule. It had some potential, true, but every surface in every room needed to be redone. It is going to take \$100,000 to get that potential, and the property would only be worth maybe \$40,000 more than the owners are asking. Leave out those pesky numbers and a less capable agent can make it seem like a great bargain. If all you're thinking of is that \$5000 rebate check, which can be fraud for reasons similar to these, you may think you got a deal.
If they had been clients of a discounter, they would have been in escrow on the first property right now. Too bad about that \$100,000 they'd have to spend to get \$40,000 benefit. On the other hand, I found the same people a property not far away that needed about \$40,000 worth of work to be worth \$120,000 more than the asking price. What does a discounter do? Write the offer on the first property. Now you've got a property you need to put \$100,000 into to make it usable, that's worth only \$40,000 more than you paid. Money the discounter would have rebated: roughly \$5000. If they didn't have a full service agent to compare with, it even looks like a great deal, because none of the value I provided these folks shows up on the HUD 1 form, or anywhere else as numbers on paper. The value is still there, as my clients know.
If you know enough about the state of the market, what problems look like and what opportunities look like, you may spend less with a discounter, or get a rebate, that doesn't cost you several times that difference. If you know everything a good agent does, there is no reason not to put that money in your pocket. But if you know everything a good agent does, why is the discounter making anything? Why aren't you doing your own transaction? Why aren't you in the business yourself and getting paid for your expertise?
A full service agent goes a long way past filling in the blanks on Winforms and faxing the offer. When I go out looking at 20 to 30 properties per week, I'm not just finding individual bargains. I'm also learning about the general state of the market, what things to look for in a given neighborhood, what common problems are with a given model of house. I know what's sold in the neighborhood recently, and I know what it looks like because I've been inside it, and I know how it compares to other stuff that's out there now. I have a pretty fair idea of what it's going to take to beautify properties, and I know what they'll be worth when the work is done, because I know what other stuff that already looks like that has sold for recently.
Real estate is a career. It may not absolutely require a college education, but many agents have one, and know many things you can't learn in college - because the professors don't know, either, unless they're active real estate agents. A good agent spends a lot of time and effort not only learning their local market, but keeping their knowledge base updated. This thing changes constantly, and it doesn't even change uniformly. How did La Jolla get to be La Jolla? I assure you it wasn't some random seagull anointing the neighborhood as having higher property values. Rancho Santa Fe doesn't even come close to the ocean, and it's the highest mean property value zip code in the nation. How did your neighborhood get to where it is? Is it likely to change, and how? What are the known and probable upcoming changes in the neighborhood? How is it likely to effect your prospective property? Wouldn't you like to know about that redevelopment zone - or the railroad tracks they intend to drive through the area?
Full service can be a very hard sale when all that you consider is the numbers on the HUD 1. There just isn't any space for "Agent kept you from making a \$60,000 mistake," let alone, "Agent showed you an \$80,000 opportunity." But people who know property know that there is a lot more to every transaction than the numbers on the HUD 1. If you're dubious, may I suggest this experiment when you're ready to buy: Find a couple full service agents willing to work with a non-exclusive buyer's agency agreements, and sign them. Then compare what happens as compared to the service of the discounter you use for properties you find yourself. There is no need to sign even a non-exclusive agreement with a discounter, by the way, as the sales contract will note the agency relationship for that transaction. Like I said in How to Effectively Shop for a Buyer's Agent, let the ineffective alternative select itself out.
I'm perfectly willing - happy, even - to do discount work for "discounter" pay. I only make half the money, but I can service a lot more than twice the clients for a much smaller level of risk and still be home in time for dinner. I'm even a better negotiator than dedicated discounters, because unlike them, I know what's really going on in the areas I serve. However, saying "full service at a discount price," does not make it so, and I refuse to pretend that it is. Furthermore, the people who approach me for discount work usually end up understanding that a real professional is worth a lot more than the extra money I make, and are happy to pay it. Most people have no problems understanding that the reason a good car commands a higher price than a bad car, let alone a skateboard, is because a good car provides more value. People will pay \$100 per seat for decent - not great - musicians in concert when you couldn't pay them enough to attend a garage band practice session. Why should this principle holds any less true for expert help in what is likely to be the biggest transaction of your life?
Caveat Emptor
## FSBO Horror and Failure to Disclose Property Defects
|
I keep getting search result hits for the string "fsbo horror." It's an amalgamation because I haven't done any postings on this specific subject.
Both buyers and sellers have problems relating to For Sale By Owner issues.
For sellers, the largest issue seems to be properly disclosing all relevant items to satisfy the liability issue. There are resources available, but the question is whether the you took proper advantage of them and made all the legally required disclosures on any issue with the property there may be. If you have an agent that fails to do this, you can sue them. If you are doing it yourself, the only one responsible is you. You are claiming to be capable of doing just as good a job as the professional, and if you didn't do it right, the buyer is going to come after you.
Now I'm going to leave the marketing and pricing questions out of the equation, because with a For Sale By Owner most folks should understand that in return for not paying a professional to help you, you've got to do it yourself. What many For Sale By Owner folks seem to fail to understand, however, is that if you haven't met legal requirements, the real nightmare may be just beginning when the property sells.
Let's say it was something fairly innocuous, like seeping water from a slow leak you didn't know about. A couple years pass, and now there's mold or settling. Perhaps the foundation cracks as a result of settling. Bills are thousands to hundreds of thousands of dollars. Your buyer goes back and finds that your water usage went up by fifteen percent in the six months before the sale. He sues, saying that even though you didn't know, you should have known based upon this evidence. Court cases are decided based upon evidence like this every day. A good lawyer paints you as maliciously selling the property as a result of this. Liability: Steep, to say the least.
Now, let's look at it from a buyer's prospective. You have a choice of two identical properties. In one, a seller is acting for themselves, in the other, they have an agent. The price may be a little cheaper on the for sale by owner one, or it may not. One of the reasons people do for sale by owner is they are greedy. But when I'm looking at a for sale by owner, the question that crosses my mind is "Are they rationally greedy, or are they just greedy?" Are they going to disclose everything wrong or that may be an issue with the property? At least here in California, the agent has pretty strong motivation to disclose if something is wrong that they know about. If they don't, they can lose their license, and even if they don't, they have potentially unlimited personal liability. If they did disclose, they're probably off the hook, and even if they aren't, their insurance will pay for the lawyers, the courts, and any liability. If there's one thing all long term agents get religion about, no matter their denomination, it's asking all of the disclosure questions.
This is not the case for many owners selling their own property. Some are every bit as good and conscientious as any agent. A good proportion, however, are intentionally concealing something about the property. What's going to happen when it comes to light? If there's an agent, there's a license number, a brokerage who was responsible for them, and insurance. The latter two are deep pockets targets for your suit, and you can find them. Once that owner gets the check, you can find them unless they're dead, but they may not have any money. Even if they do have money, it may be locked up and inaccessible via Homestead or any number of other potential reasons.
One of the reasons that I, as a buyer's agent, am always leery of a for sale by owner property is that I have to figure that first off, there's a larger than normal chance that this property has something wrong that's not properly disclosed. When that happens, my client is going to be unhappy. When my client is unhappy, they are going to sue. The first target is the seller, but if they're gone or broke, who does my erstwhile client come after? Me. So I have to figure that not only is there a larger chance of there being something wrong, I have to figure there is a larger chance of me being held responsible for something I took every step I legally could to avoid. For Sale By Owner properties usually have to be priced significantly under the market in order to persuade me that not only am I doing the right thing by my clients in trying to sell them this property, where my clients have to pay my buyer's agent fee out of their pockets rather than out of the selling agent's commission, but also that the heightened risk of future problems is worth more than the price differential to my clients. Unless the answer is a strong solid "yes" that I can document in court if I have to, I'm going to pass it by in favor of the agent-listed property next door or down the street.
Caveat Emptor (and Vendor).
## Prorated Property Taxes in California
July first marks one of the turning points in the year for California real estate, as property taxes are collected for a period running from July 1 through June 30. They are paid in two installments, the first due due November 1 (past due December 10) that covers the first six months, from July 1 to December 31. The second is due February 1st (past due April 10th), and covers the second six months, from January 1 to June 30. Other states have different set ups. For instance, Nevada property taxes are paid quarterly.
Now it happens that most folks don't actually pay their taxes until just before the "past due" date. If you have an impound account, the bank doesn't send the money until sometime around December 8th and about April 5th. But they are due and payable on the dates above, and whether you are refinancing or selling or buying, if they are due they need to paid either before the transaction is consumated, or through escrow. Prorated taxes aren't part of refinance transactions. If they're due, they have to be paid, and the current owners need to pay all of them. But for sales, what happens is the property taxes are paid past the date of the sale, or not paid up until the date of the transaction.
Let's pick a date the transaction closes. Say June 15th. The taxes were paid back in April through June 30 by the seller. But the seller didn't owe taxes past June 15th; they don't own the property any more after that. The buyer owes the other fifteen days worth. So the way it is handled is that the buyer comes up with, in addition to the purchase price, fifteen days of property taxes and pays those to the seller as part of the transaction.
If the effective date of the sale was, on the other hand, July 31st, and the seller has paid only through June 30th, then the seller will owe the buyer for taxes for the month of July, because the buyer will be paying those come November. So thirty-one days worth of taxes are taken off of the sales price by escrow and given to the buyer because they will be paying for those thirty-one days worth of taxes in November.
Prorated sales taxes are part of most sales transactions. The only exceptions are those taking place within the periods from November 1st to December 10th, and February 1st to April 10th, where taxes are paid through escrow, and not even those if the current owner already paid the taxes. Be advised that the last couple of days can be tough, especially if you have to walk them in, so if the transaction hasn't recorded at least three or four days before the end of the grace period, you want to go ahead and pay the taxes. If the transaction doesn't close, the government doesn't care why they weren't paid before the end of the grace period; you'll have to pay a penalty for being late.
Caveat Emptor.
The first thing you need to understand in reading any property advertisements is that agents write them to get people to call. They are trolling for clients. Except in the case of someone who doesn't accept dual agency advertising a listing they actually have, they are written purely with the idea of dangling something out there that clients want. Since most agents like dual agency just fine because it means they get paid twice for the same transaction, understand that only a tiny percentage of the ads that are written out there are written for any purpose other that to get potential clients to call.
Keeping this fact firmly in mind, there are two sorts of places where people go to search for property: Some that are based on MLS, and others that are not.
If it's in MLS or coming from MLS, it better be good information. I can (and do) file violations on liars in MLS. So do others - every time somebody wastes our time by saying the property has something that it doesn't. Filing violations in MLS is simple, it's effective, and after a certain low number of violations, the offender's input access gets restricted. They can't put properties in MLS and they might as well be out of business. Note that they can still "puff" a property significantly; but number of bedrooms, square footage, anything with a number or a yes/no associated with it had better be right. The big violation I'm finding most of recently is advertising it in MLS as "fee simple" when it should be either "PUD" or "Condominium". If it's got homeowner's association dues, it's not fee simple.
Everywhere else, everything that is not based on MLS, take all advertisements with a respectful amount of caution - Like at least equal in weight to the building. Once you get outside the domain of MLS, there are few sanctions possible for even the most outrageous puffery - or even advertising a property you do not, in fact, have. Or anything remotely similar. Some agents won't put anything out there that isn't as close to gospel truth as they can make it, but others are not nearly so fussy, and you really want to avoid the latter sort.
Non-MLS based property advertisements may now be pending, it may be sold, or it may in fact never have existed. The agent put that ad in trolling for buyers. What they want is your signature on an exclusive buyer broker agreement, so they can lock your business up. We know you shouldn't sign exclusive buyer's agency agreements, because that's a poor way to get a good buyer's agent, but most people don't know that and most agents are laying in wait for the ignorant.
Quite often, I have clients who should know better, as I've explained this to them - often more than once - ask me about this fantastic possibility they see somewhere else. About as surprising as gravity, they turn out to be in some way non-factual. 2 bedrooms when it's really one. 1 bedroom when it's really a studio or loft. And sometimes, they actually had it, at that price, six years ago. And then I call the agent listed on the ad, look it up on MLS, and voila! the deception becomes apparent.
There's nothing wrong with responding to such ads, and there's nothing wrong with working with the agents who advertise them, so long as you limit yourself to a nonexclusive agreement, so you can get rid of them when it become apparent they're not guarding your interests. But even with a non-exclusive agency agreement, I'd be asking myself "If they lied to get get me to call, what else are they going to lie to me about?"
I just had a client send me three prospects from one of the non-MLS based search services. All three of them were non-existent, posted by a lead generating service as bait so they could sell everyone who respondended as a lead to agents. They didn't have it, they never did have it, it wasn't even for sale, and hadn't been for over 10 years. But that's how they get leads, which they then sell to several agents.
So be aware before you respond to property advertisements that quite often, advertisements don't exist. To avoid leads services, look for specific names of the agent in the advertisement. To avoid problem agents who require an exclusive agency agreement before showing, simply refuse to sign exclusive agreements. Dual Agency is a very bad idea for buyers. If they have the listing, they're going to get paid when the property sells regardless of whether you signed that agreement. If they don't have the listing, they still risk nothing with a non-exclusive buyer's agency agreement.
Caveat Emptor
Dan Melson Amazon Author Page
The Man From Empire
A Guardian From Earth
Empire and Earth
Working The Trenches
Preparing The Ground
Building the People
The Invention of Motherhood
The Fountains of Aescalon
The Book on Mortgages Everyone Should Have!
What Consumers Need To Know About Mortgages
The Book on Buying Real Estate Everyone Should Have
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Beginner's Information: June 2007 is the next archive.
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## a couple questions on a Morgan Stanley portolio analysis
Have a question about your personal investments? No matter how simple or complex, you can ask it here.
Topic Author
Charon
Posts: 137
Joined: Thu May 03, 2018 12:08 pm
### a couple questions on a Morgan Stanley portolio analysis
My fiancée got a free portfolio analysis from someone at Morgan Stanley whom she met through business networking. She has no interest in finance details, and has (against my advice) been paying someone to manage her money for her, so I'm the one trying to make sense of the portfolio analysis. Most of it seems very standard and makes sense, but I had a couple minor questions.
1) The Morgan Stanley guy says he's a fiduciary, but his recommended investments are through Morgan Stanley. So I take it fiduciaries are required to give the best possible advice within whatever options their company offers?
2) He pointed out the advantage in fees of Morgan Stanley (1.72%) over where she has her money now (2.08%). He calculated the dollar savings under a hypothetical scenario (X dollars, Y return, Z years). He started with the dollar difference the fees led to in the first year, Δ. When I reproduce his numbers, it appears the formula he used was each year dollar_differencei = dollar_differencei-1 * (1+return_rate) + Δ.
This makes no sense to me. First, it's assuming the invested difference isn't subject to the fee, and second it's assuming Δ is constant in nominal dollars. It arises from a difference in fees that are percentages, so I assume it scales with the assets. There's no way Morgan Stanley is committing to a constant fee in nominal dollars for a 55-year time period! But this mistake makes the comparison look worse for Morgan Stanley, showing ~25% less difference over the period than there really is. So I'm questioning myself, because why would they do a calculation that put them in a worse light than reality? So are they really wrong, or am I?
For reference, my calculation was assetsi = assetsi-1 * (1+return_rate-fee_rate). I might well be wrong - I'm a good data analyst generally, but my financial knowledge is self-taught and minimal.
3) Just to throw some redder meat in here for those who read this far, my advice to my fiancée will be to shift everything to a Vanguard Target Date fund appropriate for her risk tolerance. Keep in mind that this is someone who has no interest in ever modifying asset allocations, so a set-and-forget automatic glide path seems better to me than a three- or four-fund portfolio. This is also someone who looked at a 1.72% fee and said "that seems plenty low", so I'm not worried about 0.14% vs. 0.06%. Any disagreement?
livesoft
Posts: 69764
Joined: Thu Mar 01, 2007 8:00 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
Ha! What a train wreck!
This signature message sponsored by sscritic: Learn to fish.
Taylor Larimore
Posts: 29021
Joined: Tue Feb 27, 2007 8:09 pm
Location: Miami FL
### Re: a couple questions on a Morgan Stanley portolio analysis
He pointed out the advantage in fees of Morgan Stanley (1.72%) over where she has her money now (2.08%)
Charon:
BOTH these fees are unconscionable. If you must have an advisor, consider Vanguard Personal Advisor Service (PAS) Service which charges only 0.3%. PAS certified experts are paid by salary and are without a conflict of interest.
To give you an idea of the impact of costs, consider this:
If stocks gain an average of 6% annually during the next 30 years, someone who invested \$25,000 for a 1% yearly fee will forego more than \$35,000 in gains because of the fee -- more than the original investment.
Consider a simple portfolio like this: The Three-Fund Portfolio and you will probably not need any advisor.
Best wishes.
Taylor
"Simplicity is the master key to financial success." -- Jack Bogle
investorag83
Posts: 18
Joined: Mon Feb 25, 2019 3:23 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
I'm curious to understand the real value in the PAS... USAA offers similar guidance for free even if you're not a full fledged member (I was one of the advisors at USAA for years). I can't imagine you speak to the same advisor when you call in so there really isn't that 1:1 relationship (unless I'm mistaken). Portfolio rebalancing would happen automatically or you could just do the 3 fund portfolio if you wanted to be as hands off as possible. Where am I failing to see the benefit of the PAS?
And I completely agree that those fees are most definitely on the high side. Better than before doesn't mean good. But I don't see much value in the 30 bps. Either get the fee close to 1% with a low cost implementation or handle it yourself. I'd also be curious as to the expense ratios of the new portfolio...I've seen hidden fees all over those accounts.
Mr.BB
Posts: 1117
Joined: Sun May 08, 2016 10:10 am
### Re: a couple questions on a Morgan Stanley portolio analysis
I make this really easy for you. Go online and find an expense ratio calculator. show her with her current savings and future contributions how much it will cost her at 1.72% over the next 30 years. Then do the same thing with the target index fund expense ratio that you want to use. That should open her eyes.
"We are what we repeatedly do. Excellence, then, is not an act, but a habit."
cas
Posts: 880
Joined: Wed Apr 26, 2017 8:41 am
### Re: a couple questions on a Morgan Stanley portolio analysis
Charon wrote:
Thu Mar 14, 2019 11:44 am
1) The Morgan Stanley guy says he's a fiduciary, but his recommended investments are through Morgan Stanley. So I take it fiduciaries are required to give the best possible advice within whatever options their company offers?
Is there a section of the analysis document (probably in tiny, unreadable print) somewhere way in the back (or other non-obvious place) where assorted disclaimers, including disclosure of conflicts of interest (probably buried deep in other really boring disclosures) are listed?
I'll guess there is a sentence in there somewhere saying something along the lines that, even though it is a conflict of interest and may not be in the best interest of the client, Morgan Stanley products will be given priority when products are selected for the portfolio.
At least, that is how the corporate trustee (trust department of a massive bank) for a (irrevocable) trust that
benefits one of my elderly relatives manages to meet the letter of the "fiduciary" law and still put their trust clients into their own firm's products (even when the product has obvious problems, e.g. is brand new with new managers with no track record or is older, but has been massively hemorraging assets for years). (Trustees have been legally required to act as fiduciaries for a very, very long time.)
The bank slipped up a few years back, and the relevant sentence disappeared from the disclosure for part of their clientele (discretionary clients) where they were required to act as fiduciaries. The SEC eventually (10 years after I noticed the behavior existing to egregious excess with the trust) investigated and fined them over \$300 million for breach of fiduciary duty. (A light slap on the wrist given the size of the bank.) (
https://www.sec.gov/news/pressrelease/2015-283.html )
The relevant sentence had NOT disappeared from the quarterly statements for trusts (again, page 14 or so of the statement, size 2 font, buried in the midst of other stuff), so the "breach of fiduciary duty" finding by the SEC did not apply to trust clients. Apparently is is perfectly ok (legally) for a fiduciary to act to their own benefit in a conflict of interest as long as it is disclosed ahead of time. (Having said that, I have noticed that the most egregious levels of such behavior vanished within the trust management after the SEC fine.)
(I Am Not A Lawyer, so lawyers may quibble about the exact phrasing I used in this post.)
ExitStageLeft
Posts: 1828
Joined: Sat Jan 20, 2018 4:02 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
If all the FA did was charge 1.72% including fund fees that would be bad enough. But the resulting choice of actively managed funds would most likely underperform the market, effectively adding more drag. The Dinkytown fee comparator is a good one to highlight the impact over 30 years: https://www.dinkytown.net/java/compare- ... -fees.html
Posts: 728
Joined: Tue Mar 25, 2014 8:14 pm
Location: North Alabama
### Re: a couple questions on a Morgan Stanley portolio analysis
Charon wrote:
Thu Mar 14, 2019 11:44 am
my advice to my fiancée will be to shift everything to a Vanguard Target Date fund appropriate for her risk tolerance.
Set it - and forget it!
Oh I can't, can I? That's what they said to Thomas Edison, mighty inventor, Thomas Lindberg, mighty flyer,and Thomas Shefsky, mighty like a rose.
BL
Posts: 9259
Joined: Sun Mar 01, 2009 2:28 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
I agree that a single balanced fund, especially any at Vanguard, a 3-fund portfolio at V, Fidelity, or Schwab, or, if she wants a trusted person to talk to, then Vanguard PAS may be well worth it (especially compared to the "fiduciary" at MS or wherever.) The PAS would put her into low-ER Admiral INDEX funds, AFAIK. She/he could be trusted as a fiduciary as they don't have a conflict-of-interest in what they recommend, unlike MS, etc.
montanagirl
Posts: 1242
Joined: Thu Nov 19, 2009 4:55 pm
Location: Montana
### Re: a couple questions on a Morgan Stanley portolio analysis
Got a good look at a client's MS brokerage statement last year and saw a LOT of churning. Ridiculous number of transactions but the client said he hadn't withdrawn anything, hadn't needed money for an emergency or anything like that.
And he trusted his gal there to do the right thing for him.
123
Posts: 5521
Joined: Fri Oct 12, 2012 3:55 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
The Vanguard Target Retirement funds are excellent options for those that don't want to fuss with things. They're cheap too. Since target retirement funds are becoming the default funds for many automatic enrollments in 401K plans they're a pretty safe recommendation.
The closest helping hand is at the end of your own arm.
unclescrooge
Posts: 4318
Joined: Thu Jun 07, 2012 7:00 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
montanagirl wrote:
Thu Mar 14, 2019 4:41 pm
Got a good look at a client's MS brokerage statement last year and saw a LOT of churning. Ridiculous number of transactions but the client said he hadn't withdrawn anything, hadn't needed money for an emergency or anything like that.
And he trusted his gal there to do the right thing for him.
I looked at a statement for a girl I dated once after she said her money hadn't grown over a few years. A couple of funds where proprietary non-publicly traded funds. I had to go to the SEC registration page to find information on them.
One of them was for the top 100 large cap international stocks. Basically it contained stocks like BP, Unilever, Nestle, Roche, Shell, etc.
It charged 2% upfront load, 1.5% annual fee and it self-liquidated every 3 years.
We sent in an account transfer form. They "lost" it twice. It took 3 months to get the money out.
But that still better than the IRA CD she had with Wells Fargo. Both she and my first cousin were unsuccessful at getting those funds out of Wells. Eventually, they both gave up.
Topic Author
Charon
Posts: 137
Joined: Thu May 03, 2018 12:08 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
I appreciate the interest in this post, even from those who didn't really read it
I think there was one answer to question 1, no answers to question 2, and a lot of advice against advisors and actively managed funds. To be clear, I fully agree. I'm 100% in index funds myself. That wasn't really my question, but I appreciate the charge of the Bogleheads! And to be fair, I threw question 3 in there.
My main question really was 2, if I'm crazy or if the Morgan Stanley guy actually calculated cost of fees wrong. I wrote a little Python code to calculate this and make a chart for her, to show her how unimpressive the fee difference was between the two and how impressive the fee difference was moving to Vanguard or equivalent... I think she's sold.
Topic Author
Charon
Posts: 137
Joined: Thu May 03, 2018 12:08 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
cas wrote:
Thu Mar 14, 2019 1:57 pm
Is there a section of the analysis document (probably in tiny, unreadable print) somewhere way in the back (or other non-obvious place) where assorted disclaimers, including disclosure of conflicts of interest (probably buried deep in other really boring disclosures) are listed?
I'll guess there is a sentence in there somewhere saying something along the lines that, even though it is a conflict of interest and may not be in the best interest of the client, Morgan Stanley products will be given priority when products are selected for the portfolio.
That sounds plausible to me. There were enough pages of numbers to keep me busy for a while, so I didn't delve into the legalese. I was going to go along with the fiancée and allow my portfolio to be reviewed too... until I found that stupid math error*. Not terribly interested in having a review anymore, free or not. It's not like I was going to change anything.
But the MS guy was a salesman. Definitely had her convinced on a few things. Fortunately I think she'll take my advice over his.
*Question 2 - I'm pretty sure it's a stupid math error.
Posts: 728
Joined: Tue Mar 25, 2014 8:14 pm
Location: North Alabama
### Re: a couple questions on a Morgan Stanley portolio analysis
Charon wrote:
Thu Mar 14, 2019 9:58 pm
I appreciate the interest in this post, even from those who didn't really read it
I think there was one answer to question 1, no answers to question 2, and a lot of advice against advisors and actively managed funds. To be clear, I fully agree. I'm 100% in index funds myself. That wasn't really my question, but I appreciate the charge of the Bogleheads! And to be fair, I threw question 3 in there.
My main question really was 2, if I'm crazy or if the Morgan Stanley guy actually calculated cost of fees wrong. I wrote a little Python code to calculate this and make a chart for her, to show her how unimpressive the fee difference was between the two and how impressive the fee difference was moving to Vanguard or equivalent... I think she's sold.
My guess is that the adviser is wrong. But If you already wrote a script to show the fees, what was the question again?
Oh I can't, can I? That's what they said to Thomas Edison, mighty inventor, Thomas Lindberg, mighty flyer,and Thomas Shefsky, mighty like a rose.
Topic Author
Charon
Posts: 137
Joined: Thu May 03, 2018 12:08 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
ExitStageLeft wrote:
Thu Mar 14, 2019 3:35 pm
The Dinkytown fee comparator is a good one to highlight the impact over 30 years: https://www.dinkytown.net/java/compare- ... -fees.html
Thanks for sharing. That confirms my approach - my program agrees exactly with this one, but I'm able to go out to the 55-year span (!) that the Morgan Stanley guy did.
So I guess this is an answer to question 2. MS guy did basic math wrong, or - and this is even scarier - their standard portfolio analysis software is incorrect.
Topic Author
Charon
Posts: 137
Joined: Thu May 03, 2018 12:08 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
Thu Mar 14, 2019 10:07 pm
But If you already wrote a script to show the fees, what was the question again?
I couldn't believe that MS portfolio analysis software was wrong, and thought I must be making a mistake. But apparently not. Wow.
K72
Posts: 38
Joined: Wed Dec 05, 2018 8:04 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
I hate to be un-diplomatic but I don't know how else to say this. I just left MS after 13 years of an actively managed IRA with excessive trading, high fees, and miserable results. Your mileage could vary.
Last edited by K72 on Fri Mar 15, 2019 11:29 am, edited 1 time in total.
smarcus3
Posts: 260
Joined: Sun Nov 18, 2018 9:57 pm
Location: USA
### Re: a couple questions on a Morgan Stanley portolio analysis
It's difficult to convince people to do something when they're opposed to doing it even if you can show them it's bad. My fiancee's parent pay a financial planner instead of managing the funds themselves. This has been going on for 30 years. Just think of the lost compounding due to this however trying to do anything about it is like walking around land minds. I've heard of similar difficulties other friends have in similar situations. Wish you the best as it's a tricky road. You need to show some analysis but they need to come to you for a change.
This is my personal opinion. I'm an engineer not a financial advisor.
whodidntante
Posts: 7253
Joined: Thu Jan 21, 2016 11:11 pm
Location: outside the echo chamber
### Re: a couple questions on a Morgan Stanley portolio analysis
Charon wrote:
Thu Mar 14, 2019 9:58 pm
I appreciate the interest in this post, even from those who didn't really read it
Wait, we are supposed to read posts?
Including the words Morgan Stanley in a subject line is going to elicit a reflex response around here. The boglehead's head will rotate 360 degrees, shout "costs matter" and then begin typing.
investorag83
Posts: 18
Joined: Mon Feb 25, 2019 3:23 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
No...the calculation you posted with the delta does not make sense. But I'd have to see the MS person's real numbers to confirm that's the calculation they were using. I highly doubt that a program MS uses to do the calculation is incorrect though. Only chance would be if the advisor was doing their own math on the side.
All that said, 1.78 all in is still too high.
dbr
Posts: 31502
Joined: Sun Mar 04, 2007 9:50 am
### Re: a couple questions on a Morgan Stanley portolio analysis
investorag83 wrote:
Fri Mar 15, 2019 9:22 am
No...the calculation you posted with the delta does not make sense. But I'd have to see the MS person's real numbers to confirm that's the calculation they were using. I highly doubt that a program MS uses to do the calculation is incorrect though. Only chance would be if the advisor was doing their own math on the side.
All that said, 1.78 all in is still too high.
Agreed. I don't know that his calculation is right or wrong because I don't know what the calculation actually is. Yours is right.
But the real point is that sitting down with MS to discuss whether the fee is 1.78% or 2.08% is just plain stupid (head rotating at 100 rpm). Those fees are only the top of an iceberg of money drains in dealing with MS.
A better starting point is to assume that the ideal is to manage things oneself at almost zero cost and then pay people for help when it is absolutely necessary. In that respect VPAS can be a legitimate starting point.
trustquestioner
Posts: 120
Joined: Sun Sep 03, 2017 6:50 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
1.78% AUM meets the fiduciary standard?????
investorag83
Posts: 18
Joined: Mon Feb 25, 2019 3:23 pm
### Re: a couple questions on a Morgan Stanley portolio analysis
trustquestioner wrote:
Fri Mar 15, 2019 9:42 am
1.78% AUM meets the fiduciary standard?????
Yes. Just because an alternative is lower in fee doesn't mean this wouldn't pass the fiduciary standard.
dbr
Posts: 31502
Joined: Sun Mar 04, 2007 9:50 am
### Re: a couple questions on a Morgan Stanley portolio analysis
Fiduciary may offer some protection against certain investment abuses or offer a quicker path to redress if there is an issue, but it is very weak protection and is more useful as a sales tool for the salesman than as a filter for who to do business with.
More here: https://www.forbes.com/sites/davidmarot ... 8c22d32600
including:
Consumers don't understand the difference between different types of so-called financial advisors. As Mark Schoeff Jr. writes for InvestmentNews:
Advisers must meet a fiduciary standard that requires them to act in their clients' best interests. Brokers meet a suitability standard that requires them to sell products that meet a client's objectives and risk appetite but also allows them to recommend investments that give the broker the biggest fee or commission.
"No doubt there's a great deal of confusion in the marketplace as to what standard of conduct applies to a particular relationship," Mr. Clayton said.
pkcrafter
Posts: 13783
Joined: Sun Mar 04, 2007 12:19 pm
Location: CA
Contact:
### Re: a couple questions on a Morgan Stanley portolio analysis
Charon wrote:
Thu Mar 14, 2019 11:44 am
My fiancée got a free portfolio analysis from someone at Morgan Stanley whom she met through business networking
Yes, that's their M-O.
She has no interest in finance details, and has (against my advice) been paying someone to manage her money for her, so I'm the one trying to make sense of the portfolio analysis. Most of it seems very standard and makes sense, but I had a couple minor questions.
Does friend have taxable and/or tax advantaged accounts?
1) The Morgan Stanley guy says he's a fiduciary, but his recommended investments are through Morgan Stanley. So I take it fiduciaries are required to give the best possible advice within whatever options their company offers?
2) He pointed out the advantage in fees of Morgan Stanley (1.72%) over where she has her money now (2.08%).
Good grief, where does friend have her money now? This is funny-- a high-cost advisor making a comment on the fee advantage of 1.72% vs 2.08%. Ask him what the fee advantage is between 0.05% and 1.72%. Is friend currently in class C funds? Is current advisor adding a management fee on top of the fund fees?
This makes no sense to me.
Me either!
3) Just to throw some redder meat in here for those who read this far, my advice to my fiancée will be to shift everything to a Vanguard Target Date fund appropriate for her risk tolerance. Keep in mind that this is someone who has no interest in ever modifying asset allocations, so a set-and-forget automatic glide path seems better to me than a three- or four-fund portfolio. This is also someone who looked at a 1.72% fee and said "that seems plenty low", so I'm not worried about 0.14% vs. 0.06%. Any disagreement?
My opinion is people MUST take some responsibility in understanding some fundamentals of good investing. I guess friend has no idea about the impact of costs or her reaction to losing a large sum of money in a big market crash. Using a target fund only based on age can put a person into a portfolio that is more aggressive than they are comfortable with.
Paul
When times are good, investors tend to forget about risk and focus on opportunity. When times are bad, investors tend to forget about opportunity and focus on risk.
Mr.BB
Posts: 1117
Joined: Sun May 08, 2016 10:10 am
### Re: a couple questions on a Morgan Stanley portolio analysis
whodidntante wrote:
Fri Mar 15, 2019 7:35 am
Charon wrote:
Thu Mar 14, 2019 9:58 pm
I appreciate the interest in this post, even from those who didn't really read it
Wait, we are supposed to read posts?
Including the words Morgan Stanley in a subject line is going to elicit a reflex response around here. The boglehead's head will rotate 360 degrees, shout "costs matter" and then begin typing.
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# Matrix Representation for Linear Transformation
• Feb 13th 2013, 04:27 PM
TimsBobby2
Matrix Representation for Linear Transformation
Suppose: T: R3 ----> R3 is a linear transformation whose kernel has dimension 2. Describe the form of a matrix representation for T.
We did numerous examples of linear transformations from R3 ----> R3 Doesn't this mean that the matrix has to be 3x3 since they all were in my class? If the kernel has dimension 2, the image has dimension 1. So isn't the only possibility of a matrix representation for T a 3x3?
Any help would be great.
• Feb 13th 2013, 05:05 PM
chiro
Re: Matrix Representation for Linear Transformation
Hey TimsBobby2.
For this to be 2-dimensional it means that you have one vector being a linear combination of the other two vectors.
So if you fix two vectors in the matrix, then make the other one a linear combination of the others.
• Feb 14th 2013, 09:41 AM
HallsofIvy
Re: Matrix Representation for Linear Transformation
Quote:
Originally Posted by TimsBobby2
Suppose: T: R3 ----> R3 is a linear transformation whose kernel has dimension 2. Describe the form of a matrix representation for T.
We did numerous examples of linear transformations from R3 ----> R3 Doesn't this mean that the matrix has to be 3x3 since they all were in my class? If the kernel has dimension 2, the image has dimension 1. So isn't the only possibility of a matrix representation for T a 3x3?
Any help would be great.
Yes, the matrix representation of a linear transformation from \$\displaystyle R^3\$ to \$\displaystyle R^3\$ must be a 3 by 3 matrix. But that isn't the question! The matrix representation also depends upon what basis vectors you use. If the kernel is 2 dimensional, then you can choose two independent vectors from the basis and a third vector, independent of the first two, (in fact, perpendicular to the first two) as basis vectors. What would the matrix look like in that case?
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# What is Multilayer Perceptron example?
## What is Multilayer Perceptron example?
A fully connected multi-layer neural network is called a Multilayer Perceptron (MLP). If it has more than 1 hidden layer, it is called a deep ANN. An MLP is a typical example of a feedforward artificial neural network.
### What is multi layer Perceptron used for?
The multilayer perceptron (MLP) is used for a variety of tasks, such as stock analysis, image identification, spam detection, and election voting predictions.
#### What is perceptron and Multilayer Perceptron?
Perceptron is a neural network with only one neuron, and can only understand linear relationships between the input and output data provided. However, with Multilayer Perceptron, horizons are expanded and now this neural network can have many layers of neurons, and ready to learn more complex patterns.
Is Multilayer Perceptron linear classifier?
As discussed, the perceptron is a linear classifier — an algorithm that classifies input by separating two categories with a straight line. Input is typically a feature vector x multiplied by weights w and added to a bias b : y = w * x + b .
How do you calculate multilayer perceptron?
Each layer is represented as y = f(WxT + b). Where f is the activation function (covered below), W is the set of parameter, or weights, in the layer, x is the input vector, which can also be the output of the previous layer, and b is the bias vector….
1. Forward pass.
2. Loss Calculate.
3. Backward Pass.
## What is a multi layer perceptron MLP )? Describe it in 2/3 sentences?
Multi layer perceptron (MLP) is a supplement of feed forward neural network. It consists of three types of layers—the input layer, output layer and hidden layer, as shown in Fig. 3. An arbitrary number of hidden layers that are placed in between the input and output layer are the true computational engine of the MLP.
### What is the difference between MLP and CNN?
MLP stands for Multi Layer Perceptron. CNN stands for Convolutional Neural Network. So MLP is good for simple image classification , CNN is good for complicated image classification and RNN is good for sequence processing and these neural networks should be ideally used for the type of problem they are designed for.
#### What is MLP and why is it used?
MLPs are suitable for classification prediction problems where inputs are assigned a class or label. They are also suitable for regression prediction problems where a real-valued quantity is predicted given a set of inputs.
What is multilayer perceptron classifier?
A multilayer perceptron (MLP) is a class of feedforward artificial neural network (ANN). MLP utilizes a supervised learning technique called backpropagation for training. Its multiple layers and non-linear activation distinguish MLP from a linear perceptron. It can distinguish data that is not linearly separable.
How does multilayer perceptron handle non-linear classification?
These perceptrons sum together the input linear models, and each output a non linear model. The more layers we have in our hidden layer, the more complex non linear models we can find. These models are combined at the output layer to give a final model which should be capable of classifying out input data point.
## How does Multilayer Perceptron calculate weight?
weight = weight + learning_rate * (expected – predicted) * x In the Multilayer perceptron, there can more than one linear layer (combinations of neurons).
### What are the different parts of the Multilayer Perceptron model?
Multi layer perceptron (MLP) is a supplement of feed forward neural network. It consists of three types of layers—the input layer, output layer and hidden layer, as shown in Fig.
#### What is a multilayer perceptron?
A multilayer perceptron is a neural network connecting multiple layers in a directed graph, which means that the signal path through the nodes only goes one way. Each node, apart from the input nodes, has a nonlinear activation function.
What is a multilayer perceptron (MLP)?
A Beginner’s Guide to Multilayer Perceptrons (MLP) A Brief History of Perceptrons. Multilayer Perceptrons (MLP) Subsequent work with multilayer perceptrons has shown that they are capable of approximating an XOR operator as well as many other non-linear functions. Footnotes. Further Reading Other Pathmind Wiki Posts
What is multi layer?
Multi-layer security—also known as “multi-level security” or “defense in depth”—is a suspicious-sounding phrase. Multi-layer corporate and small business internet security makes sense in the cloud, because the costs of infrastructure, bandwidth, and expertise can be shared across clients—and so can the information needed to correlate and block blended attacks.
Begin typing your search term above and press enter to search. Press ESC to cancel.
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# Cdf of multivariate normal in distributions.jl
I’m trying to use the cdf of a multivariate normal distribution using Distributions.jl.
While
``````d = Normal(0,1)
d2 = MvNormal([0.;0.], [1. 0.;0. 1.])
x = [1. ; 2.]
e1 = cdf(d,x[1])
e2 = cdf(d,x[2])
``````
works (as long as the mean vector and the var-cov matrix for the multivariate normal are Floats ), I get an error whenever I write
`e3 = cdf(d2,x)`
Here is the error message:
``````ERROR: MethodError: no method matching cdf(::Distributions.MvNormal{Float64,PDMats.PDMat{Float64,Array{Float64,2}},Array{Float64,1}}, ::Array{Float64,1})
Closest candidates are:
cdf(::Distributions.Distribution{Distributions.Univariate,S<:Distributions.ValueSupport}, ::AbstractArray{T,N}) at /Applications/JuliaPro-0.5.1.1.app/Contents/Resources/pkgs-0.5.1.1/v0.5/Distributions/src/univariates.jl:205
``````
Does the cdf function support multivariate distributions? I couldn’t find anything on this in the documentation of Distributions.jl.
No, there is not a `cdf` for the multivariate normal. Generally, it is a slightly complicated computation. We have some code evaluating the bi- and trivariate case but it hasn’t been used for a long time. For some time, I’ve wanted a dedicated package for multivariate distributions that would be using `StaticArrays` for storage. The `cdf` code for fit well in such a package.
2 Likes
That’s a shame. We really should have something in Julia for this. I’ll work around it using PyCall for the time being.
1 Like
Are you mainly interested in 2D and 3D or higher dimensions? Do you know how higher dimensions are handled in Python?
My interest is currently for 2D. I have no idea how higher dimensions are handled in Python right now but hopefully I’ll learn a little bit about it in the next few days.
If you are interested in spending a little time on it, we could probably get the 2D Julia version working again. If we do it with StaticArrays it will also be extremely fast.
Hello,
I am also interested in a julia version for the cdf for 2D normal. I might be able to spend a bit time on this, but where can I find the version that exists already? I am not familiar with StaticArrays though, but I can certainly give it a try.
Thanks!
1 Like
I think it would be very helpful to have a CDF defined for `MvNormal`, analog to MATLAB’s mvncdf. It would definitely help when dealing with truncated Gaussian priors, etc.
I’m working on it (shh…)
1 Like
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# Extracting meaning from an FFT analysis of data
My question is about Fast Fourier Transforms, since this is the first time i'm using them. So, I have a set of data by years (from 1700 - 2009) and each year corresponding to a certain value (a reading). when i plot the readings against the years it gives me the first plot below. Now, my aim is to find the dominant period with the highest readings using FFT with python (From the graph it seems that it is around 1940 - 1950). So i performed an FFT and got its amplitude and power spectra (see second plot for power spectrum). The power spectrum shows that the dominant frequencies are between 0.08 and 0.1 (cycles/year). My question is, how do i link this to the Readings vs. years ? i.e how do i know from this dominant frequency what year (or period of years) is the dominant one (or how can i use it to find it) ?
The data list can be found here: http://www.physics.utoronto.ca/%7Ephy225h/web-pages/sunspot_yearly.txt
the code i wrote is:
``````from pylab import *
from numpy import *
from scipy import *
from scipy.optimize import leastsq
import numpy.fft
#-------------------------------------------------------------------------------
# Defining the time array
tmin = 0
tmax = 100 * pi
delta = 0.1
t = arange(tmin, tmax, delta)
year, N_sunspots = loadtxt('/Users/.../Desktop/sunspot_yearly.txt', unpack = True) # years and number of sunspots
# Ploting the data
figure(1)
plot(year, N_sunspots)
title('Number of Sunspots vs. Year')
xlabel('time(year)')
ylabel('N')
# Computing the FFT
N_w = fft(N_sunspots)
# Obtaining the frequencies
n = len(N_sunspots)
freq = fftfreq(n) # dt is default to 1
# keeping only positive terms
N = N_w[1:len(N_w)/2.0]/float(len(N_w[1:len(N_w)/2.0]))
w = freq[1:len(freq)/2.0]
figure(2)
plot(w, real(N))
plot(w, imag(N))
title('The data function f(w) vs. frequency')
xlabel('frequency(cycles/year)')
ylabel('f(w)')
grid(True)
# Amplitude spectrum
Amp_spec = abs(N)
figure(3)
plot(w, Amp_spec)
title('Amplitude spectrum')
xlabel('frequency(cycles/year)')
ylabel('Amplitude')
grid(True)
# Power spectrum
Pwr_spec = abs(N)**2
figure(4)
plot(w, Pwr_spec 'o')
title('Power spectrum')
xlabel('frequency(cycles/year)')
ylabel('Power')
grid(True)
show()
``````
-
The graph below shows the data input to the FFT. The original data file contains a total of 309 samples. The zero values at the right end are added automatically by the FFT, to pad the number of input samples to the next higher power of two (2^9 = 512).
The graph below shows the data input to the FFT, with the Kaiser-Bessel a=3.5 window function applied. The window function reduces the spectral leakage errors in the FFT, when the input to the FFT is a non-periodic signal over the sampling interval, as in this case.
The graph below shows the FFT output at full scale. Without the window function. The peak is at 0.0917968 (47/512) frequency units, which corresponds to a time value of 10.89 years (1/0.0917968).
The graph below shows the FFT output at full scale. With the Kaiser-Bessel a=3.5 window applied. The peak remains in the same frequency location at 0.0917968 (47/512) frequency units, which corresponds to a time value of 10.89 years (1/0.0917968). The peak is more clearly visible above the background, due to the reduction in spectral leakage provided by the window function.
In conclusion, we can say with high certainty that the Sun spot data, provided in the original post, is periodic with a fundamental period of 10.89 years.
FFT and graphs were done with the Sooeet FFT calculator
-
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# Magnetic vector potential (Electrical Machine)
By
## Potentials and formulations
The Maxwell equations represent the physical properties of the fields. To solve them, mainly the differential form of the equations and mathematical functions, the potentials, satisfying the Maxwell equations, are used. The proper choice of a potential depends on the type of field problem. In this section, the various scalar and vector potentials are introduced.
The electric vector potential for the displacement current density will not be introduced here, because it is only important for the calculation of fields in charge-free and current-free regions such as hollow wave-guides or in surrounding fields of antennas.
Various potential formulations are possible for the different field types. Their appropriate definition ensures the accurate transition of the field problem between continuous and discrete space.
Using these artificial field quantities reduces the number of differential equations. Considering a problem described by n differential equations, a potential is chosen in such a way that one of the differential equations is fulfilled. This potential is substituted in all other differential equations, the resulting system of differential equations reduces to n-l equations. It is distinguished between magnetic and electric vector respectively scalar potentials:
Table 4.1. Definition of the potentials.
scalar potentials vector electric magnetic
By using the vector identity
and applying eq.(iii), (3.10), introduces the magnetic vector potential A. The magnetic flux density is derived as the curl of another vector field:
The magnetic vector potential is suitable in regions with and without conducting currents. The vector field A is assigned right handed to the direction of the magnetic field B (Fig. 4.1).
Fig. 4.1. Geometrical assignment of the vector potential A with the magnetic field vector B.
A static magnetic problem is described by
By using the magnetic vector potential A, the system of differential equations is reduced to
Applying the vector calculusto eq.(4.4) yields:
withand by assuming a constant permeabilityleads to the A-formulation of a magneto-static field, a Poisson equation:
To consider quasi-stationary fields, for example necessary for eddy current calculations, the magneto-dynamic formulations have to be employed. In addition to Ampere’s law, the Faraday law (i), (3.26) has to be considered to evaluate the contribution to the field by the eddy currents:
Now employing Ohm’s law to calculate the eddy currents Je yields:
Ampere’s law can now be rewritten, yielding the A-formulation for the quasi-stationary magnetic field in the time domain:
Substituting againand assuming
results in a similar A-formulation in the time domain for the transient magnetic field:
Assuming sinusoidal excitation currents with an angular frequency and thus substituting
yields the A-formulation in the frequency domain to solve eddy current problems.
This equation is the A-formulation to describe time-harmonic problems. The time dependent components of the vector potential
are expressed by:
The current is expressed in analogy in its complex representation.
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How to Read Betting Lines and Odds: A Comprehensive Guide to Sportsbooks
Rate this post
Understanding betting lines and odds is essential for anyone looking to place a bet on sports. Sportsbooks use betting lines and odds to determine the potential payout for a winning bet, as well as to reflect the likelihood of different outcomes.
For beginners, these terms can be confusing, but with a little knowledge, you can learn to read them with confidence. This guide will break down everything you need to know about betting lines and odds to help you make informed decisions at sportsbooks.
What Are Betting Lines?
1. Definition of Betting Lines
Betting lines are the different types of wagers available for a sporting event. They indicate how much you stand to win based on your bet and show which team or player is favored. Common types of betting lines include point spreads, moneylines, and totals (over/under). Each type has its own way of determining potential payouts and risks.
2. Types of Betting Lines
• Point Spread: The point spread is a popular form of betting, especially in sports like football and basketball. The spread represents the number of points a team is expected to win or lose by. The favorite team will have a negative spread (e.g., -5.5), while the underdog will have a positive spread (e.g., +5.5). For a bet on the favorite to win, they must win by more points than the spread. For a bet on the underdog, they must either win outright or lose by fewer points than the spread.
• Moneyline: The moneyline bet is straightforward: you are betting on which team or player will win the game. There is no point spread involved. Odds are expressed with a positive (+) or negative (-) number. A negative number indicates the favorite, and a positive number represents the underdog. For example, a moneyline of -150 means you must bet \$150 to win \$100, while a moneyline of +200 means a \$100 bet will win you \$200.
• Totals (Over/Under): The totals bet, also known as the over/under, involves wagering on the total number of points scored by both teams combined. The sportsbook sets a line, and you can bet on whether the actual total will be over or under that number. For example, if the line is set at 45.5 points for a football game, you can bet on whether the total points scored will be more (over) or less (under) than 45.5.
Understanding Odds in Sports Betting
1. What Are Betting Odds?
Betting odds, which are commonly found on the best sportsbooks Canada, indicate the probability of an outcome occurring and determine how much you can win from a bet. Odds can be presented in three formats: American, Decimal, and Fractional. Each format provides the same information but in a different way.
2. Types of Betting Odds
• American Odds: Also known as moneyline odds, American odds are displayed as a positive or negative number. Positive odds show how much you would win on a \$100 bet, while negative odds show how much you need to bet to win \$100. For example, +250 means you win \$250 on a \$100 bet, while -150 means you need to bet \$150 to win \$100.
• Decimal Odds: Popular in Europe, decimal odds are simple to understand. They show the total payout rather than just the profit. For example, decimal odds of 2.50 mean that for every \$1 bet, the total return is \$2.50, including your initial stake.
• Fractional Odds: Common in the UK, fractional odds are presented as fractions, such as 5/1 or 10/3. The first number represents the potential profit, and the second number represents the stake. For instance, 5/1 means you win \$5 for every \$1 wagered.
How to Read and Interpret Betting Lines and Odds
To read a point spread, look at the numbers next to each team. The team with a negative number is the favorite, and the number indicates how many points they need to win by to cover the spread. The team with a positive number is the underdog, and they must either win outright or lose by fewer points than the spread for the bet to be successful.
Example:
• Team A: -7.5
• Team B: +7.5
If you bet on Team A, they need to win by 8 or more points. If you bet on Team B, they need to either win or lose by 7 points or less.
2. Understanding Moneyline Odds
Moneyline odds are the simplest to understand. If you bet on the favorite, you will have to risk more money for a smaller return. If you bet on the underdog, you risk less money for a higher return.
Example:
• Team A: -150
• Team B: +130
A \$150 bet on Team A would win \$100 if they win, while a \$100 bet on Team B would win \$130 if they win.
When betting on totals, you decide if the combined score of both teams will be over or under the set number by the sportsbook. If the total is set at 50.5 and you bet the over, you need the combined score to be 51 or more to win. If you bet the under, the combined score must be 50 or fewer.
4. Using Decimal and Fractional Odds
To convert fractional odds to decimal, divide the first number by the second and add 1. For example, 5/1 becomes 6.00 in decimal odds. To convert American odds to decimal, for positive odds, divide by 100 and add 1. For negative odds, divide 100 by the odds and add 1.
Tips for Betting Safely and Smartly
1. Research and Analyze
Before placing any bet, do your research. Analyze teams, players, form, injuries, and other factors that can influence the outcome. Knowledge is power in sports betting.
2. Understand the Risks
Betting is inherently risky. Understand that there is no sure thing, and even bets that seem certain can lose. Manage your bankroll wisely and only bet what you can afford to lose.
3. Shop for the Best Odds
Different sportsbooks offer different odds and lines. Compare multiple sportsbooks to find the best value for your bets. Better odds can significantly increase your potential payout.
4. Consider Betting Limits and Terms
Check the betting limits and terms at your chosen sportsbook. Some sportsbooks have maximum payout limits, and others may have different rules for specific types of bets. Understanding these terms can help you avoid surprises.
Conclusion
Reading betting lines and odds is an essential skill for any sports bettor. By understanding the types of betting lines, interpreting odds correctly, and following smart betting practices, you can make more informed decisions and potentially increase your chances of success. Remember, sports betting should be fun and exciting, so always bet responsibly and enjoy the game!
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UnitConverter.netV1.2
Home / Weight Converter / Ounces To Pounds
# ounces to pounds
Enter the value that you want to convert ounces to pounds (oz to lbs) or pounds to ounces.
From: ounce To: pound
Value:
### Description
#### Ounces (oz):
An ounce is a measuring unit of mass, weight in the imperial system of measurement and US customary systems of measurement. It is denoted as "oz". One ounce is equal to 0.0625 lbs.
#### Pounds (lbs):
A pound is a unit of mass & weight measuring unit & it is denoted as "lb". It is used in the imperial system & united state customary systems. the symbolic representation of pound is lb. One lb weight is equal to 16 oz & 0.45359237 kilograms.
#### Ounces to Pounds (oz to lbs):
It is a free online Ounces to pounds (oz to lbs) weight converter. An ounce is a measuring unit of mass, weight in the imperial system of measurement and US customary systems of measurement. It is denoted as "oz". One ounce is equal to 0.0625 lbs. A pound is a unit of mass & weight measuring unit & it is denoted as lb. It is used in the imperial system & united state customary systems. the symbolic representation of pound is "lb". One lb weight is equal to 16 oz & 0.45359237 kilograms.
ounces to pounds metric conversion table
1 Ounces = 0.0625 Pounds 11 Ounces = 0.687 Pounds 30 Ounces = 1.875 Pounds
2 Ounces = 0.125 Pounds 12 Ounces = 0.75 Pounds 40 Ounces = 2.5 Pounds
3 Ounces = 0.1875 Pounds 13 Ounces = 0.812 Pounds 50 Ounces = 3.125 Pounds
4 Ounces = 0.25 Pounds 14 Ounces = 0.875 Pounds 60 Ounces = 3.75 Pounds
5 Ounces = 0.3125 Pounds 15 Ounces = 0.9375 Pounds 70 Ounces = 4.375 Pounds
6 Ounces = 0.375 Pounds 16 Ounces = 1 Pounds 80 Ounces = 5 Pounds
7 Ounces = 0.4375 Pounds 17 Ounces = 1.0625 Pounds 90 Ounces = 5.625 Pounds
8 Ounces = 0.5 Pounds 18 Ounces = 1.125 Pounds 100 Ounces = 6.25 Pounds
9 Ounces = 0.5625 Pounds 19 Ounces = 1.1875 Pounds 500 Ounces = 31.25 Pounds
10 Ounces = 0.625 Pounds 20 Ounces = 1.25 Pounds 1000 Ounces = 62.5 Pounds
### How to Convert Ounce to Pound?
how many ounces (oz) in a pound.
Example:
Convert 12 oz to lbs?
We know 1 oz = 0.0625 lbs; 1 lbs = 16 oz.
12 oz = __ lbs
12x0.0625 = 16 lbs (We know 1 oz = 0.0625 lbs)
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# Multiply By Tens Worksheets
Multiply By Tens Worksheets. This page includes powers of ten math worksheets with. The worksheets are varied including skip counting, vertical and horizontal multiplication and even color by numbers to help students learn and consolidate.
Building a strong foundation in multiplication is an important step in helping your child become proficient and confident. Multiply by multiples of ten. Multiplication by 10s (10s facts only) with these printable worksheets, students can practice multiplying by the number 10.
### The Worksheets Are Varied Including Skip Counting, Vertical And Horizontal Multiplication And Even Color By Numbers To Help Students Learn And.
This math worksheet is printable. This download includes 10 ready to go workstations for multiplication. The worksheets are varied including skip counting, vertical and horizontal multiplication and even color by numbers to help students learn and consolidate.
### Multiplying By Ten 10 With Factors 1 To 12 100 Questions A Multiplying And Dividing Decimals By 10 Worksheet, Multiplying And Dividing By Powers Of Ten Worksheets, Multiplying By Tens.
A number of examples such as 9 x 60 are demonstrated; Contribute to khufrae17/printablelesson development by creating an account on github. This provides great extra practice for kids.
### Multiplying Whole Tens By Single Digit Numbers.
Worksheets are multiplying with 10s, multiplying and dividing by 10 100 and 10, multiplication work, multiplication, 2 level a skill,. Includes all basic facts, up to 10×10, in which at least one factor. Some of the worksheets for this concept are multiplying whole, grade 3 multiplication work, multiplying by whole tens and hundreds, decimal multiplication a, multiply by whole tens.
### Equations With Single Digit Numbers Multiplying Whole Tens Are Also Included.
Download free 10 times table worksheets. Multiplication by 10s (10s facts only) with these printable worksheets, students can practice multiplying by the number 10. 12pages 3rd grade multiplication numbers 1s 2s 3s 4s 5s 6s 7s 8s 9s 10s 11s 12s multiplication worksheets learning worksheets multiplication.
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# boundary layer, very urgent
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July 18, 2006, 03:33 boundary layer, very urgent #1 nicolas Guest Posts: n/a Hi. I am a beginner in CFD. I use ICEM CFD 10.0+ cfx 5.7.1 I have a cylinder+ some others parts of a engine. I used blocking + hexa mesh for everything under ICEM. I used O grid generation, and I created a small volume around the wall, to set a fine mesh here. Someone told me that , with hexa mesh, I did'nt have to care about the boundary layer(but i need to have a fine meshin this volume). What i understood, is that I don't care about the position of the first node, and the law mesh in this volume. Is it right? wrong? I found no explanation on websites, documentations, on "how to" do a boundary layer when using hexa mesh (i found it for prism, tet)... Do I need to set exponential law in the small volume around the wall(and how to do it?) Does someone can help me to understand how to treat the boundary layer under icem, using hexa mesh (and o grid generation). If it is explained on doumentation, can you give me the references ? Regards, Nicolas.
July 18, 2006, 05:46 Re: boundary layer, very urgent #2 Kramer Guest Posts: n/a Which kind of turbulence model do you want to use : k-epsilon, k-omega, SST, or something else? SST for example needs an y+ value of approx 1 (=>very small first element on the surface). So you will need a fine mesh in the boundary layer and at least 15 hexa elements in the boundary layer.
July 18, 2006, 06:00 Re: boundary layer, very urgent #3 nicolas Guest Posts: n/a I use SST. To know the distance, where i need to set the 1st node(if i choose y+=1),I need to know the Reynolds number? In a certain way, i need to know the solution, before being able to create a good meshing for the boundary layer? other question: do you you know how to fix the distance between the wall and the 1st node, and, in the same time, set a expansion law (i have found how to set the expansion law, but i can't fix the distance). I use icem thanks, nicolas
July 18, 2006, 06:36 Re: boundary layer, very urgent #4 Kramer Guest Posts: n/a You have to make a good guess for your first node distance, its depending on the flow situation (estimate Reynolds number...) For example: i have simulated the flow around an airfoil with SST an had to set the first node at 0.01 mm to get y+ <= 1. Seconde question: there is an overall option to set the distance for the first node generally (settings an so on..., i dont remember exactly where), but independent from that you can use the spacing function (right there where u define expansion law) to set the value
July 18, 2006, 08:52 Re: boundary layer, very urgent #5 Joe Guest Posts: n/a Read the section in the turbulence modelling section entitled something like "Advice on near wall meshing".
July 18, 2006, 09:58 Re: boundary layer, very urgent #6 imad Guest Posts: n/a Dear Nicolas under blocking you find pre-mesh params select them and use edge params to select the edge near the wall and define the distance for the first and last point and if you want the mesh law
July 18, 2006, 10:23 Re: boundary layer, very urgent #7 nicolas Guest Posts: n/a Dear imad, i already try this. but i dont understand how it works. to set the distance from wall to node ->i sould use the values called in the software "spacing 1" and "spacing 2" ?? thanks for your reply. Regards, Nicolas
July 18, 2006, 13:55 Re: boundary layer, very urgent #8 imad Guest Posts: n/a DEAR Nicolas when you select the edge An arow appear, his direction define the first and the last point if you put a value and the distribution don't change, you should select copy to parallel edge or update mesh with best regards Imad
July 18, 2006, 14:00 Re: boundary layer, very urgent #9 imad Guest Posts: n/a yes spacing 1 is the first distance of edge and spacing 2 is the last but th direction of edge is the direction of arow
July 18, 2006, 22:52 Re: boundary layer, very urgent #10 Glenn Horrocks Guest Posts: n/a hi Kramer, SST does not require y+ approx =1. Using the automatic wall function approach it can switch between full boundary layer modelling (around y+=1) to wall function modelling (y+>11) and blend between. You should ask yourself whether the boundary layer needs to be modelled directly or if a wall function approach is OK. Then you generate a mesh with y+=1 or >11 based on that decision. regards, Glenn
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# Differentiate the following functions:
Question:
Differentiate the following functions:
(i) $4 \cot x-\frac{1}{2} \cos x+\frac{2}{\cos x}-\frac{3}{\sin x}+\frac{6 \cot x}{\operatorname{cosec} x}+9$
(ii) $-5 \tan x+4 \tan x \cos x-3 \cot x \sec x+2 \sec x-13$
Solution:
Formulae: -
$\frac{d}{d x} \cot x=-\operatorname{cosec}^{2} x$
$\frac{d}{d x} \cos x=-\sin x$
$\frac{d}{d x} \sec x=\sec x \tan x$
$\frac{d}{d x} \operatorname{cosecx}=-\operatorname{cosec} x \cot x$
$\frac{d}{d x} \tan x=\sec ^{2} x$
$\frac{d}{d x} \sin x=\cos x$
$\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{k}=0, \mathrm{k}$ is constant
(i) $4 \cot x-\frac{1}{2} \cos x+\frac{2}{\cos x}-\frac{3}{\sin x}+\frac{6 \cot x}{\operatorname{cosec} x}+9$
$=4 \cot x-\frac{1}{2} \cos x+2 \sec x-3 \operatorname{cosec} x+6 \cos x+9$
Differentiating with respect to $x$,
$\frac{d}{d x}\left(4 \cot x-\frac{1}{2} \cos x+2 \sec x-3 \operatorname{cosec} x+6 \cos x+9\right)$
$=4\left(-\operatorname{cosec}^{2} x\right)-\frac{1}{2}(-\sin x)+2 \sec x \times \tan x-3(-\operatorname{cosec} x \times \cot x)+6(-\sin x)+$ 0
$=-4 \operatorname{cosec}^{2} x+\frac{1}{2} \sin x+2 \sec x \tan x+3 \operatorname{cosec} x \cot x-6 \sin x$
(ii) $-5 \tan x+4 \tan x \cos x-3 \cot x \sec x+2 \sec x-13$
$=-5 \tan x+4 \sin x-3 \operatorname{cosec} x+2 \sec x-13$
Differentiating with respect to $\mathrm{x}$,
$\frac{d}{d x}(-5 \tan x+4 \sin x-3 \operatorname{cosec} x+2 \sec x-13)$
$=-5 \sec ^{2} x+4 \cos x-3(-\operatorname{cosec} x \cot x)+2 \sec x \tan x-0$
$=-5 \sec ^{2} x+4 \cos x+3 \operatorname{cosec} x \cot x+2 \sec x \tan x$
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# Perl 6 Programming/Meta Operators
## Meta OperatorsEdit
Operators do things to data. Meta operators do things to operators.
## Reduction OperatorsEdit
Reduction operators act on a list and return a scalar value. They do this by applying the reduction operator between every pair of elements in the array:
```my @nums = 1..5;
my \$sum = [+] @nums # 1 + 2 + 3 + 4 + 5
```
The `[ ]` square brackets turn any operator that normally acts on scalars into a reduction operator to perform that same operation on a list. Reductions can also be used with relational operators:
```my \$x = [<] @y; # true if all elements of @y are in ascending order
my \$z = [>] @y; # true if all elements of @y are in descending order
```
## Hyper OperatorsEdit
Reduction operators apply an operator to all the elements of an array and reduces it to a single scalar value. A hyper operator distributes the operation over all the elements in the list and returns a list of all results. Hyperoperators are constructed using the special "french quotes" symbols: « and ». If your keyboard doesn't support these, you can use the ASCII symbols `>>` and `<<` instead.
```my @a = 1..5;
my @b = 6..10;
my @c = @a »*« @b;
# @c = 1*6, 2*7, 3*8, 4*9, 5*10
```
You can also use unary operators with hypers:
```my @a = (2, 4, 6);
my @b = -« @a; # (-2, -4, -6)
```
Unary hyperoperators always return an array that is exactly the same size as the list it is given. Infix hyperoperators have different behavior depending on the sizes of its operands.
```@a »+« @b; # @a and @b MUST be the same size
@a «+« @b; # @a can be smaller, will upgrade
@a »+» @b; # @b can be smaller, will upgrade
@a «+» @b; # Either can be smaller, Perl will Do What You Mean
```
Pointing the hyper symbols in different directions affects how Perl 6 treats the elements. On the sharp side, it extends the array to be as long as the one on the dull side. If both sides are sharp, it will extend whichever is smaller.
Hypers can also be used with assignment operators:
```@x »+=« @y # Same as @x = @x »+« @y
```
## Cross OperatorsEdit
The cross is a capital `X` symbol. As an operator, the cross returns a list of all possible lists made by combining the elements of it's operands:
```my @a = 1, 2;
my @b = 3, 4;
my @c = @a X @b; # (1,3), (1,4), (2,3), (2,4)
```
The cross can also be used as a meta operator, applying the operator it's modifying against every possible combination of elements from each operand:
```my @a = 1, 2;
my @b = 3, 4;
my @c = @a X+ @b; # 1+3, 1+4, 2+3, 2+4
```
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# 17.44 Amps, volts and ohms
Before you read this post, I suggest that you read posts 16.25 and 17.17.
Everybody knows that amps and volts are electrical units. But what, exactly, do we use them to measure? And, how are they connected by Ohm’s law?
In post 16.25, we saw that amps (A) are used to measure electrical current. To have a current, something has to flow. We call whatever flows in an electrical current charge; charge is measured in coulombs (C) (post 16.25). An amp is the steady flow of 1 C in a time of 1 s (although we saw in post 16.25 that the amp is defined by the magnetic effects of a current and is then used to define a coulomb).
If you know about the mathematical techniques of differentiation (post 17.4) and integration (post 17.19), you may wish to read the rest of this paragraph – if not, ignore it. In general current, I, is the rate of change of charge, q, so that
I = dq/dt
where t represents time. Since I is defined without reference to q, the definition of q is
where the limits of integration are over the time that the current flows.
For a constant current, in which a charge Δq flows in a time interval Δt, we can write these equations as
I = Δqt and Δq = It.
Here the symbol Δ means a sample of something, as in post 17.4.
Normally we think of current flowing in metal (especially copper) wires. Electrons can move easily in many metals; we call these metals conductors. All their electrons are not necessarily associated with a single atom and so are free to move, rather like the electrons in a delocalised molecular orbital (post 16.31); electrons in a form of carbon called graphite behave similarly and it is also a conductor. In post 16.27, we saw that electrons have a negative charge; when the electrons move, a current flows in the opposite direction, because their charge is negative.
Electrons are particles with a mass (post 16.27), so to make them move, we need to apply a force (post 16.12). When the force moves an electron, it does work (post 16.20). So to move charge from a point P along a conducting wire to a point P’, we need to do work on the electrons; we say that there is a potential difference between P and P’. In this example, we say that P has a higher potential than P’, because we need to do work to cause a current to flow. If P had a lower potential than P’, the charge would spontaneously flow from P’ to P. Note that when we say charges are flowing, electrons are flowing in the opposite direction. Also, we have defined a potential difference but not an absolute value for the potential because we cannot define a true zero potential. (In practice, we can define a zero potential as being the potential in the ground around a building and call this arbitrary zero the earth.)
The potential difference between two points is measured in volts. A volt (abbreviated to V) is defined as the work done when 1 C of charge moves from one point to the other. Since we measure work in joules (J), see post 16.20, 1 V is equivalent 1 J.C-1. Don’t confuse V, the standard symbol to represent the volt, with V (in italics), the symbol I have chosen to represent postential difference (see post 16.13).We sometimes call the potential difference between a point and an arbitrary zero, its voltage.
However, when a charge flows it experiences some resistance to flow. This resistance arises because electrons are not completely free to move – for example their movement can be opposed by repulsion by other electrons, since objects with the same charge repel each other (post 16.25). The higher the potential difference, the easier it is for charge to flow. So, for a constant current, I, we define electrical resistance between two points P and P’ by
R = V/I
where V is the potential difference between P and P’. This definition means that I = V/R, so for a fixed value of V, the higher the value of R, the lower the value of I. The relationship between potential difference, current and resistance is sometimes called Ohm’s law but it isn’t really a scientific law (see post 16.2), it’s a definition. (If Ohm’s law is simply stated as “the current in a conductor is proportional to the potential difference between its ends”, this is no more than a consequence of the definition of potential difference and charge.)
Since V is measured in volts and I in amps, we might expect to measure resistance in V.A-1. But this unit has a special name, the ohm (abbreviation Ω, the Greek letter capital “omega”).
Moving charge against resistance is not like stretching a spring (post 16.49). When we stretch a spring, it stores the work done on it as potential energy. But when a charge moves against a resistance, it loses all energy that it has gained – usually in the form of heat. We call a device that is intended to have appreciable resistance a resistor. So charge moving through a resistor is analogous to an object moving through a viscous fluid (post 17.17), in that both lose all the energy they gain when work is done on them.
We can carry this analogy even further. If we move an object along a straight line, we can represent its displacement and velocity as scalars x and v, respectively (post 17.4). Hence, we can also represent the forces acting along this line as scalars. (This follows from the definition of force in the second table of post 17.39). According to post 17.17, an object moving at a constant speed through a viscous fluid experiences a drag force of
F = –M.v = –Mxt).
Here the minus sign shows that the drag force acts in the opposite direction to the force initiating movement. M depends on the dimensions of the object and the viscosity of the liquid (post 17.17). The final step in the equation above comes from the definition of v (the second table in post 17.39). When charges move through a resistor their movement is opposed by a potential difference that acts in the opposite direction to the potential difference initiating movement given by
V = –R.I = –Rqt).
So, in both cases we have equations of the same form that represent a process that dissipates energy. R depends on the dimensions of the conductor and a property of the stuff it is made from, called its resistivity. When a current flows along a conductor of length L, with a cross-sectional area A, its resistivity, ρ, is defined by the equation
R = ρL/A.
Resistivity is defined in this way because the longer the conductor, the greater the resistance that a moving charge encounters but a greater cross-sectional area means that there are more electrons available to carry the charge – so the resistance is smaller. Multiplying both sides of the previous equation by A/L gives
ρ = RA/L
Since we, measure L in m and A in m2 (post 16.13), we measure resistivity in Ω.m2/m = Ω.m.
The table above shows the resistivity of some conductors. The best conductor in the table is silver because it has the lowest resistivity. Copper has a slightly higher resistivity but is much cheaper, so is the most common material for conducting electricity. Aluminium can be used for overhead cables because it has a much lower density than copper so the cables weigh less (see posts 16.17 and 16.44).
Resistors that are mounted on electronic circuit boards (see picture above) are a few millimetres long and usually made of graphite, because it has a high resistivity, with metal connecting wires. However, graphite is not strong and it is difficult to manufacture large objects from it. So an electric kettle heating element, which is a big resistor, is made from nichrome which has a lower resistivity than graphite but is easier to handle.
In summary, current flows through conductors and is measured in amps. When a current flows, charges move. Charges move when a potential difference, measured in volts, exists between the ends of a conductor. There is some resistance, measured in ohms, in any conductor, which impedes the flow of charge.
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# Factors of 52
## |
Understanding factors and their role in mathematics is crucial for building a solid foundation in number theory. In this blog, we embark on a journey to uncover the factors of 52 and delve into the fascinating world of divisibility. So, grab your mathematical thinking caps and join us as we unlock the secrets hidden within the number 52.
### The Basics of Factors
Before we dive into the factors of 52, let’s quickly recap what factors are. Factors are the numbers that divide a given number evenly without leaving any remainder. In simpler terms, they are the building blocks that make up a number. By exploring the factors of a particular number, we gain insight into its divisibility and mathematical properties.
### Discovering the Factors of 52
To find the factors of 52, we need to determine which numbers can divide it without leaving a remainder. Let’s embark on this mathematical journey:
#### Number 1
Every whole number is divisible by 1, including 52. When we divide 52 by 1, the result is 52. Therefore, 1 is a factor of 52.
#### Number 2
Dividing 52 by 2 yields a quotient of 26. Since there is no remainder, we conclude that 2 is also a factor of 52.
#### Number 4
We are continuing our exploration, dividing 52 by 4 results in a quotient of 13, once again with no remainder. Hence, 4 is another factor of 52.
#### Number 13
Our mathematical journey takes us further as we divide 52 by 13, resulting in a quotient of 4. Therefore, 13 is a factor of 52.
### The Complete Set of Factors
Having traversed the realm of divisibility, we have discovered all the factors of 52. They are 1, 2, 4, 13, and 52. These numbers showcase the diverse ways in which 52 can be divided evenly.
### Applications and Significance
The factors of 52 find relevance in various mathematical concepts and practical applications. One important application is in determining whether a number is prime or composite. If a number has only two factors, 1 and itself, it is considered prime. However, if it has more than two factors, it is classified as composite. In the case of 52, since it has multiple factors beyond 1 and 52, it falls into the category of a composite number.
Additionally, the knowledge of factors assists in simplifying fractions, finding common denominators, and solving equations. By understanding the factors of a given number, we can unravel its properties and relationships within the mathematical landscape.
In exploring factors 52, we have gained valuable insights into its divisibility and mathematical nature. Factors 1, 2, 4, 13, and 52 contribute to the intricate web of relationships and patterns within the realm of numbers. Remember, understanding factors not only aids in problem-solving but also nurtures a deeper appreciation for the elegance and beauty of mathematics. So, continue exploring, questioning, and unlocking the secrets hidden within the vast numerical landscape.
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Solow Growth Model with Cobb-Douglas Production Function
Deriving the Law of Motion of Capital per Capita
Consider the Solow growth model with a Cobb-Douglas production function with capital share a, a constant savings rate s, a constant depreciation rate 8, and a constant population growth rate Suppose a equals 1/3, and TFP initially is A1 > 0.
1. Derive the law of motion of capital per capita.
2. Derive the steady state levels of capital and output per capita.
3. Explain intuitively why there is a steady state.
Now suppose that productivity increases by 10% to A2.
4. Derive the new steady state levels of capital and output per capita. Compare them to the old ones.
5. In three separate figures, draw the evolution through time of productivity A, capital per capita Kit, and output per capita Yit. Make sure to include several periods of the pre-shock steady state, the time of the shock, the transition to the final steady state, and several periods of the final steady state. Discuss the differences between the two figures.
6. Suppose that the economy has converged to its new steady state. Do a growth accounting exercise on output per capita. How much of the change in output per capita was due to changes in productivity, how much due to changes in capital per capita?
7. Does this result reflect well what actually occurred? 8. In the last decades, East Asian economies have grown strongly. Some studies attribute this mainly to capital accumulation. Discuss.
Assume that the Solow model provides an accurate representation of the two Germanies around the late 80s and early 90s. Although in 1990 both countries re-unified, for this problem we will treat them as two separate economies, and we will model the effects of the reunification as a one-time flow of resources (capital and/or labor) across the common border. Aside from being in steady state initially, assume that before the re-unification both economies had the same saving rate, the same rates of population growth and depreciation, and that capital was equally important for production in both countries, with a common value of = 1/2.
Let's focus first on the period before the re-unification.
1. In this initial steady state, output per capita in the West was twice the level in the East, due to its higher productivity A. Use this information and equations for the steady state to calculate the ratio of productivity between the two countries. Explain your derivations.
2. Calculate the ratio of capital per capita between the two countries. Explain your deriva-tions carefully.
3. Did the owners of capital in West Germany have any incentives to reallocate their capital from the capital abundant Western economy to the capital scarce Eastern one? Explain your answer (think about the payment per unit of capital in both countries).
4. Would East German workers have any economic incentives to migrate to the West? Explain your answer (think about wages in both countries). Now consider reunification. Let's think of it as a one-time migration of workers from East to West Germany. In the remainder of the problem, we will take the perspective of East Germany.
5. Use the phase diagram of the Solow model to trace the effects of the reunification, i.e. of a one-time decrease in the labor force of East Germany. Discuss the forces at work along the transition.
6. In three figures, sketch the evolution through time of capital per capita, investment per capita and (the logarithm of) aggregate output. Make sure your figures include the initial steady state (that is, some periods before the re-unification), the shock, several periods after the shock, and the final steady state.
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## The Solar System
Remember our discussion about the scale of the Solar System? (see our EBook "Solar System").
We noticed how difficult it is to get a true sense of distances as compared to the sizes of Sun, planets and moons, and that it isn’t possible to build a scale model within a normal building that is true to scale in both distance and size.
The picture at left shows the Sun and planets to the same scale as regards size, not their distance. Is it possible to draw a picture of the Solar system where both and size and distance are at the same scale?
In our Observatory we have a wall space of about 5 metre length available to build a scale model of the Solar System. This needs to show at least the distances of the eight planets from the Sun at the true scale.
We calculated that if Neptune would have to fit within the 5 metre, the Sun would have a diameter of 1.4 mm. We used an ordinary pinhead of about that size. We realised that the Sun is the only visible object at this scale, and that Neptune, the furthest planet, is 4.5 metre away from the Sun. See the table below for details.
Scale Model Solar System
Actual (km)Scaled (mm)
ObjectDiameterDistance
from Sun
DiameterDistance
from Sun
Sun 1,392,000 - 1.4 -
Mercury 4,853 57,950,000 0.005 58
Venus 12,132 108,110,000 0.012 108
Earth 12,771 149,570,000 0.013 150
Mars 6,768 227,840,000 0.007 228
Jupiter 143,031 778,140,000 0.14 778
Saturn 120,683 1,427,000,000 0.12 1,427
Uranus 51,083 2,870,300,000 0.051 2,870
Neptune 49,550 4,499,900,000 0.050 4,500
What we have learned from this scale model?
The first four rocky planets are all very close to the Sun and the gas planets are, relatively speaking, very far away. Especially Uranus and Neptune are at a lonely distance.
The other thing that we have learned is that the planets are unimaginably small in comparison to their mutual distance. They are mere specks of dust at this scale. Just look at Neptune, a speck of only 0.05 mm orbiting a pinhead of 1.4 mm at a distance of 4.5 metre.
Hence the Solar System is a very empty place, in contrast to what is suggested on many diagrams of the Solar System.
The difficulty is to draw both distance and size to a scale so that you can see something.
You cannot build such a scale model of the Solar System inside a normal building.
The other thing we must realise is that the planets are never lined up all in a straght line away from the Sun. In reality each planet could be anywhere in the plane of the Solar System (the ecliptic). So in this scale model we have invisibly small planets that could be anywhere in a circle at the given distance from the pinhead. This really hits it home that the Solar System is a very empty place, and that pictures such as the one above are very unrealistic. In this picture Neptune would have to be at a distance of more than 400 metres from the Sun.
Actually, the Solar System extends all the way to the Oort Cloud, which is up to about half a light year from the Sun. In this scale model that is 4.7 km. So that is the real size of the Solar System at this scale, in which the Sun is the size of a pinhead.
Above we show another scale model that fits within the text area of this page. The Sun and planets are now even much smaller than in the previous model (see the details in the diagram). Even the Sun cannot be made visible at this scale,
but you can appreciate the true relative distances here.
In this EBook we are going to use this scale model in which our Sun is the size of a pinhead of 1.4 mm.
We will see how this scale model increases in size when we are leaving our Solar System, on a trip to the far reaches of the observable Universe.
## Make your own scale model of the Solar System
Maybe you have space in a hall or corridor that is much longer than the 4.5 metres we use.
Can you double that, or even more?
Alternatively, you can decide that Mercury must be e.g. 1 mm in diameter. Then calculate all the other dimensions of your scale model and see if you can fit it inside somewhere (or not). At least you will get a good idea about the true scale of the Solar System.
Use the values for size and distance in the table above.
Good luck!
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sum-of-the-maclaurin-series-kristakingmath
# Interactive video lesson plan for: Sum of the maclaurin series (KristaKingMath)
#### Activity overview:
► My Sequences & Series course: https://www.kristakingmath.com/sequences-and-series-course
Learn how to find the sum of the maclaurin series, which is the taylor series centered at 0, or with a=0. You can easily find the sum of the series if you can compare the given series to a well known maclaurin series.
● ● ● GET EXTRA HELP ● ● ●
If you could use some extra help with your math class, then check out Krista’s website // http://www.kristakingmath.com
● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: http://www.kristakingmath.com
INSTAGRAM // https://www.instagram.com/kristakingmath/
PINTEREST // https://www.pinterest.com/KristaKingMath/
QUORA // https://www.quora.com/profile/Krista-King
Tagged under: sinx,taylor,compare,calculus,expert,maclaurin,sum, ,series,^,educational
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12 Feb 2019
# Penrose Tiles
### Mathematics, Penrose Tiles, Roger Penrose
———————
In 1974, Roger Penrose, a British mathematician, created a revolutionary set of tiles that could be used to cover an infinite plane in a pattern that never repeats. In 1982, Daniel Shechtman, an Israeli crystallographer, discovered a metallic alloy whose atoms were organized unlike anything ever observed in materials science. Penrose garnered public renown on a scale rarely seen in mathematics. Shechtman won the Nobel Prize. Both scientists defied human intuition and changed our basic understanding of nature’s design, revealing how infinite variation could emerge within a highly ordered environment.
At the heart of their breakthroughs is “forbidden symmetry,†so-called because it flies in the face of a deeply ingrained association between symmetry and repetition. Symmetry is based on axes of reflection—whatever appears on one side of a line is duplicated on the other. In math, that relationship is reflected in tiling patterns. Symmetrical shapes such as rectangles and triangles can cover a plane with neither gap nor overlap, and in an ever-repeating pattern. Repeated patterns are called “periodic†and are said to have “translational symmetry.†If you move a pattern from place to place, it looks the same.
Penrose, a bold, ambitious scientist, was interested less in identical patterns and repetition, and more in infinite variation. To be precise, he was interested in “aperiodic†tiling, or sets of tiles that can cover an infinite plane with neither gap nor overlap, without the tiling pattern ever repeating itself. That was a challenge because he couldn’t use tiles with two, three, four, or six axes of symmetry—rectangles, triangles, squares, and hexagons—because on an infinite plane they would result in periodic or repeated patterns. That meant he had to rely on shapes believed to leave gaps in the tiling of a plane—those with forbidden symmetries.
Penrose turned to five-axis symmetry, the pentagon, to create his plane of non-repeating patterns, in part, he has said, because pentagons “are just nice to look at.†What was remarkable about Penrose tiles was that even though he derived his tiles from the lines and angles of pentagons, his shapes left no awkward gaps. They snugged together perfectly, twisting and turning across the plane, always coming close to repetition, but never quite getting there.
Penrose tiling captured public attention for two major reasons. First, he found a way to generate infinitely changing patterns using just two types of tiles. Second, and even more spectacular, his tiles were simple, symmetrical shapes that on their own betrayed no sign of their unusual properties.
Penrose made several versions of his aperiodic tile sets. One of his most famous is known as the “kite†and the “dart.†The kite looks like the kids’ toy of the same name, and the dart looks like a simplified outline of a stealth bomber. Both divide cleanly along axes of symmetry and each has two simple, symmetrical arcs on their surface. Penrose established one placing rule: for a “legal†tile placement these arcs must match up, creating contiguous curves. Without this rule, kites and darts can be placed together in repeating patterns. With this rule, repetition never comes. The kite and the dart tile forever, dancing around their five axes, creating starbursts and decagons, winding curves, butterflies and flowers. Shapes recur but new variations keep creeping in.
Edmund Harriss, an assistant clinical professor in mathematical studies at the University of Arkansas, who wrote his Ph.D. thesis on Penrose tiles, offers a comparison. “Imagine you’re on a world that is just made up of squares,†Harriss says. “You start walking, and when you get to the edge of the square, and the next square is exactly the same, you know what you’re going to see if you walk forever.†Penrose tiling has the exact opposite nature. “No matter how much information you have, how much you’ve seen of the tiling, you’ll never be able to predict what happens next. It will be something that you’ve never seen before.â€
RTWT
4 Feedbacks on "Penrose Tiles"
### Anon
OK, I get the uniqueness and it is mildly interesting. But of what value? What practical value does it have to anyone?
### another 'non
“What is the good of a newborn baby?â€
Ben Franklin, I think.
### Anon
But that is a non sequitur that only serves to beg the question. If it’s only value is math it rates far below other important math such as compound interest or Circumference = 2 Ï€ r. If it’s value is artistic it rates right up there with modern art. If it has a practical value then I don’t see it. It is mildly interesting and has a nerd value but other than that, what?
### Surellin
It’s an aspect of chaos theory.
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# Python numpy.diagonal() Examples
The following are 30 code examples for showing how to use numpy.diagonal(). These examples are extracted from open source projects. You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example.
You may check out the related API usage on the sidebar.
You may also want to check out all available functions/classes of the module , or try the search function .
Example 1
def log_multivariate_normal_density(X, means, covars, min_covar=1.e-7):
"""Log probability for full covariance matrices. """
if hasattr(linalg, 'solve_triangular'):
# only in scipy since 0.9
solve_triangular = linalg.solve_triangular
else:
# slower, but works
solve_triangular = linalg.solve
n_samples, n_dim = X.shape
nmix = len(means)
log_prob = np.empty((n_samples, nmix))
for c, (mu, cv) in enumerate(zip(means, covars)):
try:
cv_chol = linalg.cholesky(cv, lower=True)
except linalg.LinAlgError:
# The model is most probabily stuck in a component with too
# few observations, we need to reinitialize this components
cv_chol = linalg.cholesky(cv + min_covar * np.eye(n_dim),
lower=True)
cv_log_det = 2 * np.sum(np.log(np.diagonal(cv_chol)))
cv_sol = solve_triangular(cv_chol, (X - mu).T, lower=True).T
log_prob[:, c] = - .5 * (np.sum(cv_sol ** 2, axis=1) + \
n_dim * np.log(2 * np.pi) + cv_log_det)
return log_prob
Example 2
def _log_multivariate_normal_density_full(X, means, covars, min_covar=1.e-7):
"""Log probability for full covariance matrices.
"""
n_samples, n_dim = X.shape
nmix = len(means)
log_prob = np.empty((n_samples, nmix))
for c, (mu, cv) in enumerate(zip(means, covars)):
try:
cv_chol = linalg.cholesky(cv, lower=True)
except linalg.LinAlgError:
# The model is most probably stuck in a component with too
# few observations, we need to reinitialize this components
cv_chol = linalg.cholesky(cv + min_covar * np.eye(n_dim),
lower=True)
cv_log_det = 2 * np.sum(np.log(np.diagonal(cv_chol)))
cv_sol = linalg.solve_triangular(cv_chol, (X - mu).T, lower=True).T
log_prob[:, c] = - .5 * (np.sum(cv_sol ** 2, axis=1) +
n_dim * np.log(2 * np.pi) + cv_log_det)
return log_prob
Example 3
def CRE(size):
r"""Circular real ensemble (CRE).
Parameters
----------
size : tuple
(n, n), where n is the dimension of the output matrix.
Returns
-------
U : ndarray
Orthogonal matrix drawn from the CRE (=Haar measure on O(n)).
"""
# almost same code as for CUE
n, m = size
assert n == m # ensure that mode in qr doesn't matter.
A = np.random.standard_normal(size)
Q, R = np.linalg.qr(A)
# Q-R is not unique; to make it unique ensure that the diagonal of R is positive
# Q' = Q*L; R' = L^{-1} *R, where L = diag(phase(diagonal(R)))
L = np.diagonal(R)
Q *= np.sign(L)
return Q
Example 4
def CUE(size):
r"""Circular unitary ensemble (CUE).
Parameters
----------
size : tuple
(n, n), where n is the dimension of the output matrix.
Returns
-------
U : ndarray
Unitary matrix drawn from the CUE (=Haar measure on U(n)).
"""
# almost same code as for CRE
n, m = size
assert n == m # ensure that mode in qr doesn't matter.
A = standard_normal_complex(size)
Q, R = np.linalg.qr(A)
# Q-R is not unique; to make it unique ensure that the diagonal of R is positive
# Q' = Q*L; R' = L^{-1} *R, where L = diag(phase(diagonal(R)))
L = np.diagonal(R).copy()
L[np.abs(L) < 1.e-15] = 1.
Q *= L / np.abs(L)
return Q
Example 5
def O_close_1(size, a=0.01):
r"""return an random orthogonal matrix 'close' to the Identity.
Parameters
----------
size : tuple
(n, n), where n is the dimension of the output matrix.
a : float
Parameter determining how close the result is on O;
:math:\lim_{a \rightarrow 0} <|O-E|>_a = 0 (where E is the identity).
Returns
-------
O : ndarray
Orthogonal matrix close to the identiy (for small a).
"""
n, m = size
assert n == m
A = GOE(size) / (2. * n)**0.5 # scale such that eigenvalues are in [-1, 1]
E = np.eye(size[0])
Q, R = np.linalg.qr(E + a * A)
L = np.diagonal(R) # make QR decomposition unique & ensure Q is close to one for small a
Q *= np.sign(L)
return Q
Example 6
def cov(self):
r"""The covariance matrix describing the Gaussian state.
The diagonal elements of the covariance matrix correspond to the
variance in the position and momentum quadratures:
.. math::
\mathbf{V}_{ii} = \begin{cases}
(\Delta x_i)^2, & 0\leq i\leq N-1\\
(\Delta p_{i-N})^2, & N\leq i\leq 2(N-1)
\end{cases}
where :math:\Delta x_i and :math:\Delta p_i refer to the
position and momentum quadrature variance of mode :math:i respectively.
Note that if the covariance matrix is purely diagonal, then this
corresponds to squeezing :math:z=re^{i\phi} where :math:\phi=0,
and :math:\Delta x_i = e^{-2r}, :math:\Delta p_i = e^{2r}.
Returns:
array: the :math:2N\times 2N covariance matrix.
"""
return self._cov
Example 7
def test_diagonal(self):
a = np.arange(12).reshape((3, 4))
assert_equal(a.diagonal(), [0, 5, 10])
assert_equal(a.diagonal(0), [0, 5, 10])
assert_equal(a.diagonal(1), [1, 6, 11])
assert_equal(a.diagonal(-1), [4, 9])
b = np.arange(8).reshape((2, 2, 2))
assert_equal(b.diagonal(), [[0, 6], [1, 7]])
assert_equal(b.diagonal(0), [[0, 6], [1, 7]])
assert_equal(b.diagonal(1), [[2], [3]])
assert_equal(b.diagonal(-1), [[4], [5]])
assert_raises(ValueError, b.diagonal, axis1=0, axis2=0)
assert_equal(b.diagonal(0, 1, 2), [[0, 3], [4, 7]])
assert_equal(b.diagonal(0, 0, 1), [[0, 6], [1, 7]])
assert_equal(b.diagonal(offset=1, axis1=0, axis2=2), [[1], [3]])
# Order of axis argument doesn't matter:
assert_equal(b.diagonal(0, 2, 1), [[0, 3], [4, 7]])
Example 8
def sp_to_adjency(C, threshinf=0.2, threshsup=1.8):
""" Thresholds the structure matrix in order to compute an adjency matrix.
All values between threshinf and threshsup are considered representing connected nodes and set to 1. Else are set to 0
Parameters
----------
C : ndarray, shape (n_nodes,n_nodes)
The structure matrix to threshold
threshinf : float
The minimum value of distance from which the new value is set to 1
threshsup : float
The maximum value of distance from which the new value is set to 1
Returns
-------
C : ndarray, shape (n_nodes,n_nodes)
The threshold matrix. Each element is in {0,1}
"""
H = np.zeros_like(C)
np.fill_diagonal(H, np.diagonal(C))
C = C - H
C = np.minimum(np.maximum(C, threshinf), threshsup)
C[C == threshsup] = 0
C[C != 0] = 1
return C
Example 9
def diagflat(v, k=0):
"""Returns a 2-d array with flattened v as diagonal.
Args:
v: array_like of any rank. Gets flattened when setting as diagonal. Could be
an ndarray, a Tensor or any object that can be converted to a Tensor using
tf.convert_to_tensor.
k: Position of the diagonal. Defaults to 0, the main diagonal. Positive
values refer to diagonals shifted right, negative values refer to
diagonals shifted left.
Returns:
2-d ndarray.
"""
v = asarray(v)
return diag(tf.reshape(v.data, [-1]), k)
Example 10
def get_like_from_mats(ky_mat, l_mat, alpha, name):
""" compute the likelihood from the covariance matrix
:param ky_mat: the covariance matrix
:return: float, likelihood
"""
# catch linear algebra errors
labels = _global_training_labels[name]
# calculate likelihood
like = (-0.5 * np.matmul(labels, alpha) -
np.sum(np.log(np.diagonal(l_mat))) -
math.log(2 * np.pi) * ky_mat.shape[1] / 2)
return like
#######################################
##### KY MATRIX FUNCTIONS and gradients
#######################################
Example 11
def zz(matrix, nb):
r"""Zig-zag traversal of the input matrix
:param matrix: input matrix
:param nb: number of coefficients to keep
:return: an array of nb coefficients
"""
flipped = np.fliplr(matrix)
rows, cols = flipped.shape # nb of columns
coefficient_list = []
for loop, i in enumerate(range(cols - 1, -rows, -1)):
anti_diagonal = np.diagonal(flipped, i)
# reversing even diagonals prioritizes the X resolution
# reversing odd diagonals prioritizes the Y resolution
# for square matrices, the information content is the same only when nb covers half of the matrix
# e.g. [ nb = n*(n+1)/2 ]
if loop % 2 == 0:
anti_diagonal = anti_diagonal[::-1] # reverse anti_diagonal
coefficient_list.extend([x for x in anti_diagonal])
# flattened = [val for sublist in coefficient_list for val in sublist]
return coefficient_list[:nb]
Example 12
def _fit(self, X, model_dump_path=None, verbose=True):
"""
:param X: shape (n_videos, n_frames, n_descriptors_per_image, n_dim_descriptor)
:param model_dump_path: (optional) path where the fitted model shall be dumped
:param verbose - boolean that controls the verbosity
:return: fitted Fisher vector object
"""
assert X.ndim == 4
self.feature_dim = X.shape[-1]
X = X.reshape(-1, X.shape[-1])
# fit GMM and store params of fitted model
self.gmm = gmm = GaussianMixture(n_components=self.n_kernels, covariance_type=self.covariance_type, max_iter=1000).fit(X)
self.covars = gmm.covariances_
self.means = gmm.means_
self.weights = gmm.weights_
# if cov_type is diagonal - make sure that covars holds a diagonal matrix
if self.covariance_type == 'diag':
cov_matrices = np.empty(shape=(self.n_kernels, self.covars.shape[1], self.covars.shape[1]))
for i in range(self.n_kernels):
cov_matrices[i, :, :] = np.diag(self.covars[i, :])
self.covars = cov_matrices
assert self.covars.ndim == 3
self.fitted = True
if verbose:
print('fitted GMM with %i kernels'%self.n_kernels)
if model_dump_path:
with open(model_dump_path, 'wb') as f:
pickle.dump(self,f, protocol=4)
if verbose:
print('Dumped fitted model to', model_dump_path)
return self
Example 13
def init_amps(self, eris):
time0 = time.clock(), time.time()
mo_e = eris.fock.diagonal()
nocc = self.nocc
eia = mo_e[:nocc,None] - mo_e[None,nocc:]
eijab = lib.direct_sum('ia,jb->ijab',eia,eia)
t1 = eris.fock[:nocc,nocc:] / eia
eris_oovv = np.array(eris.oovv)
t2 = eris_oovv/eijab
self.emp2 = 0.25*einsum('ijab,ijab',t2,eris_oovv.conj()).real
logger.info(self, 'Init t2, MP2 energy = %.15g', self.emp2)
logger.timer(self, 'init mp2', *time0)
return self.emp2, t1, t2
Example 14
def make_cholesky_unique(chol):
"""Make a lower triangular cholesky factor unique.
Cholesky factors are only unique with the additional requirement that all diagonal
elements are positive. This is done automatically by np.linalg.cholesky.
Since we calucate cholesky factors by QR decompositions we have to do it manually.
It is obvious from that this is admissible because:
chol sign_swither sign_switcher.T chol.T = chol chol.T
"""
sign_switcher = np.sign(np.diag(np.diagonal(chol)))
return chol @ sign_switcher
Example 15
def test_diagonal():
b1 = np.matrix([[1,2],[3,4]])
diag_b1 = np.matrix([[1, 4]])
array_b1 = np.array([1, 4])
assert_equal(b1.diagonal(), diag_b1)
assert_equal(np.diagonal(b1), array_b1)
assert_equal(np.diag(b1), array_b1)
Example 16
def test_diagonal(self):
a = [[0, 1, 2, 3],
[4, 5, 6, 7],
[8, 9, 10, 11]]
out = np.diagonal(a)
tgt = [0, 5, 10]
assert_equal(out, tgt)
Example 17
def get_correlations(cca, data_0, data_1):
"""
:param cca:
:param data_0:
:param data_1:
:return:
"""
x_scores, y_scores = get_cca_projection(cca, data_0, data_1)
corrs_tmp = np.corrcoef(x_scores.T, y_scores.T)
corrs = np.diagonal(corrs_tmp, offset=data_0.shape[1])
return corrs
Example 18
def compute_nrg(xframes):
n_frames = xframes.shape[1]
return np.diagonal(np.dot(xframes,xframes.T))/float(n_frames)
Example 19
def test_diagonal():
b1 = np.matrix([[1,2],[3,4]])
diag_b1 = np.matrix([[1, 4]])
array_b1 = np.array([1, 4])
assert_equal(b1.diagonal(), diag_b1)
assert_equal(np.diagonal(b1), array_b1)
assert_equal(np.diag(b1), array_b1)
Example 20
def test_diagonal(self):
a = [[0, 1, 2, 3],
[4, 5, 6, 7],
[8, 9, 10, 11]]
out = np.diagonal(a)
tgt = [0, 5, 10]
assert_equal(out, tgt)
Example 21
def metrics(confusions, ignore_unclassified=False):
"""
Computes different metrics from confusion matrices.
:param confusions: ([..., n_c, n_c] np.int32). Can be any dimension, the confusion matrices should be described by
the last axes. n_c = number of classes
:param ignore_unclassified: (bool). True if the the first class should be ignored in the results
:return: ([..., n_c] np.float32) precision, recall, F1 score, IoU score
"""
# If the first class (often "unclassified") should be ignored, erase it from the confusion.
if (ignore_unclassified):
confusions[..., 0, :] = 0
confusions[..., :, 0] = 0
# Compute TP, FP, FN. This assume that the second to last axis counts the truths (like the first axis of a
# confusion matrix), and that the last axis counts the predictions (like the second axis of a confusion matrix)
TP = np.diagonal(confusions, axis1=-2, axis2=-1)
TP_plus_FP = np.sum(confusions, axis=-1)
TP_plus_FN = np.sum(confusions, axis=-2)
# Compute precision and recall. This assume that the second to last axis counts the truths (like the first axis of
# a confusion matrix), and that the last axis counts the predictions (like the second axis of a confusion matrix)
PRE = TP / (TP_plus_FN + 1e-6)
REC = TP / (TP_plus_FP + 1e-6)
# Compute Accuracy
ACC = np.sum(TP, axis=-1) / (np.sum(confusions, axis=(-2, -1)) + 1e-6)
# Compute F1 score
F1 = 2 * TP / (TP_plus_FP + TP_plus_FN + 1e-6)
# Compute IoU
IoU = F1 / (2 - F1)
return PRE, REC, F1, IoU, ACC
Example 22
def IoU_from_confusions(confusions):
"""
Computes IoU from confusion matrices.
:param confusions: ([..., n_c, n_c] np.int32). Can be any dimension, the confusion matrices should be described by
the last axes. n_c = number of classes
:param ignore_unclassified: (bool). True if the the first class should be ignored in the results
:return: ([..., n_c] np.float32) IoU score
"""
# Compute TP, FP, FN. This assume that the second to last axis counts the truths (like the first axis of a
# confusion matrix), and that the last axis counts the predictions (like the second axis of a confusion matrix)
TP = np.diagonal(confusions, axis1=-2, axis2=-1)
TP_plus_FN = np.sum(confusions, axis=-1)
TP_plus_FP = np.sum(confusions, axis=-2)
# Compute IoU
IoU = TP / (TP_plus_FP + TP_plus_FN - TP + 1e-6)
# Compute mIoU with only the actual classes
counts = np.sum(1 - mask, axis=-1, keepdims=True)
mIoU = np.sum(IoU, axis=-1, keepdims=True) / (counts + 1e-6)
# If class is absent, place mIoU in place of 0 IoU to get the actual mean later
return IoU
Example 23
def test_diagonal():
b1 = np.matrix([[1,2],[3,4]])
diag_b1 = np.matrix([[1, 4]])
array_b1 = np.array([1, 4])
assert_equal(b1.diagonal(), diag_b1)
assert_equal(np.diagonal(b1), array_b1)
assert_equal(np.diag(b1), array_b1)
Example 24
def test_diagonal(self):
a = [[0, 1, 2, 3],
[4, 5, 6, 7],
[8, 9, 10, 11]]
out = np.diagonal(a)
tgt = [0, 5, 10]
assert_equal(out, tgt)
Example 25
def expected_durations(self):
"""
(array) Expected duration of a regime, possibly time-varying.
"""
return 1. / (1 - np.diagonal(self.regime_transition).squeeze())
Example 26
def logpdf_obs(self, x):
x = x - self.mean
x_whitened = self.whiten(x)
#sigmainv = linalg.cholesky(sigma)
logdetsigma = np.log(np.linalg.det(sigma))
sigma2 = 1. # error variance is included in sigma
llike = 0.5 * (np.log(sigma2)
- 2.* np.log(np.diagonal(self.cholsigmainv))
+ (x_whitened**2)/sigma2
+ np.log(2*np.pi))
return llike
Example 27
def mvn_nloglike_obs(x, sigma):
'''loglike multivariate normal
assumes x is 1d, (nobs,) and sigma is 2d (nobs, nobs)
brute force from formula
no checking of correct inputs
use of inv and log-det should be replace with something more efficient
'''
#Sturla: sqmahal = (cx*cho_solve(cho_factor(S),cx.T).T).sum(axis=1)
#Still wasteful to calculate pinv first
sigmainv = linalg.inv(sigma)
cholsigmainv = linalg.cholesky(sigmainv)
#2 * np.sum(np.log(np.diagonal(np.linalg.cholesky(A)))) #Dag mailinglist
# logdet not needed ???
#logdetsigma = 2 * np.sum(np.log(np.diagonal(cholsigmainv)))
x_whitened = np.dot(cholsigmainv, x)
#sigmainv = linalg.cholesky(sigma)
logdetsigma = np.log(np.linalg.det(sigma))
sigma2 = 1. # error variance is included in sigma
llike = 0.5 * (np.log(sigma2) - 2.* np.log(np.diagonal(cholsigmainv))
+ (x_whitened**2)/sigma2
+ np.log(2*np.pi))
return llike, (x_whitened**2)
Example 28
def mvn_loglike_chol(x, sigma):
'''loglike multivariate normal
assumes x is 1d, (nobs,) and sigma is 2d (nobs, nobs)
brute force from formula
no checking of correct inputs
use of inv and log-det should be replace with something more efficient
'''
#Sturla: sqmahal = (cx*cho_solve(cho_factor(S),cx.T).T).sum(axis=1)
sigmainv = np.linalg.inv(sigma)
cholsigmainv = np.linalg.cholesky(sigmainv).T
x_whitened = np.dot(cholsigmainv, x)
logdetsigma = np.log(np.linalg.det(sigma))
nobs = len(x)
from scipy import stats
print('scipy.stats')
print(np.log(stats.norm.pdf(x_whitened)).sum())
llf = - np.dot(x_whitened.T, x_whitened)
llf -= nobs * np.log(2 * np.pi)
llf -= logdetsigma
llf *= 0.5
return llf, logdetsigma, 2 * np.sum(np.log(np.diagonal(cholsigmainv)))
#0.5 * np.dot(x_whitened.T, x_whitened) + nobs * np.log(2 * np.pi) + logdetsigma)
Example 29
def mvn_nloglike_obs(x, sigma):
'''loglike multivariate normal
assumes x is 1d, (nobs,) and sigma is 2d (nobs, nobs)
brute force from formula
no checking of correct inputs
use of inv and log-det should be replace with something more efficient
'''
#Sturla: sqmahal = (cx*cho_solve(cho_factor(S),cx.T).T).sum(axis=1)
#Still wasteful to calculate pinv first
sigmainv = np.linalg.inv(sigma)
cholsigmainv = np.linalg.cholesky(sigmainv).T
#2 * np.sum(np.log(np.diagonal(np.linalg.cholesky(A)))) #Dag mailinglist
# logdet not needed ???
#logdetsigma = 2 * np.sum(np.log(np.diagonal(cholsigmainv)))
x_whitened = np.dot(cholsigmainv, x)
#sigmainv = linalg.cholesky(sigma)
logdetsigma = np.log(np.linalg.det(sigma))
sigma2 = 1. # error variance is included in sigma
llike = 0.5 * (np.log(sigma2) - 2.* np.log(np.diagonal(cholsigmainv))
+ (x_whitened**2)/sigma2
+ np.log(2*np.pi))
return llike
Example 30
def mean_photon(self, mode, **kwargs):
return mean, var `
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