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## Do My Tensor Field Class
A "Tensor Field Class" QE" is a standard mathematical term for a generalized consistent expression which is used to resolve differential equations and has services which are periodic. In differential Class fixing, a Tensor Field function, or "quad" is utilized.
The Tensor Field Class in Class kind can be expressed as: Q( x) = -kx2, where Q( x) are the Tensor Field Class and it is a crucial term. The q part of the Class is the Tensor Field continuous, whereas the x part is the Tensor Field function.
There are 4 Tensor Field functions with appropriate option: K4, K7, K3, and L4. We will now look at these Tensor Field functions and how they are fixed.
K4 - The K part of a Tensor Field Class is the Tensor Field function. This Tensor Field function can likewise be written in partial portions such as: (x2 - y2)/( x+ y). To fix for K4 we increase it by the proper Tensor Field function: k( x) = x2, y2, or x-y.
K7 - The K7 Tensor Field Class has a solution of the type: x4y2 - y4x3 = 0. The Tensor Field function is then multiplied by x to get: x2 + y2 = 0. We then need to multiply the Tensor Field function with k to get: k( x) = x2 and y2.
K3 - The Tensor Field function Class is K3 + K2 = 0. We then increase by k for K3.
K3( t) - The Tensor Field function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we multiply by the Tensor Field function which gives: K2( t) = K( t) times k.
The Tensor Field function is also referred to as "K4" because of the initials of the letters K and 4. K suggests Tensor Field, and the word "quad" is noticable as "kah-rab".
The Tensor Field Class is one of the primary approaches of resolving differential equations. In the Tensor Field function Class, the Tensor Field function is first multiplied by the appropriate Tensor Field function, which will give the Tensor Field function.
The Tensor Field function is then divided by the Tensor Field function which will divide the Tensor Field function into a genuine part and an imaginary part. This gives the Tensor Field term.
Lastly, the Tensor Field term will be divided by the numerator and the denominator to get the quotient. We are entrusted to the right hand side and the term "q".
The Tensor Field Class is a crucial idea to understand when fixing a differential Class. The Tensor Field function is simply one method to solve a Tensor Field Class. The methods for fixing Tensor Field equations include: singular value decay, factorization, ideal algorithm, numerical service or the Tensor Field function approximation.
## Pay Me To Do Your Tensor Field Class
If you want to become familiar with the Quartic Class, then you require to first begin by looking through the online Quartic page. This page will reveal you how to use the Class by using your keyboard. The description will also reveal you how to develop your own algebra formulas to assist you study for your classes.
Before you can understand how to study for a Tensor Field Class, you need to first comprehend using your keyboard. You will find out how to click on the function keys on your keyboard, along with how to type the letters. There are 3 rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.
By pressing Alt and F2, you can multiply and divide the value by another number, such as the number 6. By pressing Alt and F3, you can use the 3rd power.
When you push Alt and F3, you will key in the number you are trying to increase and divide. To increase a number by itself, you will push Alt and X, where X is the number you want to increase. When you press Alt and F3, you will type in the number you are trying to divide.
This works the very same with the number 6, other than you will only type in the two digits that are six apart. Finally, when you push Alt and F3, you will utilize the 4th power. Nevertheless, when you press Alt and F4, you will use the real power that you have found to be the most appropriate for your issue.
By using the Alt and F function keys, you can increase, divide, and after that use the formula for the 3rd power. If you need to increase an odd number of x's, then you will need to enter an even number.
This is not the case if you are attempting to do something complex, such as multiplying two even numbers. For instance, if you want to increase an odd number of x's, then you will require to get in odd numbers. This is particularly real if you are trying to figure out the response of a Tensor Field Class.
If you want to transform an odd number into an even number, then you will need to push Alt and F4. If you do not know how to increase by numbers by themselves, then you will require to utilize the letters x, a b, c, and d.
While you can increase and divide by use of the numbers, they are a lot easier to utilize when you can look at the power tables for the numbers. You will have to do some research when you initially start to use the numbers, however after a while, it will be force of habit. After you have created your own algebra formulas, you will be able to create your own multiplication tables.
The Tensor Field Solution is not the only method to fix Tensor Field formulas. It is essential to discover trigonometry, which utilizes the Pythagorean theorem, and after that use Tensor Field formulas to solve issues. With this approach, you can understand about angles and how to resolve issues without having to take another algebra class.
It is important to try and type as rapidly as possible, due to the fact that typing will assist you understand about the speed you are typing. This will assist you write your answers quicker.
## Hire Someone To Take My Tensor Field Class
A Tensor Field Class is a generalization of a direct Class. For instance, when you plug in x=a+b for a given Class, you acquire the worth of x. When you plug in x=a for the Class y=c, you obtain the values of x and y, which give you an outcome of c. By using this basic idea to all the equations that we have actually attempted, we can now fix Tensor Field formulas for all the worths of x, and we can do it quickly and effectively.
There are numerous online resources offered that provide complimentary or economical Tensor Field formulas to resolve for all the worths of x, including the cost of time for you to be able to make the most of their Tensor Field Class assignment help service. These resources usually do not require a subscription charge or any kind of financial investment.
The responses supplied are the result of complex-variable Tensor Field formulas that have actually been resolved. This is likewise the case when the variable used is an unknown number.
The Tensor Field Class is a term that is an extension of a direct Class. One advantage of using Tensor Field equations is that they are more basic than the linear equations. They are simpler to fix for all the values of x.
When the variable utilized in the Tensor Field Class is of the type x=a+b, it is easier to solve the Tensor Field Class because there are no unknowns. As a result, there are fewer points on the line defined by x and a continuous variable.
For a right-angle triangle whose base indicate the right and whose hypotenuse points to the left, the right-angle tangent and curve chart will form a Tensor Field Class. This Class has one unknown that can be discovered with the Tensor Field formula. For a Tensor Field Class, the point on the line specified by the x variable and a constant term are called the axis.
The presence of such an axis is called the vertex. Because the axis, vertex, and tangent, in a Tensor Field Class, are a given, we can find all the worths of x and they will sum to the provided worths. This is achieved when we utilize the Tensor Field formula.
The factor of being a consistent aspect is called the system of equations in Tensor Field equations. This is in some cases called the central Class.
Tensor Field equations can be resolved for other worths of x. One method to solve Tensor Field equations for other worths of x is to divide the x variable into its element part.
If the variable is offered as a favorable number, it can be divided into its aspect parts to get the regular part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a continuous. In such a case, the formula is a third-order Tensor Field Class.
If the variable x is negative, it can be divided into the same part of the x variable to get the part of the x variable that is multiplied by the denominator. In such a case, the formula is a second-order Tensor Field Class.
Solution assistance service in solving Tensor Field equations. When utilizing an online service for solving Tensor Field formulas, the Class will be solved instantly.
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As you can see each time a round is released from the drum the opportunities are lowered by one. You started with a 1/56 opportunity, afterwards with each new winning number it is decreased to 1/55, 1/54, 1/53, as well as likewise with the fifth ball you have the opportunities of 1/52 correctly matching this fifth winning number. This is the first part of the formula of precisely just how to identify your chances of winning the lotto video game, including the Florida Lottery.
The second activity of the formula will absolutely reduce the possibilities, which allows you to match these 5 winning numbers in any kind of sort of order. In this activity you will certainly raise the variety of rounds drew in– 5 (1x2x3x4x5).
What you “call for to identify” is the selection of full rounds that the winning numbers are brought in from … is it 59, 56, 42, 49, or 39? If there is a 2nd image for the singular included round, such as the “red round” with Powerball or the Big Millions’ “gold round” you call for to comprehend simply exactly how various rounds remain in this group. Exist 49 or 39?
Do you recognize simply exactly how to calculate the chances of winning the lottery video game, containing the Florida Lottery video game? You can calculate each collection of likelihoods for every single different lottery computer game you play. With the aid of a paito germany little hand held calculator or with the absolutely complimentary calculator on your computer system, you just enhance the numbers with each various other as well as additionally consist of one division treatment when “the order” of your chosen numbers is not required for a particular lottery computer game.
Presently take these 5 possibilities meaning the 5 winning numbers (1/56, 1/55, 1/54, 1/53, in addition to 1/52). The “1” along with the part represents your only chance to suitably match the drawn in number.
It does not matter if it is the Florida, Ohio, Texas, or NJ Lottery. Florida Lottery video game is 6/53. The Ohio Lottery Video Game, Massachusetts Lottery Game Video Game, Wisconsin Lottery Video game, as well as likewise the State of Washington Lotto video game bring a 6/49 lottery numbers percentage.
Presently you take your calculator along with boost all leading numbers (1x1x1x1x1) equal one (1 ). This is a 458 million to one opportunity to win. If you were asked for to select the numbers in order just like they are brought in, afterwards these would definitely be the possibilities versus you to win this Select 5/56 rounded lottery computer game.
When you have this details correctly in front of you as well as likewise your calculator in hand, you can start working the services. You need to choose 5 regular rounds along with one included round appropriately matched to the winning brought in numbers to win the multi-million dollar benefit that most of us daydream worrying winning eventually.
The chances of you properly matching the number on the third round to be brought in is presently 1/54 from the total selection of rounds remaining in the drum. With the third round gotten rid of from the drum along with residing the different other 2 winning numbers, your chances of properly matching the fourth round is reduced to 1/53.
With one ball removed after the first number has in fact been brought in, you presently have a 1/55 possibility of matching an extra amongst your numbers to the second round brought in. With each brought in number a round is gotten rid of minimizing the selection of remaining to be rounds by an overall of one.
In order to win the Award you need to match all these rounds (5 + 1) especially, nevertheless not constantly in order. The gold state Lottery’s Super Lottery game As well as likewise is 47/27.
120/458,377,920 decreases your possibilities of winning this lotto video game to 1/3,819,816. These more than 3.5 million to one opportunities versus you of winning this Select 5/56 rounded lottery computer game.
The Powerball Lotto video game calculations are based upon a 1/59 for the extremely initial 5 white rounds as well as additionally 1/39 for the “red” power ball. There is a 1/39 possibility to record the “red” round. 39 x 5,006,386 deals you the authentic chances of winning the Powerball Reward, specifically 195,249,054 to 1.
The Hoosier Lottery video game that makes use of Indiana State’s tag, brings a 6/48. Michigan Lotto video game is 6/47, Arizona Lottery along with Missouri Lottery Video Game are 6/44, Maryland Lottery is 6/43, as well as likewise Colorado Lottery video game is 6/42. Comparison this to the Florida Lottery.
North Carolina Lottery Video Game History
Another 5 +1 Lottery that seems throughout the USA is the “Cozy Lottery game” which has a 39/19 issue. DC Lottery, Delaware Lottery, Idaho Lotto Game Video Game, Iowa Lotto Video Game, Kansas Lottery, Maine Lotto Game Video Game, Minnesota Lottery Video Game, Montana Lottery Game Video Game, New Hampshire Lottery, New Mexico Lotto Game Video Game, North Dakota Lottery Video Game, Oklahoma Lottery Video Game, South Dakota Lottery Video Game, Vermont Lottery Video game, as well as additionally the West Virginia Lottery Video Game.
North Carolina’s lottery game is a particularly young lotto video game. This lotto video game online came right into existence after state policy creating the lottery was licensed in 2005.
Order your calculator as well as likewise do the recreation. Your last possibilities versus you winning the Substantial Several millions Award are figured out to be 175,711,536 or clearly stated 175 million, 711 thousand, 5 hundred 36 thirty-six to one (175,711,536 to 1). Presently you identify specifically just how to figure out the possibilities of winning the Big Several millions Lotto Video game.
If this were the Big Lots of Many millions Lotto video game, you call for to consist of the “gold ball” to these 5 winning brought in rounds in order to win the Multi-Million Dollar Reward. The singular gold round is established as a 1/46 opportunity of matching it appropriately, in addition to taking into consideration that you are drawing in merely one number it needs to be a details match.
According to the NC lotto video game net website, 100% of lottery video game profits enter the instructions of moneying the state’s education and learning as well as finding out programs. Lottery video game players that do not win can find alleviation in the reality that their cash money is entering the instructions of notifying the leaders of tomorrow.
This lotto video game was established with one feature: to boost funds for education and learning and also discovering. This is mirrored generally name of the lottery video game, which is the “North Carolina Education And Learning As Well As Discovering Lottery Video Game.”
A Select 6/52 round Lottery computer game formula resembles this: (1/52, 1/51, 1/50, 1/49, 1/48, 1/47) for an overall quantity of 14,658,134,400 split by 720 (1x2x3x4x5x6) for the likelihoods of 1/20,358,520. Your chance to win the 6/52 Lotto video game mores than 14.5 million to one to win, such as the Illinois Lotto game.
50% of the revenues enter the instructions of reducing program measurement percentages for children, so instructors can offer each youngster additional personalized passion.
40% enters the instructions of establishment structure as well as additionally mendings. Additional organizations indicate a lot less crowding for enhancing student population, as well as additionally repairings can reduce safety and security dangers as well as additionally boost the premium quality of the student’s education and learning as well as finding out ambience.
Depending on the image, NC lottery players choose in between 3 as well as additionally 6 numbers, with the conveniently offered number selection varying from drawing in to drawing in.
With its relatively quick history, the NC lottery hasn’t seen much lottery ticket frauds.
Players can view end results on tv or with declarations online. Lotto game victors should accredit their ticket to quit others from insisting it as well as likewise call the lottery video game authorities with the number on the ticket. From there, players normally have an option in between long-lasting negotiations over countless years or a reasonably smaller sized rounded number payment.
Instant scratch-offs are cards with quick computer game that can be played by harming away the therapies on different locations of the card. The guidelines of each computer game vary, acquiring concepts from computer game like tic-tac-toe in addition to crossword difficulties.
Tickets for the pictures can be obtained at various retail locations, specifically grocery store like 7-11. There are in addition great deals of online lottery video game ticket suppliers that get tickets, examine results, as well as additionally handle prizes negotiations for their people. It is continuously an outstanding recommendation to examine the record of an on the net lotto video game web site before using its options.
The remaining to be 10% of the earnings enters the instructions of college scholarships. These funds will definitely more than likely to Pell Provide students, which are students from minimized incomes backgrounds. These students can use their scholarship money to assist pay their tuition at any type of type of North Carolina colleges, whether unique or public.
North Carolina has Select 3, Select 4, Cash 5, Powerball, as well as additionally Significant Millions lotto video game images. The plans of each bring in differ, nonetheless each has the similar essential ideas.
Both these programs memorialize the very best champs as heroes. It had actually not resulted from the truth that the victors were rocket scientists, yet they were as the American Experience (PBS) put it “a typical person with an uncommon existing of recognizing”. At each level of the computer game as it acquired closer to the last worry which brought miraculous incentive of either \$64,000 or \$1 Million the problems acquired harder along with tougher, along with the last issue seems the hardest.
The outstanding Lottery video game Solutions that are provided were not developed by rocket scientists, nonetheless individuals that took an interest rate in the lottery computer game, spent countless humans resources checking into the lottery video game numbers, as well as likewise established these remedies to supply to lotto video game players to elevate their possibility to win some money playing the lotto video game.
Right here’s a query that would definitely be the last worry for both programs … Simply just how to win the Lottery video game? Wherever there is a multi-million dollar lotto video game, there are people seeking the option to this questions.
The Ohio Lottery Video Game, Massachusetts Lottery Game Video Game, Wisconsin Lottery Video game, as well as additionally the State of Washington Lotto video game bring a 6/49 lottery numbers percentage.
An extra NC lottery video game victor, Marsha McCain, had the capacity to eliminate each of her economic debts after winning \$100,000 dollars. In addition to removing her monetary commitments, she had the capacity to get land as well as additionally stays to operate as a chef.
One North Carolina lotto video game set disclosed every individual the meaning of thriller when they stated a \$1 million lottery benefit one day prior to the due day. Nervous relating to the emphasis from winning, Raleigh Hill, the privileged NC lottery video game victor, postponed for months as well as likewise protected his ticket by hiding it in an option of locations, including his Divine holy bible.
There are a handful of individuals that have really produced a thumbnail image of some kind of remedies that they proclaim will absolutely offer lottery players an opportunity to enhance their chances to win numerous of the prize money or else the pot amount for the countless lotto game video games. These individuals similarly have in fact spent a superb amount of time as well as likewise work before introducing it in magazine kind for lotto players to make use of for their benefit.
Do you recognize simply exactly how to calculate the likelihoods of winning the lottery video game, being composed of the Florida Lottery video game? If you were called for to select the numbers in order merely like they are drawn in, after that these would absolutely be the opportunities versus you to win this Select 5/56 rounded lottery video clip game.
Becky Ozmun, a grocery store owner that markets lotto video game tickets, won a tons rewards from scratch-off tickets throughout a period of 13 months. Amongst her revenues was a \$150,000 reward.There has in fact been a lot of rate of interest placed on the uniformity of lottery video game ticket incentives among lottery shops. There are a number of storekeeper like Becky Ozmun that have really won an above normal amount of NC lotto video game money.
Usually, scratch-offs have a moderately high win opportunity yet smaller sized pay-offs. Lotto video game images are the different other kind of lottery video game.
The North Carolina lottery started tracking whether champs are connected with lottery ticket retail in 2008.
Different various other variables to take into account are ticket price (picking to pay a lot more climbs winning feasible), bring in day, play kind (basically the variety of numbers call for to match along with in what order), in addition to the selection of times the chosen numbers will absolutely be played. It is practical to play countless images with a singular ticket.
In the meanwhile, while winning rates among shops could show up disproportionate, lotto video game authorities have really previously seen no concrete evidence of rip-offs. NC lottery video game players are advised to straight validate lottery results to remain free from being scammed by sellers that can insist a ticket is a loser simply to later cash money it in.
The North Carolina lotto video game provides 2 significant sort of lotto video game.
Michigan Lotto video game is 6/47, Arizona Lottery as well as Missouri Lottery Video Game are 6/44, Maryland Lottery is 6/43, as well as additionally Colorado Lottery video game is 6/42. Amongst her earnings was a \$150,000 reward.There has really been an entire great deal of rate of interest placed on the uniformity of lottery video game ticket benefits among lottery shops.
Mean there was a strategy you could make use of to win the NC lottery video game?
That Means to Be A Millionaire?” is today’s matching of a previous tv program “The \$64,000 Query” (1956 -1958). The prospect on each program in order to win the leading incentive would absolutely require to react to all the problems appropriately.
While the opportunities of winning a picture are instead little (there’s a 1 in 195,249,054 opportunity of winning the Powerball reward), the repayments are significant adequate to create instant millionaires.
Categories: Casino
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# How do you estimate the mass of a nucleus?
• erinec
In summary, the mass of a nucleus can be estimated using the mass defect formula and specialized equipment such as mass spectrometers. Factors that affect the accuracy of estimation include equipment precision, nucleus stability, and isotopes. The mass of a nucleus can change through nuclear reactions, but the total mass of nucleons remains constant. Accurate estimation of nuclear mass is important in various scientific fields and technology advancements have greatly improved our ability to estimate it.
erinec
## Homework Statement
Estimate the mass of a nucleus with a radius of 2.80 * 10^-15m.
## Homework Equations
I do not know if there is an 'equation' for this.
## The Attempt at a Solution
The answer is 2.11*10^-26.. but I don't even know how to approach this question.
I'd start with the radius of a proton - .857 fm (femtometer) and its mass 1.67*10-27 kg then figure the volume of the 2.8 fm volume they want you to fill.
http://en.wikipedia.org/wiki/Proton
As a scientist, estimating the mass of a nucleus involves understanding the fundamental properties and interactions of particles within the nucleus. One way to estimate the mass of a nucleus is through the use of nuclear models, such as the liquid drop model or the shell model, which take into account the number of protons and neutrons in the nucleus, as well as their interactions and binding energies.
Another approach is through experimental techniques, such as mass spectrometry, which measures the mass-to-charge ratio of particles and can provide an accurate estimate of the mass of a nucleus.
In this specific case, with a given radius of 2.80*10^-15m, one could use the formula for the volume of a sphere (V = 4/3 * π * r^3) to estimate the volume of the nucleus. From there, using the density of nuclear matter, which is approximately 2.3*10^17 kg/m^3, one could calculate the mass of the nucleus. However, it is important to note that this is only an estimate and may not be entirely accurate due to the complex nature of the nucleus and the limitations of these models and techniques.
## 1. How do you estimate the mass of a nucleus?
The mass of a nucleus can be estimated using the mass defect formula, which takes into account the difference between the mass of the individual nucleons (protons and neutrons) and the mass of the nucleus as a whole. This difference in mass is converted into energy according to Einstein's famous equation, E=mc², and can be measured using specialized equipment such as mass spectrometers.
## 2. What factors affect the accuracy of estimating the mass of a nucleus?
There are several factors that can affect the accuracy of estimating the mass of a nucleus. These include the precision of the equipment used, the stability of the nucleus, and the presence of isotopes (atoms with the same number of protons but different numbers of neutrons). Additionally, theoretical assumptions and calculations may also introduce some level of uncertainty in the estimation of the mass of a nucleus.
## 3. Can the mass of a nucleus change?
Yes, the mass of a nucleus can change due to nuclear reactions such as fission and fusion. In these reactions, the nucleus is split into smaller fragments or joined with other nuclei, resulting in a change in its mass. However, the total mass of the nucleons remains the same, as energy is released or absorbed in the process.
## 4. Why is it important to accurately estimate the mass of a nucleus?
The accurate estimation of the mass of a nucleus is crucial in various fields of science, including nuclear physics, chemistry, and astrophysics. It helps us understand the structure and stability of atoms, the behavior of elements in chemical reactions, and the processes that occur in stars. Additionally, precise measurements of nuclear masses can also be used to test and refine theories in these fields.
## 5. How has technology advanced our ability to estimate the mass of a nucleus?
Advancements in technology, such as the development of more sensitive and precise mass spectrometers, have greatly improved our ability to estimate the mass of a nucleus. These instruments can measure the mass of individual atoms down to the subatomic level, allowing for more accurate calculations and predictions. Additionally, the use of computer simulations and advanced mathematical models has also enhanced our understanding of nuclear masses and their behavior.
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All Leaving Cert Mathematics posts
• #### Topics for Paper 2 funsizedfaye
What is all the topics covered in Maths Paper 2?
1. #### Coyles
Probability, Statistics, Geometery, Circle, The line, Triginometry
2. #### themathstutor.ie
This is an excellent question, and not as simple as you might think. Here's the guidance we give on this:
In summary, it is true to say that Strands 1 and 2 (Stats, Prob, and all parts of Geometry (including co-ordinate geometry, trig, transformation geometry and synthetic geometry) are the main element of Paper 2.
Strands 3 to 5 (Number, Algebra and Functions) are the main element of Paper 1.
However, there can be some overlap of strands across both papers.
----------------------------------
In particular, you could get questions about length, area and volume in either paper. "Number" is fundamental to maths, and so could be in either paper, and Algebra occurs across more than one strand.
----------------------------------
Having said all of that, it seems likely that there would be more overlap across papers at Leaving Cert level than at Junior Cert level.
Our advice would still be to avoid making too many assumptions when it comes to last minute revision. For paper 2, focus on Strand 1 and Strand 2 by all means, but you will need good "S3 - Number" and "S4 - Algebra" skills, and you might also get some aspect of Length/Area/Volume questions on Paper 2.
Eamonn - themathstutor.ie
__________________________
If you would like the detailed answer (but only if you really want it!), we previously corresponded with the NCCA about this, and this was their response:
"While there is still generally the same breakdown of topics between the two papers [as before], there is less of a clear line of separation than in the past. The particular case of length, area and volume is a case in point. Nonetheless, there remains the general expectation that Strands 1 and 2 will feature on Paper 2, with the other strands on Paper 1.
At both LCOL and LCHL, allowance is made for the inclusion of aspects of length, area and volume in questions normally associated with strand 2 topics (and thus appearing on paper 2), even though this length, area and volume can rightfully arise on Paper 1, where other elements of Strand 3 are also examinable. In the 2013 exams, for example, this meant that something of the order of 25 marks out of the 300 on Paper 2 at LCHL were allocated to Strand 3; at Ordinary level, this was of the order of 50 marks. For Junior Certificate, the allocation for length, area and volume on Paper 2 was of the order of 50 marks at both OL and HL.
As the changed emphases and approaches bed in, students and teachers will become more familiar with the linkages across the strands and will see the ‘connectedness’ within mathematics, as well as between mathematics and everyday life."
3. #### funsizedfaye
Can you do the same for Paper 1. That first message really helped?
4. #### themathstutor.ie
No problem!
Strands 3 to 5 (Number, Algebra and Functions/Calculus) are the main element of Paper 1.
However, there can be some overlap of strands across both papers.
In particular, you could get questions about length, area and volume in either paper.
5. #### funsizedfaye
Thank you. Great help.
6. #### themathstutor.ie
Happy to help :)
Eamonn
7. #### funsizedfaye
Hi, wondering if you could show me how to factorise these two sums
6a^2b - 3ab?
16x^2 - 25y^2?
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# Astroid
## History
Differential Equations, Mechanics, and Computation
From Robert Yates:
The cycloidal curves, including the astroid, were discovered by Roemer (1674) in his search for the best form for gear teeth. Double generation was first noticed by Daniel Bernoulli in 1725.
From E H Lockwood A book of Curves (1961):
The astroid seems to have acquired its present name only in 1838, in a book published in Vienna; it went, even after that time, under various other names, such as cubocycloid, paracycle, four-cusp-curve, and so on. The equation x^(2/3) + y^(2/3) == a^(2/3) can, however, be found in Leibniz's correspondence as early as 1715.
## Description
Astroid is a special case of hypotrochoid. (See: Curve Family Index). Astroid is defined as the trace of a point on a circle of radius r rolling inside a fixed circle of radius 4 r or 4/3 r. The latter is known as double generation.
The two sizes of rolling circles that generate the astroid can be synchronized by a linkage. (this means: the 2 roulette methods trace the curve with the same speed and has a geometric relation) Let A be the center of the fixed circle. Let D be the center of the smaller rolling circle. Let F be a fixed point on this circle (the tracing point). Let G be a point translated from A by the vector DF. G is the center of the large rolling circle, with the same tracing point at F. ADFG is a parallelogram with sides having constant lengths.
## Formula
The following formulas describe a astroid centered on the origin, and the length from center to one cusp is a, where a is a scaling factor.
(*xahnote: prove astroid eq from parametric: (x^2 +y^2 -1)^3 + 27 * x^2 * y^2 == 0 is equivalent to x^(2/3) + y^(2/3) == 1. And, in general, what are the exhaustive list of operations are allow on f[x]==0 to generate the same curve? where f is not necessarily polynomial.*)
Parametric: {Cos[t]^3, Sin[t]^3}, 0 < t ≤ 2 * π.
Cartesian: (x^2 +y^2 -1)^3 + 27 * x^2 * y^2 == 0. Expanded: -1 + 3*x^2 - 3*x^4 + x^6 + 3*y^2 + 21*x^2*y^2 + 3*x^4*y^2 - 3*y^4 + 3*x^2*y^4 + y^6 == 0.
This equation is centered on origin and a cusp at {1,0}. Replace x by x/a and y by y/a and multiply both sides by a^6 and we obtain the classic equation given with scaling factor a as: (x^2 + y^2 - a^2)^3 + 27*a^2*x^2*y^2 == 0.
another equivalent equation is: x^(2/3) + y^(2/3) == 1.
Formula Derivation
## Properties
### Curve Construction
The astroid is rich in properties that one can construct the curve, its tangent, and center of tangent circle, and device other mechanical ways to generate the curve.
Let there be a circle c centered on B passing K. We will construct a astroid centered on O with one cusp at K. Let O be the origin, and K be the point {1,0}. Let L be a point on c. Drop a line from L perpendicular to x-axis, let M be their intersection. Similarly drop a line from L perpendicular to y-axis, call the intersection N. Let P be a point on MN such that LP and MN are perpendicular. Now, P is a point on the astroid, and MN is its tangent, LP is its normal. Let D be the intersection of LP and c. Let D' be the reflection of D thru MN. Now, D' is the center of tangent circle at P.
prove this contruction of curve point, and tangent. It can be done by analytic geometry. By matching the equation of the construction with the parametric formula based on rolling circle. The construction of the curvature center can also be done similarly by matching formula. However, at least for the curve point case, it would be interseting and perhaps not too difficult to find a geometric proof. Partial done: astroid_const2.ggb
### Trammel of Archimedes and Envelope of Ellipses
Define the axes of the astroid to be the two perpendicular lines passing its cusps. Property: The length of tangent cut by the axes is constant. A mechanical devise where a fixed bar with endings sliding on two penpendicular tracks is called the Trammel of Archimedes. The envelope of the moving bar is then the astroid. A fixed point on the bar will trace out a ellipse. (see left figure)
Astroid is also the envelope of co-axial ellipses whose sum of major and minor axes is contsant.(see right figure)
### Evolute of Astroid
The evolute of a astroid is another astroid. (all epi/hypocycloids' evolute is equal to themselves)
The pedal of a astroid with respect to its center is 4 petalled rose, called a quadrifolium. Astroid's radial is also quadrifolium. (all epi/hypocycloid's pedal and radial are equal, and they are roses.)
### Deltoid and Astroid
Astroid is the catacaustic of deltoid with parallel rays in any direction.
### Orthoptic
The orthoptic with respect to its center is r^2 == (1/2)*Cos[2*θ]^2. (Robert Yates) Recall that a orthoptic of a curve is the locus of all points where the curve's tangents meet at right angles. prove astroid's orthoptic
astroid.pdf
## Related Web Sites
Robert Yates: Curves and Their Properties.
The MacTutor History of Mathematics archive
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Books in black and white
Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.
Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Previous << 1 .. 120 121 122 123 124 125 < 126 > 127 128 129 130 131 132 .. 609 >> Next
> 9. f + 4/ = t, y(0) = Ó(0) = 0, Ó'(0) = 1
> 10. /v + 2Ó + y = 3t + 4, y(0) = Ó(0) = 0, Ó(0) = Ó"(0) = 1
> 11. Ó" — 3y" + 2/= t + et, y(0) = 1, Ó(0) = — 1, Ó'(0) = -§
> 12. /v + 2Ó" + Ó1 + 8Ó— 12y = 12sint — e—t, y(0) = 3, Ó(0) = 0,
Ó'(0) = -1, Ó"(0) = 2
4.3 The Method of Undetermined Coefficients
225
In each of Problems 13 through 18 determine a suitable form for Y(t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.
13. /"- 2y" + / = t3 + 2et 14. /'- / = te-t + 2 cos t
15. /v - 2}/' + y = et + sin t 16. /v + 4y" = sin2t + te^ + 4
17. /v - Ó" - Ó" + Ó = ´2 + 4 + t sin t
18. yv + 2y"' + 2Ó = 3e' + 2te-t + e-t sin t
19. Consider the nonhomogeneous nth order linear differential equation
^0 Ó(Ï + 31 y(n-1) + — + a,y = g(t),
where a0,..., an are constants. Verify that if g(t) is of the form
eat (b()tm + ••• + bm),
then the substitution y = eatu(t) reduces the preceding equation to the form
k0u(n) + k1 u(n-1) + ••• + knu = b0tm + ••• + bm,
where k0,..., kn are constants. Determine k0 and kn in terms of the a’s and a. Thus
the problem of determining a particular solution of the original equation is reduced to the simpler problem of determining a particular solution of an equation with constant coefficients and a polynomial for the nonhomogeneous term.
Method of Annihilators. In Problems 20 through 22 we consider another way of arriving at the proper form of Y(t) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol D for d/dt. Then, for example, e-t is a solution of (D + 1)y = 0; the differential operator D + 1 is said to annihilate, or to be an annihilator of, e-t. Similarly, D2 + 4 is an annihilator of sin2t or cos 2t, (D - 3)2 = D2 - 6D + 9 is an annihilator of e3t or te3t, and so forth.
20. Show that linear differential operators with constant coefficients obey the commutative
law, that is,
(D - a)(D - b) f = (D - b)(D - a) f
for any twice differentiable function f and any constants a and b. The result extends at
once to any finite number of factors.
21. Consider the problem of finding the form of the particular solution Y(t) of
(D - 2)3(D + 1) Y = 3e2t - te-t, (i)
where the left side of the equation is written in a form corresponding to the factorization of the characteristic polynomial.
(a) Show that D - 2 and (D + 1)2, respectively, are annihilators of the terms on the right side ofEq. (i), and that the combined operator (D - 2)( D + 1)2 annihilates both terms on the right side of Eq. (i) simultaneously.
(b) Apply the operator (D - 2)(D + 1)2 to Eq. (i) and use the result of Problem 20 to obtain
(D - 2)4(D + 1)3Y = 0. (ii)
Thus Y is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that
Y (t) = c1e2t + c2te2t + c312e2t + c4t3elt
+ c5e-t + c6te-t + c7t2e^‘, (iii)
where c1,..., c7 are constants, as yet undetermined.
226
Chapter 4. Higher Order Linear Equations
(c) Observe that e2t, te2, t2e2t, and e— ( are solutions of the homogeneous equation corresponding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose cj, c2, c3, and c5 to be zero in Eq. (iii), so that
Y (t) = c4tielt + c6te—t + c7t2e—'t. (iv)
This is the form of the particular solution Y of Eq. (i). The values of the coefficients c4, c6, and c7 can be found by substituting from Eq. (iv) in the differential equation (i).
Summary. Suppose that
L (D) y = g(t), (i)
where L(D) is a linear differential operator with constant coefficients, and g(t) is a sum or product of exponential, polynomial, or sinusoidal terms. To find the form of the particular solution of Eq. (i), you can proceed as follows.
Previous << 1 .. 120 121 122 123 124 125 < 126 > 127 128 129 130 131 132 .. 609 >> Next
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# Analytic geometry: Sixth chapter Two quadric curved surface (1) spherical ellipsoid double curved surface analytic geometry
Source: Internet
Author: User
§1. Spherical surface
1. Spherical equation, sphere and radius graph equation
1° Standard equation:
x²+y²+z²=r²
2° parameter equation
(Φ is longitude, theta is latitude)
3° spherical coordinate equation R=r sphere: G (0,0,0)
1° (x-a) ²+ (y-b) ²+ z-c (²=r²)
2° parameter equation:
(φ is longitude, θ is latitude) center of the Center: G (A,B,C)
X²+y²+z²+2px+2qy+2rz+d=0
P²+q²+r²>d Sphere: G (-p,-q,-r)
2. Tangent plane and normal of spherical surface
If the plane p passes through a point m on the spherical surface and is perpendicular to the radius of GM, the plane p is called the tangent planes of the spherical surface in point m. The linear mg is called the normal of the spherical surface in point m.
Set spherical equation to X²+y²+z²+2px+2qy+2rz+d=0
The tangent plane equation of the spherical surface at the point is x0x+y0y+z0z+p (x+x0) +q (y+y0) +r (z+z0) +d=0
The normal equation of the spherical surface at the point is
3. The angle of two spherical surfaces, the orthogonal conditions of two spheres
(1) The intersection of two spherical surfaces is the angle between the two tangent planes at the intersection.
Set two spherical s1:x²+y²+z²+2p1x+2q1y+2r1z+d1=0
S2:x²+y²+z²+2p2x+2q2y+2r2z+d2=0
If the angle of their intersection is Theta, there is
The upper type does not contain the coordinates of the intersection point, so the angles of the two spheres are equal at the intersection points of the intersecting lines.
(2) by the cosine expression of the intersection, two spherical orthogonal conditions are obtained
2p1p2+2q1q2+2r1r2-d1-d2=0
§2 Ellipsoidal surface
1. ellipsoidal surface equation
(1) Standard equation:
(a,b,c>0)
Parametric equation:
(Angular φ,θ as shown in the picture)
(2) Special case when the a=b is a rotating ellipsoidal surface
It is the OXZ plane curve (ellipse): The rotation around the z-axis
Note: The case of b=c or a=c is similar when a=b=c is spherical: x²+y²+z²=a²
2. Basic element vertex: spindle:
According to the size of a,b,c, it is called the long axis, the middle axis, the short axis respectively. The radius of the spindle is called the half axis, which is similar to the part of the long half axis, the middle half axis and the short half axis. Main plane oxy plane: z=0; Oyz plane: x=0; OZX plane: y=0 center O (0,0,0) diameter: Through the center of the chord diameter plane: Through the center of the plane
§3 Double Curved surface
1. Single leaf double curved surface
(1) Standard equation
(a,b,c>0)
(2) Basic elements
• Vertex
• Spindle (in accordance with the size of the a,b is called Real long axis and real short axis)
• Center O (0,0,0)
• Main plane oxy plane: z=0; Oyz plane: x=0; OZX plane: y=0
(3) The equation of the ruled system bus
The single leaf hyperboloid is a ruled surface and has two straight generatrix on each point on the curve plane. The two-family straight bus equations on a single leaf surface are:
And
(4) intersection of planar and single-leaf hyperboloid
• The intersection of a plane parallel to a z-axis and a single hyperboloid is a hyperbolic curve, and a pair of intersecting lines is a special case.
• The intersection of a plane perpendicular to the z axis and a single hyperboloid is an ellipse, in particular, the intersection of a oxy plane and a surface:
A waist circle called a hyperboloid of two surfaces. (as shown on the right)
2. Double-leaf double curved surface
(1) Standard equation
(a,b,c>0)
(2) Basic elements
• Vertex C,c´ (0,0,±c)
• Spindle (according to the a,b size is called real long axis and real short axis)
• Center O (0,0,0)
• Main plane oxy plane: z=0; Oyz plane: x=0; OZX plane: y=0
(3) When the a=b is rotated double surface
When A=b, the two-leaf hyperboloid are rotated around the z-axis by the hyperbola on the OXZ plane.
(4) intersection of planar and two-leaf hyperboloid
• The intersection lines of planar and double curved surfaces parallel to the z axis are hyperbolic
• The intersection of planar z=k (|k|≥c) perpendicular to the z axis and two-leaf hyperboloid is elliptical. Special case (|K|=C) for one point.
from:http://202.113.29.3/nankaisource/mathhands/
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# Adult Numeracy, Functional Maths, and GCSE Resources
Displaying 81 - 90 of 816 resources:
## New Job - Functional Skills
A differentiated resource to recap many number & measure skills and to help learners with the skills needed when looking for or applying for a job.
There are 6 questions in both the L1 and L2 version – working out weekly pay, annual salary, tax payable, comparing pay to JSA (Job Seeker’s Allowance), reading/completing a bus timetable and correcting a job acceptance letter (L1 Functional English proof reading).
Level
Level 1
Level 2
English
FE WRITING Spelling
Maths
FM Straightforward problem(s) with more than 1 step
FM E3.10 Calculate with money using decimal notation & express money correctly in writing in pounds and pence
FM E3.13 Read time from analogue & 24 hour digital clocks in hours & minutes
FM L1.14 Calculate percentages of quantities, inc. simple percentage increase / decrease by 5% & multiples of
FM L1.20 Convert between units of length, weight, capacity, money and time, in the same system
AN N2/L1.5 Calculate with decimals up to 2dp
AN MSS1/L2.2 Calculate, measure and record time
AN MSS1/L1.3 Calculate with and convert between units of time
AN MSS1/L1.1 Add, subtract, multiply & divide sums of money and record
AN MSS1/L1.2 Read & measure time and use timetables
AN N2/L2.8 Find percentage parts of quantities and measurements
AN N2/L1.9 Find simple percentage parts of quantities and measurements
Context
Employment skills & Public services
## E2-E3 Train journeys Functional Maths
This functional resource gets the learner to answers some questions about a depicted rail ticket and then look up costs for two journeys and carry out further calculations: Woking to London (E2), London to Glasgow (E3 extension questions).
Addition is used to find the cost of two tickets and subtraction to find the difference between two tickets and the cost of two tickets. Multiplication is used to find the cost of an hourly car park ticket.
Level
Entry Level 3
Entry Level 2
Maths
Functional Maths - numbers and the number system
Functional Maths - measures, shape & space
ICT
Functional Skills ICT: Finding and selecting information
Context
Leisure, Hobbies, Travel & Tourism
## Functional Maths skills checks E2/E3 and E3/L1
A range of questions that I have designed for my Functional Skills Maths learners.The focus is on the four operations (+ – x ÷) with wide coverage of measures and money.
There are two sets of similar, but differentiated, questions so that learners at both both levels can complete them in the same lesson. Each set of questions cover several themes (dog show and music festival) followed by a set of mixed questions and a final page of stretch and challenge tasks.
Editor’s note
With curriculum mapping.
Level
Level 1
Entry Level 3
Entry Level 2
Maths
Functional Maths - numbers and the number system
Functional Maths - measures, shape & space
Number
Context
Art Film Media Music Radio TV
Animals, animal care, farming & equine
## Money in, money out
This a functional resource. This resource uses addition, subtraction, multiplication and rounding numbers.
Sam and Tom have weekly incomes from various sources and lots of expenditure!
Subtraction is needed to find out how much money they have left, and multiplication is used to work out money needed for lunches for each day. Each section ends with a rounding task showing items the students have saved for.
Level
Entry Level 3
Entry Level 2
Maths
Functional Maths - numbers and the number system
Functional Maths - measures, shape & space
Context
Independent living
## Number patterns & sequences for Functional Maths
Although many learners find patterns and sequences (including odd & even numbers) straightforward, there is a sizeable proportion that really struggle – particularly those with dyslexia and/or dyscalculia.
This resource grew from my frustration at the lack of practice material available for this topic at Entry Levels 2 and 3 in my one to one support sessions. There is generally only one question on this topic per FS assessment, and I found it difficult to come up with real life functional problems off the top of my head.
Level
Entry Level 2
Entry Level 3
Maths
AN N1/E3.1 place value <1000, odd / even, count in 10s /100s
AN N1/E2.2 Read, write, order & compare numbers to 100; odd even nos.
AN N1/E2.1 Count reliably to 100, count on in 2s and 10s
N1/E1.8
FM E2.3 Recognise & sequence odd & even numbers up to 100
FM E3.6 Recognise & continue linear sequences of numbers up to 100
Context
Sport
Retail Hospitality Customer service
Leisure, Hobbies, Travel & Tourism
Gardening & Horticulture
## Functional Skills Easter Quiz
I designed this quiz as revision for my learners.
Trying to disguise the Functional Maths by putting it together with other rounds as a quiz to make it more interesting!
Editor’s note
Covers many aspects of E3 and L1 Maths – along with interesting and fun general knowledge questions for all levels.
The answer sheet and curriculum mapping is only available to site contributors. Please see details within the resource
Level
Level 1
Entry Level 3
Maths
Functional Maths
General
Generic resources for literacy, numeracy and beyond
## Solar panels L2 Functional Maths questions
5 questions, increasing in difficulty, to get learners to work out how many solar panels can fit on different size roofs. Involves metric conversions between mm, cm and m. Learners have to work with the panels being placed in 2 different orientations to work out the maximum number of panels that can fit.
Level
Level 2
Maths
MSS2/L2.3
MSS2/L2.2
Functional Maths - measures, shape & space
Context
Painting Decorating & DIY
Employment skills & Public services
Construction Carpentry Plumbing
## Probability questions for L2 Functional Maths
5 questions based on functional skills exam questions where learners find the probability of an event happening.
Level
Level 2
Maths
HD2/L2.1
Functional Maths - handling information and data
## Cutting shapes out of materials L2 Functional Maths
Four challenging Functional Skills maths questions to encourage learners to use diagrams to cut shapes out of different materials. The questions are based on past Edexcel Functional skills exam questions that learners find difficult. Also involves metric conversions.
Editor’s note
Includes answer sheet and curriculum mapping. Excellent coverage of a topic that almost all learners find tricky.
Level
Level 2
Maths
MSS2/L2.3
MSS2/L2.2
Functional Maths - measures, shape & space
Context
Painting Decorating & DIY
Construction Carpentry Plumbing
## Scale drawing questions involving plans and elevations
Three challenging Functional skills questions to get learners to draw 2D representations of 3D objects using a given scale. The questions are based on an exam question from an Edexcel L2 paper last year that learners found difficult.
Editor’s note
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+0
# sqrt(2x+3)+2= sqrt(6x+7) I have two radicals that equal each other and i need to find x, the only thing your calculator does is take my typed in variables and
0
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Guest Mar 5, 2017
#1
0
Solve for x:
2 + sqrt(2 x + 3) = sqrt(6 x + 7)
Raise both sides to the power of 2:
(2 + sqrt(2 x + 3))^2 = 6 x + 7
Subtract 6 x + 7 from both sides:
-7 - 6 x + (2 + sqrt(2 x + 3))^2 = 0
-7 - 6 x + (2 + sqrt(2 x + 3))^2 = 4 sqrt(2 x + 3) - 4 x:
4 sqrt(2 x + 3) - 4 x = 0
Simplify and substitute y = sqrt(2 x + 3):
4 sqrt(2 x + 3) - 4 x = 6 + 4 sqrt(2 x + 3) - 2 (sqrt(2 x + 3))^2 = -2 y^2 + 4 y + 6 = 0:
-2 y^2 + 4 y + 6 = 0
The left hand side factors into a product with three terms:
-2 (y - 3) (y + 1) = 0
Divide both sides by -2:
(y - 3) (y + 1) = 0
Split into two equations:
y - 3 = 0 or y + 1 = 0
y = 3 or y + 1 = 0
Substitute back for y = sqrt(2 x + 3):
sqrt(2 x + 3) = 3 or y + 1 = 0
Raise both sides to the power of two:
2 x + 3 = 9 or y + 1 = 0
Subtract 3 from both sides:
2 x = 6 or y + 1 = 0
Divide both sides by 2:
x = 3 or y + 1 = 0
Subtract 1 from both sides:
x = 3 or y = -1
Substitute back for y = sqrt(2 x + 3):
x = 3 or sqrt(2 x + 3) = -1
Raise both sides to the power of two:
x = 3 or 2 x + 3 = 1
Subtract 3 from both sides:
x = 3 or 2 x = -2
Divide both sides by 2:
x = 3 or x = -1
2 + sqrt(2 x + 3) ⇒ 2 + sqrt(3 + 2 (-1)) = 3
sqrt(6 x + 7) ⇒ sqrt(7 + 6 (-1)) = 1:
So this solution is incorrect
2 + sqrt(2 x + 3) ⇒ 2 + sqrt(3 + 2 3) = 5
sqrt(6 x + 7) ⇒ sqrt(7 + 6 3) = 5:
So this solution is correct
The solution is:
Guest Mar 5, 2017
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https://socratic.org/questions/what-is-the-distance-between-3-5-2-and-8-5-4
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# What is the distance between (3,5,-2) and (-8,5,4)?
Nov 29, 2015
$d = \sqrt{157} \approx 12.53$
#### Explanation:
Recall the very useful formula to calculate the distance in 2 dimensions i.e: between 2 points:$\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
In 3 dimensional space the distance between 3 points is calculated by adding 3rd dimension to the above formula, so now the distance between points:$\left({x}_{1} , {y}_{1} , {z}_{1}\right) , \left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
In this case the points are: (3,5,−2),(−8,5,4) so we have:
$d = \sqrt{{\left(- 8 - 3\right)}^{2} + {\left(5 - 5\right)}^{2} + {\left(4 - \left(- 2\right)\right)}^{2}}$
$d = \sqrt{{\left(- 11\right)}^{2} + {\left(0\right)}^{2} + {\left(6\right)}^{2}}$
$d = \sqrt{121 + 0 + 36}$
$d = \sqrt{157}$
$d \approx 12.53$
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# Solving Rational Inequalities
• Aug 11th 2011, 05:43 PM
Dragon08
Solving Rational Inequalities
Solve the inequality: 2x-10/x >x+5
I tried it and got an unfactorable polynomial: x^2+3x+10, but the answer says it's x<-5 and -2<x<0.
• Aug 11th 2011, 05:57 PM
e^(i*pi)
Re: Solving Rational Inequalities
What is the question and what have you tried?
$\dfrac{2x-10}{x} > x+5$ or is it $2x - \dfrac{10}{x} > x+5$
• Aug 11th 2011, 07:24 PM
SammyS
Re: Solving Rational Inequalities
Quote:
Originally Posted by Dragon08
Solve the inequality: 2x-10/x >x+5
I tried it and got an unfactorable polynomial: x^2+3x+10, but the answer says it's x<-5 and -2<x<0.
To get the book answer, the problem would have to be equivalent to solving: $\dfrac{-2x-10}{x} > x+5\,.$
• Aug 11th 2011, 07:42 PM
Dragon08
Re: Solving Rational Inequalities
It's the first one, and I made the moved x+5 onto the left side and made a common denominator. The resulting polynomial numerator is unfactorable.
• Aug 11th 2011, 07:43 PM
Dragon08
Re: Solving Rational Inequalities
To SammyS: Oh, then the the book's answer may be wrong.
• Aug 12th 2011, 01:19 AM
Siron
Re: Solving Rational Inequalities
First of all, like SammyS said it has to be -2x in the numerator of the RS, because if it's 2x then you'll get a negative discriminant and so no roots in $\mathbb{R}$.
Rational inequality
$\frac{-2x-10}{x}>x+5$ and $x\neq 0$
Solve:
$\frac{-2x-10}{x}>x+5 \Leftrightarrow -2x-10>x(x+5) \Leftrightarrow -2x-10>x^2+5x \Leftrightarrow -x^2-7x-10>0 \Leftrightarrow -(x+5)(x+2)>0$
Now you can make a sign table to determine the solution.
• Aug 12th 2011, 03:57 AM
e^(i*pi)
Re: Solving Rational Inequalities
Quote:
Originally Posted by Siron
First of all, like SammyS said it has to be -2x in the numerator of the RS, because if it's 2x then you'll get a negative discriminant and so no roots in $\mathbb{R}$.
Rational inequality
$\frac{-2x-10}{x}>x+5$ and $x\neq 0$
Solve:
$\frac{-2x-10}{x}>x+5 \Leftrightarrow -2x-10>x(x+5) \Leftrightarrow -2x-10>x^2+5x \Leftrightarrow -x^2-7x-10>0 \Leftrightarrow -(x+5)(x+2)>0$
Now you can make a sign table to determine the solution.
Good spot on the -2x instead of 2x (Clap)
Alternatively, if $x < 0$ then we need to change the direction of the inequality when multiplying through by x.
$-2x-10 < x^2+5x \Longleftrightarrow x^2+7x+10 > 0 \Longleftrightarrow (x+2)(x+5) > 0$
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V22.0101 ......Homework Assignment 1 ...... Spring 2007
Binary Representation of Numbers
Assigned: WED, Jan 24.
First Version Due (by email to etutor): (Marateck) TUES, Feb 6, at midnight
Final Version Due (by email to etutor): 7 days after receiving comments from etutor
Please read the assignment carefully. If you need help, don't hesitate to contact your E-TUTOR, but send the email EARLY, NOT just before the deadline. You can also see the TA or the Professor during their office hours. Even though this is the first assignment, it uses concepts from all of Chapters 2 through 5. However, you do not need to read all of these chapters yet. Read the parts you need, especially Sections 3.3.1 and 3.3.2 for the if statement and Section 4.4 for the for loop.
Decimal numbers from 0 through 255 can be represented in the binary number system using 8 bits. Here are some examples:
DecimalBinary
0 00000000
1 00000001
2 00000010
3 00000011
7 00000111
15 00001111
16 00010000
25 00011001
128 10000000
129 10000001
254 11111110
255 11111111
Thus, for example, the decimal number 25 is represented in binary as 00011001, meaning 0 times 128 plus 0 times 64 plus 0 times 32 plus 1 times 16 plus 1 times 8 plus 0 times 4 plus 0 times 2 plus 1 times 1. Notice that EACH BIT CORRESPONDS TO A POWER OF 2. Make sure you understand how this works before you go any further.
This assignment asks you to write a Java class called Binary. The decimal number will be stored in the variable numVal. Another variable binStr will store the binary string corresponding to numVal. The class Binary will have two methods that you must write. Here are the details.
• the variable binStr (type String) is where the string of bits corresponding to the binary representation of the number is stored by one of the methods (see below).
• The main method should check to make sure the decimal value is between 0 and 255. If it is not, it should print an informative error message (using System.out.println) and store the number 0 instead.
• the method getBin (public static String getBin(int num )) should take the value that is passed to it, generate the corresponding binary string and returns it. YOU MUST WRITE THIS METHOD BY YOURSELF. YOU MAY NOT CALL A METHOD FROM THE JAVA LIBRARY TO DO THIS JOB. The method should consist of a single for loop. This loop generates the 8 bits of the bit string, one at a time. There are two possible ways to do this. The easiest is to generate the bits from LEFT to RIGHT, by successively subtracting powers of 2 from numVal. Take the number 25 for example. If we subtract the biggest relevant power of 2, the number 128, from this, we get a negative answer so we know the first bit must be 0. The first power of 2 which is smaller than 25 is 16, so that tells us that the bit corresponding to that power (that's actually the fourth bit from the left) is 1. Once 16 is taken away from 25, we have 9 left. The next power of 2 is 8, which is less than 9, so the next bit (the fifth from the left) is also 1. Now we only have 1 left, so the sixth and seventh bits are 0, and the final eights bit is 1. You should not type all the powers of 2 into the program. You should start with 128, and then repeatedly divide this by 2 in the loop. Don't use any of the numbers in the table above for an example. Choose another number, say between 70 and 250 for your example. Don't use the same example as a classmate, if you can avoid it. (Of course, some people may choose the same example by accident -that's OK.)
There's an alternative way to code this method, also using a for loop but generating the bits from RIGHT to LEFT, using repeated division. See p. 11 of the text. See the Powerpoint presentation at the last link for the homework http://cs.nyu.edu/courses/spring02/V22.0102-002/binary.ppt. Look at the slides up to the the ASCII table.
As each bit is generated, either left to right or right to left, you have to concatenate it to the current value of binStr. Start with the empty string.
• the method checkBin (public static boolean checkBin(String binStr, int numVal)) should take the binary string in binStr and check to see if it correctly represents the number in numVal, essentially reversing the operation of getBin. This is easy, again using a loop: you need to extract the first bit of the string binStr, multiply it by 128, then add this to the second bit multiplied by 64, and so on. Again, don't type all the powers of 2 into the program: just start with 128 and repeatedly divide it by 2. Another variation is to first multiply the first bit by 2, then add the second bit and multiply all that by 2, and then add the third bit, and so on. Again, make sure you use a for loop and document it with COMMENTS. The method checkBin should return a Boolean value, indicating whether the result matches the number in numVal. To extract each bit, you need to use the String method charAt (see p. 265) to get the relevant char value and you can then test this against the char values '0' and '1' to complete the job.
You can "hard-wire" the example numbers into the main method, or you can read them from the the keyboard using the Scanner class described in the text book (page 53). You must have JDK1.5 or later, and import java.util.Scanner to use the Scanner class. To read an integer you would write Scanner kbd = new Scanner(System.in); followed by int num = kbd.nextInt().
After your program runs properly, alter it slightly so that it can read the data from the command line. If you have a mac, the terminal program gives you the command line. If you have a PC, see http://www.computerhope.com/issues/ch000549.htm (the link is on the web page under the assignment) for how to include the java and javac programs in your path. However, the easiest way is to place your program in the Bin directory of the jdk directory. Hand this version into your etutor. If you can't get this version to work, 0.2 points will be deducted from a possible high of 4.0 points.
Please get started on the assignment immediately, and contact your etutor as soon as possible with any questions. Don't expect a response the day before the due date. Late homeworks will be accepted up to 7 days late, but they will be penalized 25%.
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## Pages
Bankers Discount and True Discount Questions and Answers: Banker's discount questions and answers is one of the rarely asked question in the competitive exams and the interviews (off campus & walk ins ). There are many true discount aptitude questions are found in both online and offline mode. Our Indiagrade team members selected the repeatedly asked true discount bankers discount aptitude questions from all the competitive exam question papers. Candidates are advised use this opportunity and they are asked to start the preparation for their bank exams and other examinations. You can download the essential formulas for the bankers discount and true discount as PDF file.
### True Discount and Bankers Discount example question and Solutions:
Q1: Calculate the The rate percentage where the bankers discount on a certain sum due two years, hence is 11/10 of the true discount.
(a) 5%
(b) 10%
(c) 15%
(d) 2%
Q2: The interest on rupees 750 for two years is the same as the true discount on rupees 960 due two years. The rate of interest is the same in both cases.
(a) 10%
(b) 8%
(c) 14%
(d) 18%
Q3: Find the true discount. The banker's gain on bill due one year at 12% per year is six rupees.
(a) 40
(b) 25
(c) 30
(d) 50
Q4: The true discount and simple interest the on a certain sum for a given time and at a given rate are Rs.80 and Rs.85 respectively. Find the sum?
(a) 1450
(b) 1360
(c) 1854
(d) 1550
Q5: The certain worth of a certain sum due sometime is rupees 600 and the true gain is rupees 160. Calculate banker's discount?
(a) 16
(b) 15
(c) 12
(d) 17
Q6: The true gain on a bill due nine months at 16% per year is rupees 189. Calculate the amount of the bill?
(a) 1850
(b) 1578
(c) 1764
(d) 1650
Q7: The true discount on 1760 rupees due after a certain time at 12% per year is 160 rupees. Calculate the time after which it is due?
(a) 5 months
(b) 4 months
(c) 12 months
(d) 10 months
Q8: Ten rupees be allowed as true discount on a bill of 110 rupees due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time?
(a) 15.54
(b) 16.30
(c) 18.33
(d) 16.50
Q9: The bankers discount on 1600 rupees at 15% per year is the same as true discount on 1680 rupees for the same time and at the same rate. Find the time?
(a) 4 months
(b) 6 months
(c) 10 months
(d) 8 months
Q10: A trader owes a merchant 10,028 INR due one year. The trader wants to settle the amount after three months. How much money should he pay, If the rate of interest 12% per year.
(a) 8000
(b) 9200
(c) 7500
(d) 8560
Q1: (a)
Q2: (c)
Q3: (d)
Q4: (b)
Q5: (a)
Q6: (c)
Q7: (d)
Q8: (c)
Q9: (a)
Q10:(b)
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# Search by Topic
#### Resources tagged with SM - Advanced probability similar to Any Win for Tennis?:
Filter by: Content type:
Stage:
Challenge level:
### There are 16 results
Broad Topics > Stage 5 Statistics Mapping Document > SM - Advanced probability
### The Derren Brown Coin Flipping Scam
##### Stage: 5 Challenge Level:
Calculate probabilities associated with the Derren Brown coin scam in which he flipped 10 heads in a row.
### Why Do People Find Probability Unintuitive and Difficult?
##### Stage: 2, 3, 4 and 5
Uncertain about the likelihood of unexpected events? You are not alone!
### Knock-out
##### Stage: 5 Challenge Level:
Before a knockout tournament with 2^n players I pick two players. What is the probability that they have to play against each other at some point in the tournament?
### Ante Up
##### Stage: 5 Challenge Level:
Use cunning to work out a strategy to win this game.
### Taking Chances Extended
##### Stage: 4 and 5
This article, for students and teachers, is mainly about probability, the mathematical way of looking at random chance.
##### Stage: 5 Challenge Level:
Use combinatoric probabilities to work out the probability that you are genetically unique!
### Succession in Randomia
##### Stage: 5 Challenge Level:
By tossing a coin one of three princes is chosen to be the next King of Randomia. Does each prince have an equal chance of taking the throne?
### Bet You a Million
##### Stage: 4 Challenge Level:
Heads or Tails - the prize doubles until you win it. How much would you pay to play?
### Weekly Challenge 37: Magic Bag
##### Stage: 5 Challenge Level:
A weekly challenge concerning combinatorical probability.
### FA Cup
##### Stage: 5 Challenge Level:
In four years 2001 to 2004 Arsenal have been drawn against Chelsea in the FA cup and have beaten Chelsea every time. What was the probability of this? Lots of fractions in the calculations!
### Snooker
##### Stage: 5 Challenge Level:
A player has probability 0.4 of winning a single game. What is his probability of winning a 'best of 15 games' tournament?
### Spinners
##### Stage: 5 Challenge Level:
How do scores on dice and factors of polynomials relate to each other?
### Dicey Decisions
##### Stage: 5 Challenge Level:
Can you devise a fair scoring system when dice land edge-up or corner-up?
### Teams
##### Stage: 5 Challenge Level:
Two brothers belong to a club with 10 members. Four are selected for a match. Find the probability that both brothers are selected.
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Saltar al contenido principal
# 7.1: Resolver ecuaciones trigonométricas con identidades
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##### Objetivos de aprendizaje
• Verificar las identidades trigonométricas fundamentales.
• Simplifica las expresiones trigonométricas usando álgebra y las identidades.
En las películas de espionaje, vemos espías internacionales con múltiples pasaportes, cada uno reclamando una identidad diferente. No obstante, sabemos que cada uno de esos pasaportes representa a la misma persona. Las identidades trigonométricas actúan de manera similar a múltiples pasaportes; hay muchas formas de representar la misma expresión trigonométrica. Así como un espía elegirá un pasaporte italiano cuando viaje a Italia, elegimos la identidad que aplica al escenario dado al resolver una ecuación trigonométrica.
En esta sección, comenzaremos un examen de las identidades trigonométricas fundamentales, incluyendo cómo podemos verificarlas y cómo podemos utilizarlas para simplificar las expresiones trigonométricas.
#### Verificación de las Identidades Trigonométricas Fundamentales
Comenzaremos con las identidades pitagóricas (Tabla$$\PageIndex{1}$$), que son ecuaciones que involucran funciones trigonométricas basadas en las propiedades de un triángulo rectángulo. Ya hemos visto y utilizado la primera de estas identificaciones, pero ahora también usaremos identidades adicionales.
$${\sin}^2 \theta+{\cos}^2 \theta=1$$ $$1+{\cot}^2 \theta={\csc}^2 \theta$$ $$1+{\tan}^2 \theta={\sec}^2 \theta$$
La segunda y tercera identidades se pueden obtener manipulando la primera. La identidad$$1+{\cot}^2 \theta={\csc}^2 \theta$$ se encuentra reescribiendo el lado izquierdo de la ecuación en términos de seno y coseno.
Demostrar:$$1+{\cot}^2 \theta={\csc}^2 \theta$$
\begin{align*} 1+{\cot}^2 \theta&= (1+\dfrac{{\cos}^2}{{\sin}^2})\qquad \text{Rewrite the left side}\\ &= \left(\dfrac{{\sin}^2}{{\sin}^2}\right)+\left (\dfrac{{\cos}^2}{{\sin}^2}\right)\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\sin}^2+{\cos}^2}{{\sin}^2}\\ &= \dfrac{1}{{\sin}^2}\\ &= {\csc}^2 \end{align*}
De igual manera, se$$1+{\tan}^2 \theta={\sec}^2 \theta$$ puede obtener reescribiendo el lado izquierdo de esta identidad en términos de seno y coseno. Esto da
\begin{align*} 1+{\tan}^2 \theta&= 1+{\left(\dfrac{\sin \theta}{\cos \theta}\right )}^2\qquad \text{Rewrite left side}\\ &= {\left (\dfrac{\cos \theta}{\cos \theta}\right )}^2+{\left (\dfrac{\sin \theta}{\cos \theta}\right)}^2\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\cos}^2 \theta+{\sin}^2 \theta}{{\cos}^2 \theta}\\ &= \dfrac{1}{{\cos}^2 \theta}\\ &= {\sec}^2 \theta \end{align*}
Recordemos que determinamos qué funciones trigonométricas son impares y cuáles son pares. El siguiente conjunto de identidades fundamentales es el conjunto de identidades pares e impares. Las identidades pares impares relacionan el valor de una función trigonométrica en un ángulo dado con el valor de la función en el ángulo opuesto (Tabla$$\PageIndex{2}$$).
$$\tan(−\theta)=−\tan \theta$$ $$\sin(−\theta)=−\sin \theta$$ $$\cos(−\theta)=\cos \theta$$ $$\cot(−\theta)=−\cot \theta$$ $$\csc(−\theta)=−\csc \theta$$ $$\sec(−\theta)=\sec \theta$$
Recordemos que una función impar es aquella en la que$$f(−x)= −f(x)$$ para todos$$x$$ en el dominio off. f. La función seno es una función impar porque$$\sin(−\theta)=−\sin \theta$$. La gráfica de una función impar es simétrica sobre el origen. Por ejemplo, considere las entradas correspondientes de$$\dfrac{\pi}{2}$$ y$$−\dfrac{\pi}{2}$$. La salida de$$\sin\left (\dfrac{\pi}{2}\right )$$ es opuesta a la salida de$$\sin \left (−\dfrac{\pi}{2}\right )$$. Por lo tanto,
\begin{align*} \sin\left (\dfrac{\pi}{2}\right)&=1 \\[4pt] \sin\left (-\dfrac{\pi}{2}\right) &=-\sin\left (\dfrac{\pi}{2}\right) \\[4pt] &=-1 \end{align*}
Esto se muestra en la Figura$$\PageIndex{2}$$.
Recordemos que una función par es aquella en la que
$$f(−x)=f(x)$$para todos$$x$$ en el dominio de$$f$$
La gráfica de una función par es simétrica alrededor del eje y. La función coseno es una función par porque$$\cos(−\theta)=\cos \theta$$. Por ejemplo, considere las entradas correspondientes$$\dfrac{\pi}{4}$$ y$$−\dfrac{\pi}{4}$$. La salida de$$\cos\left (\dfrac{\pi}{4}\right)$$ es la misma que la salida de$$\cos\left (−\dfrac{\pi}{4}\right)$$. Por lo tanto,
\begin{align*} \cos\left (−\dfrac{\pi}{4}\right ) &=\cos\left (\dfrac{\pi}{4}\right) \\[4pt] &≈0.707 \end{align*}
Ver Figura$$\PageIndex{3}$$.
Para todos$$\theta$$ en el dominio de las funciones seno y coseno, respectivamente, podemos afirmar lo siguiente:
• Ya que$$\sin(−\theta)=−\sin \theta$$, seno es una función impar.
• Ya que$$\cos(−\theta)=\cos \theta$$, el coseno es una función par.
Las otras identidades pares-impares se derivan de la naturaleza par e impar de las funciones seno y coseno. Por ejemplo, considere la identidad tangente,$$\tan(−\theta)=−\tan \theta$$. Podemos interpretar la tangente de un ángulo negativo como
$\tan (−\theta)=\dfrac{\sin (−\theta)}{\cos (−\theta)}=\dfrac{−\sin \theta}{\cos \theta}=−\tan \theta. \nonumber$
Tangente es por lo tanto una función impar, lo que significa que$$\tan(−\theta)=−\tan(\theta)$$ para todos$$\theta$$ en el dominio de la función tangente.
La identidad cotangente,$$\cot(−\theta)=−\cot \theta$$, también se desprende de las identidades seno y coseno. Podemos interpretar la cotangente de un ángulo negativo como
$\cot(−\theta)=\dfrac{\cos(−\theta)}{\sin(−\theta)}=\dfrac{\cos \theta}{−\sin \theta}=−\cot \theta.\nonumber$
Cotangente es por lo tanto una función impar, lo que significa que$$\cot(−\theta)=−\cot(\theta)$$ para todos$$\theta$$ en el dominio de la función cotangente.
La función cosecante es la recíproca de la función sinusoidal, lo que significa que la cosecante de un ángulo negativo se interpretará como
$\csc(−\theta)=\dfrac{1}{\sin(−\theta)}=\dfrac{1}{−\sin \theta}=−\csc \theta. \nonumber$
Por lo tanto, la función cosecante es impar.
Finalmente, la función secante es la recíproca de la función coseno, y la secante de un ángulo negativo se interpreta como
$\sec(−\theta)=\dfrac{1}{\cos(−\theta)}=\dfrac{1}{\cos \theta}=\sec \theta. \nonumber$
La función secante es, pues, parejo.
En resumen, sólo dos de las funciones trigonométricas, coseno y secante, son parejos. Las otras cuatro funciones son impares, verificando las identidades pares-impares.
El siguiente conjunto de identidades fundamentales es el conjunto de identidades recíprocas, que, como su nombre lo indica, relacionan funciones trigonométricas que son recíprocas entre sí. (Tabla$$\PageIndex{3}$$). Recordemos que por primera vez encontramos estas identidades al definir funciones trigonométricas desde ángulos rectos en trigonometría de ángulo recto.
$$\sin \theta=\dfrac{1}{\csc \theta}$$ $$\csc \theta=\dfrac{1}{\sin \theta}$$ $$\cos \theta = \dfrac{1}{\sec \theta}$$ $$\sec \theta=\dfrac{1}{\cos \theta}$$ $$\tan \theta=\dfrac{1}{\cot \theta}$$ $$\cot \theta=\dfrac{1}{\tan \theta}$$
El conjunto final de identidades es el conjunto de identidades de cocientes, que definen las relaciones entre ciertas funciones trigonométricas y pueden ser muy útiles para verificar otras identidades (Tabla$$\PageIndex{4}$$).
$$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$$ $$\cot \theta=\dfrac{\cos \theta}{\sin \theta}$$
Las identidades recíprocas y cocientes se derivan de las definiciones de las funciones trigonométricas básicas.
Las identidades pitagóricas se basan en las propiedades de un triángulo rectángulo.
${\cos}^2 \theta+{\sin}^2 \theta=1$
$1+{\cot}^2 \theta={\csc}^2 \theta$
$1+{\tan}^2 \theta={\sec}^2 \theta$
Las identidades pares impares relacionan el valor de una función trigonométrica en un ángulo dado con el valor de la función en el ángulo opuesto.
$\tan(−\theta)=−\tan \theta$
$\cot(−\theta)=−\cot \theta$
$\sin(−\theta)=−\sin \theta$
$\csc(−\theta)=−\csc \theta$
$\cos(−\theta)=\cos \theta$
$\sec(−\theta)=\sec \theta$
Las identidades recíprocas definen recíprocas de las funciones trigonométricas.
$\sin \theta=\dfrac{1}{\csc \theta}$
$\cos \theta=\dfrac{1}{\sec \theta}$
$\tan \theta=\dfrac{1}{\cot \theta}$
$\csc \theta=\dfrac{1}{\sin \theta}$
$\sec \theta=\dfrac{1}{\cos \theta}$
$\cot \theta=\dfrac{1}{\tan \theta}$
Las identidades del cociente definen la relación entre las funciones trigonométricas.
$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$
$\cot \theta=\dfrac{\cos \theta}{\sin \theta}$
##### Ejemplo$$\PageIndex{1}$$: Graphing the Equations of an Identity
Grafica ambos lados de la identidad$$\cot \theta=\dfrac{1}{\tan \theta}$$. En otras palabras, en la calculadora gráfica, gráfica$$y=\cot \theta$$ y$$y=\dfrac{1}{\tan \theta}$$.
Solución
Ver Figura$$\PageIndex{4}$$.
Análisis
Solo vemos una gráfica porque ambas expresiones generan la misma imagen. Uno está encima del otro. Esta es una buena manera de probar cualquier identidad. Si ambas expresiones dan la misma gráfica, entonces deben ser identidades.
1. Trabajar en un lado de la ecuación. Por lo general, es mejor comenzar por el lado más complejo, ya que es más fácil de simplificar que de construir.
2. Busque oportunidades para factorizar expresiones, cuadrar un binomio o agregar fracciones.
4. Si estos pasos no dan el resultado deseado, intente convertir todos los términos en senos y cosenos.
##### Ejemplo$$\PageIndex{2}$$: Verifying a Trigonometric Identity
Verificar$$\tan \theta \cos \theta=\sin \theta$$.
Solución
\begin{align*} \tan \theta \cos \theta &=\left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta \\[4pt] &=\sin \theta \end{align*}
Análisis
Esta identidad era bastante sencilla de verificar, ya que solo requería escribir$$\tan \theta$$ en términos de$$\sin \theta$$ y$$\cos \theta$$.
##### Ejercicio$$\PageIndex{1}$$
Verificar la identidad$$\csc \theta \cos \theta \tan \theta=1$$.
Contestar
\begin{align*} \csc \theta \cos \theta \tan \theta=\left(\dfrac{1}{\sin \theta}\right)\cos \theta\left(\dfrac{\sin \theta}{\cos \theta}\right) \\[4pt] & =\dfrac{\cos \theta}{\sin \theta}(\dfrac{\sin \theta}{\cos \theta}) \\[4pt] & =\dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[4pt] &=1 \end{align*}
##### Ejemplo$$\PageIndex{3A}$$: Verifying the Equivalency Using the Even-Odd Identities
Verifique la siguiente equivalencia usando las identidades pares-impares:
$$(1+\sin x)[1+\sin(−x)]={\cos}^2 x$$
Solución
Trabajando en el lado izquierdo de la ecuación, tenemos
$$(1+\sin x)[1+\sin(−x)]=(1+\sin x)(1-\sin x)$$
Desde
\begin{align*} \sin(-x)&= -\sin x \\ [5pt] &=1-{\sin}^2 x\qquad \text{Difference of squares} \\ [5pt] &={\cos}^2 x \\ {\cos}^2 x&= 1-{\sin}^2 x \\ \end{align*}
##### Ejemplo$$\PageIndex{3B}$$: Verifying a Trigonometric Identity Involving $${\sec}^2 \theta$$
Verificar la identidad$$\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta}={\sin}^2 \theta$$
Solución
\ [\ begin {alinear*}
\ dfrac {{\ seg} ^2\ theta-1} {{\ seg} ^2\ theta} &=\ dfrac {({\ tan} ^2\ theta +1) -1} {{\ seg} ^2\ theta}\\
{\ seg} ^2\ theta&= {tan} ^2\ theta +1\\
=\ dfrac {{\ tan} ^2\ theta} {{\ seg} ^2\ theta}\\
&= {\ tan} ^2\ theta\ izquierda (\ dfrac {1} {{\ seg} ^2\ theta}\ derecha)\\
&= {\ tan} ^2\ theta\ izquierda ({\ cos} ^2\ theta\ derecha)\\
{\ cos} ^2\ theta&=\ dfrac {1} {{\ seg} ^2\ theta}\\
&=\ izquierda (\ dfrac {{\ sin} ^2\ theta} {{cos} ^2\ theta}\ derecha)\\
{\ tan} ^2\ theta&=\ dfrac {{\ sin} ^2\ theta} {{\ cos} ^2\ theta }\\
&= {\ sin} ^2\ theta
\ final {alinear*}\]
Hay más de una manera de verificar una identidad. Aquí hay otra posibilidad. Nuevamente, podemos comenzar por el lado izquierdo.
\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{{\sec}^2 \theta}{{\sec}^2 \theta}-\dfrac{1}{{\sec}^2 \theta}\\ &= 1-{\cos}^2 \theta\\ &= {\sin}^2 \theta \end{align*}
Análisis
En el primer método, utilizamos la identidad$${\sec}^2 \theta={\tan}^2 \theta+1$$ y seguimos simplificando. En el segundo método, dividimos la fracción, poniendo ambos términos en el numerador sobre el denominador común. Este problema ilustra que hay múltiples formas en las que podemos verificar una identidad. Emplear algo de creatividad a veces puede simplificar un procedimiento. Siempre y cuando las sustituciones sean correctas, la respuesta será la misma.
##### Ejercicio$$\PageIndex{2}$$
Demostrar eso$$\dfrac{\cot \theta}{\csc \theta}=\cos \theta$$.
Contestar
\begin{align*} \dfrac{\cot \theta}{\csc \theta}&= \dfrac{\tfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\ &= \dfrac{\cos \theta}{\sin \theta}\cdot \dfrac{\sin \theta}{1}\\ &= \cos \theta \end{align*}
##### Ejemplo$$\PageIndex{4}$$: Creating and Verifying an Identity
Crear una identidad para la expresión$$2 \tan \theta \sec \theta$$ reescribiendo estrictamente en términos de seno.
Solución
Hay varias formas de comenzar, pero aquí usaremos el cociente y las identidades recíprocas para reescribir la expresión:
\begin{align*} 2 \tan \theta \sec \theta&= 2\left (\dfrac{\sin \theta}{\cos \theta}\right )\left(\dfrac{1}{\cos \theta}\right )\\ &= \dfrac{2\sin \theta}{{\cos}^2 \theta}\\ &= \dfrac{2\sin \theta}{1-{\sin}^2 \theta}\qquad \text{Substitute } 1-{\sin}^2 \theta \text{ for } {\cos}^2 \theta \end{align*}
Por lo tanto,
$$2 \tan \theta \sec \theta=\dfrac{2 \sin \theta}{1−{\sin}^2 \theta}$$
##### Ejemplo$$\PageIndex{5}$$: Verifying an Identity Using Algebra and Even/Odd Identities
$$\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)}=\cos \theta−\sin \theta$$
Solución
Empecemos por el lado izquierdo y simplifiquemos:
\begin{align*} \dfrac{{\sin}^2(-\theta)-{\cos}^2(-\theta)}{\sin(-\theta)-\cos(-\theta)}&= \dfrac{{[\sin(-\theta)]}^2-{[\cos(-\theta)]}^2}{\sin(-\theta)-\cos(-\theta)}\\ &= \dfrac{{(-\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta} \;\; \; , \sin(-x) = -\sin\space x\text { and } \cos(-x)=\cos \space x\\ &= \dfrac{{(\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta}\qquad \text{Difference of squares}\\ &= \dfrac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{-(\sin \theta+\cos \theta)}\\ &= \cos \theta-\sin \theta \end{align*}
##### Ejercicio$$\PageIndex{3}$$
Verificar la identidad$$\dfrac{{\sin}^2 \theta−1}{\tan \theta \sin \theta−\tan \theta}=\dfrac{\sin \theta+1}{\tan \theta}$$.
Contestar
\begin{align*} \dfrac{{\sin}^2 \theta-1}{\tan \theta \sin \theta-\tan \theta}&= \dfrac{(\sin \theta +1)(\sin \theta -1)}{\tan \theta(\sin \theta -1)}\\ &= \dfrac{\sin \theta+1}{\tan \theta} \end{align*}
##### Ejemplo$$\PageIndex{6}$$: Verifying an Identity Involving Cosines and Cotangents
Verificar la identidad:$$(1−{\cos}^2 x)(1+{\cot}^2 x)=1$$.
Solución
\begin{align*} (1-{\cos}^2 x)(1+{\cot}^2 x)&= (1-{\cos}^2 x)\left(1+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right)\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x}{{\sin}^2 x}+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right )\qquad \text{Find the common denominator}\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x +{\cos}^2 x}{{\sin}^2 x}\right)\\ &= ({\sin}^2 x)\left (\dfrac{1}{{\sin}^2 x}\right )\\ &= 1 \end{align*}
## Uso del álgebra para simplificar las expresiones trigonométricas
Hemos visto que el álgebra es muy importante en la verificación de identidades trigonométricas, pero es igual de crítico para simplificar las expresiones trigonométricas antes de resolverlo. Estar familiarizado con las propiedades y fórmulas básicas del álgebra, como la fórmula de diferencia de cuadrados, la fórmula cuadrada perfecta, o la sustitución, simplificará el trabajo involucrado con expresiones y ecuaciones trigonométricas.
Por ejemplo, la ecuación$$(\sin x+1)(\sin x−1)=0$$ se asemeja a la ecuación$$(x+1)(x−1)=0$$, que utiliza la forma factorizada de la diferencia de cuadrados. El uso del álgebra hace que encontrar una solución sea sencillo y familiar. Podemos establecer cada factor igual a cero y resolver. Este es un ejemplo de reconocimiento de patrones algebraicos en expresiones o ecuaciones trigonométricas.
Otro ejemplo es la fórmula de diferencia de cuadrados$$a^2−b^2=(a−b)(a+b)$$, que es ampliamente utilizada en muchas áreas distintas a las matemáticas, como la ingeniería, la arquitectura y la física. También podemos crear nuestras propias identidades expandiendo continuamente una expresión y haciendo las sustituciones adecuadas. El uso de propiedades y fórmulas algebraicas hace que muchas ecuaciones trigonométricas sean más fáciles de entender y resolver.
##### Ejemplo$$\PageIndex{7A}$$: Writing the Trigonometric Expression as an Algebraic Expression
Escribe la siguiente expresión trigonométrica como expresión algebraica:$$2{\cos}^2 \theta+\cos \theta−1$$.
Solución
Observe que el patrón mostrado tiene la misma forma que una expresión cuadrática estándar,$$ax^2+bx+c$$. Dejando$$\cos \theta=x$$, podemos reescribir la expresión de la siguiente manera:
$$2x^2+x−1$$
Esta expresión se puede factorizar como$$(2x+1)(x−1)$$. Si se pusiera igual a cero y quisiéramos resolver la ecuación, usaríamos la propiedad del factor cero y resolveríamos cada factor para$$x$$. En este punto, reemplazaríamos$$x$$ con$$\cos \theta$$ y resolveríamos para$$\theta$$.
##### Ejemplo$$\PageIndex{7B}$$: Rewriting a Trigonometric Expression Using the Difference of Squares
Reescribir la expresión trigonométrica usando la diferencia de cuadrados:$$4{cos}^2 \theta−1$$.
Solución
\begin{align*} 4{\cos}^2 \theta-1&= {(2\cos \theta)}^2-1\\ &= (2\cos \theta-1)(2\cos \theta+1) \end{align*}
Análisis
Si esta expresión se escribiera en forma de una ecuación establecida igual a cero, podríamos resolver cada factor usando la propiedad de factor cero. También podríamos usar la sustitución como hicimos en el problema anterior y dejar$$\cos \theta=x$$, reescribir la expresión as$$4x^2−1$$, y factor$$(2x−1)(2x+1)$$. Luego reemplace$$x$$ con$$\cos \theta$$ y resuelva para el ángulo.
##### Ejercicio$$\PageIndex{4}$$
Reescribir la expresión trigonométrica usando la diferencia de cuadrados:$$25−9{\sin}^2 \theta$$.
Contestar
Esta es una fórmula de diferencia de cuadrados:$$25−9{\sin}^2 \theta=(5−3\sin \theta)(5+3\sin \theta)$$.
##### Ejemplo$$\PageIndex{8}$$: Simplify by Rewriting and Using Substitution
Simplifica la expresión reescribiendo y usando identidades:
$${\csc}^2 \theta−{\cot}^2 \theta$$
Solución
Podemos comenzar con la identidad pitagórica.
\begin{align*} 1+{\cot}^2 \theta&= {\csc}^2 \theta\\ \text{Now we can simplify by substituting } 1+{\cot}^2 \theta \text{ for } {\csc}^2 \theta\\ {\csc}^2 \theta-{\cot}^2 \theta&= 1+{\cot}^2 \theta-{\cot}^2 \theta\\ &= 1 \end{align*}
##### Ejercicio$$\PageIndex{5}$$
Utilizar técnicas algebraicas para verificar la identidad:$$\dfrac{\cos \theta}{1+\sin \theta}=\dfrac{1−\sin \theta}{\cos \theta}$$.
(Pista: Multiplica el numerador y denominador del lado izquierdo por$$1−\sin \theta$$.)
Contestar
\begin{align*} \dfrac{\cos \theta}{1+\sin \theta}\left(\dfrac{1-\sin \theta}{1-\sin \theta}\right)&= \dfrac{\cos \theta (1-\sin \theta)}{1-{\sin}^2 \theta}\\ &= \dfrac{\cos \theta (1-\sin \theta)}{{\cos}^2 \theta}\\ &= \dfrac{1-\sin \theta}{\cos \theta} \end{align*}
##### Medios
Acceda a estos recursos en línea para una instrucción adicional y práctica con las identidades trigonométricas fundamentales.
## Ecuaciones Clave
Identidades pitagóreas $${\cos}^2 \theta+{\sin}^2 \theta=1$$ $$1+{\cot}^2 \theta={\csc}^2 \theta$$ $$1+{\tan}^2 \theta={\sec}^2 \theta$$ Identidades pares e impares $$\tan(−\theta)=-\tan \theta$$ $$\cot(-\theta)=-\cot \theta$$ $$\sin(-\theta)=-\sin \theta$$ $$\csc(-\theta)=-\csc \theta$$ $$\cos(-\theta)=\cos \theta$$ $$\sec(-\theta)=\sec \theta$$ Identidades recíprocas $$\sin \theta=\dfrac{1}{\csc \theta}$$ $$\cos \theta=\dfrac{1}{\sec \theta}$$ $$\tan \theta=\dfrac{1}{\cot \theta}$$ $$\csc \theta=\dfrac{1}{\sin \theta}$$ $$\sec \theta=\dfrac{1}{\cos \theta}$$ $$\cot \theta=\dfrac{1}{\tan \theta}$$ Identidades de cocientes $$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$$ $$\cot \theta=\dfrac{\cos \theta}{\sin \theta}$$
## Conceptos clave
• Existen múltiples formas de representar una expresión trigonométrica. Verificar las identidades ilustra cómo se pueden reescribir las expresiones para simplificar un problema.
• Graficar ambos lados de una identidad la verificará. Ver Ejemplo$$\PageIndex{1}$$.
• Simplificar un lado de la ecuación para igualar al otro lado es otro método para verificar una identidad. Ver Ejemplo$$\PageIndex{2}$$ y Ejemplo$$\PageIndex{3}$$.
• El enfoque para verificar una identidad depende de la naturaleza de la identidad. A menudo es útil comenzar en el lado más complejo de la ecuación. Ver Ejemplo$$\PageIndex{4}$$.
• Podemos crear una identidad y luego verificarla. Ver Ejemplo$$\PageIndex{5}$$.
• Verificar una identidad puede implicar álgebra con las identidades fundamentales. Ver Ejemplo$$\PageIndex{6}$$ y Ejemplo$$\PageIndex{7}$$.
• Las técnicas algebraicas se pueden utilizar para simplificar las expresiones trigonométricas. Utilizamos técnicas algebraicas a lo largo de este texto, ya que consisten en las reglas fundamentales de las matemáticas. Ver Ejemplo$$\PageIndex{8}$$$$\PageIndex{9}$$, Ejemplo y Ejemplo$$\PageIndex{10}$$.
This page titled 7.1: Resolver ecuaciones trigonométricas con identidades is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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Samples and ProbabilityType of Unit: ConceptualPrior KnowledgeStudents should be able to:Understand the concept of a ratio.Write ratios as percents.Describe data using measures of center.Display and interpret data in dot plots, histograms, and box plots.Lesson FlowStudents begin to think about probability by considering the relative likelihood of familiar events on the continuum between impossible and certain. Students begin to formalize this understanding of probability. They are introduced to the concept of probability as a measure of likelihood, and how to calculate probability of equally likely events using a ratio. The terms (impossible, certain, etc.) are given numerical values. Next, students compare expected results to actual results by calculating the probability of an event and conducting an experiment. Students explore the probability of outcomes that are not equally likely. They collect data to estimate the experimental probabilities. They use ratio and proportion to predict results for a large number of trials. Students learn about compound events. They use tree diagrams, tables, and systematic lists as tools to find the sample space. They determine the theoretical probability of first independent, and then dependent events. In Lesson 10 students identify a question to investigate for a unit project and submit a proposal. They then complete a Self Check. In Lesson 11, students review the results of the Self Check, solve a related problem, and take a Quiz.Students are introduced to the concept of sampling as a method of determining characteristics of a population. They consider how a sample can be random or biased, and think about methods for randomly sampling a population to ensure that it is representative. In Lesson 13, students collect and analyze data for their unit project. Students begin to apply their knowledge of statistics learned in sixth grade. They determine the typical class score from a sample of the population, and reason about the representativeness of the sample. Then, students begin to develop intuition about appropriate sample size by conducting an experiment. They compare different sample sizes, and decide whether increasing the sample size improves the results. In Lesson 16 and Lesson 17, students compare two data sets using any tools they wish. Students will be reminded of Mean Average Deviation (MAD), which will be a useful tool in this situation. Students complete another Self Check, review the results of their Self Check, and solve additional problems. The unit ends with three days for students to work on Gallery problems, possibly using one of the days to complete their project or get help on their project if needed, two days for students to present their unit projects to the class, and one day for the End of Unit Assessment.
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In this group task students collect data and analyze from the class to answer the question "is there an association between whether a student plays a sport and whether he or she plays a musical instrument? "
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This lesson provides students with an introduction to exponential functions. The class first explores the world population since 1650. Students then conduct a simulation in which a population grows at a random yet predictable rate. Both situations are examples of exponential growth.
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A work in progress, this FlexBook is an introduction to theoretical probability and data organization. Students learn about events, conditions, random variables, and graphs and tables that allow them to manage data.
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This lesson unit is intended to help teachers assess how well students are able to translate between graphs and algebraic representations of polynomials. In particular, this unit aims to help you identify and assist students who have difficulties in: recognizing the connection between the zeros of polynomials when suitable factorizations are available, and graphs of the functions defined by polynomials; and recognizing the connection between transformations of the graphs and transformations of the functions obtained by replacing f(x) by f(x + k), f(x) + k, -f(x), f(-x).
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Place and Location are two of the five themes of geography and a natural starting point for a study of the Arctic and Antarctica. Location answers the question, "Where am I?" while the study of place asks, "What kind of a place is it?" and, "How does this place connect to my hometown?" This issue of Beyond Penguins and Polar Bears examines how you can introduce the Arctic and Antarctica and use science, geography, literacy, and technology to help your students compare and contrast these two dramatically different areas as well as their own home. Get ready for an adventure as you and your students develop your polar sense of place!
Beyond Penguins and Polar Bears is an online professional development magazine for elementary teachers which focuses on preparing teachers to teach polar science concepts in an already congested curriculum by integrating inquiry-based science with literacy teaching. Such an integrated approach can increase students' science knowledge, academic language, reading comprehension, and written and oral discourse abilities.
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This lesson is from Tools 4 NC Teachers. This lesson involves determining whether data is categorical, numerical, or changes over time. This is remixable.
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This lesson is from Tools4NCTeachers.In this lesson, students build pattern block animals inspired by literature, describe and graph the shapes used, and interpret the data by asking and answering questions to develop the concepts of geometry and graphing. Remix this lesson to include extension ideas or samples of student work.
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This article provides ideas, lessons and resources on how elementary teachers can integrate map skills, math, and art into lessons about the geography of the Arctic and Antarctica.
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07/30/2019
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This is a simple task about interpreting the graph of a function in terms of the relationship between quantities that it represents.
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Review all graphs included in Data Packet in order to understand juncos migrating patterns.
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Biology
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Analyze the graphs. The top graph is the buffalo population graph you have previously examined. The bottom graph is the wildebeest population in the Serengeti during the same time period. What patterns do you notice in the table and graph?
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Biology
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In this task students interpret two graphs that look the same but show very different quantities. The first graph gives information about how fast a car is moving while the second graph gives information about the position of the car. This problem works well to generate a class or small group discussion. Students learn that graphs tell stories and have to be interpreted by carefully thinking about the quantities shown.
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Illustrative Mathematics
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Illustrative Mathematics
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# Can Principal Moments of Inertia Exceed the Sum of the Others?
• knightpraetor
In summary, the principal moment of inertia is a physical property that describes an object's resistance to rotational motion and is calculated by summing the products of the object's mass and distance from the axis of rotation. Factors that affect it include mass distribution, distance from the rotation axis, and orientation. It is used in engineering and physics to analyze stability and motion, and is different from the more general term "moment of inertia."
knightpraetor
How do i go about proving that none of the principal moments of inertia can exceed the sum of the other two?
Someone suggested the triangle inequality, but i don't understand how to use it.
Just write out the definitions:
$$I_x=\int (y^2+z^2)dm$$
$$I_y=\int (x^2+z^2)dm$$
$$I_z=\int (x^2+y^2)dm$$
and fiddle around with that.
## What is the principal moment of inertia and why is it important?
The principal moment of inertia is a physical property of an object that describes its resistance to rotational motion. It is important because it helps determine the object's stability, how it will respond to external forces, and how it will rotate around different axes.
## How is the principal moment of inertia calculated?
The principal moment of inertia is calculated by summing the products of the mass of each particle in the object and the square of its distance from the axis of rotation. This calculation is typically done using integrals in calculus.
## What factors affect the principal moment of inertia?
The principal moment of inertia is affected by the mass distribution of the object, the distance of the mass from the axis of rotation, and the orientation of the object. Objects with a larger mass and/or mass that is farther from the axis of rotation will have a larger moment of inertia.
## How is the principal moment of inertia used in engineering and physics?
The principal moment of inertia is used in engineering and physics to analyze the stability and motion of objects, such as buildings, bridges, and vehicles. It is also used in the design of rotating machinery, such as turbines and engines.
## What is the difference between principal moment of inertia and moment of inertia?
The principal moment of inertia is a specific type of moment of inertia that describes the object's resistance to rotational motion around its principal axes. The moment of inertia, on the other hand, is a more general term that describes an object's resistance to rotational motion around any axis. The principal moment of inertia is typically used when analyzing objects with a defined orientation, while the moment of inertia can be used for any type of rotational motion.
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## Beginner's Marathon
For students of class 6-8 (age 12 to 14)
Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm
### Re: Beginner's Marathon
They start, from Mirpur, one with the bike($A$) and the other($B$) on foot, and from Lalmatia($C$) on foot of course. After one hour $A$ will give the bike to $C$, and $B$ will stop, and wait for the bike. $A$ will reach Lamatia after 2 hours. $C$ rides the bike until he meets $B$ , and gives it to him. After this they'll reach their targets in one hour, so $2$:$40$ in total.
Frankly, my dear, I don't give a damn.
Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm
### Re: Beginner's Marathon
Problem $23$
Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.
Frankly, my dear, I don't give a damn.
dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm
### Re: Beginner's Marathon
$\text{Problem 23}$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm
### Re: Beginner's Marathon
$\text{Problem 24}$
Let $n$ be a positive integer and let $a_1, a_2,.....a_k$(here $k$ > 1) be distinct integers in the set {${1,2.....n}$} such that $n$ divides $a_i(a_{i+1}-1)$ for $i = 1,2,.....k-1$. Prove that $n$ does not divide $a_k(a_1 - 1)$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm
### Re: Beginner's Marathon
Solution to problem $24$ (OVERKILL)
Assume that $n$ divides $a_k(a_1-1)$.
Assume that $(a_i, n)=1$. Then $n| a_i(a_{i+1}-1)$ implies $n|a_{i+1}-1$ which implies $a_{i+1}=1$. Then $n|1(a_{i+2}-1)$ which is only possible if $a_{i+2}=1$ but this contradicts the distinction.
So there exist some prime $p$ such that $p|n$ and $p|a_i$.
Now we inductively prove that $p$ divides all $a_j$ such that $1\leq j \leq n$. We have already shown the base case. Now assume that $p|a_m$. We have, $p |n |a_{m-1}(a_m-1)$. So, we must have $p|a_{m-1}$. Here, the indices are taken modulo $k$. So, $p$ divides all $a_j$ such that $1\leq j \leq n$. Note that the induction was done by starting from $a_i$. We get that if some prime divides at least one of the $a_j$'s, that prime divides all the $a_j$'s.
We will strengthen the statement more with dealing with prime powers.
Assume that $p^x||n$. Then,as $p^x||n|a_j(a_{j+1}-1)$ for all $j$, $p^x|a_j$ for all $j$.
So, we get that $(n,a_1)=(n,a_2)=.......=(n,a_k)= C$ where C is an integer greater than $1$.
Let, $b_j=\frac {a_j}{C}$ for all $j$. Also let, $n_1=\frac {n}{C}$. Then, we have $(n_1, C)=(n_1,b_j)=1$ for all $j$.
Now, as $n_1C|b_jC(b_{j+1}C-1)$, we get $n_1|b_{j+1}C-1$ for all $j$. So, $n_1|b_xC-1-b_yC+1=C(b_x-b_y)$.
We get that $n_1|b_x-b_y$. Which is impossible since $b_x, b_y$ are different integers less than $n_1$. So we get a contradiction. So our first assumption is false. Now, this solution is really unclear. Someone tell me where does the logic breaks if our first assumption is false.
Last edited by ahmedittihad on Mon Jun 26, 2017 4:00 am, edited 1 time in total.
Frankly, my dear, I don't give a damn.
Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm
### Re: Beginner's Marathon
Solution to Problem $24$ (NOT OVERKILL)
As the previous solution we solve by contradiction.
So, assume that $a_1a_k \equiv a_k (mod n)$.
$a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k-2}\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k$.
Therefore, $0 < |a_1-a_k| < n$ is divisible by $n$. Contradiction.
EID MUBARAK
Frankly, my dear, I don't give a damn.
Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm
### Re: Beginner's Marathon
As this is the Beginner's Marathon, I request everyone to not give shortlist problems.
Problem $25$
Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $\Omega$ is a circle passing through $A,B,C,D$. Let $\omega$ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
Frankly, my dear, I don't give a damn.
dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm
### Re: Beginner's Marathon
Let $AD \cap \Omega = P$. Now, note that $AP = BB_1, AA_1 = CA_2$ and $BB_1 = CB_2$. Now, power of point implies that $AP.AD = AC.AA_1 \Rightarrow BB_1.BC = CA_2.AC \Rightarrow CA_2.AC=CB_2.BC$. Therefore, $A,B,A_2,B_2$ are cyclic.
Somebody post the next problem.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
- Charles Caleb Colton
aritra barua
Posts: 57
Joined: Sun Dec 11, 2016 2:01 pm
### Re: Beginner's Marathon
Problem $26$
Given an acute angled triangle $\bigtriangleup ABC$.Inscribe in it a triangle $UVW$ of least possible perimeter.
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# How can obtain the relative orientation between two quaternions?
How do I obtain the relative orientation given two orientations (represented by quaternions q0 and q1)?
• not sure what you are asking here, since you are not telling how did you calculate the quaternion in the first place. but my answer here might be related gamedev.stackexchange.com/questions/67199/… and btw you are mixing euler angles with quaternions which is sth I also addressed in the answer. – concept3d Jan 1 '14 at 23:09
• which API are you using? it is supposed to be stated in the documentation of how do you interpret orientation. – concept3d Jan 1 '14 at 23:37
• from what I understand you can't extract euler angles in the traditional sense. Try converting the quaternion to a matrix and there is a way of analyzing that matrix to extract the euler angles which may not work. But TBH I feel you are asking the wrong question. why do you want to extract pitch in the first place ? – concept3d Jan 1 '14 at 23:48
Quaternion is another representation of axis angle. The solution is to create a new quaternion from the original quaternion that only has the needed components.
An axis angle representation can be converted to a quaternion using the following formula
q[0] = cos(R/2);
q[1] = sin(R/2)*x;
q[2] = sin(R/2)*y;
q[3] = sin(R/2)*z;
Where R is the angle in radians, and (x,y,z) represents the axis, and quaternion is (R,x,y,z).
So in order to create a new quaternion with only the pitch component you just zero out the other components and normalize the quaternion:
Quaternion q; // this is your original quaternion
q.x = 0.0;
q.z = 0.0;
q.Normalize();
Sensors AFAIK calculate orientation relative to gravity, in other words Y (or Z) is the direction of the gravity. You need take that into consideration. And I think you don't need to multiply it with the inverse initial orientation.
• This works great if I know that the motion is pitch. In general, that is unknown. – ilp Jan 2 '14 at 0:32
• @ilp well, what you can do, is extract the component with the highest contribution to the rotation and use that. So instead of taking a fixed axis. Use the one with the highest contribution and zero out the others. – concept3d Jan 2 '14 at 0:35
• @ilp at this point I can't answer your question, because I don't know the neccessary details :-) maybe you should edit the question and give it a context so we are able to help. – concept3d Jan 2 '14 at 0:43
• @ilp edited the answer. – concept3d Jan 2 '14 at 1:12
The relative orientation is obtained simply by division:
q = q0 / q1
Or, if division is not available:
q = q0 * inverse(q1)
Note that since the quaternions used to represent rotations are unit quaternions, the inverse of q1 is simply its conjugate q1*, and is obtained by flipping the sign of x, y, z but not w.
• if you checked the edits, he is already doing that q = q0 * inverse(q1), so am not particularly sure what he is trying to do. – concept3d Jan 2 '14 at 10:54
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https://proofwiki.org/wiki/22
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# 22
Previous ... Next
## Number
$22$ (twenty-two) is:
$2 \times 11$
The $1$st positive integer which can be expressed as the sum of $2$ odd primes in $3$ ways:
$22 = 19 + 3 = 17 + 5 = 11 + 11$
The $2$nd integer after $1$ which equals the number of digits in its factorial:
$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$
which has $22$ digits
The $2$nd Smith number after $4$:
$2 + 2 = 2 + 1 + 1 = 4$
The $3$rd pentagonal number after $1$, $5$ which is also palindromic:
$22 = 1 + 4 + 7 + 10 = \dfrac {4 \left({3 \times 4 - 1}\right)} 2$
The $3$rd hexagonal pyramidal number after $1$, $7$:
$22 = 1 + 6 + 15$
The $3$rd number after $1$, $3$ whose $\sigma$ value is square:
$\sigma \left({22}\right) = 36 = 6^2$
The $4$th pentagonal number after $1$, $5$, $12$:
$22 = 1 + 4 + 7 + 10 = \dfrac {4 \left({3 \times 4 - 1}\right)} 2$
The $7$th generalized pentagonal number after $1$, $2$, $5$, $7$, $12$, $15$:
$22 = \dfrac {4 \left({3 \times 4 - 1}\right)} 2$
The $8$th semiprime after $4$, $6$, $9$, $10$, $14$, $15$, $21$:
$22 = 2 \times 11$
The $8$th of $35$ integers less than $91$ to which $91$ itself is a Fermat pseudoprime:
$3$, $4$, $9$, $10$, $12$, $16$, $17$, $22$, $\ldots$
The $10$th even number after $2$, $4$, $6$, $8$, $10$, $12$, $14$, $16$, $20$ which cannot be expressed as the sum of $2$ composite odd numbers.
The $11$th positive integer which is not the sum of $1$ or more distinct squares:
$2$, $3$, $6$, $7$, $8$, $11$, $12$, $15$, $18$, $19$, $22$, $\ldots$
$22$ is a palindromic number whose square is also palindromic:
$22^2 = 484$
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http://www.brighthubeducation.com/lesson-plans-grades-1-2/27382-how-to-read-a-ruler/
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written by: Kathy Foust • edited by: Trent Lorcher • updated: 2/16/2012
All items on a ruler can be daunting to young learners. This activity lets students use a hands on method that is fun and educational
• slide 1 of 1
### Introduction to Rulers
By now you may have already done the previous lesson plans in this series (see the links below). If so, your students have seen you use a ruler and have been introduced to how to use a ruler. Now it's time to get more detailed about rulers and similar instruments. For this lesson plan you will need the items listed below.
• Ruler
• Folding Yard Stick
• Tape measure
This lesson is based in part on how information is processed in our brains. Americans are still using the ruler as a standard tool of measurement because whether we realize it or not, it is easier for us to imagine the units because they are based on our bodies, where the metric system was once based on the distance from the earth to the sun. I don't know about you, but that's a measurement I can't easily wrap my mind around and I don't expect my students to be able to either.
Explain the history of measuring tools. Let them know that a foot is based on the size of a man's foot, so that when there were no measuring tools available, a man could walk toe to toe across an area to get a measurement. A yard is based on the space between the average man's normal footsteps. Display this with the ruler and the yardstick. Of course, the measurements may not be exact, but it gives your students a good visual concept. As a point of interest, let them know that horses still today are measured in hands. explain this concept to your students, but don't focus too long on it.
Display the 3 types of measuring tools you have. Explain applications for each one. Take measurements around the classroom of various objects to demonstrate the use. When measuring, write on the board what the class estimates something to measure. Keep it simple. Start off with estimate measurements. This means you may measure a book and say it is about a foot long. Then, get more precise.
Write the phrase "out of 16" on the board. When you measure something and you come up with say 2 3/4 inches, write the number 2 above the "out of 16". Next, write 9 in front of the out of sixteen. You should be measuring with the students and counting the notches on the ruler or other measuring tool so that students can understand how you came up with this number.
It's best to not reduce the fraction unless you have already gone over that material with your students. Or, if you want the reduced number, then count with the students the size line corresponding with that number. For 3/4, you should have the students focus on counting the larger lines that correspond with the division of four on the ruler.
Have the students think of 3 items in their home that they can measure with a ruler. Write them down. Have students write down what they think the item measures, then go home and measure the items with a ruler.
#### Second Grade Math Lesson Plans for Measuring
This series is dedicated to lesson plans that begin with an introduction to measuring and extend through using specific measuring tools.
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https://math.stackexchange.com/questions/1657001/argmax-of-integral-under-log-transformation
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# argmax of integral under log transformation
I am trying the find $x^*$ defined as: $$x^* = argmax_x \int f(x,y)p(y)dy$$ where $f(x,y)$ can be an arbitrary positive function and $p(y)$ is a probability density.
Does the location of $x^*$ hold true under a log transformation of $f$? Specifically, does the previous equation imply: $$x^* = argmax_x \int log \{f(x,y)\} p(y)dy$$
I doubt that this is true, because of the $p(y)$ term in the integral, but I can't quite pin down why not. A formal proof is not necessary - some intuitive explanation (or counterexample) would suffice.
• $\log(f(x, y))$ needs not be integrable to begin with. – user251257 Feb 16 '16 at 1:17
• True, but in this case lets assume that the log transformed integrand has an analytical solution (which is one of the reasons we want to do a log transform), while the original formulation may or may not have one. – Supratik Paul Feb 16 '16 at 11:51
On $[\frac12, 1]^2$ let $f(x,y) = y$ for $\frac12\le y \le x \le 1$, $f(x,y) = 0$ else. Let $p(y) = 2$ for $\frac12\le y \le 1$. Then, we have $$g(x) := \int_{1/2}^1 f(x, y) p(y) dy = 2\int_{1/2}^x \underbrace{y}_{\ge 0}\, dy$$ and $\operatorname{argmax} g = 1$. On other hand, we have $$h(x) := \int_{1/2}^1 \log(f(x,y)) p(y) dy = 2\int_{1/2}^x \underbrace{\log(y)}_{\le 0} dy$$ and thus $\operatorname{argmax} h = \frac{1}{2}$.
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http://hotmath.com/hotmath_help/topics/multiplying-and-dividing-negatives.html
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# Multiplying and Dividing with Negatives
## Multiplying with Negatives
There's a simple rule here:
• (positive) × (positive) = (positive)
• (positive) × (negative) = (negative)
• (negative) × (positive) = (negative)
• (negative) × (negative) = (positive)
Example:
• (3)(7) = 21
• (3)(–7) = –21
• (–3)(7) = –21
• (–3)(–7) = 21
Important Thing To Notice:
Since a positive times a positive is positive, and a negative times a negative is also positive, it follows that any number times itself is either zero or positive.
• (12)(12) = 144
• (–12)(–12) = 144
## Dividing with Negatives
The same rule works with division:
• (positive) ÷ (positive) = (positive)
• (positive) ÷ (negative) = (negative)
• (negative) ÷ (positive) = (negative)
• (negative) ÷ (negative) = (positive)
Example:
• 56 ÷ 7 = 8
• 56 ÷ (–7) = –8
• –56 ÷ 7 = –8
• –56 ÷ (–7) = 8
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http://www.math.niu.edu/~beachy/abstract_algebra/study_guide/32.html
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Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996
## § 3.2 Subgroups
Definition 3.2.1. Let G be a group, and let H be a subset of G. Then H is called a subgroup of G if H is itself a group, under the operation induced by G.
Example 3.2.1. Q× and R× are subgroups of C×, the multiplicative group of complex numbers.
Example 3.2.2. SLn(R), the set of all n × n matrices over R with determinant 1, is a subgroup of GLn(R).
Proposition 3.2.2. Let G be a group with identity element e, and let H be a subset of G. Then H is a subgroup of G if and only if the following conditions hold:
(i) for all a,b in H, the product ab is in H;
(ii) e belongs to H;
(iii) for all a in H, the inverse a-1 is in H.
Corollary 3.2.3. Let G be a group and let H be a subset of G. Then H is a subgroup of G if and only if H is nonempty and ab-1 is in H for all a,b in H.
Corollary 3.2.4. Let G be a group, and let H be a finite, nonempty subset of G. Then H is a subgroup of G if and only if ab is in H for all a,b in H.
Definition 3.2.5. Let G be a group, and let a be any element of G. The set
<a> = { x in G | x = an for some n in Z }
is called the cyclic subgroup generated by a.
The group G is called a cyclic group if there exists an element a in G such that G=<a>. In this case a is called a generator of G.
Proposition 3.2.6. Let G be a group, and let a be an element of G.
(a) The set <a> is a subgroup of G.
(b) If K is any subgroup of G such that a is an element of K, then <a> K.
Examples 3.2.7 - 3.2.9. Z and Zn are cyclic groups, but Zn× may not be cyclic. (See Corollary 7.5.11.)
Definition 3.2.7. Let a be an element of the group G. If there exists a positive integer n such that an = e, then a is said to have finite order, and the smallest such positive integer is called the order of a, denoted by o(a).
If there does not exist a positive integer n such that an = e, then a is said to have infinite order.
Proposition 3.2.8. Let a be an element of the group G.
(a) If a has infinite order, and ak = am for integers k,m, them k=m.
(b) If a has finite order and k is any integer, then ak = e if and only if o(a) | k.
(c) If a has finite order o(a)=n, then for all integers k, m, we have
ak = am if and only if k m (mod n).
Furthermore, | <a> | = o(a).
Lemma 3.2.9. Let H be a subgroup of the group G. For a,b in G define a ~ b if ab-1 is in H. Then ~ is an equivalence relation.
Theorem 3.2.10 (Lagrange). If H is a subgroup of the finite group G, then the order of H is a divisor of the order of G.
Corollary 3.2.11. Let G be a finite group of order n.
(a) For any a in G, o(a) is a divisor of n.
(b) For any a in G, an = e.
Example 3.2.12. (Euler's theorem) Let G be the multiplicative group of congruence classes modulo n. The order of G is given by (n), and so by Corollary 3.2.11, raising any congruence class to the power (n) must give the identity element.
Corollary 3.2.12. Any group of prime order is cyclic.
## § 3.2 Subgroups: Solved problems
If the idea of a subgroup reminds you of studying subspaces in your linear algebra course, you are right. If you only look at the operation of addition in a vector space, it forms an abelian group, and any subspace is automatically a subgroup. Now might be a good time to pick up your linear algebra text and review vector spaces and subspaces.
Lagrange's theorem is very important. It tells us that in a finite group the number of elements in any subgroup must be a divisor of the total number of elements in the group. This is a useful fact to know when you are looking for subgroups in a given group.
It is also important to remember that every element a in a group G defines a subgroup < a > consisting of all powers (positive and negative) of the element. This subgroup has o(a) elements, where o(a) is the order of a. If the group is finite, then you only need to look at positive powers, since in that case the inverse a-1 of any element can be expressed in the form an, for some n > 0.
For a group G, the centralizer C(a) of an element a in G is defined in Exercise 3.2.14 of the text as
C(a) = { x in G | xa = ax } .
Exercise 3.2.14 shows that C(a) is a subgroup of G that contains <a>.
The center of the group G, denoted by Z(G), is defined in Exercise 3.2.16 as
Z(G) = { x in G | xg = gx for all g in G } .
Exercise 3.2.16 shows that Z(G) is a subgroup of G.
23. Find all cyclic subgroups of Z24×. Solution
24. In Z20×, find two subgroups of order 4, one that is cyclic and one that is not cyclic. Solution
25. (a) Find the cyclic subgroup of S7 generated by the element (1,2,3)(5,7).
(b) Find a subgroup of S7 that contains 12 elements. You do not have to list all of the elements if you can explain why there must be 12, and why they must form a subgroup.
Solution
26. In G = Z21×, show that
H = { [x]21 | x 1 (mod 3) } and K = { [x]21 | x 1 (mod 7) }
are subgroups of G. Solution
27. Let G be an abelian group, and let n be a fixed positive integer. Show that
N = { g in G | g = an for some a in G }
is a subgroup of G. Solution
28. Suppose that p is a prime number of the form p = 2n + 1.
(a) Show that in Zp× the order of [2]p is 2n.
(b) Use part (a) to prove that n must be a power of 2.
Solution
29. In the multiplicative group C× of complex numbers, find the order of the elements
= and = .
Solution
30. Let K be the following subset of GL2 (R).
K =
Show that K is a subgroup of GL2 (R). Solution
31. Compute the centralizer in GL2 ( R) of the matrix . Solution
32. Let G be the subgroup of GL2 (R) defined by G = .
Let A = and B = .
Find the centralizers C(A) and C(B), and show that C(A) C(B) = Z(G), where Z(G) is the center of G. Solution
## § 3.2 Lab questions
To answer these questions experimentally, you can use the Groups15 Java applet written by John Wavrik of UCSD.
Lab 1. For each group G of order 8 on the list of groups given by Groups15, find the elements of order 2. Do these elements (together with the identity element) form a subgroup of G?
Lab 2. For each group G of order n < 16, find the largest possible order of an element a in G.
Lab 3. Find all groups G of order n < 16 for which there is a divisor m of n, but no corresponding subgroup of G of order m. (These are the groups of order 15 or less for which the converse of Lagrange's theorem fails.)
Lab 4. If |G| = n, where n < 16, and p is a prime divisor of n, can you always find a subgroup H of G with |H| = p? What can you say if you ask the question for prime power? (You only need to worry about n = 8 and n = 12.)
Lab 5. List the groups G in Groups15 for which the center
Z(G) = { x in G | xg = gx for all g in G }
contains only the identity element.
Lab 6. For a group G, the centralizer of an element a in G is defined as
C(a) = { x in G | xa = ax } .
In the group of order 12 called A4 in Groups15, find the centralizers C(B) and C(D) of the elements listed as B and D.
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https://www.cram.com/flashcards/work-and-power-212526
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### 30 Cards in this Set
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What takes more power, to lift an object 1 meter high in 2 minutes or in 2 seconds? 2 seconds - because the same amount of work is done in less time What is the formula for power rearranged for work? work = power x time W = P x t If time and force remain the same, power is increased when the __________ increases. (HINT: Write the formula for power, then substitute in the formula for work. Evaluate each variable.) work What is POWER? The rate at which work is done, or the amount of work per unit time. Forty watts of power is required to do work on an object for ten seconds. How much power is required to do the same amount of work in 5 seconds? 80 watts It takes twice as much power to do the same amount of work in half the time. For a given amount of work, power and time behave in opposite ways. (They have an inverse relationship. If one increases the other decreases by the same factor.For example, time was divided by two so power was multiplied by two to keep the amount of work the same.) For work to be done, the force applied to an object and the distance over which it acts must be in the same ______________. direction How much work is done when a 1 kg mass is raised a vertical distance of 1 meter? 1 joule What UNIT is WORK measured in? joule (J) A pulley is used to lift a 100 kg mass 5 m. How much work is done? W = m x g (This W is weight.) W = 100 kg x 9.8 m/s2 W = 980 N W = F x d (This W is work.) W = 980 N x 5 m W = 4,900 Nm ANSWER: W = 4,900 J What is the FORMULA for WORK? work = force x distance OR W = F x d A 15 kg rock is lifted 4 m in 6 sec. How much power is consumed? W = m x g = 15 kg x 9.8 m/s2 = 142.5 N P = W/t = (F x d)/t = (142.5 N x 4 m)/6 s = 579 J / 6 s ANSWER: 95 watts (W) What is WORK? Energy transferred through motion or a force acting through a distance. A teacher pushed a 12 kg desk across the floor for a distance of 10 m. She exerted a force of 45 N. How much work was done? W = F x d = 45 N x 10 m = 450 Nm ANSWER: W = 450 J * Note: The weight of the desk is not the resistance because an equal and opposite force is exerted on it by the floor (Newton's third law.) Instead, the teacher must overcome the inertia of the desk and friction. How much work is done when a crane lifts a 30,000 N object 25 m ? W = F x d W = 30,000 N x 25 m W = 750,000 Nm ANSWER: W = 750,000 J A weight lifter lifts a 200 kg barbell from the floor to above his head a distance of 2.6 m. He holds the barbell for 5 s. How much work is done during the 5 second interval? No work is done (0 J). The barbell is not moving. What is the formula for power rearranged for time? time = work / power t = W / P What are the SI UNITS of POWER? Watts (W) In terms of the SI units of force and distance, what is one joule of work equivalent to? One joule is the amount of work that is done in moving an object a distance of one meter with a force of one Newton. joule = Newton x meter (J = N x m) In terms of the SI units of work and time, what is one watt of power equivalent to? One watt is the power needed to do one joule of work in one second. watt = joule / second ( W = J/s ) What is the formula for work rearranged for force? force = work / distance F = W / d A woman carries a bag of groceries home. Is work being done on the groceries? Why or why not? No, work is not being done on the groceries because the direction of the force on the bags is upward and the direction of motion is forward. Since theses directions are not the same, work is not being done. A cable car at Stone Mountain moves along an incline to transport passengers up the mountain. Is work being done if the direction of force exerted on the passengers (upward) is not the same as the direction of motion (at an incline)? Explain. Yes, work is being done because part of the force being applied to lift the passengers up is in the direction of motion of the car. Or, conversely, part of the displacement vector acts along the direction of force. If force and distance are ________________ to eachother, then no work is done. perpendicular A forklift is used to move two boxes that weigh and 20 N and 30 N, respectively, from the floor to a shelf that is 10 m high. How much work is done? The combined weight of the boxes is 50 N. Therefore, the forklift must apply an upward force of 50 N to overcome the downward weight of the two boxes and move them at a constant speed. W = F x d W = 50 N x 10 m W = 500 Nm ANSWER: W = 500 J What is the formula for work rearranged for distance? distance = work / force d = W / F How much work is done to carry a bookbag weighing 30 N on your shoulders home a distance of 500 meters? None. Force and distance are not in the same direction. Because they are perpendicular to eachother, no work is being done. As long as part of a force acts in the direction of motion, some work is done. However, to maximize the amount of work that is done using a given force, the direction of force and the distance must be _____________ to eachother. parallel What is the FORMULA for POWER? power = work / time OR P = W / t A girl climbs a tree by carefully pulling herself up the trunk. Is she doing work? Why or why not? Yes, work is being done because the direction of force and the distance over which it acts are the same. Both are upward. What is the SI unit of time used to calculate power? seconds (s)
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## Algebra and Trigonometry 10th Edition
$x=0$ $x=\pi$ $x=\frac{\pi}{2}$
$cos^2x+sin^2x=1$ $cos^2x=1-sin^2x$ $cos^2x+sin~x=1$ $1-sin^2x+sin~x=1$ $-sin^2x+sin~x=0$ $-sin~x(sin~x-1)=0$ $-sin~x=0$ $sin~x=0$ $x=0$ or $x=\pi$ $sin~x-1=0$ $sin~x=1$ $x=\frac{\pi}{2}$
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# log
• Nov 15th 2012, 08:21 AM
Petrus
log
ima solve al none negative solution för this equation ehmm idk how to write it but its ^(4) *logx*^(8)*log x = ^(16)*log x idk how to solve this:(
• Nov 15th 2012, 12:21 PM
HallsofIvy
Re: log
I'm not sure what the "^4" means but I will guess that it is the base of the logarithm. I am used to that being written as a subscript, after. If that is correct then the equation is $\displaystyle log_4(x)(log_8(x))= log_16(x)$.
Now, "$\displaystyle y= log_a(x)$" is the same as "$\displaystyle a= e^y$". Do you notice that all of those bases are powers of 2? $\displaystyle 4= 2^2$, $\displaystyle 8= 2^3$ and $\displaystyle 16= 2^4$. So $\displaystyle y= log_4(x)$ is the same as $\displaystyle x= 4^y= (2^2)^y= 2^{2y}$ and so $\displaystyle log_2(x)= 2y$, thus $\displaystyle y= log_{4}(2)= \frac{log_2(x)}{2}$. $\displaystyle y= log_8(x)$, so $\displaystyle x= 8^y= (2^3)^y= 2^{3y}$, thus $\displaystyle 3y= log_2(x)$ and $\displaystyle y= log_8(x)= \frac{log_2(x)}{3}$. Similarly, $\displaystyle log_{16}(x)= \frac{log_2(x)}{4}$ so the equation can be written
$\displaystyle \frac{log_2(x)}{2}\frac{ln_2(x)}{3}= \frac{ln_2(x)}{16}$
If we let $\displaystyle y= log_2(x)$ that is the same as $\displaystyle \frac{y}{2}\frac{y}{3}= \frac{y^2}{6}= \frac{y}{4}$.
Can you solve that?
• Nov 15th 2012, 12:25 PM
MarkFL
Re: log
Do you mean:
$\displaystyle \log_4(x)\cdot\log_8(x)=\log_{16}(x)$ ?
If so, try the identity:
$\displaystyle \log_{b^n}(a)= \frac{{\log_b(a)}}{n}$
• Nov 15th 2012, 12:27 PM
Petrus
Re: log
Quote:
Originally Posted by MarkFL2
Do you mean:
$\displaystyle \log_4(x)\cdot\log_8(x)=\log_{16}(x)$ ?
If so, try the identity:
$\displaystyle \log_{b^n}(a)= \frac{{\log_b(a)}}{n}$
nvm did not see halloy..
• Nov 15th 2012, 12:45 PM
skeeter
Re: log
Quote:
Originally Posted by Petrus
ima solve al none negative solution för this equation ehmm idk how to write it but its ^(4) *logx*^(8)*log x = ^(16)*log x idk how to solve this
please do not use "text talk" abbreviations ... it makes the post confusing and difficult to read.
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# Two Step Equations With Fractions And Decimals Worksheet
Two Step Equations With Fractions And Decimals Worksheet – Excellent, fun and free of charge math worksheets should be able to existing a numerical issue in a different way. Math is all things considered simply a numeric expression of a number of life’s most basic queries: How much money do I have left basically if i get a soft drinks? In the end of the week, how much of my every day allowance can i be able to conserve should i don’t?
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# A Course of Pure Geometry by E. Askwith
Similar geometry books
Geometry and Trigonometry for Calculus: A Self-Teaching Guide
If you would like geometry and trigonometry as a device for technical paintings … as a refresher direction … or as a prerequisite for calculus, here’s a brief, effective approach that you should study it! With this publication, you could educate your self the basics of aircraft geometry, trigonometry, and analytic geometry … and learn the way those issues relate to what you know approximately algebra and what you’d prefer to find out about calculus.
Independent Axioms for Minkowski Space-Time
The first objective of this monograph is to elucidate the undefined primitive thoughts and the axioms which shape the root of Einstein's thought of precise relativity. Minkowski space-time is constructed from a collection of autonomous axioms, said by way of a unmarried relation of betweenness. it's proven that each one versions are isomorphic to the standard coordinate version, and the axioms are constant relative to the reals.
Extra info for A Course of Pure Geometry
Sample text
Euclidean and Hyperbolic Geometry Lemma 10. Let a = 0 be an element of X and τ > 1 be a real number. 17) 1 + a2 . τ 2α = (τ − 1) Proof. Since {x√∈ X | x −√ce− + x − ce get with c := a τ , := ln τ , obviously, a = ce− , τ a = ce , 2α = (e − e− ) = 2 sinh · √ 1 + c2 } is B (c, ), we 1 + c2 = (τ − 1) 1 + a2 . τ Proposition 11. Suppose that B (c, ), B (c , ) are hyperbolic balls satisfying and B (c, ) ⊆ B (c , ). 18) . Proof. 18), c = c and > 0. a, a motion µ such that µ (c) = 0, µ (c ) = λj, λ > 0.
5. Balls, hyperplanes, subspaces 49 a euclidean hyperplane of X. If e ∈ X satisfies e2 = 1, if t ∈ R and ω1 , ω2 ∈ O (X), then ω1 Tt ω2 (e⊥ ) = {ω1 Tt ω2 (x) | x ∈ e⊥ } will be called a hyperbolic √ hyperplane, where {Tt | t ∈ R} is based on the axis e and the kernel sinh · 1 + h2 . Of course, mutatis mutandis, also the euclidean hyperplanes can be described this way. In Proposition 17 parametric representations of hyperbolic hyperplanes will be given. Proposition 12. If H (a, α) and H (b, β) are euclidean hyperplanes with H (a, α) ⊆ H (b, β), then H (a, α) = H (b, β) and there exists a real λ = 0 with b = λa and β = λα.
E. 42). 42). e. 41), α (1 + δβ) − β (1 + δα) 2 = ϕ2 (ξ − η). e. α (1 + δβ) > β (1 + δα). 45) 30 Chapter 1. Translation Groups holds true for all ξ > η ≥ 0. 45) also holds true for ξ = η ≥ 0. a), we will distinguish two cases, namely δ = 0 and δ > 0. 45) yields ϕ (ξ − η) = ϕ (ξ) − ϕ (η) for all ξ ≥ η ≥ 0. Given arbitrarily t, s ∈ R≥0 , put ξ := t + s and η := s. Hence ξ ≥ η ≥ 0 and thus ϕ (t + s) = ϕ (t) + ϕ (s). 46) for all t ∈ R≥0 with a constant l > 0, in view of ϕ (1) > ϕ (0) = 0. a we get ψ (h) = 1 for all h ∈ H.
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# Weak inductive argument. LESSON # 4 2019-01-21
Weak inductive argument Rating: 6,5/10 1561 reviews
## How to Write an Effective Inductive Argument
And the conclusion may turn out to be true John doubts it. Inducing is 'bringing about', while reasoning is 'arriving at a conclusion'. This argument is inductively weak even if the two premises are true. Therefore, if Olivia purchases a pair of glasses identical to Dorothy's, her vision should be as good as Dorothy's is now. So, you are allergic to peanuts. This strikes to the heart of what determines the strength of an inductive argument. Considering the next example: All humans are reptiles.
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## Examples of Inductive Reasoning
And, sometimes when new information becomes available, an argument that used to look sound doesn't look so good anymore. Reasoning that the mind must contain its own categories organizing , making experience of space and time possible, Kant concluded a priori. The chair in the dining room is red. Such statements are logically correct. Deductive Arguments A argument is either or.
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## 2. Inductive Arguments and Strong Reasoning
Comparison chart Deductive versus Inductive comparison chart Deductive Inductive Introduction from Wikipedia Deductive reasoning, also called deductive logic, is the process of reasoning from one or more general statements regarding what is known to reach a logically certain conclusion. Some additional notes: an argument that misuses a form what we will call a formal fallacy may not be valid but then we need to look at it in terms of inductive strength. For example: All reptiles are mammals. But this is quite unlikely to happen if the lottery is indeed a fair one. If you must differentiate what is a strong or weak argument, it is critics that you carefully see through what is written by a writer.
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## [A08] Inductive Reasoning
The deductive nature of mathematical induction is based on the non-finite number of cases involved when using mathematical induction, in contrast with the finite number of cases involved in an enumerative induction procedure with a finite number of cases like. Deductive reasoning applies general rules to make conclusions about specific cases. The goal of a causal argument is to help determine the most probable cause of a specific effect. But this argument is not good support for its conclusion. The arguments resulting from such thinking are called inductive arguments. In any case, the conclusion may well end up being invalid because inductive reasoning does not guarantee validity of the conclusions.
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## What is a cogent argument?
As this 's premises, even if true, do not entail the conclusion's truth, this is a form of inductive inference. We would have a much stronger case for this if we also knew the test was comparatively easy. Like inductive generalizations, causal arguments can be applied to empirical evidence. This is a formal inductive framework that combines with the Bayesian framework. The conclusion is induced in these types of statements. Similar to the concept of soundness for deductive arguments, a strong inductive argument with true premises is termed. Most birds can fly and a penguin is a bird.
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## Critical Thinking: Understanding Inductive Arguments
In other words, it is impossible for the premises to be true but the conclusion false. If the earth was flat, then ships sailing on the ocean would fall off. There are four kinds of inductive arguments: 1. Humans are similar to chimpanzees, and therefore they tend to get violent when exposed to rage. This occurs when they can be replaced and they have to work with other causes to bring about an effect.
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## Inductive Reasoning Examples ~ To Accelerate your English Grammar
However, if we increase X to say 2000, then the inductive strength of the argument will of course increase. He is 99 and is in a coma. And last, to quantify the level of probability in any mathematical form is problematic. Going through some examples of this form of reasoning will help you get a better understanding of the concept. An invalid argument is always unsound. Second, the concluding All is a very bold assertion. Rather, explanations take the veracity of those things they are trying to explain for granted, and instead work to clarify the how and why it came to be.
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## Examples of Inductive Reasoning
~Every time you get a call from some unknown number, you find a telemarketer on the other side of the line. All arguments are either valid or invalid, and either sound or unsound; there is no middle ground, such as being somewhat valid. John Nolt, Dennis Rohatyn, Archille Varzi. The Nobel prize-winning biologist Herbert Ralls has stated that chlorinated hydrocarbons in our water supply constitute a major threat to the public health. Therefore, A is an obtuse angle. This article considers conductive arguments to be a kind of inductive argument. Bob has bought the diamond ring to give to Joan.
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## Examples of Inductive Reasoning
Both Dorothy and Olivia have trouble seeing distant objects clearly. For example, we see dark clouds in the sky and think it is likely to rain so we bring an umbrella. For example, if it is hypothesized that Sally is a sociable individual, subjects will naturally seek to confirm the premise by asking questions that would produce answers confirming that Sally is, in fact, a sociable individual. For all natural numbers n, if P holds of n then P also holds of n + 1. Here is a stronger inductive argument based on better evidence: Two independent witnesses claimed John committed the murder.
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## What is a cogent argument?
Therefore, John won't be able to attend our meeting today. An inductive argument can be affected by acquiring new premises evidence , but a deductive argument cannot be. A weak inductive argument will not do this. Just what you'd expect--you need to find other premises that support the one in question, and build an argument that proves it, and hopefully those premises will be common knowledge and acceptable. Quote of the page Problems are to the mind what exercise is to the muscles; they toughen and make strong.
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# math
posted by on .
(2/3-5/12)^2+(-5/6-2/3)^2
• math - ,
2/3 - 5/12 = 8/12 - 5/12 = 3/12 = 1/4
-5/6 - 2/3 = -5/6 - 4/6 = -9/6 = -3/2
(1/4)^2 + (-3/2)^2 = 1/16 + 9/4 = 1/16 + 36/16 = 37/16
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# Prove this is the only solution
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Nov 9th 2006, 10:22 AM
Ruichan
Prove this is the only solution
$\displaystyle a^2 + 2 = b^3$
a=5 and b=3
Prove that a=5 and b=3 are the only solutions.
Any ideas?
• Nov 9th 2006, 11:09 AM
CaptainBlack
Quote:
Originally Posted by Ruichan
$\displaystyle a^2 + 2 = b^3$
a=5 and b=3
Prove that a=5 and b=3 are the only solutions.
Any ideas?
There must be some restriction on what constitutes a solution missing here,
as otherwise a=-5, b=3 is also a solution. Perhaps you are interested in
solutions in natural numbers?
RonL
• Nov 9th 2006, 12:53 PM
Ruichan
Quote:
Originally Posted by CaptainBlack
There must be some restriction on what constitutes a solution missing here,
as otherwise a=-5, b=3 is also a solution. Perhaps you are interested in
solutions in natural numbers?
RonL
I have no idea, my prof gave us this game to play, this is the only detail he gave us. He said those are the only solutions to the above. a = 5 and b = 3
• Nov 9th 2006, 01:42 PM
CaptainBlack
Quote:
Originally Posted by Ruichan
I have no idea, my prof gave us this game to play, this is the only detail he gave us. He said those are the only solutions to the above. a = 5 and b = 3
What were you studying in class just before he set this?
(it could be a clue to what restrictions on solutions are intended)
RonL
• Nov 9th 2006, 03:23 PM
Ruichan
Quote:
Originally Posted by CaptainBlack
What were you studying in class just before he set this?
(it could be a clue to what restrictions on solutions are intended)
RonL
Actuarial Math but i don't think it's related.
Other stuffs he sort of provided are:
$\displaystyle (a^2 -25) = b^3 - 27$
$\displaystyle (a + 5)(a - 5) = (b-3)(b^2+3b+9)$
a (subscript k) +5 = $\displaystyle k^2 +10k$
1.(0,37) = 37
2.(37,98) = 61
3. (98,189) = 91
4. (189,316) = 127
2.-1. = 24
3.-2. = 30
4.-3. = 36
He did say that he proved it long time ago, nobody else, at least those he knows could prove it.
I actually have no idea what's he talking about though.
• Nov 9th 2006, 06:10 PM
ThePerfectHacker
In the 17th Century my Favorite mathemation Pierre de Fermat wrote that 26 is the only number contained in between a square and a cube.
$\displaystyle x^2+2=y^3$.
Fermat never wrote down the proof (like usual) and many people failed to prove it (including me :mad: ) but about 100 years later Leonard Euler proved it. Though (I think) it he spend 7 years on this problem! (Just goes to show the Fermat was the master of Diophantine Equations). I myself have no idea how to show that.
I happen to know that $\displaystyle x^3+1=y^2$ also has no solutions accept $\displaystyle x=2,y=3$ which is (special case of Catalan Conjecure--- now proven). But was shown to be true also by Euler I think and again all texts on number theory omit the prove because it is way too long (though clever). However, I think there is a simpler proof due to Gauss based on the Quadradic fields and complex factorization of the cube which leads to an impossibility. But I do not think Gauss' method works here because the expression $\displaystyle x^3+2$ is not factorable while $\displaystyle x^3+1=(x+1)(x^2-x+1)$ which is the entire argument. Finally there is certainly a way to show this using the theory of Elliptic curve because this leads to a the square root of a cubic polynomial, but sadly I am an noob :( so I cannot know of one.
By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied.
• Nov 9th 2006, 09:00 PM
topsquark
Quote:
Originally Posted by ThePerfectHacker
In the 17th Century my Favorite mathemation Pierre de Fermat wrote that 26 is the only number contained in between a square and a cube.
$\displaystyle x^2+2=y^3$.
Fermat never wrote down the proof (like usual) and many people failed to prove it (including me :mad: ) but about 100 years later Leonard Euler proved it. Though (I think) it he spend 7 years on this problem! (Just goes to show the Fermat was the master of Diophantine Equations). I myself have no idea how to show that.
I happen to know that $\displaystyle x^3+1=y^2$ also has no solutions accept $\displaystyle x=2,y=3$ which is (special case of Catalan Conjecure--- now proven). But was shown to be true also by Euler I think and again all texts on number theory omit the prove because it is way too long (though clever). However, I think there is a simpler proof due to Gauss based on the Quadradic fields and complex factorization of the cube which leads to an impossibility. But I do not think Gauss' method works here because the expression $\displaystyle x^3+2$ is not factorable while $\displaystyle x^3+1=(x+1)(x^2-x+1)$ which is the entire argument. Finally there is certainly a way to show this using the theory of Elliptic curve because this leads to a the square root of a cubic polynomial, but sadly I am an noob :( so I cannot know of one.
By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied.
Crap. I was really hoping TPH would come through on this one (as he likes these) and I've now spent several hours working on it. All I can say is that the solution must be of the form:
a = 210x + 5, b = 210y + 3
or
a = 210x + 5, b = 210y + 33
(x and y integers, not necessarily positive)
Obviously the first pair contains the (presumably only) solution and I have yet to find any solutions using the second. Unfortunately the method I am using will never generate only one solution, so it's a dead end (though an enjoyable piece of work.)
-Dan
• Nov 9th 2006, 09:04 PM
Quick
Quote:
Originally Posted by ThePerfectHacker
26 is the only number contained in between a square and a cube.
What's that supposed to mean :confused:
• Nov 9th 2006, 09:06 PM
Ruichan
Quote:
Originally Posted by ThePerfectHacker
By the way, I really do not think your professor proved it using Elementary number theory. I can promise you that this problem is a problem only for the gods. I think he just lied.
I don't think he's lying though. His solution was published in reader's digest or whatever magazine. I wasn't paying attention to what he was saying coz I was practically sleeping with my eyes open by 8.30pm last night.
Guess I just need to ask him next week on which magazine that solution was published in. However, it was published almost 2 decades ago though.
He didn't proved the theory, just proved that a = 5 and b = 3 are the only solutions.
• Nov 10th 2006, 05:17 AM
topsquark
Quote:
Originally Posted by Quick
What's that supposed to mean :confused:
$\displaystyle 25 = 5^2 < 26 < 3^3 = 27$: 26 is the only such positive integer that is in succession between a square and a cube.
-Dan
• Nov 10th 2006, 05:20 AM
Quick
Quote:
Originally Posted by topsquark
$\displaystyle 25 = 5^2 < 26 < 3^3 = 27$: 26 is the only such positive integer that is in succession between a square and a cube.
-Dan
Oh, now I see. I was thinking things like 3^2<10<3^3
• Nov 10th 2006, 05:22 AM
topsquark
I don't know, maybe someone will gain some inspiration from what I was trying and succeed where I failed. I was trying an inductive type proof that a and b could only be expressed as:
$\displaystyle a = \left (2 \prod_{i = 1}^n(2i+1) \right )m + 5$
$\displaystyle b = \left (2 \prod_{i = 1}^n(2i+1) \right )q + 3$
where m and q are integers.
If this is true and you can do an induction, then we can show that n is without bound. If this is true then the only solution is for m, q = 0. That is, a = 5, b = 3. But I couldn't manage to prove the inductive step. (I kept on finding other forms that I couldn't prove were incorrect.)
-Dan
• Nov 10th 2006, 06:18 AM
ThePerfectHacker
Quote:
Originally Posted by Ruichan
I don't think he's lying though. His solution was published in reader's digest or whatever magazine. I wasn't paying attention to what he was saying coz I was practically sleeping with my eyes open by 8.30pm last night.
.
I would love to see it.
---
Topsquark just drop it. You are not going to solve it. I spend hours on these diophantine equations including this one. Trust me when I say that they are not easy, they might seem easy. Especially when you introduce cubes. I have seen proofs of several diophantine equations for example Fermat's last theorem for n=3, by Euler, it is huge, just pages and pages and pages all using manipulations divisibility arguments. They also involve beyond belive algebraic manipulations and the most unusual substitution. Which is why Euler solved most of them, becuase he was one of the few that was able to do this.
Solving diophantine equations in this type is called an "elementary" solution. My problem of the week used such an argument. There are also other non-elementary methods, one became famous in 19th century using field theory. And today there is a another one using elliptic curves. I think your professor solve it using the two latter methods, not the classical, but it would be nice if you can post it here. I would like to see how it is approach using modern math.
• Nov 16th 2006, 01:04 AM
Ruichan
Mathematics or Mathematical Monthly. It's from Mathematics Association of America, Journal form. That issue was published between 1980-1981.
Prof doesn't remember how to prove it, HEHEHE, he's looking for the hard copy of his proof. He forgot where he place his hardcopy.
That's why he's asking if we know how to prove it.
He said he only remembered that he used the modular formula or whatever.
I'm too lazy to go search for through the last 20 years of that issue which his proof was accredited.
• Nov 16th 2006, 06:22 AM
ThePerfectHacker
Quote:
Originally Posted by Ruichan
He said he only remembered that he used the modular formula or whatever.
Exactly, what I said, a non-elementary proof. Modular forms are 20th century mathematics. If he really knows them that is something to be proud of.
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# 2.6 – Similar Triangles
## Objectives
• Define similar as it relates to geometric figures and list the key characteristics of similar triangles (congruent angles and proportional sides).
• Dilate geometric figures to determine if they are similar.
• Use correct notation to indicate similarity.
• Calculate the scale factor of triangles by setting up ratios of corresponding sides.
## Key Terms
• Dilate – To change the size but not the shape of a geometric figure.
• The dilated figure is similar to the original but is not congruent to it.
• A dilation stretches or shrinks the original figure by a certain scale factor in relation to a point called the center of dilation
• The center of dilation is the point from which the scale factor between the original image points and dilation points is determined.
• In a dilation, segments that go through the center of dilation remain unchanged and segments that do not will become stretched.
• Proportional – Having equal ratios.
• Ratio – A comparison that shows the relative size of one quantity with respect to another.
• The ratio a to b is often written with a colon (a:b) or as a fraction (a/b).
• Scale Factor – The ratio of the lengths of corresponding sides in similar figures.
• In a dilation, this is the factor by which the original figure is multiplied.
• Multiply by a proper fraction (a number smaller than 1) to compress (shrink, or make smaller)
• Multiply by a number larger than 1 to enlarge (grow, get make bigger)
• Whole numbers and decimals work, as long as the value is > 1
• Similar – Having exactly the same shape.
• When figures are similar, corresponding angles are equal (congruent) in measure and corresponding sides are proportional in length.
• The symbol ~ means “is similar to.”
• Notation: $\Delta ABC \sim \Delta DEF$
• Similar Triangles – Triangles that have exactly the same shape.
• Their corresponding angles are equal (congruent)
• Their corresponding sides are proportional in length (same ratio).
## Notes
Similar Triangles
• Similar Triangles
• Triangles have equal (congruent) angle measures, but may not have congruent side measures.
• Triangles have proportional side measures
• ALL congruent triangles are similar
• SOME similar triangles are congruent
• The ratios of all three pairs of corresponding sides are equal
• The corresponding angles are equal (congruent)
• Determining Similarity
• Are the side lengths proportional?
• Do the fractions reduce down to the same value?
• Are all three angle measures equal (congruent)?
• If you answer yes to all of these questions, then you have similar triangles!
Dilating Triangles
• Dilating Triangles will compress (shrink) or enlarge (grow) them, without changing the shape
• If a triangle gets bigger, the scale factor is > 1 (greater than 1)
• If a triangle gets smaller, the scale factor is < 1 (less than 1, which is a fraction)
Scale Factor
• To Determine Scale Factor
• Step 1: List ALL of the corresponding sides of a triangle
• If the first triangle gets larger (grows, or enlarges), put the corresponding sides of the larger triangle top of the fraction (numerator) and put the corresponding sides of the smaller triangle on the bottom of the fraction (denominator)
• If the first triangle gets smaller (compresses, or shrinks), put the corresponding sides of the smaller triangle top of the fraction (numerator) and put the corresponding sides of the larger triangle on the bottom of the fraction (denominator)
• Step 2: Reduce the fraction
• Did the triangles enlarge (grow) or compress (shrink) from the 1st to the 2nd?
• Is the result larger than 1 or smaller than 1?
• If the triangles compress (shrink), the fraction should be less than 1
• If the triangles enlarge (grow), the fraction should be greater than 1
Examples
• Example of Dilation (Compression)
• Example of Similar Triangles, NOT Congruent
• All angles measures of the 1st triangle are equal (congruent) to all angle measures on the 2nd triangle
• $\frac{8}{16}=\frac{10}{20}=\frac{6}{12}$
• The ratio for all three triangles is reduced to $\frac{1}{2}$
• The first triangle is 1/2 the size of the second one
• The dilation has doubled from the first triangle to the 2nd
• So, the scale factor = 2 (because the triangle got larger)
• Example of Similar Triangles that ARE Congruent
• All angles measures of the 1st triangle are equal (congruent) to all angle measures on the 2nd triangle
• All side lengths of the 1st triangle are congruent to all side lengths on the 2nd triangle (shown with hashmarks)
• The triangle side lengths have a proportional ration of 1:1
• Written as a fraction $\frac{1}{1}$ or just 1.
• The scale factor = 1
• Example of Finding the Scale Factor
• What is the scale factor of $\Delta ABC \sim \Delta DEF$?
• The triangles dilate larger (enlarge or grow), so:
• $\frac{Large}{Small}$
• $\frac{\overline{EF}}{\overline{BC}}=\frac{\overline{AB}}{\overline{DE}}=\frac{\overline{DF}}{\overline{AC}}$
• $\frac{48}{6}=\frac{48}{6}=\frac{32}{4}$
• Reduce the fractions:
• They all equal $\frac{8}{1}=8$
• Scale Factor = 8
• Lesson Video – Dilation & Scale
• Click icon in the bottom right to view in “Full Screen” mode
[ All images are licensed under Creative Commons unless otherwise stated. ]
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# An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and nature of the image.
Here, Object distance, u = -27 cm (To the left side of mirror)
Image distance, v = ? (To be calculated)
Focal length, f = -18 cm (It is a concave mirror)
Height of object, h1 = 7.0 cm
And, Height of image, h2 = ? (To be calculated)
Now, For a mirror:
Image distance, v = -54 cm
Since the image distance is minus 54 cm, therefore, the screen should be placed at a distance of 54 cm in front of the concave mirror (on its left side). The nature of image obtained on the screen is real and inverted.
Also, For a mirror,
= -14.0 cm
Thus, the size of image is 14.0 cm.
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# How Many Tablespoons In 1/4 Cup
If you’re reading this article, you probably found yourself in a situation where a recipe called for 1/4 cup of a specific ingredient, but you only have tablespoons to measure it out. Don’t worry – we’ve got you covered. In this article, we’ll explain how many tablespoons are in 1/4 cup and provide some additional information about measuring ingredients in the kitchen.
Baca Cepat
## Converting Cups to Tablespoons
When measuring ingredients in the kitchen, it’s important to have a basic understanding of volume conversions. In the United States, the standard measurement for volume is the cup. One cup is equal to 16 tablespoons. Therefore, if you need to measure out 1/4 cup of an ingredient, you can simply multiply 1/4 by 16 to get the number of tablespoons you need.
1/4 x 16 = 4 tablespoons
So, 1/4 cup is equal to 4 tablespoons.
## Measuring Ingredients Precisely
While knowing how many tablespoons are in 1/4 cup is useful, it’s important to note that exact measurements can be the difference between a successful recipe or a disaster. Measuring precisely is especially important in baking recipes, where a small deviation in ingredient amounts can lead to a flat cake or a soupy batter.
When measuring ingredients, be sure to use measuring tools designed for the task. A liquid measuring cup should be used for liquids, while dry measuring cups should be used for dry ingredients. Measuring spoons are also useful for smaller amounts of ingredients.
When measuring dry ingredients, use a spoon to fill the measuring cup with the ingredient, then level off the top with a straight edge. Be sure not to pack the dry ingredient down, as this can lead to an excess amount being used.
When measuring liquids, fill the measuring cup to the desired amount at eye level. Keep in mind that some ingredients may need to be measured by weight rather than volume for the most precise results.
## Common Ingredient Conversions
Knowing how to convert between units of measurement can be helpful in the kitchen. Here are some common conversions to keep in mind:
• 1 tablespoon = 3 teaspoons
• 1/4 cup = 4 tablespoons
• 1/3 cup = 5 tablespoons + 1 teaspoon
• 1/2 cup = 8 tablespoons
• 1 cup = 16 tablespoons
• 1 fluid ounce = 2 tablespoons
• 1 quart = 4 cups
• 1 gallon = 16 cups
### In Conclusion
Knowing how many tablespoons are in 1/4 cup is just one piece of the puzzle when it comes to measuring ingredients in the kitchen. By using the correct measuring tools and being precise, you can ensure that your recipes will turn out perfectly. By keeping common ingredient conversions in mind, you can easily convert between different units of measurement. Happy cooking!
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## Wednesday, June 30, 2021
Published June 30, 2021 by with 1 comment
# If 10 Vaccinated and 10 Unvaccinated People Die, Can We Still Say Vaccines Work?
You will almost certainly be seeing headlines about vaccinated people dying and might even see that more vaccinated than unvaccinated die. Here's one from the week that I wrote this post. Why do we still say vaccines work if this is happening?
Imagine as an example that you see '10 vaccinated and 10 unvaccinated doctors died from COVID-19 today'. Your brain probably thinks 'well...the vaccine didn't work I guess.' We see those numbers, then just assume that the populations were similar. They're all doctors right?
Digging more, say that it turns out that 90% of the doctors were vaccinated. To make it easy, assume that there are 1,000 total doctors. 90% vaccinated means there were 900 vaccinated and 100 unvaccinated. If 10 died from each group, that means:
• 10 / 100, or 10% of unvaccinated doctors died
• 10 / 900, or 1.1% of vaccinated doctors died
Unvaccinated doctors were 9 times as likely to die as vaccinated ones. Another way of phrasing that is that the vaccine's efficacy was:
vaccine efficacy = 1 - (vaccinated risk/unvaccinated risk) = 1 - (0.011/0.1) = 89%
This is how you have to think about things like this. Vaccines, masks, seat belts, helmets, etc. aren't 100% effective. Use the calculation above whenever you see headlines like this and want to know the actual story.
You can even have more vaccinated deaths than unvaccinated. Imagine for the 89% efficacy vaccine above, you have 99% of the population vaccinated. For 10,000 doctors in that example, you'd expect to have 10% of the 100 unvaccinated die and 1.1% of the vaccinated 9900 die, so that's 10 unvaccinated deaths and 120 vaccinated deaths. A highly effective vaccine can still have more vaccinated people die than unvaccinated ones.
In case a visual helps, here is the initial example's distribution as a colored grid (red = dead and green = alive):
#### 1 comment:
1. 💢 ลงหลัก100 ได้หลัก #พัน งบน้อย อย่ากลัว เล่นง่าย ได้จริง 🔥
🌀ไม่ต้องทำเทรินให้เสียเวลา ไม่ล็อคยูส เเตกง่าย
🙀ครบจบเว็บเดียว🙀
🎰บอล มวย บๅคๅs่ๅ สล็oต แคuดี้ ยิงปลๅ
🔥 ฝากไม่มีขั้นต่ำแค่ 1 บาทก็ฝากได้เล่นได้ 🔥
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# Project #4487 - College Algebra
Solve the problem
1)The growth in the population of a certain rodent at a dump site fits the exponential function A(t)=708e^0.024t, where t is the number of years since 1988. Estimate the population in the year 2000.
Solve the equation
2) 4^(x-3)=25 Round to three decimal places.
Use a change of base formula to evaluate the given logarithm. Approximate to three decimal places.
3) \$2500 is invested at 5% compounded quarterly. In how many years will the account have grown to \$4000? Round your answer to the nearest tenth of a year.
Solve the problem
4) Wind speed varies in the first twenty meters above the ground. For a particular day, let f(x)=3.6 ln x +7.4 model the wind speed x meters about the ground. At what height is the wind speed 9 meters per second? Round results to the nearest hundredth.
Predict the end behavior of the graph of the function.
5) f(x)= -1.42x^4-x^3+x^2+7
Use factoring by grouping to solve the equation
6) x^3+2x^2-36x-72=0
Use synthetic division to find the quotient and remainder
7) (2x^3+3x^2+4x-10) / (x+1)
Determine all possible rational solutions of the polynomial equation
8) f(x)=2x^3-5x^2+7x-7
Solve the equation exactly in the complex number system
9) x^3+7x^2+36x+252=0
Give the equations of any vertical asymptotes for the graphs of the rational functions
10) f(x)= 5x^3+4x-9
--------------
x^2+3x-54
Provide an appropriate response.
11) If Q varies inversely as the square root of R and Q=3 when R=4, what is Q when R is 16?
Use algebraic and/or graphical methods to solve the inequality
12) (x+7)(x^2-16)>0
13) x+15
------- <4
x+5
Subject Mathematics Due By (Pacific Time) 04/16/2013 12:00 am
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# Partial derivative problem
1. Aug 7, 2012
### geekba
Hi!
Here is my function:
I think I know how to find ∂u/∂x, but I have no idea how to find ∂/∂z(∂u/∂x). Here is how I found ∂u/∂x:
http://oi48.tinypic.com/prsly.jpg
Does someone know how to find ∂/∂z(∂u/∂x)?
I appreciate any help :)
Last edited by a moderator: Aug 8, 2012
2. Aug 7, 2012
### Sourabh N
Before the second partial derivative, you should fix the error in your calculation of ∂u/∂x, specifically ∂($\frac{xy}{z}$)/∂x.
3. Aug 7, 2012
### geekba
What's wrong with ∂(xy/z)/∂x? I checked it and it seems correct to me...
4. Aug 8, 2012
### geekba
It's very important so all suggestions are welcome :)
5. Aug 8, 2012
### Sourabh N
Never mind. I hadn't scrolled all the way down, it is correct.
I believe you are having trouble calculating $\frac{∂}{∂z}$($∂\rho/∂s$) and $\frac{∂}{∂z}$($∂\rho/∂t$) (Let me know if this is not the case).
To simplify this, get rid of s and t by writing $∂\rho/∂s$ and $∂\rho/∂t$ as partial derivatives of $\rho$ w.r.t. x, y and z, using the chain rule. Since you know how s and t depend on x, y and z, this can be done.
Once you have done this, calculating $\frac{∂}{∂z}$($∂\rho/∂s$) and $\frac{∂}{∂z}$($∂\rho/∂t$) would be straightforward.
6. Aug 8, 2012
### geekba
I got it finally Thaks a lot!
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# Thread: probability
1. ## probability
i am not sure where my mistake is . Can someone pls check for me .
Q : In a test , the math marks if candidates at a certain centre are normally distributed with a mean of 55 marks and a standard deviation of 5 marks . If a sample of 5 students is chosen at random , find the probability that at least one student obtains marks of less than 60 .
My working :
Let X be the random variable of the math marks of the candidates ..
X - N ( 55 , 25 )
P ( X < 60 ) = 0.84134
then , let Y be the random variable of the number of students obtaining marks less than 60 .
X - B ( 5 , 0.84134 )
therefore P ( Y ≥ 1 ) = 0.999 ----- my answer
But the answer given is 0.5784
Where did i go wrong ??
2. Originally Posted by mathaddict
i am not sure where my mistake is . Can someone pls check for me .
Q : In a test , the math marks if candidates at a certain centre are normally distributed with a mean of 55 marks and a standard deviation of 5 marks . If a sample of 5 students is chosen at random , find the probability that at least one student obtains marks of less than 60 .
My working :
Let X be the random variable of the math marks of the candidates ..
X - N ( 55 , 25 )
P ( X < 60 ) = 0.84134
then , let Y be the random variable of the number of students obtaining marks less than 60 .
X - B ( 5 , 0.84134 )
therefore P ( Y ≥ 1 ) = 0.999 ----- my answer
But the answer given is 0.5784
Where did i go wrong ??
Your working is excellent. Your answer is correct.
The given answer of 0.5784 is wrong. It's the probability that at least one student obtains marks of greater than 60.
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# Limit problem.
Noncircular proof that the limit:
$$\lim_{h \to 0}\frac{e^{h}-1}{h}$$
Tends to 1, which is $$(ln(e))$$ as $${h \to 0}$$
Been trying logarithms and other kind of stuff but always seem to get 0/0 :(
Last edited:
L'Hospital gives the desired result.
HallsofIvy
Homework Helper
Yes, but he asked for a "non-circular" proof! The point is that the standard proof of the derivative of ex requires taking that limit. Thus, using the derivative to get that limit is "circular".
However, it is not necessary to use the "standard proof" of the derivative. One thing commonly done in modern Calculus books is to define the logarithm first:
Define
$$ln(x)= \int_1^x \frac{1}{t} dt$$
All of the usual properties of the natural logarithm can be derived from that including the fact that it is a one-to-one function from the postive real numbers to the real numbers and so has an inverse function from the real numbers to the positive real numbers. Call that inverse function "exp(x)". Since it is clear from the definition above that the derivative of ln(x) is 1/x, knowing that ln(exp(x))= x tells us that (1/exp(x))(exp(x))'= 1 so the derivative of exp(x) is exp(x) itself. Once we have that, then we can use the L'Hospital as Matthollyw00d says.
(Also, if y= exp(x), then x= ln(y) and, if $x\ne 0$ 1= (1/x)ln(y)= ln(y1/x. Going back to the exp form, exp(1)= y1/x so y= (exp(1))x. That is, the "exp" function, defined as the inverse to ln(x), really is ex where e is defined as exp(1), the number whose natural logarithm is 1.)
But if you want a more direct proof, you have to go back to the definition of "e". e can be defined by $e= \lim_{h\to 0}(1+ h)^{1/h}$. So we can say that, for h close to 0, we have $e= (1+h)^{1/h}$, approximately. Then $e^h= 1+h$ so $e^h-1 = h$ and then $(e^{h}-1)/h= 1$. That is, as I said, approximate. taking the limit, as h goes to 0, makes it exact.
I love math. Im a pilot but can still sit and study math for several hours. Problem is that I don't have anyone to share my thoughts with and discuss. However you guys seem to be really bright in math. Best math/physics forum so far!
Thank you for the reply guys!
I seem to suggest using series all the time but the Taylor series expansion of e^h might be useful.
Then (e^h-1)/h = 1+h(some terms) tends to 1 as h goes to 0. But this may again be circular.
lurflurf
Homework Helper
This limit is the definition (Newton quotient) of exp'(0).
Thus L'Hospital is overkill.
What we need is a definition of exp where either the limit or exp'(0) is given or easy to deduce.
In fact exp'(0)=1 is what is special about e.
limit means not simply substituting the values but it merely means that at its neighbourhood
Ok something is wrong here.
$$\frac{d}{dx}{a^x}=ln({a}){a^x}$$
Tis is by proving that the limit of (ah-1)/h tends to ln(a).
The limit above with e instead: (eh-1)/h tends to 1. But it should actually tend to ln(e) wich happens to be 1. This according to "(ah-1)/h tends to ln(a)" statement which is true.
Does this mean that L'Hospitals rule was just a coincidence and doesent actually work with (ah-1)/h since ln(a)$$\neq$$ 1 if ofcourse a$$\neq$$e
The derivative of ex is actually ln(e)ex, same applies for ax, then dy/dx(ax) = ln(a)ax.
Its kind of a contradiction there if you use L'hospital to prove that these limits tend to ln(e) respectively ln(a).
Does anyone understand what I mean or is my reasoning idiotic?
Actually as someone suggested, it is very easy using the definition of e^x as the infinite power series:
$$\displaystyle\sum_{k=0}^\infty \frac{x^k}{k!}$$ . Try taking the derivative and you will see it is the same sum, so that D(e^x) = e^x. Then you may use L'hopital's rule, whose derivation does not depend in any way on e^x (uses generalized mean value theorem), so no circular argument is used.
HallsofIvy
Homework Helper
Ok something is wrong here.
$$\frac{d}{dx}{a^x}=ln({a}){a^x}$$
Tis is by proving that the limit of (ah-1)/h tends to ln(a).
The limit above with e instead: (eh-1)/h tends to 1. But it should actually tend to ln(e) wich happens to be 1. This according to "(ah-1)/h tends to ln(a)" statement which is true.
Does this mean that L'Hospitals rule was just a coincidence and doesent actually work with (ah-1)/h since ln(a)$$\neq$$ 1 if ofcourse a$$\neq$$e
The derivative of ex is actually ln(e)ex, same applies for ax, then dy/dx(ax) = ln(a)ax.
Its kind of a contradiction there if you use L'hospital to prove that these limits tend to ln(e) respectively ln(a).
Does anyone understand what I mean or is my reasoning idiotic?
I don't understand what "contradiction" you are talking about. IF you have proven the derivative of ex without using that limit itself, then, yes, L'Hospital's method works nicely for either ex or ax.
Assume that we already know that d(ex)/dx= ex and that d(ax/dx= ln(a)ax. Then Applying L'Hospital to $lim_{h\to 0}(e^h- 1)/h$, differentiating both numerator and denominator, with respect to h, separately, we get $\lim_{h\to 0}e^h/1= 1$. Doing the same with $\lim_{h\to 0}(a^h-1)/h$ we get $\lim_{h\to 0}ln(a)a^h/1= ln(h)$, both of which are correct.
What does exactly 'non-circular' proof mean?
What does exactly 'non-circular' proof mean?
Means that you shouldnt use things you know in forehand to proof something. Sounds weird but here is an example:
Lets say you want to proove that the derivative of the function f(x)=ax is f'(x)ln(a)ax by using the definition of the derivative
f'(x)=(a(x+h)-ax)/h as h tends to 0.
After simplifying you will get ax times the limit as h tends to 0 of (ah-1)/h.
You can prove that this limit tends to ln(a) by using L'Hopitals rule, but then you have to differentiate ah, however, how can you differentiate it, if you havent prooven its derivative yet?
see my point?
jjjjj
I love math. Im a pilot but can still sit and study math for several hours. Problem is that I don't have anyone to share my thoughts with and discuss. However you guys seem to be really bright in math. Best math/physics forum so far!
Thank you for the reply guys!
Means that you shouldnt use things you know in forehand to proof something. Sounds weird but here is an example:
Lets say you want to proove that the derivative of the function f(x)=ax is f'(x)ln(a)ax by using the definition of the derivative
f'(x)=(a(x+h)-ax)/h as h tends to 0.
After simplifying you will get ax times the limit as h tends to 0 of (ah-1)/h.
You can prove that this limit tends to ln(a) by using L'Hopitals rule, but then you have to differentiate ah, however, how can you differentiate it, if you havent prooven its derivative yet?
see my point?
Yeah...I got it...Best Regards...
did we get an answer? is $\lim_{x \rightarrow 0} \frac{ln(1+x)}{x} = 1$ allowed or not? because if that's allowed write $a^x = e^{xlna}$ & make the substitution $e^{xlna} - 1 = y$ so that $x = \frac{lna}{ln(y+1)}$. Then $$\lim_{x \rightarrow 0} \frac{a^x - 1}{x} = \lim_{y \rightarrow 0} \frac{ylna}{ln(y+1)} = lna$$
Last edited:
Yes it is.
L'Hopitals rule gives you:
$$\lim_{x \rightarrow 0}\frac{1/(1+x)}{1} = 1$$
HallsofIvy
Homework Helper
did we get an answer? is $\lim_{x \rightarrow 0} \frac{ln(1+x)}{x} = 1$ allowed or not? because if that's allowed write $a^x = e^{xlna}$ & make the substitution $e^{xlna} - 1 = y$ so that $x = \frac{lna}{ln(y+1)}$. Then $$\lim_{x \rightarrow 0} \frac{a^x - 1}{x} = \lim_{y \rightarrow 0} \frac{ylna}{ln(y+1)} = lna$$
Well, the point was that whether or not that is "allowed" depends upon exactly how you have defined the logarithm. If you used the exponential to define the logarithm (ln(x) is the inverse function to ex), and derived the derivative of the logarithm from the derivative of the exponential, then you cannot use properties of the logarithm to find that limit. Of course, as I said before, you don't have to use the exponential to define the logarithm. Many texts do it the other way: define the logarithm function to be
$$ln(x)= \int_1^x \frac{1}{t} dt$$
and then define the exponential to be the inverse function to the logarithm. That way you do not use $\lim_{x\to 0}(e^x- 1)/x$ to define the derivative of the exponential (and then the derivative of the logarithm) so there is no "circular reasoning" involved in using the logarithm to prove that limit.
Actually as someone suggested, it is very easy using the definition of e^x as the infinite power series:
$$\displaystyle\sum_{k=0}^\infty \frac{x^k}{k!}$$ . Try taking the derivative and you will see it is the same sum, so that D(e^x) = e^x. Then you may use L'hopital's rule, whose derivation does not depend in any way on e^x (uses generalized mean value theorem), so no circular argument is used.
To differentiate term by term you first need to prove the series converges uniformly (I think it's easier in the case of power series).
Another way to use this power series (and btw the series itself derives from the original definition of e) is to just substitute it into the limit, and since you know it converges you can work with it with no worries. You will end up with a series which's first term is 1 and all of the rest are powers of h, so when h tends to zero the series tends to 1.
Well, the point was that whether or not that is "allowed" depends upon exactly how you have defined the logarithm. If you used the exponential to define the logarithm (ln(x) is the inverse function to ex), and derived the derivative of the logarithm from the derivative of the exponential, then you cannot use properties of the logarithm to find that limit. Of course, as I said before, you don't have to use the exponential to define the logarithm. Many texts do it the other way: define the logarithm function to be
$$ln(x)= \int_1^x \frac{1}{t} dt$$
and then define the exponential to be the inverse function to the logarithm. That way you do not use $\lim_{x\to 0}(e^x- 1)/x$ to define the derivative of the exponential (and then the derivative of the logarithm) so there is no "circular reasoning" involved in using the logarithm to prove that limit.
is that what was meant by no circular reasoning? I thought it just meant don't use l'Hopital's rule
When I took Calculus we learned the power rule for obtaining the derivative. d(X^n)/dX = nX^(n-1). This carries over to negative exponents. So that now we knew, in general how to integrate X^-a, for integers and then fractions.
BUT the whole matter breaks down in attempting to integrate 1/X. Thus my class proceeded in the way written above by HallsofIvy defining the log of X.
is that what was meant by no circular reasoning? I thought it just meant don't use L'Hopital's rule
We could proceed to use the Taylor expansion e^X = 1+X/1 +X^2/2! if we were to get this by expanding the difinition of e^x = lim(1+x/n)^n as n goes to infinity.
L'Hospital's rule employs the derivative, so that if we assume that derivative and at no previous point prove it, then it is circular.
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Comment on page
1231.Divide-Chocolate
题目描述
You have one chocolate bar that consists of some chunks. Each chunk has its own sweetness given by the array sweetness.
You want to share the chocolate with your K friends so you start cutting the chocolate bar into K+1 pieces using K cuts, each piece consists of some consecutive chunks.
Being generous, you will eat the piece with the minimum total sweetness and give the other pieces to your friends.
Find the maximum total sweetness of the piece you can get by cutting the chocolate bar optimally.
Example 1:
Input: sweetness = [1,2,3,4,5,6,7,8,9], K = 5
Output: 6
Explanation: You can divide the chocolate to [1,2,3], [4,5], [6], [7], [8], [9]
Example 2:
Input: sweetness = [5,6,7,8,9,1,2,3,4], K = 8
Output: 1
Explanation: There is only one way to cut the bar into 9 pieces.
Example 3:
Input: sweetness = [1,2,2,1,2,2,1,2,2], K = 2
Output: 5
Explanation: You can divide the chocolate to [1,2,2], [1,2,2], [1,2,2]
Constraints:
0 <= K < sweetness.length <= 10^4
1 <= sweetness[i] <= 10^5
代码
class Solution {
public int maximizeSweetness(int[] sweetness, int K) {
int l = 0, r = Arrays.stream(sweetness).sum() / (K + 1);
while (l < r) {
int mid = l + (r - l + 1) / 2;
int cnt = 0;
int sum = 0;
for (int i = 0; i < sweetness.length; i++) {
sum += sweetness[i];
if (sum >= mid) {
sum = 0;
cnt++;
}
}
if (cnt >= K + 1) {
l = mid;
} else {
r = mid - 1; // ???
}
}
return l;
}
}
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# Help with Steady State Heat Transfer Problem
• Diego Saenz
In summary, the conversation is about a steady state heat transfer problem with a flux source of 400 W/m². The goal is to calculate the heat flux passing through a surface that is placed and oriented in space with air in between. The proposed method is to use ∫∫(-k∇T)*(n dS), but the person is unsure about what ∇T represents. The suggested solution is to find T by solving the heat equation with additional boundary conditions. If the geometry is simple, it can be solved analytically, but if not, software is needed. The conversation also mentions the use of energy conservation and a constant temperature distribution to find the heat flux. The person requests a simple example to better understand the
#### Diego Saenz
Hello everyone,
I have this setady state, heat transfer problem; I hope you can help me with it.
I have flux source of 400 W/m² (a lamp), and i want to calculate the heat flux passing through a surface arbitrarily placed and oriented in the space. There is air in between. How can i do this?
I thought that I could use ∫∫(-k∇T)*(n dS)
But i don't know what ∇T is...
Thanks.
That will give you the answer once you've found T. Finding T involves solving the heat equation with a local heat source. You'll need some more boundary conditions to do this, e.g. what are the temperatures of the walls?
If your geometry is simply you can solve for T analytically then differentiate to find your heat flux through a surface.
If not, you'll need some software.
Hi mikeph,
My surface is simple, is a square. Is it possible to use this information to solve for T? Can you explain how can I do this?
Thanks
If the heat source is inside the surface then it's just 400 W/m^2, this is from energy conservation using the steady state assumption.
If you have a constant temperature distribution then the heat being added inside the box (400) must equal the heat leaving through the walls, otherwise the temperature of the box would have to increase which would introduce a time variation.
1 person
I really appreciate your help mikeph.
But I have problems visualizing the solution, could you provide simple example?
## 1. How do I set up a steady state heat transfer problem?
To set up a steady state heat transfer problem, you will need to define the boundary conditions, material properties, and heat source/sink in your system. Additionally, you will need to create a mathematical model using governing equations like Fourier's Law of Heat Conduction.
## 2. What is the difference between steady state and transient heat transfer?
Steady state heat transfer refers to a condition where the temperature in a system remains constant over time. Transient heat transfer, on the other hand, occurs when there is a change in temperature over time. In other words, steady state heat transfer is a state of thermal equilibrium, while transient heat transfer is a dynamic process.
## 3. How do I solve a steady state heat transfer problem numerically?
To solve a steady state heat transfer problem numerically, you will need to use a numerical method such as the Finite Difference Method or Finite Element Method. These methods involve discretizing the system into smaller elements and solving the governing equations for each element. The results can then be combined to obtain a solution for the entire system.
## 4. What factors affect the steady state heat transfer rate?
The steady state heat transfer rate is affected by several factors, including the temperature difference between the hot and cold ends, the thermal conductivity of the material, the size and shape of the system, and the boundary conditions. The presence of insulation or heat sinks/sources can also affect the heat transfer rate.
## 5. How can I improve the efficiency of heat transfer in a steady state system?
To improve the efficiency of heat transfer in a steady state system, you can increase the temperature difference between the hot and cold ends, use materials with higher thermal conductivity, optimize the shape and size of the system, and minimize heat losses through insulation or other methods. Additionally, using advanced techniques like multi-physics simulations can help optimize the design for efficient heat transfer.
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# Stepper Motors Part 2
Hopefully some of you are still with me, and you haven’t blown up your supply, switching device or coil.
Before we continue I’d like to ask a question; what current can your supply handle? Let’s take a brief look at Ohm’s Law, or at least the mathematical equation describing Ohm’s Law. The equation V=IR is incredibly important in electronics, and so you should have it burned into the back of your skull. Assuming a 5V supply (not uncommon), and a coil resistance of 1.6ohms (again, not unheard of) – this gives a current of 3.125 Amps PER COIL. Please also bear in mind that by using the Full Stepping method we would be drawing 6.25 Amps and you could be facing some current issues. I was using an old PC power supply as they have reasonably hefty current ratings, and although the 5V rail was fine, running off the 12V kept tripping the thermal fuse when trying to switch just a single coil.
Stepping a single step is easy enough, but you will invariably try to spin your stepper fast. You will hit a certain point where the stepper just stalls and lets out a whine (corresponding to the switching frequency). This is not necessarily a limitation of the motor, but of your driver circuitry. You have reached a point where you can’t energise the coils quickly enough, but there is a solution. By increasing the drive voltage it is possible to decrease the time taken for the current to saturate the coil, and so increase your switching speed. We have already ascertained that an increased voltage leads to an increased current and a blown supply, so we must find a way to increase voltage and keep current the same. There are a number of ways to do this, but I am going to cover the two simplest; linear resistive current limiting, and a chopper drive circuit.
Linear resistive current limiting is far simpler than it sounds. In fact, it is just adding a resistor in series with each coil. Similar to a current limiting resistor when driving an LED, this resistor limits the current going through the coil. Most motor manufacturers actively encouraged this technique up to the mid-80’s, but it has fallen out of fashion – either way this is probably the easiest and cheapest method available. The major disadvantage of this method is that the current is getting dumped in this resistor, and so the resistor will get HOT. This is NOT a job for the 1/4W resistor.
The chopper drive circuit uses PWM to limit the current in the coil. Either this can be open-loop control where the electronics has no knowledge of the current levels, and the duty cycle is determined either through experimentation or calculation. Alternatively, closed-loop control can be implemented through the use of more complex electronics measuring coil current. The disadvantage of this method is based purely around the PWM frequency. Higher frequencies will cause RF issues, and lower frequencies (i.e. audio frequencies) the motor will literally “squeel”.
Next article I will be dealing with actual schematics, and including some code used to step the motor using a PIC18F.
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# how much is 10 trillion, like how does compare with something else?
i heard neanderthals walked the earth 1 trillion SECONDS ago!...so how much is 10 trillion?
Relevance
• 1 decade ago
try to picture 1000 of something. What was the earth like 1000 years ago? The Byzantine Empire was in power, and Leif Ericson the viking first landed on North America (Greenland).
10 times 1000 years ago was roughly 8000BC, or the end of the Ice Age.
So 1000 is pretty large as far as years are concerned (with a human perspective). Now, there are 3,600 seconds in an hour, 86,400 seconds in a day, and 31,536,000 seconds in a year. 1,000,000,000,000 seconds is a long time ago.
When I try to comprehend large numbers, it sometimes helps to divide them. For example, 10,000,000,000,000 is 10,000 billions or 10 million millions. Just comprehending a million is difficult though.
Imagine you have a puzzle with 1,000 pieces. Just picture an ordinary jigsaw puzzle, and what the pieces look like in a pile before you put the puzzle together. It takes up about 75 to 100 cubic inches of space.
Now you have 1,000 jigsaw puzzles (1 million pieces). If you took them all out of the boxes, you could approximately fill a large kiddie pool with pieces.
Now you have 1,000 kiddie pools filled with puzzle pieces (1 billion pieces). You could now fill 60 18-wheeler truck loads of puzzle pieces, or about the equivalent of a freight barge or small water tower.
With 1000 of those (60,000 truck loads) or (1 trillion pieces), you could fill your local water reservoir with puzzle pieces. That's pretty impressive, but we're still not at 10 trillion.
10 reservoirs worth of pieces is about a medium-sized lake. An eighth of a cubic kilometer. That's an unbelievable amount of puzzle pieces.
Now if you started building a massive puzzle with all of those pieces at a rate of 10 pieces PER SECOND at the time the Neanderthals walked the earth, you would just be finishing about now.
The puzzle would be about the size of India.
• 1 decade ago
10 trillion dollar bills touching end to end would encircle the earth at its equator (27,000 miles) 85,000 times, or it would stretch to the sun (93 million miles away) almost 54 times. It would take the speed of light 215 minutes (3.58 hours) to get to the end of it.
And the cash register just keeps on clicking. It is 3 times more than the total currency that we have printed for circulation.
• 1 decade ago
Add Obama's \$1 trillion in new spending to it and then compare that \$11 trillion to the tears of all the hungry children there will be when he drives us further in to the abyss and that's about right.
• 1 decade ago
You could pay Al Gore's electric bill with \$10 trillion.
• bob z
Lv 5
A bit less than 2000x the population of earth.
-or-
100x the number of stars in the our entire galaxy
-or-
4000x the number of heart beats in an average lifetime.
-or-
10x more than it takes to bribe the president, house, and senate of the united states.
• 1 decade ago
Who didn't see this coming? Congress passes a huge give away and there is fraud and corruption from day one.
Shrink Washington. Bring programs back to the state, city, county level where there is more accountability.
• 1 decade ago
Its... basically... Imagine having a trillion toothpicks in a box.
Ten of those.
• Anonymous
It is precisely the number of molecules in a certain unit. Oompa oompa!
• Anonymous
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Quiz3sol - NAME Quiz 3 7.6 A sample of methane gas of mass...
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Unformatted text preview: NAME: Quiz 3 7.6. A sample of methane gas of mass 25 g at 250 K and 18.5 atm expands isothermally until its pressure is 2.5 atm. Calculate the change in entropy of the gas given that Pile= Vf/Vi, AS = n R In (Vp/Vi), and R = 8.3145 J/mol K (where Pi and Pf are the final and initial pressures of the gas and Vf and Vi are the final and initial volumes of the same gas). The molecular weight of methane gas is 16.04 g/mol. AS = n R In (Pi/Pf) = (25 g/16.04 g/mol) x (8.3145 J/ mol K) x In (18.5 atm/2.5 atrn) = + 26 J/k. 7.15. With the knowledge that AHsystem for a reversible process is 100 kJ/mol, find the entropy change at 298K at constant pressure. A: constant P: as", = ‘M—m = —-" I T \ Note: also constant J T of 298 K 100, 000 — J Thergflnm, as“, =—"""= 335.57 29811" mol 8.15. Find the AGf at T=298K and P2=3atm. AGf at STP = 27%. Use me Ang = —Rln [g] and AG = AH — TAS . State any simplifications or assumptions (hint: I examine pressure dependence contributions of the different terms). AS“ = —Rh1[%] and M} = AH -Tfi.S . State any simplifications or assumptions (hint: 1 examine pressure depemience contributions of the different terms). In AG = 5H —TASJ the :15 term is much more pressure dependent than the 111-! term, so AG : -TAS _ Using thafirsr 1:155” equation; one gets a final equation of so). = soft“) + RTln[%]. Plugging in the numbersfinr P1=Inrm, P;=3a:m, T=29£Kj I R=3.31'45 Jr’molfi', and AGIW = -smanit gives afina! answer of 35:24.33! me I ma | ...
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Limits - x^n formula
Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Concept wise
This video is only available for Teachoo black users
### Transcript
Ex 13.1, 10 Evaluate the Given limit: limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) = (γ(1)γ^(1/3) β 1)/(γ(1)γ^(1/6) β 1) = (1 β 1)/(1 β 1) = 0/0 Since it is form 0/0, We can solve it by using (πππ)β¬(π₯βπ) (π₯^π β π^π)/(π₯ β π) = nan β 1 Hence, limβ¬(zβ1) (π§^(1/3) β 1)/(π§^(1/6) β 1) = limβ¬(zβ1) π§^(1/3) β 1 Γ· limβ¬(zβ1) π§^(1/6) β 1 = limβ¬(zβ1) π§^(1/3) β γ(1)γ^(1/3) Γ· limβ¬(zβ1) π§^(1/6) β γ(1)γ^(1/6) Multiplying and dividing by z β 1 = limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β 1) Γ· limβ¬(zβ1) (π§^(1/6) βγ (1)γ^(1/6))/(π§ β 1) Using (πππ)β¬(π₯βπ) ( π₯^π β π^π)/(π₯ β π) = nan β 1 limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β1) = 1/3 γ(1)γ^(1/3 β 1) = 1/3 Γ 1 = 1/3 limβ¬(zβ1) (π§^(1/6) β γ(1)γ^(1/6))/(π§ β1) = 1/6 γ(1)γ^(1/6 β 1) = 1/6 Γ 1 = 1/6 Hence our equation becomes = limβ¬(zβ1) (π§^(1/3) β γ(1)γ^(1/3))/(π§ β 1) Γ· limβ¬(zβ1) (π§^(1/6) β 6)/(π§ β 1) = 1/3 Γ·1/6 = 1/3 Γ 6/1 = 2
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# Roots and Powers of Algebraic Expressions
### Percents: Definition, Application & Examples
Maybe you know that 95% is an A and 75% is a C. But what do those percents really mean? In this lesson, we'll learn about percents, inclpercent-–-definitionuding how to convert them to fractions and decimals.
### Percent – Definition
OK, team, I need you to give all you've got. I want you to go out there and give 110%. I know it sounds impossible. And, well, it is. Before we work on plays, let's talk about percents.
The word percent literally means 'per hundred.' We use this symbol - % - for percents.
Let's take the word apart. It's per and cent. Where have you seen 'cent' before? Well, it's the word for a penny. It's also in the word century. What's a century? A hundred years. And then there's centennial; that's the 100-year anniversary. A centipede has 100 legs. Well, I think it does. I've never tried to count. And a woman who has centuplets is going to be crazy tired.
Let's talk about what percents mean. When you're hitting 99% of your shots, what did you do? If you took 100 shots, you made 99 of them. 99 per one hundred, per-cent. But what if you took 1,000 shots? Whoa. I bet your arms are tired. 99% means that you got 99 out of each hundred. So 99% of 1,000 is 990. Hitting 99% of your shots would also make you the best basketball player in the history of the world.
### Fractions & Decimals
This is a stats-driven game, so let's talk about what we can do with percents. We can convert percents to fractions quite easily. For example, our team makes 43% of its free throws. Let's say we want to convert 43% to a fraction. That's 43 per one hundred. As a fraction, it's 43/100. That's it!
And then there's Fred the Flying Monkey, our team mascot. He jumps off a trampoline to make crazy dunks during halftime. He only makes 8% of his dunks. That sounds bad, but it's actually one of the best percentages among flying monkeys, with or without trampolines.
Anyway, if we had 8%, it'd be 8/100. No matter what your percent, just put it over 100, and you've made it into a fraction. With 8/100, we can simplify that to 2/25, which still doesn't sound great.
What about decimals? What is 43% as a decimal? Just drop the percent sign and move the decimal two places to the left. So 43% becomes .43. Why? Because .43 is 43 one-hundredths.
I said we make 43% of our free throws. What if we wanted to know what 43% of 17 is. We had 17 free throws in the last game. If we multiply 17 times .43, we get 7.31. The team made 8 of 17 free throws, so we were slightly above our average percentage.
What about 8%? I know, I know. Fred doesn't like to talk about it. But still, just drop the sign and move the decimal two places to the left. So 8% becomes .08. The math is the same. To figure out his success in 50 attempts, we'd multiply 50 times .08, which is 4. Hey, 4 is better than 0!
### Practice Problems
Let's try some practice problems involving percents. Just as there are different ways to win a basketball game, there are different ways to solve a percent problem. As we go through these, let's try a few different methods for solving them.
At a home game, 84% of the seats are filled. If there are 5,200 seats, how many seats are filled? To solve this, let's set up two fractions: 84/100 = x/5,200. Remember, 84% as a fraction is just 84/100. If we cross multiply, we get 100x = 436,800. Divide by 100, and we find out that 4,368 fans showed up.
We also could have converted 84% to a decimal. 84% would become .84. And then we just multiply .84 times 5,200, which is, again, 4,368.
Here's another one: If a team has 15 players and 9 travel for a road game, how many players stay home? This one has a little trick to it. Note that the question is asking how many players stay home. So instead of 9, we want 15 - 9, or 6. So what is 6 of 15?
If we set this up as a fraction, we have x/100 = 6/15. Cross multiply to get 15x = 600. 600 divided by 15 is 40. So 40/100 students stayed home. What is 40/100 as a percent? 40%. The bigger question is this: Where's the dedication on that team? 40% stayed home? Not cool.
Here's another one: In a game, a team makes 36 shots and misses 42. What percent of shots were made? The trick here is that we're not given the total. The fraction for the made shots isn't 36/42. It's 36/(36 + 42), so a total of 78 shots were attempted. To solve this one, let's try something different. Let's just divide 36 by 78. That gets us a decimal, .46. We can convert that to a percent by moving the decimal two places to the right. So .46 is 46%. The team made 46% of its shots.
### Lesson Summary
To summarize, we learned about percents. 'Percent' means per hundred.
To convert a percent to a fraction, we just put the number over 100. 75% becomes 75/100. 2% becomes 2/100.
To convert a percent to a decimal, we drop the sign and move the decimal two places to the left. 15% becomes .15. 9% becomes .09.
Oh, and 110%? If 100% is the maximum you can give, how do you give 110%? Well, see, it's a metaphor. But that's a whole other topic...
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Square Root 144
Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Square root of a number implies finding the value of the given number which is raised to an exponent of ‘1/2’. Square root is the radical sign used to represent the notation for a number or an expression. Therefore square root of 144 is represented as ‘√144’. Its value can be calculated by finding the numbers which when multiplied by itselfgives 144. In this case, when 12 is multiplied by 12 we get 144 and similarly, -12 multiplied by -12 also gives ‘144’. Hence square root of 144 = ± 12.
Example 1: What is the square root of 225?
Square root of ‘225’ can be alsorepresented with the radical sign as ‘√225’.
In order to find its value, we need to find the numbers which when multiplied by itself gives ‘225’.
15 * 15 = 225 and we also have -15 * -15 = 225.
This implies that 15 multiplied by itself or -15 multiplied by itself gives 225 as the answer.
Hence, square root of 225, which implies √225 = ±15.
Example 2: What is the square root of 256?
Square root of ‘256’ can be also represented with the radical sign as ‘√256’.
In order to find its value, we need to find the numbers which when multiplied by itself gives ‘256’.
16 * 16 = 256 and we also have -16 * -16= 256.
Therefore 16 multiplied by itself or -16 multiplied by itself gives 256 as the answer.
Hence, square root of 256, which implies √256 = ±16.
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# Rope Sliding off a Peg
1. Oct 4, 2014
### SenatorAstro
1. The problem statement, all variables and given/known data
A limp rope with a mass of 2.4 kg and a length of 1.5 m is hung, initially at rest, on a frictionless peg that has a negligible radius, as shown in the Figure. y1 is equal to 0.48 m. What is the vertical velocity of the rope just as the end slides off the peg?
2. Relevant equations
PE = mgh
KE = 1/2mv^2
3. The attempt at a solution
Because the kinematics of this system seemed incredibly complicated, I figured it best to use conservation of energy in the system. Knowing that the center of mass will fall 0.48 meters, I assumed:
mgΔH = 1/2mv^2
2.4*9.81*0.48 = 1/2*2.4*v^2
v = 3.06881084461 m/s
Unfortunately, this seems to be incorrect.
2. Oct 4, 2014
### Simon Bridge
Welcome to PF;
How do you know the answer is incorrect?
Some of the rope is falling, but some of the rope is lifting. How did you account for the work done lifting part of the rope?
The behavior is not kinematic - since the acceleration will not be constant.
You can, however, use Newton's laws or conservation of energy to set up a differential equation.
Last edited: Oct 4, 2014
3. Oct 5, 2014
### ehild
I do not see any figure.
ehild
4. Oct 5, 2014
### haruspex
As I understand the term, kinematics concerns the geometry of motion, such as the movement of linkages. It does not concern itself with forces, energy, etc.
5. Oct 5, 2014
### Simon Bridge
Fersure - constant acceleration is a subset of kinematics, which is, strictly, the geometry of motion.
Kinematics at the secondary education level, is usually given as the geometry of motion where acceleration is a constant.
That is the level I was answering at - leaving OP to contradict me if I got it wrong. I suspected that OP did not want to use the suvat (or kinematic) equations. Probably should have been more careful.
6. Oct 5, 2014
### ehild
from Encyclopedia Britannica.
ehild
7. Oct 5, 2014
### SenatorAstro
My apologies. Here's the image.
File size:
11.1 KB
Views:
172
8. Oct 5, 2014
### Simon Bridge
Cool - that's what I figured: what about the questions in post #2?
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# Area Between Curves
## Homework Statement
Find the area bounded by the curves y=x^2 and y= 2 - x^2 for 0 ≤ x ≤ 2.
∫top - ∫bottom
## The Attempt at a Solution
∫(2-x^2)dx - ∫x^2dx
What I'm confused about is that the two equations only cross on [-1,1] so within the interval of the problem I only have an enclosed area on [0,1]. But the problem asks for the area on [0,2]. How do I reconcile the differing intervals?
LCKurtz
Homework Helper
Gold Member
If you draw the vertical line x = 2 it gives a right boundary just like x = 0 gives the left boundary. Your curves cross so you have to do it in two parts.
OH! I think I see that now.
So I'll have:
[∫(0→1)(2-x^2)dx - ∫(0→1)x^2dx] + [∫(1→2)(2-x^2)dx - ∫(1→2)x^2dx]
Basically the area between the curves on [0,1] plus the bits hanging off on [1,2].
A = 4/3 un^2
I knew there was something I was missing and it's been a couple of weeks since we did that.
Thanks for the helps!
LCKurtz
Homework Helper
Gold Member
OH! I think I see that now.
So I'll have:
[∫(0→1)(2-x^2)dx - ∫(0→1)x^2dx] + [∫(1→2)(2-x^2)dx - ∫(1→2)x^2dx]
Basically the area between the curves on [0,1] plus the bits hanging off on [1,2].
A = 4/3 un^2
I knew there was something I was missing and it's been a couple of weeks since we did that.
Thanks for the helps!
Your integrand is always y-upper - y-lower. Check that on the interval [1,2].
Oh! Yep. I forgot that my lines crossed.
One step at a time.... :)
A = 4 un^2
Thanks again.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Rational Numbers in Applications
## Story problems including fractions
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Practice Rational Numbers in Applications
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Rational Numbers in Applications
What if you sent out invitations to a birthday party to 64 of your friends? One week later 12\begin{align*}\frac{1}{2}\end{align*} have responded that they will attend. Two weeks later, another 14\begin{align*}\frac{1}{4}\end{align*} have responded that they will attend. At the last minute, 18\begin{align*}\frac{1}{8}\end{align*} of those who originally said they would attend change their mind. How could you determine how many of your friends will be in attendance? After completing this Concept, you'll be able to solve real-world problems like this one.
### Watch This
Watch this video for help with the Examples above.
### Guidance
Let's use the skills we learned in the last concept to solve some real-world problems.
#### Example A
Peter is hoping to travel on a school trip to Europe. The ticket costs 2400. Peter has several relatives who have pledged to help him with the ticket cost. His parents have told him that they will cover half the cost. His grandma Zenoviea will pay one sixth, and his grandparents in Florida will send him one fourth of the cost. What fraction of the cost can Peter count on his relatives to provide? The first thing we need to do is extract the relevant information. Peter’s parents will provide 12\begin{align*}\frac{1}{2}\end{align*} the cost; his grandma Zenoviea will provide 16\begin{align*}\frac{1}{6}\end{align*}; and his grandparents in Florida 14\begin{align*}\frac{1}{4}\end{align*}. We need to find the sum of those numbers, or 12+16+14\begin{align*}\frac{1}{2} + \frac{1}{6} + \frac{1}{4}\end{align*}. To determine the sum, we first need to find the LCD. The LCM of 2, 6 and 4 is 12, so that’s our LCD. Now we can find equivalent fractions: 121614=6162=612=2126=212=3134=312 Putting them all together: 612+212+312=1112\begin{align*}\frac{6}{12} + \frac{2}{12} + \frac{3}{12} = \frac{11}{12}\end{align*}. Peter will get 1112\begin{align*}\frac{11}{12}\end{align*} the cost of the trip, or2200 out of \$2400, from his family.
#### Example B
A property management firm is buying parcels of land in order to build a small community of condominiums. It has just bought three adjacent plots of land. The first is four-fifths of an acre, the second is five-twelfths of an acre, and the third is nineteen-twentieths of an acre. The firm knows that it must allow one-sixth of an acre for utilities and a small access road. How much of the remaining land is available for development?
The first thing we need to do is extract the relevant information. The plots of land measure 45,512,\begin{align*}\frac{4}{5}, \frac{5}{12},\end{align*} and 1920\begin{align*}\frac{19}{20}\end{align*} acres, and the firm can use all of that land except for 16\begin{align*}\frac{1}{6}\end{align*} of an acre. The total amount of land the firm can use is therefore 45+512+192016\begin{align*}\frac{4}{5} + \frac{5}{12} + \frac{19}{20} - \frac{1}{6}\end{align*} acres.
We can add and subtract multiple fractions at once just by finding a common denominator for all of them. The factors of 5, 9, 20, and 6 are as follows:
5 512223202256 23
We need a 5, two 2’s, and a 3 in our LCD. 2235=60\begin{align*}2 \cdot 2 \cdot 3 \cdot 5 = 60\end{align*}, so that’s our common denominator. Now to convert the fractions:
45512192016=124125=4860=55512=2560=319320=5760=101106=1060
We can rewrite our sum as 4860+2560+57601060=48+25+571060=12060\begin{align*}\frac{48}{60} + \frac{25}{60} + \frac{57}{60} - \frac{10}{60} = \frac{48 + 25 + 57 - 10}{60} = \frac{120}{60}\end{align*}.
Next, we need to reduce this fraction. We can see immediately that the numerator is twice the denominator, so this fraction reduces to 21\begin{align*}\frac{2}{1}\end{align*} or simply 2. One is sometimes called the invisible denominator, because every whole number can be thought of as a rational number whose denominator is one.
Solution
The property firm has two acres available for development.
Evaluate Change Using a Variable Expression
When we write algebraic expressions to represent a real quantity, the difference between two values is the change in that quantity.
#### Example C
The intensity of light hitting a detector when it is held a certain distance from a bulb is given by this equation:
Intensity=3d2
where d\begin{align*}d\end{align*} is the distance measured in meters, and intensity is measured in lumens. Calculate the change in intensity when the detector is moved from two meters to three meters away.
We first find the values of the intensity at distances of two and three meters.
Intensity (2)Intensity (3)=3(2)2=34=3(3)2=39=13
The difference in the two values will give the change in the intensity. We move from two meters to three meters away.
Change=Intensity (3)Intensity (2)=1334\begin{align*}\text{Change} = \text{Intensity} \ (3) - \text{Intensity} \ (2) = \frac{1}{3} - \frac{3}{4}\end{align*}
To find the answer, we will need to write these fractions over a common denominator.
The LCM of 3 and 4 is 12, so we need to rewrite each fraction with a denominator of 12:
1334=4143=412=3334=912
So we can rewrite our equation as 412912=512\begin{align*}\frac{4}{12} - \frac{9}{12} = -\frac{5}{12}\end{align*}. The negative value means that the intensity decreases as we move from 2 to 3 meters away.
Solution
When moving the detector from two meters to three meters, the intensity falls by 512\begin{align*}\frac{5}{12}\end{align*} lumens.
Watch this video for help with the Examples above.
### Vocabulary
• Subtracting a number is the same as adding the opposite (or additive inverse) of the number.
• To add fractions, rewrite them over the lowest common denominator (LCD). The lowest common denominator is the lowest (or least) common multiple (LCM) of the two denominators.
• When adding fractions: ac+bc=a+bc\begin{align*}\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}\end{align*}
• When subtracting fractions: acbc=abc\begin{align*}\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}\end{align*}
• Commutative property: the sum of two numbers is the same even if the order of the items to be added changes.
• Associative Property: When three or more numbers are added, the sum is the same regardless of how they are grouped.
• Additive Identity Property: The sum of any number and zero is the original number.
• The number one is sometimes called the invisible denominator, as every whole number can be thought of as a rational number whose denominator is one.
• The difference between two values is the change in that quantity.
### Guided Practice
Elsa baked a small cake for her family. First her sister ate one quarter and her mom ate one third. How much was left for Elsa?
Solution:
The whole cake is represented by 1. To solve this problem, we subtract the fraction that each person ate.
11413\begin{align*}1-\frac{1}{4}-\frac{1}{3}\end{align*}.
To complete this problem, we must give the terms common denominators. Since the denominators do not share any factors, we simply multiply them together: 43=12\begin{align*} 4\cdot 3=12\end{align*}.
There is 512\begin{align*}\frac{5}{12}\end{align*} of the original cake left for Elsa.
### Practice
Which property of addition does each situation involve?
1. Whichever order your groceries are scanned at the store, the total will be the same.
2. However many shovel-loads it takes to move 1 ton of gravel, the number of rocks moved is the same.
3. If Julia has no money, then Mark and Julia together have just as much money as Mark by himself has.
1. 715+29\begin{align*}\frac{7}{15} + \frac{2}{9}\end{align*}
2. 519+227\begin{align*}\frac{5}{19} + \frac{2}{27}\end{align*}
3. 512918\begin{align*}\frac{5}{12} - \frac{9}{18}\end{align*}
4. 2314\begin{align*}\frac{2}{3} - \frac{1}{4}\end{align*}
5. Ilana buys two identically sized cakes for a party. She cuts the chocolate cake into 24 pieces and the vanilla cake into 20 pieces, and lets the guests serve themselves. Martin takes three pieces of chocolate cake and one of vanilla, and Sheena takes one piece of chocolate and two of vanilla. Which of them gets more cake?
6. Nadia, Peter and Ian are pooling their money to buy a gallon of ice cream. Nadia is the oldest and gets the greatest allowance. She contributes half of the cost. Ian is next oldest and contributes one third of the cost. Peter, the youngest, gets the smallest allowance and contributes one fourth of the cost. They figure that this will be enough money. When they get to the check-out, they realize that they forgot about sales tax and worry there will not be enough money. Amazingly, they have exactly the right amount of money. What fraction of the cost of the ice cream was added as tax?
7. The time taken to commute from San Diego to Los Angeles is given by the equation time=120speed\begin{align*}time = \frac{120}{speed}\end{align*} where time is measured in hours and speed is measured in miles per hour (mph). Calculate the change in time that a rush hour commuter would see when switching from traveling by bus to traveling by train, if the bus averages 40 mph and the train averages 90 mph.
### Vocabulary Language: English
Equivalent Fractions
Equivalent Fractions
Equivalent fractions are fractions that can each be simplified to the same fraction. An equivalent fraction is created by multiplying both the numerator and denominator of the original fraction by the same number.
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# Can someone help me calculate big O? c++
• I'm only advancing the right side of the window by 1 or more places every iteration. So I was thinking my solution is O(n) but then I grow the window X amount of times depending on how many palindromes are found so O(2n). Can someone explain the big O for my solution, thanks!
``````class Solution {
public:
string longestPalindrome(string s) {
if(s.size() < 2)
{
return s;
}
int eos = s.size() - 1; // end of string
int r(0), l(0), loc(0), size(0); //right and left window, substring location, substring size
for(r = l + 1; r < s.size(); l = r++){
if(s[l] == s[r] || (l > 0 && s[l-1] == s[r])) //palindrom test for aa or aba
{
if(s[l] != s[r]) //assume aba type palindrom found
{
l--;
} else {
//find multiples in a row aaa, aaaaa, aaaaaa
while(r < eos && s[l] == s[r + 1]){
r++;
}
}
//find size of palindrom by growing window in both directions
while(l > 0 && r < eos && s[l-1] == s[r+1])
{
l--;
r++;
}
//if found palindrom bigger than current, record it's place
if(r-l+1 > size)
{
size = r-l+1;
loc = l;
}
//if found palindrom is bigger than 2, reset right to halfway
//between left and right side of the window
if(r-l > 2){
r = 1+r-((r-l)/2);
}
}
}
return size > 0 ? s.substr(loc,size) : s.substr(0,1);
}
};
``````
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https://learnool.com/c2h2f2-lewis-structure/
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# C2H2F2 Lewis structure
The information on this page is ✔ fact-checked.
C2H2F2 (1,2-difluoroethylene) has two carbon atoms, two hydrogen atoms, and two fluorine atoms.
In the C2H2F2 Lewis structure, there is a double bond between the two carbon atoms, and each carbon is attached with one hydrogen atom and one fluorine atom, and on each fluorine atom, there are three lone pairs.
Contents
## Steps
Use these steps to correctly draw the C2H2F2 Lewis structure:
#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure
Let’s discuss each step in more detail.
### #1 First draw a rough sketch
• First, determine the total number of valence electrons
In the periodic table, carbon lies in group 14, hydrogen lies in group 1, and fluorine lies in group 17.
Hence, carbon has four valence electrons, hydrogen has one valence electron, and fluorine has seven valence electrons.
Since C2H2F2 has two carbon atoms, two hydrogen atoms, and two fluorine atoms, so…
Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of two hydrogen atoms = 1 × 2 = 2
Valence electrons of two fluorine atoms = 7 × 2 = 14
And the total valence electrons = 8 + 2 + 14 = 24
• Second, find the total electron pairs
We have a total of 24 valence electrons. And when we divide this value by two, we get the value of total electron pairs.
Total electron pairs = total valence electrons ÷ 2
So the total electron pairs = 24 ÷ 2 = 12
• Third, determine the central atom
Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.
Now we have to choose the central atom from carbon and fluorine. Place the least electronegative atom at the center.
Since carbon is less electronegative than fluorine, assume that the central atom is carbon.
Here, there are two carbon atoms, so we can assume any one as the central atom.
Let’s assume that the central atom is left carbon.
Therefore, place carbons in the center and hydrogen and fluorine on either side.
• And finally, draw the rough sketch
### #2 Mark lone pairs on the atoms
Here, we have a total of 12 electron pairs. And five bonds are already marked. So we have to only mark the remaining seven electron pairs as lone pairs on the sketch.
Also remember that carbon is a period 2 element, so it can not keep more than 8 electrons in its last shell. Hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And fluorine is a period 4 element, so it can keep more than 8 electrons in its last shell
Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines, hydrogens, and right carbon. But no need to mark on hydrogen, because each hydrogen has already two electrons.
So for each fluorine, there are three lone pairs, for right carbon, there is one lone pair, and for left carbon, there is zero lone pair because all seven electron pairs are over.
Mark the lone pairs on the sketch as follows:
### #3 Calculate and mark formal charges on the atoms, if required
Use the following formula to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
For left carbon atom, formal charge = 4 – 0 – ½ (6) = +1
For right carbon atom, formal charge = 4 – 2 – ½ (6) = -1
For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0
For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0
Here, both carbon atoms have charges, so mark them on the sketch as follows:
The above structure is not a stable Lewis structure because both carbon atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.
### #4 Convert lone pairs of the atoms, and minimize formal charges
Convert a lone pair of the right carbon atom to make a new C — C bond with the left carbon atom as follows:
In the above structure, you can see that the central atom (left carbon) forms an octet. The outside atoms (right carbon and fluorines) also form an octet, and both hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.
Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of C2H2F2.
Deep
Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.
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# What does a Riemann sum represent?
A Riemann Sum begins with the question of how to find the area under a curve (i.e., between a positive curve and the x-axis, which is essentially a Geometry question). It turns out, though, that as with many seemingly limited questions in mathematics, we can expand the application of the answer to this question to other areas (e.g., finding distance travelled by a moving object in physics). In order to do this, we must generalize the initial question using what is called a Riemann Sum.
We begin with finding the area under a positive curve (i.e., a curve with positive y-values). For example, suppose we wish to find the area under the curve ##f(x) = x^2## from ##x = 1## to ##x = 3##. We divide this interval into “subintervals.” For example, we might choose four subintervals of equal width (1-1.5, 1.5-2, 2-2.5, 2.5-3).
We then approximate the area of each subinterval using a rectangle. The width of the rectangle is the the change in x which is 0.5 (or ##Delta x = 0.5##). The height is a selected y-value for each subinterval. In this case, we shall choose the right-hand endpoints. Thus we have the following rectangular areas:
##f(1.5)*0.5+f(2)*0.5+f(2.5)*0.5+f(3)*0.5 = 10.75##.
Note that we can abbreviate this using summation (sigma) notation as:
##sum_(i=1)^4(f(1+i*0.5)*(0.5))## or more generally as ##sum_(i=1)^4(f(1+i*Deltax)*(Deltax))##.
We now finally come to the idea of a Riemann Sum. A Riemann Sum is encountered when we relax our rules for situations like the previous one to allow for a more general view of the problem. That is, we could use different values for each ##Delta x## and select various points for each ##f(x)##. We could also include functions that are not always positive.
The Riemann Sum would then look like this: ##sum_(i=1)^n(f(barx_i)*(Deltax_i))##, where ##barx_i## is some x value in the ith. subinterval and ##Deltax_k## is the width of that subinterval.
Some of our products might now be negative, if some of the y-values are negative. These would represent “negative areas” (i.e., areas below the x-axis). The Riemann Sum would be an approximation of the difference between the “positive” and “negative” areas.
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# Density of primes among the first $k$ numbers formed by the digits of $\pi$?
Consider the digits numbers formed by the first $$k$$ digits in the decimal representation of $$\pi, k \ge 1$$
$$3\\31\\ 314 \\3141 \\31415\\314159 \\ 3141592 \\31415926 \\314159265 \\ ...$$
Out of the first $$10^4$$ such numbers $$4000$$ (approximately $$40\%$$) end in $$1,3,7$$ or $$9$$. Since all primes $$> 5$$ end in one of these four digits I checked how many of these $$4000$$ numbers were primes and I could find only corresponding to $$k = 2,6,38$$ which is much lower than what I anticipated.
Question: In general, assuming $$0 < \alpha < 1$$ to be a normal in base $$10$$, what is the expected density of primes among the first $$k$$ numbers formed by the digits of $$\alpha$$ as explained above?
• If we assume random digits, we can estimate the number of primes with the $1/ln(n)$-approach. – Peter Jan 16 '20 at 13:54
• I think, a high search limit has already been established. Maybe, you look it up at OEIS. – Peter Jan 16 '20 at 13:56
• digits in order ? there's a thread on mersenne forum by david eddy you might like. – user645636 Jan 16 '20 at 18:48
• oh and the digits prior to a 1 or 7 will never form a number that is 2 mod 3. all digits prior to 3 or 9 will never form a multiple of 3. – user645636 Jan 16 '20 at 19:08
• you forgot 31 but got 13... – user645636 Jan 17 '20 at 15:48
The Prime Number Theorem states, essentially, that the number of primes less than $$N$$ is approximately $$\frac{N}{\ln{N}}$$. That means (roughly) that the probability of $$N$$ itself being prime is about $$\frac{1}{\ln{N}}$$. Remember, though, that $$N$$ is the number itself, not the number of digits it has. Since $$\ln{N}$$ is roughly equal to the number of digits of $$N$$ multiplied by $$\ln{10}$$, and since a number's index in your sequence is just the number of digits it has, that suggests that the probability that the $$k$$th number is prime should be about $$\frac{1}{k\ln{10}}$$. Assuming that whether each number is prime is independent of the primality of the ones before it, the expected number of primes should be
$$\frac{1}{\ln{10}}\left(1 + \frac12 + \frac13 + \frac14 + \cdots + \frac1k\right)$$
By a result of Euler and Mascheroni, this happens to be almost exactly $$\frac{\ln{k}}{\ln{10}} = \log_{10}k$$. $$\log_{10}{10000} = 4$$, which is exactly what your data indicates!
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Blogg | Combine
Blogg
Or ”can daedalean words actually help make more accurate descriptions of your random variable? Part 1: Kurtosis
Is a common belief that gaussians and uniform distributions will take you a long way.
Which is understandable if one considers the law of large numbers: with a large enough number of trials, the mean converges to the expectation. Depending on your reality, you may not have a large enough number of trials. You could always calculate an expected value out of that (limited) number of trials, and for certain conditions on your random variable, you could get a bound on the variation of that expectation, together with a probability of that bound. And that alone is a lot of information on how your random variable might behave.
But you could do a bit more with those expected value calculations! Especially if you have a reason to believe that the best fit for your random variable might not be normal, yet you simply don’t have enough samples to commit to a model right now. Could it be that the apparent misbehavior of a random variable is actually a feature instead of a bug?
Kurtosis: Peakyness, Broad-shoulderness and Heavy-tailedness
Kurtosis (we can use κ interchangeably to denote it) describes both the peakyness and the tailedness in a distribution. κ can be expressed in terms of the expectation:
β_2 = E(X-μ)^4 / (E(X-μ)^2)^2
Or in terms of the sample means (a.k.a. sample κ):
b_2 = (Sum((Xi- μ{hat})^4)/n) / (Sum((Xi-μ{hat})^2)/n)^2
Kurtosis represents a movement of mass that does not affect the variance and reflects the shape of a distribution apart from its variance. Is defined in terms of an even power of the standard score, so is invariant under linear transformations. Normal distributions have the nice property of having κ = 3. Take the following re-enactment of the wikipedia illustration for normal distributions:
As you may have guessed from the labels, we have plotted three gaussians with different σ^2 (0.2, 1, 5), all with mean zero. If we draw samples from these distributions and apply the sample κ expressions, we will in all cases get κ ~3. All these three distributions have the same Kurtosis. You can try with sample sizes as small as 9, and move all the way to 1000, 3000 and more. You will still get κ ~3. This is why there is a relative measure called Excess Kurtosis, which is the kurtosis of a distribution with respect that of a normal distribution. And as you may have guessed, is built with κ – 3.
Distributions with negative κ – 3 are known as platykurtic. The term ”Platykurtosis” refers to the case in which a distribution is Platykurtic. Relative to the normal, platykurtosis occurs with light tails, relatively flatter center and heavier shoulders.
Distributions with excess kurtosis (with a positive κ – 3) are known as leptokurtic. Just as with Platykurtosis, the term ”Leptokurtosis” refers to the case in which a distribution is Leptokurtic. When Leptokurtosis occurrs, heavier tails are often accompanied by a higher peak. Is easier to think of hungry tails eating variance from the shoulders/tips or ”thinning” them (the greek word ”Lepto” means ”thin, slender”).
So excess kurtosis on most cases captures how much of the variance would go from the shoulders to the tails and to grow the peak (leptokurtosis) or from the tails and the peak height into the shoulders (platykurtosis). Leptokurtosis can either occur because the underlying distribution is not normal, or because outliers are present. So if you are sure that your underlying phenomenon is normal yet you experience leptokurtosis, then you can either re-evaluate your assumptions, or consider the presence of outliers.
Detecting Bimodality
According to De Carlo (first reference of this post) Finucan in his 1964 ”A note on Kurtosis” noted that, since bimodal distributions can be viewed as having heavy shoulders, they should tend to have negative kurtosis. Moreover, Darlington in ”Is kurtosis really ’peakedness’?” (1970) argued that excess kurtosis can be interpreted as a measure of unimodality versus bimodality, in which with large negative kurtosis is related to bimodality, with the uniform distribution (κ = -1.2) setting a dividing point.
Discussing Darlington’s results and Finucan’s note would probably require another post… Let’s see how we go for that later :). For now, I think the following plot from wikipedia shows all the former behaviours really nicely:
Kurtosis for assessing normality
Gaussians are like sedans: you can drive them in the city, you can drive them in the highway, you can drive them to the country. They will take you thru many roads. But sometimes you need a proper truck. Or fabulous roller-skates. And knowing when would you need either can save you from toiling.
Thanks to the higher moments of a distribution, is possible to make relatively cheap tests for normality, even with really small sample sizes. If you wanted to compare the shape of a (univariate) distribution relative to the normal distribution, you can by all means use Excess Kurtosis. Just calculate sample kurtosis for the data you are studying. If it deviates significantly from 3, even for a small number of samples, then you are either not dealing with a gaussian distribution, or there is too much noise for such a sample size.
Multivariate normality & multivariate methods
The assumption of normality in the multivariate case prevails in many application fields, often without verifying how reasonable it would be in each particular case. There is a number of other conditions to check for multivariate normality. A property of multivariate normality is that all the marginals are also normal, so this should first be checked (this can be quickly performed with excess kurtosis). Also, linear combinations of the marginals should also be normal, squared Mahalanobis distances have an approximate chi-squared distribution (often q-q plots are used for this purpose), and we can go on. You could even use this web interface to an R package dedicated exclusively to check multivariate normality.
Kurtosis can affect your study more than you think. When applying a method for studying your random variable, keep in mind that some methods can be more or less affected by the higher moments of the distribution. Some methods are better at dealing with some skewdness (for another post) while some tests are more affected by it. Similarly with Kurtosis. Tests of equality of covariance matrices are known to be affected by kurtosis. Analyses based on covariance matrices (shout out PC Analysis!) can be affected by kurtosis. Kurtosis can affect significance tests and standard errors of parameter estimates.
Final words
The contents of this post have been influenced by:
Is too bad that I could not find a way to read or buy Finucan’s note on kurtosis, because it seemed interesting. If anyone has access to it, please comment and let me know how to get it.
There is more than Monte Carlo when talking about randomized algorithms. It is not uncommon to see the expresions ”Monte Carlo Approach” and ”randomized approach” used interchangeably. More than once you start reading a paper or listening to a presentation, in which the words ”Monte Carlo” appear on the keywords and even on the title, and as you keep reading/listening, you notice in the algorithm description a part in which the runs with incorrect outputs are discarded until only correct outputs are given… Which effectively turns a Monte Carlo into a Las Vegas. Running time is no longer deterministic-ish, and the algorithm will only provide correct answers. So, they say/write ”Monte Carlo” here, ”Monte Carlo” there, and when it comes to what actually happens, you are not really in Monte Carlo, and you might be in Vegas, baby.
You can do a small check yourself. Use your favorite search engine with ”monte carlo simulation” and ”las vegas simulation”. Depending on your search engine, you may get something looking more or less like this:
aprox 84 200 000 results (0,32 seconds) – monte carlo simulation
aprox 9 560 000 results (0,56 seconds) – las vegas simulation
Almost a whole order of magnitude more in relevant results from Monte Carlo, in almost half the time. Now, if you used something like google, you may or may not get relevant results in your first screen, depending on how well google knows you. Of course, there are the cases of ”monte carlo simulation” that most likely refer to other topics e.g. Monte Carlo integration and the like, the same with ”las vegas simulation”. And depending on how much your search engine knows your habits, you might not get any results of Las Vegas simulations right away.
And probably the next thing you may do, is to do a quick wikipedia search of ”Las Vegas algorithm”. You may possibly read it, and find in the end ”Atlantic City algorithm”. And maybe you may want to follow that little wikipedia rabbit hole, and end up reading about it. And then no one can save you.
The featured image in this post is from here.
Remember your friend from our very first post? . Well, I am sorry to say that he never really reached French Guyana. He ended up in Carcass, one of the Malvinas/Falkland islands. And his boat was (peacefully) captured by overly friendly pirate penguins. Now he spends his days counting penguins and sheep. He did keep a coin and this calculator as a memento from his past life. You know, just in case.
As for you, you are still at the office and now your task is to get some samples from a distribution that strangely resembles a cropped half co-sine. Your problem is that such distribution is not in NumPy (gasp!). Don’t fret, we got you covered. NumPy not having a distribution should not cause weeping or gnashing of teeth. In fact, your friend can easily build pretty much any distribution with the help of some math, his coin and his trusty office memento (provided that the coin is unbiased and that he has enough patience). This is possible because of the Probability Integral Transform (PIT). As part of this tutorial, we will do the exercise in slide 11 of this lecture, so you might want to open the slides to do the steps together.
One consequence of the PIT is that as long as you can have a random variable in which a cumulative distribution function (CDF) exists, you can then transform the CDF to express the random variable in terms of the uniform distribution. Which means that for a very long list of distributions it would be possible for you to build your own random number generator, using nothing but some simple functions and a uniform distribution sampler!
So, let’s begin with the exercise of the tutorial. We will use our trusty calculator node, so you can place 3 calculator nodes and a dummy input (which can easily be built with a random table generator). We will not use the actual contents of the dummy input, since we will generate our own signals in the calculator. Your start flow may look like this:
You can change the labels in the node by double clicking the text. Now, right click on the first calculator node (from left to right) and click ”configure”. Make a variable that will be your independent axis through the calculations. If you look in the exercise, the range of the pdf goes from (-pi/2 to pi/2]. Note that the lower bound is an open bound. I will name it x, and you can build it in the calculation field with: np.linspace(-np.pi/2,np.pi/2, num=100)[1:]. You can then double click on the node to run it, in order to have ”x” available as an input for the next node, which you will use to build the pdf.
In the calculation field of the node to build the pdf, you simply need to enter the same constraints that appear in the exercise: The pdf is 1/2 cos (x) for x (-pi/2,pi/2]. We could easily deal with that by removing the last sample on every variable.
np.array([0 if x > np.pi/2. or x <= -np.pi/2. else 0.5*np.cos(x) for x in ${x}]) I was first trying out how would the plots look like if I allowed more than the non-zero range to be generated. But you don’t need to do that. You can just do this instead: np.array([0.5*np.cos(x) for x in${x}])
Since we will build a CDF from this PDF, and we will want to plot them together, then is convenient to let the previous signal pass through. So you can simply write a variable named ”x”, and drag the variable name in ”Signals” of ”x” into the calculations field. Your configuration screen will look something like this:
We will use densFunc and x to evaluate how well our approximation behaves in the end.
Now, building a CDF from this easily integrable analytic function is quite simple. You can go the analytic or the numeric route. For this tutorial we will also show the numeric route because not all pdfs you may run into will have an easily integrable analytic form. We use the trapezoidal rule just to show that the PIT will approximate the original random variable X even with a not-so-smooth approximation of the CDF. To build the CDF, we take each value of the integration and sum it with the previous value, like so:
np.array( [ ( (${x}[index]-${x}[index-1]) * (${densFunc}[index]+${densFunc}[index-1])/2.) + np.sum([ ( (${x}[indexK]-${x}[indexK-1]) * (${densFunc}[indexK]+${densFunc}[indexK-1])/2.) if indexK>0 else 0 for indexK in range(index) ]) if index>0 else 0 for index in range(0,len(${x})) ]) To analytically build a the CDF, you can start by integrating 1/2 cos(x) dx. Which will give you 1/2 sin(x) + C. To find C, solve the initial condition sin(pi/2) = 1, and that gives you 1/2 sin(x) + 1/2 as the function from which you can generate a CDF. Now, if we plot the PDF and the CDF together (you can use the plot tables node, and choose x as the independent axis), they will look like this: Ok, NOW we have all we need to generate some data with the same distribution as X. And we will both use the numeric and the analytic ways. The general way to obtain a function of the uniform distribution is in slide 10 of the presentation, so solving for x: u = 1/2 sin(x) + 1/2 -> x= arcsin(2u-1) You can add a new calculator node, and feed the output of CDF to it. Let us take as many samples as elements exist in our initial independent axis (just for being able to use the same axis across different graphics), and name that simRandVarFromU. You can enter in the calculation field: np.array([np.arcsin(-1+2.*np.random.uniform(0,1)) for b in${x}])
Now, your friend in Carcass Island may actually find it easier to use the numeric way. He probably has a lot of stones and writings on the sand (drawing a lot of slots for his coin to emulate different positions between 0 and 1), and he has been spending a lot of energy scaring away the sheep and the penguins, so let’s make his life a little bit easier. He can calculate numericSim by doing the following:
np.array([${x}[np.where(${cdf}<=np.random.uniform(0,1))[0][-1]] for inx in ${x}]) See what he did there? he rolled his coin until he got a uniformly distributed number ”u” between 0 and 1. Since he knows that the range of the CDF is between 0 and 1, it is safe to see in which value of x was the CDF less than or equal to ”u”, and take that as an index to obtain ”x”. After building the two estimates of X, let us find a way to compare them. So, what we are ideally constructing is a way of building a variable with a distribution as close as possible to X. So, let’s compare the distribution of: • Original X • PIT generated X (analytical) • PIT generated X (numeric) For both PIT – generated cases, we first obtained a histogram, applied a smoothing over the histogram, and then re-sampled the smooth histogram for a linear space like the one in the independent variable of Original X. To build the histograms: Notice that we removed a bin since numpy will generate 1 extra bin. To smooth the histograms, we use a convolution with a window size of 10 samples, like this: np.convolve(${numHist},np.ones(10)/10,mode=’same’)
For both PIT generated histograms. Remember to add all the other signals to be able to make proper plots. So here is a plot of both smooth histograms and their bins:
To be able to compare the smooth histograms and the real PDF in the same plot, we made an interpolation (to change sample size). We simply did:
And after putting together all signals on the same table, we were able to produce this plot:
So, one can see that at least with 100 samples and a smoothing window of 10 samples, with 50 bins, we got this approximation. What would happen if we start playing with this parameters? First, changing the number of bins at this small amount of samples would not make much difference. Second, getting more samples is very cheap for you in your desktop, but changing a convolution window would be cheaper for your friend in Carcass. An increase of a smoothing window to 23 samples will give us:
But if we really wanted to see if the signals converge together, throw a bigger number of samples. Say, throw 1000 samples, and leave the smoothing window at size 10:
I hope that you can add the PIT to your back of tricks, and use it whenever you have the need. Until the next post!
This post borrowed the featured image from here.
Or ”Martingales are awesome!”.
In a previous post, we talked about bounds for the deviation of a random variable from its expectation that built upon Martingales, useful for cases in which the random variables cannot be modeled as sums of independent random variables (or in the case in which we do not know if they are independent or not, or even their distribution). We briefly explained that a martingale sequence is a sequence X_0, X_1,… so that for all i>0:
E[X_i | X_0,…,X_{i-1}] = X_{i-1}
and also:
E[X_i] = E[X_0] for all i≥0.
We discussed two bounds over variation of expectation when we have a Martingale (Azuma and Kolmogorov-Doob). However, the point of treating random variables as Martingales was not touched.
And, how exactly can I treat any random variable as a Martingale?
Easy, cowboy. Consider a probability space (Ω,F,Pr), in which Pr is a probability measure assigned to a field (Ω,F), from which Ω is a sample space, and F is a collection of subsets of the sample space Ω. One can see a sample space as an arbitrary set containing all possible outcomes in a probabilistic experiment. Events are subsets of the sample space. For instance, if the experiment is a sequence of two coin-flips, then Ω contains {hh,ht,th,tt}. An event ε ⊆ Ω for the experiment of two coin-flips can be the cases in which we get different values on each coin flip, containing {ht,th}. The collection of events in F must satisfy:
1. ∅ ∈ F
2. ε ∈ F ⇒ complement of ε also ∈ F (closure under complement)
3. ε1, ε2,… ∈ F ⇒ ε1 ∪ ε2 ∪ … ∈ F (closure under countable union)
Closure under complement together with closure under countable union also imply closure under countable intersection. For a field in which F = 2^Ω, is possible to build sequences of nested subsets of F so that F_0 ⊆ F_1 ⊆ … ⊆ F_n. From that, we can see that F_0 contains the empty event, and F_n contains 2^Ω. Those sequences of nested subsets of F is what we call a filter or filtration in the context of probability spaces. Given an arbitrary F_i in a filter, we can say that an event ε ∈ F is F_i-measurable when ε ∈ F_i. This basically means that an event can only be measurable in a filtration, if the subset of the sample space associated to that event is also contained in the filtration. As a consequence of filtration being nested subsets, if an event is F_i-measurable, then the event is also F_{i+1}, F_{i+2},… and definitely F_n measurable.
Now consider a random variable X over a probability space (Ω,F,Pr). X can be viewed as a function X: Ω → R, meaning that for a given sample ω ∈ Ω, the random variable will take the value X(ω) ∈ R. In other words, we have a random variable that is used to associate a random sample in the sample space with a number in the domain of the real numbers. Considering the F_i-measurability concept, we can say that the random variable X is F_i-measurable if for each x ∈ R, the event {X = x} is contained in F_i. That also means that X is F_j measurable for all j≥i.
Starting from this random variable X, we can build our Martingale. What can be said about X with respect to F_{i-1}? Since X is a function over the values of Ω, we know that the values X will take will be constants over each event. X is a function and its value does not have to be a constant for a sequence of events. It can have a different behavior. However, X is well defined enough to have an expected value over a sequence of events, and such expected value is:
E[ X | F_{i-1} ]
Which is the expected value of our random variable X over the events contained in F_{i-1}. And this expected value is in itself another random variable. A random variable that constitutes a mapping from F_{i-1} into the reals. An interesting property of this random variable, is that its value is constant if X is also F_{i-1}-measurable. This means that the expected value of X given F_{i-1} will be constant as long as there are events in F_{i-1} for all values X can take. And of course, there is the case in which E[ X | F_{i-1} ] can be constant if E[X] is constant, when X is independent of F_{i-1} (since then, from conditional expectation rules, E[ X | F_{i-1} ] will simply be E[X]).
Putting everything together: for a probability space (Ω,F,Pr) with a filtration F_0 ⊆ F_1 ⊆ …, and a sequence of random variables X_0, X_1, …, so that for all i≥0, X_i is F_i-measurable. Then, the sequence X_0,… X_n is a martingale, as long as for all i≥0,
E[ X_{i+1} | F_i ] = X_i
Any subsequence of a martingale is also a martingale, relative to the corresponding subsequence of the filter. Considering that, and knowing that X is a random variable over the probability space, then is possible to define:
X_i = E[ X | F_i ]
and it will follow that the sequence X_0,… X_n is also a martingale. This is how you construct a martingale from a random variable, and is also known as a Doob martingale. The key point is that each for each 0≤i≤n, X_i is F_{i-1}-measurable.
So, I can treat a random variable as a Martingale. What can I do with that?
You could do this. And while you are at it, you can consider the following:
• All martingale properties are applicable to this martingale sequence.
• The expectation variation bounds with martingales discussed in the previous post are applicable. Those are powerful bounds since they are not so restrictive.
• Constructions of martingale differences is possible by having Y_i = X_i – X_{i-1} and requiring E[Y_{i+1} | F_i] =0
But if you want some examples, we can give you some ideas to build a relaxation of your real application.
Polya’s Urn Scheme:
Consider an urn that initially contains b black balls and w white balls. We perform a sequene of random selections from this urn, where at each step the chosen ball is replaced with c balls of the same colour. Let X_i denote the fraction of black balls in the urn after the ith trial. Show that the sequence X_0, X_1… is a martingale.
To prove this, we will use a property of martigales, by which for all i≥0, E[X_i] = E[X_0], and we will first take the trivial case, and use it to build a general case.
First, let’s take the trivial case. The value of X_o is a constant depending on b and w, which is which is b/(b+w). Since the expected value of a constant is the constant, then E[X_o] = X_o .
Now, let’s build an expression for the general case of X_i. Since at each trial we replace the chosen ball with c balls of the same colour, then on every ith step we add in total (c-1) balls into the urn. This means that in any given point i, we will have in total:
b+w+i*(c-1)
balls. X is the fraction of black balls in a given time i, so that numerator becomes itself a random variable depending on the expected value of black balls at point i. Then, we can say that at a certain point i we will add (c-1) black balls with probability E[X_{i-1}]. We use E[X_{i-1}], since the fraction of balls in the urn (X) is a probability, but since the value of such probability is a random variable itself, then we talk expected values of that random variable. In this way, the numerator becomes:
b + expected value of black balls at time i
which is:
b + Σ (number of black balls added on each i) * (probability of adding a black ball at i)
which for all i>0 is:
b + Σ (from 0 to k=i-1) (c-1) * E[X_k]
Now putting numerator and denominator together:
E[Xi] = (b + Σ (from 0 to k=i-1) (c-1) * E[X_k]) / (b+w+i*(c-1))
This expression holds for i>0. Let us first verify that this expression holds for i=1:
E[X_1] = (b + Σ (from 0 to k=0) (c-1) * E[X_0]) / (b+w+(c-1))
E[X_1]= (b+ (c-1) * b/(b+w) ) / (b+w+c-1) = ((b*(b+w)+b*(c-1)) / (b+w)) / (b+w+c-1)
= (b*(b+w+c-1)) / ((b+w)*(b+w+c-1)) = b/(b+w)
So, E[X_1] = E[X_0]. For the nth case:
E[X_n] = b*(b+w+n*(c-1)) / (b+w)*(b+w+n(c-1)) = b/(b+w)
Again, E[X_n] = E[X_0]. Which is what we intended to demonstrate. Oh yeah. You can look at other examples for inspiration in 18.2.1 of here, the whole presentation here, and something a bit more hardcore yet well explained here.
The featured image of the title comes from here.
In the previous post we looked at Chebyshev’s, Markov’s and Chernoff’s expressions for bounding (under certain conditions) the divergence of a random variable from its expectation. Particularly, we saw that the Chernoff bound was a tighter bound for the expectation, as long as your random variable was modeled as sum of independent Poisson trials. In this post, we will expand to the cases in which our random variables cannot be modeled as sums of independent random variables (or in the case in which we do not know if they are independent or not, or even their distribution). For that, this post will be on two inequalities involving Martingales: Kolmogorov-Doob and Azuma.
Martingales have attracted more than one, especially those related to gambling (for some anecdotal cases, check here). The martingales we refer to in this post, are sequences of random variables. So, for a sequence X_0, X_1,… if for all i>0:
E[X_i | X_0,…,X_{i-1}] = X_{i-1}
And this is known as a martingale sequence. Consequently, in martingale sequences, E[X_i] = E[X_0] for all i≥0. Taking of X_i as the capital of a gambler playing a fair game after the ith bet. In a fair game, both players have the same chance of winning, so the expected gain or loss from each bet is zero. So even if sometimes you win and sometimes you lose, gains and losses should cancel each other and converge at the initial value. Hence the phrase ”quit while you are ahead”.
When we get a sequence with the following property:
E[X_i |X_0,…, X_{i-1}] ≥ X_{i-1}
then the expected value of our capital would either remain or grow with each bet. It doesn’t mean we will never lose, it only means that time is in our side. And such sequences are known (unsurprisingly) as super-martingales.
The concept becomes even more appealing to gamblers when we define martingales in terms of net gains or losses from the ith bet. So, being Y_i=X_i-X_{i-1} (or the difference between the capital of the gambler in bet i and in bet i-1) and X_i = X_0 + ∑Yj (from j=1 until j=i), we can get something called a sequence of differences (also named martingale differences). In martingale differences, it holds that for all i≥1:
E[Y_i | Y_1,…,Y_{i-1}] = 0
There is a way to construct martingale sequences from any random variable (also known as Doob martingales). It is also possible to convert super-martingales into martingales. But maybe that is a matter to talk about in another post. Maybe titled ”Martingales are awesome”.
Now, lets get into the matter of our expectation bounds. Now, how would a person deciding where to invest its capital feel if he/she could have an idea of when to ”pull out” of the investment (a.k.a the time i of expected maximum capital during the cycle). Or if he/she has provisioned enough to survive the roughest time of the investment?.
Kolmogorov-Doob bound:
Let X_0, X_1,… be a martingale sequence. Then for any λ > 0 :
Pr[max_{0≤i≤n} X_i ≥ λ] ≤ E[|X_n|] / λ
This form of Kolmogorov-Doob tells us that for a capital X, the probability of having maximum capital ≥ λ in any bet i of the whole game (a.k.a. that moment in which you are on fire) is bounded by E[|X_n|]/λ.
Notice that E[|X_n|] is the sum of the product of all possible positive AND negative values times their probabilities (shamelessly citing wikipedia here). Following that, if there was an equal probability of X_n taking any value between X_n and -X_n, then E[|X_n|] would become 0 and supporting myself on |E[X]| ≤ E[|X|], and being |E[X]| non zero if X0 is nonzero, then it must apply that in a martingale, the probabilities of having any specific value of X_n between X_n and -X_n are not the same for every value, unless (possibly) X_0 was 0 -interestingly, I have not found explicit restrictions over the value of X_0-. If you wanted to be pessimistic and lazy, you could use |E[X_n]|/ λ = |X_0|/λ as your bound instead -if the worst case scenario does not look so bad and you have no other information, you may as well dwell with care into it-. You may optionally scream ”YOLO” for courage.
Azuma’s bound:
Can be found in other textbooks as Hoeffdings’s inequality. Let X_0, X_1,… be a martingale sequence so that for each for each k:
|X_k – X_{k-1}| ≤ c_k
where c is independent of k. Then, for all t≥0 and any λ >0,
Pr[|X_t-X_0| ≥ λ] ≤ 2e^( (-λ^2) /(2*∑(c_k)^2 (for k=1 to k=t) ) )
This bound over martingales resembles the exponential bounds in Chernoff’s inequality. It basically tells us that the probability of X deviating more than λ from its initial value in a bet t, is bounded by an exponential function, given that the successive differences in each bet are less than or equal to a certain quantity c. The process of applying this bound to a martingale sequence is also known as the method of bounded differences.
To apply any of the former methods you would have to of course do your homework and find out as much information as you can on the system from which the martingale is generated. Sometimes it can be useful to build the martingale from a relaxation of the system, make your calculations over it, and see if your relaxation is harder than how the system is more likely to behave. And there are many other useful bounds over martingales that are not in this post. In other words, if you can possibly model a system’s behaviour with martingales, you can get a much better idea of the system without necessarily relying on experimental data (see Polya’s urn problem in our next post, ”Martingales are Awesome”).
Most information in this post was taken from section 4 of the Randomized Algorithms book from R. Motwani and P. Raghavan, and from sections 7 and 12 from the Probability and Random Processes book from Grimmett and Stirzaker.
In our previous post, we briefly explored the plotting capabilities built in Sympathy, and also the enormous flexibility that the calculator node brings into our flow. Today we will use those two nodes to build and plot one of the simplest, most widely used models of data: a linear regression.
Our flow will simply have four parts:
1. A data capturing/manipulation part (that we use to obtain/import/filter the measurements and sample number / time information),
2. A part for obtaining the model coefficients
3. A part for generating a plot of the model together with the input data
4. A part for exporting the data and plots.
In the following figure, parts 2-4 are visible.
For some help on how to do part 1, you can read here.
Regarding part 2, we will use the calculator node. More specifically in the flow in here, is the node in the leftmost bottom part that says ”Calculate linear coefficients”. Our configuration screen looks as follows:
The two input signals have the same size and were generated with the same time axis. As you can see, we are using polyfit in numpy for obtaining the model coefficients in each signal (see numpy documentation). This means that if you know your data can be better fit to another polynomial, you could request a different degree for the coefficients. In this example, for the two measurements, we have requested a first degree. Measurement 1 is more or less well behaved, while Measurement 2 is much noisier.
Now, the output of this calculation operation will be the two coefficients, ordered by the degree of the coefficient, i.e. if the model is mX^1 + bX^0, then we will have that linearCoeffs1[0] = b and linearCoeffs1[1] = m.
With this in mind, we can go to the configuration of the next calculator node (the one that says build linear model) in order to generate the data for plotting the model:
So, since our linear model is Y = mX+b, then we generate a numpy array taking m and b from the linearCoeffs that we would like to plot, and we do that over the length of the original measurements (in this example it was 50 samples). In this way, the output signal from the plottable model is as long as the input data, and we can easily apply ”Hjoin Tables”, in order to have all the information we need to plot it.
After joining the original data with the models, we can go to the configuration of the plot node, and put together the model and the input data in the same canvas. You can play with the colors, the types of lines and markers, add a grid, place proper names, and even add the mean of each signal in the label (as the featured image in this post).
The final step is exporting. A common use of the plotter in sympathy is as an inspection tool for signals that we have in our data, in that way we may notice if we need to filter the data a bit better or if the signals are behaving as expected. Then, to export the plots in the plotting node, we feed the signal into an exporter, and we choose the type of file we want to export the graphic in ”plot output extension”. We can choose between png, pdf, and a couple more. This will export both the data and the plot.
And you are set! I hope you have some nice holidays. I will do, for sure!
In a previous post we were discussing the pros and cons of parametric and non-parametric models, and how they can complement each other. In this post, we will add a little more into the story. More specifically, we are going to talk about bounds to the probability that a random variable deviates from its expectation. In these two posts we will talk about the following well-known techniques:
1. Chebyshev’s Inequality
2. Markov’s Inequality
3. Chernoff’s Bounds
They exploit the knowledge on some feature of the random variable e.g. the variance of the distribution, or the random variable being obtained from a poisson process, or knowing that the random variable does not have negative values. You can find extensions for other properties of the random variable under ”tail inequalities” or ”deviation inequalities”. Techniques 1 and 2 were developed in the last part of the 1800’s. The latest is a bit more modern (the author is still alive!).
Why would you want to estimate bounds over the probability that a random variable deviates from its expectation? Some applications rely on expected values to make estimates/predictions. And on many cases, this is a reasonable assumption to make. If you want to use expected values to represent a random variable, or as part of another technique to provide an output in a decision making process, then is sensible to provide some bounds on the probability that the random variable may deviate from the expectation.
If you are lucky enough to find a problem (or a simplification of a problem) in so that it satisfies all conditions necessary for all techniques, the order of ”tightness” of the bounds is Chernoff-Chebyshev-Markov, being Chernoff the tightest. This is possibly the reason why, in spite of Chebyshev’s inequality being older, many textbooks choose to talk about Markov´s inequality first. It is not rare to see authors use Markov’s inequality in the proof for Chebyshev’s. Actually, is not rare at all to see Markov’s inequality while going thru proofs, simply because it requires so little from your random variable, so it became a staple in the pantry of statisticians.
Chebyshev’s Inequality
Take-home message: ”The probability that a random variable is at least t standard deviations away from its expectation is at most 1/t^2”
Condition: We know the variance of the distribution.
Boring footnote: Some notes use interchangeably the distance from the expectation and the distance from the mean, which for a big enough number of samples becomes reasonable. I chose to use the expectation instead because the demostration of Chebyshev via Markov uses a random variable composed of the absolute value of the difference between your random variable and its expectation, so is easy to remember. But both are probably just as fine.
Markov’s Inequality
Take-home message: ”The probability that a random variable takes values bigger than or equal to t times its expectation, is less than or equal to 1/t”.
Condition: The random variable must not take negative values.
Boring footnote: It would make more pedagogical sense to start with Markov’s inequality, but I feel I need to start making some historical justice. If I find something older and just as useful as Chebyshev’s, I will probably make a newer post and scorn myself. Like this.
Ok, sorry about that link. Here, have some eyebleach.
Chernoff’s Bounds
Take-home messages:
”The probability that a random variable X built from the sum of independent Poisson Trials deviates to less than (1-δ) times their expectation μ, is less than exp(-μ δ ^2 / 2)”
”For the same random variable X, the probability of X deviating to more than (1-δ) times μ, is less than (exp(δ) / (1+δ)^(1+δ) ) ^ μ.
Conditions: The random variable must be a sum of independent Poisson trials.
This post will hopefully motivate you to study this topic on your own. If you do, some books you can check out:
The beautiful image for the preview is from here.
This post is my interpretation of Chapter 10 of the book ”Advanced Data Analysis from an Elementary point of view”. It is one of the most interesting reads I have found in quite some time (together with this).
Actually, the original title for the post was ”Book Chapter review: Using non-parametric models to test parametric model mis-specifications”. But shorter titles tends to attract more viewers.
The first question that one might ask is ”If we are going to use a non-parametric model to test a parametric model, why not going straight to non-parametric modelling instead?”. Well, there are advantages to having a parametric model if you can build one. It could be of interest for your application to express your process in terms of known mathematical models. Also, a well specified parametric model can converge faster to the true regression function than a non-parametric model. Finally, if you only have a small number of samples, you could have better predictions by using a parametric model (even if slightly mis-specified). The reason is simply because parametric models tend to have a significantly smaller bias than non-parametric models. Also, for the same number of samples, a well specified parametric model is likely to have less variance in its predictions than a non-parametric model. Now, if you have a cheap and fast way to obtain many more samples, a non-parametric model can make better predictions than a mis-specified parametric model.
This is a consequence of the bias-variance trade-off that the author explains in a way that a person without a background in statistics can understand (in chapter 1.1.4 of the same book).
Non-parametric models ”have an open mind” when building a model, while parametric models ”follow a hunch”, if you will. One can find some similarities between modelling (parametric, non-parametric) and search algorithms, more specifically in uninformed search (BFS, DFS, Iterative deepening and variants) vs informed search (say, A* and variants). Even with a relaxed (and admissible) version of the optimal heuristic, A* is expected to traverse a shorter path than any uninformed search algorithm. However, it will traverse a larger path if the heuristic is poorly constructed, and most likely be outperformed by uninformed search. With this in mind, another question that may tickle you is: Can one translate a modelling problem into a search problem and have a machine automatically and optimally find the best parametric model for a given problem? Oh, yes (you can leave that as background music for the day if you like). You will of course need to express the problem in a way that the program terminates before our sun dies, and an admissible heuristic also helps. But yeah, you can do that. Humans often solve harder problems than they give themselves credit for.
If you were to build such a machine, the author can give you some suggestions to check for errors in your model. For instance, he suggests that if you have a good reason to think that the errors in a model can ONLY come from certain mis-specifications (say, that instead of being Y= θ1 X + ε it can only be Y=θ1 X + θ2 X + ε or maybe a couple other forms) then it may probably be faster and less sample-hungry for you to simply check whether the estimated θ2 is significantly different from 0, or whether the residuals from the second model are significantly smaller than those from the first. However, when no good reason is available to argue for one or other source of mis-specification, you can use non-parametric regression to check for all sources by doing either one of the following:
• If the parametric model is right, it should predict as well as, or even better than the non-parametric one. So, you can check if the difference between Mean Squared errors of the two estimators is small enough.
• If the parametric model is right, the non-parametric estimated regression curve should be very close to the parametric one. So, you can check that the distance between the two regression curves is approximately zero in all points.
• If the parametric model is right, then its residuals should be patternless and independent of input features. So, you can apply non-parametric smoothing to the parametric residuals and see if their expectation is approximately zero everywhere.
For the last method, I can elaborate on the explanation from the author. If the residuals are Y-f(x;θ), then the expected value of the residuals given an input X is E[Y-f(x;θ)|X] (the author did not made it explicit, but I assume that the residuals must be calculated with x ⊆ X. I could be wrong on this, so please correct me if I am). Now, being our typical regression model something in the lines of Y=f(X;θ)+ε, we substitute this in the expectation, and we end up with E[f(x;θ)+ε-f(x;θ) | X]. In this expression, we have that f(x;θ)+ε-f(x;θ) = ε, so you end up with E[ε|X]. Since the constant is independent from X, then the expression becomes E[ε|X] = E[ε]. Since the expected value of a constant ε will always be the constant, then E[ε]=ε. And with a significantly small enough ε, we can say that ε ≅ 0, so no matter what input X we have, the expected value of the residuals of the predictor should be approximately equal to zero.
So yes, under some circumstances (too little samples) you can actually be better off with a slightly mis-specified (i.e. relaxed) model, than with a full non-parametric model. And yes, you can indeed check if the assumptions for your model are actually valid.
Have fun, and see you in the next post!
Allow me to introduce you to your new best friend from Sympathy 1.2.x: The improved calculator node. The node takes a list of tables, from which you can establish a new signal with the output for a calculation. There is already a menu with the most popular calculations and a list of signals from the input tables, all in the same configuration window.
This is the configuration window
To do this tutorial, it is recommended to use some data in csv format, and the latest version of Sympathy for Data (1.2.5 at the time of writing this post).If you would like to more complex formats (say, INCA-files) here are some pointers that can help you on that.
Now, with your data set and ready, follow the steps in here and locate the nodes to build the following flow:
Try to build this flow in sympathy 1.2!
And then right-click on the ”Calculator” node to open the configuration window for the node. Now, lets get our first calculation going!
Our First Calculation: A Column operation
1. Write a Signal Name (you don’t have to do it as the first step, but it helps you keep track of your work). In this example, we have chosen to write ”sumA” since we intend to obtain the sum of the values of a signal whose name starts with A.
2. Drag-and-drop a function from ”Available functions” into ”Calculation”. In this example, we have chosen to drag ”Sum”.
3. Drag-and-drop a signal name into approximately the point in the calculation in which the variable name is supposed to go.
You may need to delete some remaining text in the calculation until the preview in the top right shows you a correct number. It should look similar to the following screen:
First calculation. Note that the preview in the top right shows you a valid value
Second Calculation: A Row-wise operation
Now lets create a new calculation: click on ”New” (on top of ”Edit Signal”). We will now make a new signal that will be the sum of two signals, elementwise. As you can see in the figure below, this signal will be called ”sumTwoSignals”. So, simply drag-and-drop the names of the signals you want to sum, and make sure you put the operator between the two names. In this case, the operator was a sum (+).
Second calculation: elementwise sum two signals
What happens if the calculation outputs differ in shape?
In many cases (like in this example), the calculations may not have the same dimensions, and you may want to separate them into different output tables, because then Sympathy will inform you that it can not put together two columns with different lengths on the same table. For those cases, uncheck the ”Put results in common outputs” box, and then the results will go to different columns. In newer versions of sympathy the checkbox ”Put results in common output” is checked by default.
Now, lets go into something more interesting…
You can use the principles of the previous two sections together with the operators in ”available functions” in order to accomplish many day-to-day tasks involving calculations over your data.
But there is more fun from where all of that came from!. The real hidden superpower of your new best friend relies on the fact that you can access numpy from it. Yes, you can have the best of both worlds: the power of numpy from a graphic interface! Your colleague can stay in French Guyana and watch some space launches, you have almost everything you can ever dream of!
For our next example, lets build a binomial distribution. All you need to do is to write in the calculation field:
np.random.binomial(number_of_tests,success_pr,size=(x))
In which ”x” is the number of instances for the trials. This will give you a signal of size ”x”, from which each element in the output signal is the number of successful outcomes out of ”number_of_tests” for an event with probability ”success_pr” of succeeding. You can now manually assign a value to those elements. Notice in the figure below that you can either hand-type a set of values in there for quick estimations, or use variable names from input signals as inputs to the function.
Examples for using np.random.binomial with hand-written parameters and with signals from input files
Building and plotting a PDF (Probability Density Function)
A nice feature of the histograms in numpy, is that you can use them to build probability density functions by setting the parameter density to True (see numpy documentation entry here). Since np.histogram will return two arrays (one with the PDF, and one with the bin_edges) then you may want to also add a [0] in the end. So, it can look like this:
np.histogram(${yourVariable},bins=10,density=True)[0] Now lets plot the distribution!. To be able to plot anything, since the plotting in sympathy is based upon matplotlib, you will require an independent axis, form which your PDF will be the dependent axis. You could take the other returning value of the histogram for that purpose, by writing a new calculation which will serve as the new independent axis for plotting: np.histogram(${NVDA-2006-MA3day},bins=10,density=False)[1][:-1]
Note that the length of the bin edges are always length(hist)+1, and here we have chosen the left as the start of the distribution. A simple way to be clean about what you intend to plot, is to extract the signals and then hjoin then in a table, which will be fed to the ”Plot Table” module in sympathy. It can look like this:
An example on selecting the signals for plotting
In the configuration menu of the ”Plot Table” module, you can select the labels for the independent (X) and dependent (Y) axis, the scale, ticks and if you would like to add a grid into the plot. In the ”Signal” Tab, you can select which signal goes to which axis, and the colours and style of the plot. See below an example of a plot with 10 bins, a plot with 100 bins, and a plot with 1000 bins.
A plot of a PDF with only 10 bins
Same data, but 100 bins
Same data, 1000 bins
We hope that with this tutorial you can have some tools to study your data. And if you get bored and start daydreaming about your colleague at French Guyana, you can always watch this video. Have fun!
When your data is incomplete, somewhat corrupted or you simply need to use a black-box tool, you can help yourself by using statistics. Statistics build upon probability theory to draw an inference from our data.
Now, let’s get something out of the way: Not everyone has the same interpretation of the ”Probability” concept, and I will try my best to not to impose my view on yours. If you would like a nip on that discussion, check out this Wikipedia entry.
No matter what your view on it is, the quality of your inference depends mainly on the quality and quantity of data, the methods used for the inference and HOW the methods were chosen. How much do you really know about your data? Information on your domain can help you make assumptions to use techniques that do more with less data. While is entirely possible to infer something when you hardly know the domain, then you may need to use more ”data hungry” techniques. How much error/bias will you be introducing by using a technique when the assumptions do not entirely hold on the data? (and how much are you willing to accept for a clue?). You may hit a wall when you have a very small number of samples and little knowledge on the data. Then you should either get more samples, or get a better understanding on the data from other sources.
So here is what you can expect: Statistics are not magic. Getting good data is hard. Formatting data to make it useful requires work (which we hope tools like Sympathy can help you with). Choosing a set of techniques for your data is not a kitchen recipe, even though some sciences through the years have devised methods for specific situations. A method can get you information that is simply not there (bias), so be very careful and double check everything you find.
Econometrics theory is like an exquisitely balanced French recipe, spelling out precisely with how many turns to mix the sauce, how many carats of spice to add, and for how many milliseconds to bake the mixture at exactly 474 degrees of temperature. But when the statistical cook turns to raw materials, he finds that hearts of cactus fruit are unavailable, so he substitutes chunks of cantaloupe; where the recipe calls for vermicelli he uses shredded wheat; and he substitutes green garment dye for curry, ping-pong balls for turtle’s eggs and, for Chalifougnac vintage 1883, a can of turpentine.”
This quote from Cosma Rohilla Shalizi’s wonderful 800-page draft for his textbook ”Advanced Data Analysis from an Elementary Point of View”, in which he was quoting Stefan Valavanis, quoted in Roger Koenker, “Dictionary of Received Ideas of Statistics” s.v. “Econometrics”. When the circumstances surrounding your data are hard to control you may turn into the Swedish chef, but make do.
Now, lets get into the matter in the next post!
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Time remaining:
# Multiply and simplify 36*7/4
label Algebra
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5
Multiply and simplify 36*7/4
Nov 17th, 2017
The easiest way to do this is to imagine as a fraction as well. So, 36/1 X 7/4
It's hard to picture in this text, but multiply the top numbers together, then multiply the bottom number together. Up top we have 36 x 7 = 252
On the bottom we have 1 x 4 = 4
252/4 = 63
Oct 21st, 2014
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Nov 17th, 2017
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Nov 17th, 2017
Nov 18th, 2017
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posted by on .
Given f(x) = 2root x+7 and g(x) = 5x^2 – 9, evaluate (f/g)(2) . I don't understand how to do this, can someone please explain the process??!! thanks!
• grade 11 math - ,
I assume that you mean the function of a function, f[g(x)]
First calculate g(2)
g(2) = 5*4 - 9 = 11
Then calculate f(11)
Is f(x) = (2sqrtx)+7 or 2 sqrt(x+7) ?
I can't tell from the way your wrote it.
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# Symbolic Logic
Topics: Logic, Semantics, Truth Pages: 3 (542 words) Published: January 21, 2013
Midterm Examination
In
-------------------------------------------------
SYMBOLIC LOGIC
-------------------------------------------------
“Great knowledge comes to those who are willing to learn.”
Test I. Identification. Choose the correct answer from the choices provided inside the box. Hypothetical PropositionBroad Disjunctive~
Conditional PropositionConjunctive PropositionV
Disjunctive PropositionSymbolic Logic .
Strict Disjunctive=Ɔ
Conditional PropositionConjunctive PropositionV
Disjunctive PropositionSymbolic Logic .
Strict Disjunctive=Ɔ
_____________1. This symbol is used to refer the word of negation “not”. _____________2. This symbol is used to refer to the hypothetical proposition “if . . . then”. _____________3. This symbol is used to connote that p and q are both true or if p and q are both false. _____________4. This symbol is used for strict disjunctives and broad disjunctives. _____________5. This symbol means that if p and q are both true, all other instances are false.
_____________6. It establishes an artificial language, which eliminates the vagueness of ordinary language. _____________7. It is usually a compound proposition, which contains a proposed or tentative explanation. _____________8. It is a compound proposition in which one clause asserts something as true provided that the other clause is true.
_____________9. It is a compound proposition in which one member or more than one member may be true. _____________10. It asserts that two alternatives cannot be true at the same time or both alternative may be false.
Test II. Determine the following Hypothetical Propositions as a. Strict Disjunctive,
c. Conditional Proposition,
d. Categorical Proposition,
e. Conjunctive Proposition or
f. Not a Hypothetical Proposition.
_____________ 1. You are against the administration or you are not....
Please join StudyMode to read the full document
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# Algebra 2 – Set Equality and Subsets
Hello. I’m Professor Von Schmohawk and welcome to Why U. In the previous lecture
we saw how to define sets and their members through the use of set-builder notation. In this lecture, we will learn about several
types of relations between sets. One such relation is “equality”. If two sets contain exactly the same elements
they are said to be equal. For example, set A, which consists of
the elements one, two and three and set B consisting of
the elements two, three, and one are equal since they both contain
the same elements. Notice that it doesn’t matter that their elements
are listed in different orders. This is because the order in which the elements
of a set are listed is irrelevant. The set consisting of one, two, and three is exactly the same as the set
consisting of two, three, and one. The list does not imply that there is any
particular order to the elements. It only says that every element listed
is a member of the set. So if sets A and B contain the same elements
then they are equal. Another important relation between sets
is the relation of “subset”. If all the elements of set A
are also contained in set B then we say that set A
is a subset of set B. This relation is denoted using
the subset symbol. If two sets, A and B, are equal
then they are subsets of each other. This is because all the elements of A
are members of B and all the elements of B
are also members of A. So any two sets that are equal
are subsets of each other. And since every set is equal to itself
every set is also a subset of itself. So the subset symbol can be used whenever all the elements of A
are also members of B whether A has the same number of elements
as B or has fewer elements. But when A has fewer elements, we can be more
specific and call A a proper subset of B. We denote this using the proper subset symbol which is the subset symbol
without the line underneath. In addition to the relation of subset,
a set can also be a “superset”. In any case where set A
is a subset of set B we can also say that set B
is a superset of set A. This relation is denoted using the superset
symbol, which is the subset symbol reversed. Likewise, if A is a proper subset of B
then B is a proper superset of A. But what about the empty set,
the set containing no elements? The empty set is considered to be a subset
of every other set including itself. In fact, the empty set is the only set
which is a subset of every set. So far we have represented sets
by listing their members or using set-builder notation or by drawing little ovals
with the elements inside. In the next lecture, we will see how to
visualize sets using Venn Diagrams.
### 26 thoughts on “Algebra 2 – Set Equality and Subsets”
• July 28, 2012 at 7:06 am
i like your revolutioninzed way of teaching….
• October 15, 2012 at 11:21 pm
only 2 comments and only 2500, subscribers, he deserves a lot more people. Oh yeah, only 16 likes (I just liked đ
• November 7, 2012 at 11:32 am
Very useful for teaching middle school. Thanks NyWhyU đ
• November 7, 2012 at 1:35 pm
• February 26, 2013 at 11:25 am
hi .. can i download ur video … it so very helpful for student here to learn …
• June 12, 2013 at 10:57 pm
A very nice representation of set, subsets, and supersets! (Discrete Math-VCU).
• September 12, 2013 at 9:12 am
thanks man very innovative way of teaching
• April 22, 2014 at 5:20 pm
woot, math…
• May 9, 2015 at 7:59 pm
great way to explain
• October 13, 2015 at 6:53 pm
on the proper subset, what If A has fewer elements than B, also A has an element that B does NOT. Would this still make A a proper subset of B?
• February 7, 2016 at 9:21 pm
Unbelievable these videos should have way more subscribers and views!
• March 8, 2016 at 12:53 pm
• May 25, 2016 at 6:57 am
03:30 And this makes a problem when we try to count the number of elements in a set: if an empty set is always a subset of every set, that means that it also needs to be counted as an element when we have a set of sets. And this is the case with the set-theoretical definition of numbers, which is based on nesting empty sets one inside another and counting them.
• May 25, 2016 at 7:00 am
+quantumsingularityup not really. In order to reason about sets, you need the rules of reasoning in the first place. Therefore, logic is more basic than set theory (and perhaps the most basic in all Mathematics).
• October 16, 2016 at 6:16 pm
then what's the difference between equality and subset ?
• November 25, 2016 at 11:45 am
That into is like powerpuff girls
• March 5, 2017 at 4:20 pm
Well Explained. Thanks.
However, I'm wondering about the tobacco smoking pipes.
tobacco smoking pipes!? Are you promoting tobacco !?
• March 7, 2017 at 3:25 pm
lol love it i love whyu
• May 16, 2017 at 7:40 pm
Can you make videos in Combinatorics?
• July 25, 2017 at 6:09 am
This is a lot easier and more fun than being shoved into a classroom at 7:00am while people throw rubber bands and insult you across the room while the teacher corrects homework all day instead of teaching and then fail because you forget to turn in your work :).
• October 7, 2017 at 5:25 am
I want to make a set of all fractions.
Can you please tell me how to say that a fraction has to be simplified in notation? (note: the fraction is one variable, not two seperate variables)
• February 2, 2019 at 2:31 pm
That video is fire
• March 25, 2019 at 10:58 am
Why is the empty set considered to be a subset of every set? A set other than an empty set is not empty (has some number of elements), how are they even related?
• May 1, 2019 at 12:19 am
Ummmm the lady has 1,48 foods she ate 1,67 how many DOSE she have leftover heheheheheh MATH ppl
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## Why It Matters: Rational Expressions and Equations
#### Why study rational expressions and equations?
If you were around in 2011, you probably heard about Occupy Wall Street, the protest against social and economic inequality in the US. According to Wikipedia, the main issues raised by Occupy Wall Street were social and economic inequality, greed, corruption, and the perceived undue influence of corporations on government—particularly from the financial services sector. The OWS slogan, “We are the 99%”, refers to income inequality and wealth distribution in the U.S. between the wealthiest 1% and the rest of the population. To achieve their goals, protesters acted on consensus-based decisions made in general assemblies which emphasized direct action over petitioning authorities for redress.
Economists are also interested in how wealth is distributed in economies. The Lorenz curve is a mathematical representation of the distribution of income or of wealth. Max O. Lorenz developed it in 1905 to represent inequality of the wealth distribution.
The curve in the graph below shows the proportion of overall income or wealth assumed by the bottom x% of the people. It is often used to represent income distribution, where it shows for the bottom x% of households, what percentage (y%) of the total income they have. The percentage of households is plotted on the x-axis, the percentage of income is on the y-axis. It can also be used to show distribution of assets. In such use, many economists consider it to be a measure of social inequality.
Lorenz Curve
Points on the Lorenz curve represent statements like “the bottom 20% of all households have 10% of the total income.” A perfectly equal income distribution would be one in which every person has the same income. In this case, the bottom N% of society would always have N% of the income. This can be depicted by the straight line y = x; called the “line of perfect equality.”
By contrast, a perfectly unequal distribution would be one in which one person has all the income and everyone else has none. In that case, the curve would be at y = 0% for all x < 100%, and y = 100% when x = 100%. This curve is called the “line of perfect inequality.”
The Gini coefficient is the ratio of the area between the line of perfect equality and the observed Lorenz curve to the area between the line of perfect equality and the line of perfect inequality. The higher the coefficient, the more unequal the distribution is. This is given by the ratio $G=\frac{A}{A+B}$, where A and B are the areas of regions as marked in the graph below.
Gini Index
The closer the Gini index index is to 1, the less equally wealth is distributed amongst an economy. The closer it is to 0, the more equally wealth is distributed. Below is a map of the world with Gini Index color coded.
World map color coded by Gini Index
The Gini Index is a rational expression, or ratio. In this module, we will define and apply mathematical operations to rational expressions. We will also solve rational equations.
### Learning Outcomes
• Operations With Rational Expressions
• Define and simplify rational expressions
• Multiply and divide rational expressions
• Add and subtract rational expressions
• Rational Equations
• Solve rational equations
• Solve proportional problems
• Solve work equations
• Solve variation problems
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# ON A CLASS OF PRESENTATIONS OF MATRIX ALGEBRAS
1 Definitions and a positive result
For a brief history of the problem of how to identify a complete matrix ring,
and of how to present a matrix ring by generators and relations, we refer to
the introduction of [1]. In that paper it was established that a k- algebra R
is an
(m+n) ×(m+n) matrix algebra if and only if there are three elements
a, b, c ∈ R satisfying the relations and . There it
was also shown that the k-algebra presented by the two generators a and b
and (1) be low is nonzero if i/m = j/n. This paper deals further with these
two-element presentations (1), whose theory is by no means as complete as
that of the three-element presentation studied in [1].
Unless otherwise stated k is an arbitrary associative commutative ring with
1. All algebras will here be as sumed associative and with 1.
Let us make
Definition 1.1 Let i, j, m and n be positive integers. Let R(k; i, j, m, n)
denote the k-algebra presented by two generators a and b and the relations
We will show that the question whether R(k; i, j, m, n) is nonzero, and
whether it admits a homomorphism to a matrix algebra over k, will only
depend on the ordered pair of positive rational numbers (p, q) = (i/j, m/n).
For a field k we will show that R(k; i, j, m, n) can be mapped into
for some N and hence is non zero if p is arbitrary and q = 1. On the other
hand in the next section we will show that for all k, R(k; i, j, m, n) is trivial
if q ≠ 1 and p = 1.
We will begin with the following lemma which will be useful in proving the
next theorem.
Lemma 1.2 Let R be a ring, and d a positive integer, and regard R as
embedded in by the diagonal embedding. Then every element of R
has a d-th root in .
Proof: A d-th root of x is given by
We can now show that the behavior of R(k; i, j, m, n) only depends on the
ordered pair (i/j, m/n).
Theorem 1.3 For positive integers c and d we have the following:
1. R(k; i, j, m, n) is nonzero if and only if R(k; ci, cj, dm, dn) is nonzero.
2. R(k; i, j, m, n) can be mapped into for some positive integer N,
if and only if R(k; ci, cj, dm, dn) can be mapped into for some
N'.
Proof: Let c and d be positive integers. It is clear that if R(k; ci, cj, dm, dn)
is nonzero then so is R(k; i, j, m, n), also if R(k; ci, cj, dm, dn) can be mapped
into for some N then so can R(k; i, j, m, n).
Assume R(k; i, j, m, n) to be nonzero. If we let e be a common multiple
of c and d, then is a nonzero k-algebra in which
the relations defining R(k; ei, ej, em, en) have a solution by Lemma 1.2.
R(k; ei, ej, em, en) is therefore nonzero and hence also R(k; ci, cj, dm, dn).
Let us finally assume that R(k, i, j, m, n) can be mapped into for
some N. Equivalently there are matrices a and b in satisfying (1).
Let again e be a common multiple of c and d . Then by Lemma 1.2 we
can find e-th roots of a and b in . Hence R(k; ei, ej, em, en) can be
mapped into , and therefore so can R(k; ci, cj, dm, dn).
Definition 1.4 Let be the set of ordered pairs of positive rational numbers
(i/j, m/n) such that R(k; i, j, m, n) is nonzero, and let be the
set of (i/j, m/n) such that R(k; i, j, m, n) can be mapped into for
some positive integer N.
Theorem 1.3 shows that the sets and are indeed well defined.
There are some basic relations worth noticing. We see by left-right symmetry
that for X ∈ { B, C} we have
If there exists a homomorphism of commutative rings k → k' then
If k and k' contain fields of the same characteristic, then .
We will prove the next theorem in a few steps .
Theorem 1.5 If k is a field and p a positive rational number then (p, 1) ∈
Let us first notice a few things that will ease the task. When seeking a map
R(k; i, j, 1, 1) → for some N it suffices by Theorem 1.3 to consider
the case gcd(i, j) = 1.
Lemma 1.6 If is the algebraic closure of the field k, then .
Proof: Clearly if k is a sub field of k'. If k'/k is a finite field
extension then . Since where the union is taken over
all finite field extensions k'/k, the lemma follows.
Theorem 1.5 will clearly follow once we have proved the following lemma:
Lemma 1.7
If k is an algebraically closed field and i, j are relatively prime
positive integers then one can always find a nonzero k-algebra homomorphism
R(k; i, j, 1, 1) → .
Proof: We will consider the following three cases:
Case 1. i + j is even (hence i, j are odd): Here we clearly have a surjection
R(k; i, j, 1, 1) → given by .
Case 2. char(k) : By left-right symmetry of R(k; i, j, 1, 1) we may
assume j > i to hold. Consider the polynomials .
Since char(k) does not divide i + j we have that is separable and
has therefore a root which is not a root of . Hence in this case there
is an r ∈ k such that and .
Now for this r ∈ k the k-algebra homomorphism R(k; i, j, 1, 1) →
given by
is clearly well defined.
Case 3. char(k) l i+j and i+j is odd: Let char(k) = p. Since gcd(i, j) = 1,
we have gcd(i, p) = 1 and thus, i is invertible in the prime field of k with an
inverse c. Consider the k-algebra homomorphism R(k; i, j, 1, 1) →
defined as follows:
We easily see that:
and computation using the fact that i+j is odd shows that we get
1 and . Hence the above homomorphism is well defined.
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# 0110ge. Geometry Regents Exam Which expression best describes the transformation shown in the diagram below?
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1 0110ge 1 In the diagram below of trapezoid RSUT, RS TU, X is the midpoint of RT, and V is the midpoint of SU. 3 Which expression best describes the transformation shown in the diagram below? If RS = 30 and XV = 44, what is the length of TU? 1) 37 ) 58 3) 74 4) 118 In ABC, m A = x, m B = x +, and m C = 3x + 4. What is the value of x? 1) 9 ) 31 3) 59 4) 61 1) same orientation; reflection ) opposite orientation; reflection 3) same orientation; translation 4) opposite orientation; translation 1
2 4 Based on the construction below, which statement must be true? 6 Which transformation is not always an isometry? 1) rotation ) dilation 3) reflection 4) translation 7 In ABC, AB BC. An altitude is drawn from B to AC and intersects AC at D. Which conclusion is not always true? 1) ABD CBD ) BDA BDC 3) AD BD 4) AD DC 1) m ABD = 1 m CBD ) m ABD = m CBD 3) m ABD = m ABC 8 In the diagram below, tangent PA and secant PBC are drawn to circle O from external point P. 4) m CBD = 1 m ABD 5 In the diagram below, ABC is inscribed in circle P. The distances from the center of circle P to each side of the triangle are shown. If PB = 4 and BC = 5, what is the length of PA? 1) 0 ) 9 3) 8 4) 6 Which statement about the sides of the triangle is true? 1) AB > AC > BC ) AB < AC and AC > BC 3) AC > AB > BC 4) AC = AB and AB > BC
3 9 Which geometric principle is used to justify the construction below? 1 Lines j and k intersect at point P. Line m is drawn so that it is perpendicular to lines j and k at point P. Which statement is correct? 1) Lines j and k are in perpendicular planes. ) Line m is in the same plane as lines j and k. 3) Line m is parallel to the plane containing lines j and k. 4) Line m is perpendicular to the plane containing lines j and k. 1) A line perpendicular to one of two parallel lines is perpendicular to the other. ) Two lines are perpendicular if they intersect to form congruent adjacent angles. 3) When two lines are intersected by a transversal and alternate interior angles are congruent, the lines are parallel. 4) When two lines are intersected by a transversal and the corresponding angles are congruent, the lines are parallel. 13 In the diagram below of parallelogram STUV, SV = x + 3, VU = x 1, and TU = 4x Which equation represents the circle whose center is (, 3) and whose radius is 5? 1) (x ) + (y + 3) = 5 ) (x + ) + (y 3) = 5 3) (x + ) + (y 3) = 5 4) (x ) + (y + 3) = 5 11 Towns A and B are 16 miles apart. How many points are 10 miles from town A and 1 miles from town B? 1) 1 ) 3) 3 4) 0 What is the length of SV? 1) 5 ) 3) 7 4) 4 14 Which equation represents a line parallel to the line whose equation is y 5x = 10? 1) 5y x = 5 ) 5y + x = 10 3) 4y 10x = 1 4) y + 10x = 8 3
4 15 In the diagram below of circle O, chords AD and BC intersect at E, mac = 87, and mbd = What is an equation of the line that contains the point (3, 1) and is perpendicular to the line whose equation is y = 3x +? 1) y = 3x + 8 ) y = 3x 3) y = 1 3 x 4) y = 1 3 x What is the degree measure of CEA? 1) 87 ) 61 3) ) 6 19 In the diagram below, SQ and PR intersect at T, PQ is drawn, and PS QR. 16 In the diagram below of ADB, m BDA = 90, AD = 5, and AB = 15. What is the length of BD? 1) 10 ) 0 3) 50 4) 110 What technique can be used to prove that PST RQT? 1) SAS ) SSS 3) ASA 4) AA 17 What is the distance between the points ( 3, ) and (1, 0)? 1) ) 3 3) 5 4) 5 4
5 0 The equation of a circle is (x ) + (y + 4) = 4. Which diagram is the graph of the circle? 1 In the diagram below, ABC is shown with AC extended through point D. 1) If m BCD = 6x +, m BAC = 3x + 15, and m ABC = x 1, what is the value of x? 1) 1 ) ) 16 4) ) 3) Given ABC DEF such that AB DE = 3. Which statement is not true? BC 1) EF = 3 m A ) m D = 3 3) area of ABC area of DEF = 9 4 4) perimeter of ABC perimeter of DEF = 3 4) 5
6 3 The pentagon in the diagram below is formed by five rays. 6 What is the image of point A(4, ) after the composition of transformations defined by R 90 r y = x? 1) ( 4, ) ) (4, ) 3) ( 4, ) 4) (, 4) 7 Which expression represents the volume, in cubic centimeters, of the cylinder represented in the diagram below? What is the degree measure of angle x? 1) 7 ) 96 3) 108 4) 11 4 Through a given point, P, on a plane, how many lines can be drawn that are perpendicular to that plane? 1) 1 ) 3) more than 4) none 5 What is the slope of a line that is perpendicular to the line whose equation is 3x + 4y = 1? 3 1) 4 ) ) 3 4) 4 3 1) 16π ) 34π 3) 97π 4) 3, 888π 8 What is the inverse of the statement If two triangles are not similar, their corresponding angles are not congruent? 1) If two triangles are similar, their corresponding angles are not congruent. ) If corresponding angles of two triangles are not congruent, the triangles are not similar. 3) If two triangles are similar, their corresponding angles are congruent. 4) If corresponding angles of two triangles are congruent, the triangles are similar. 6
7 9 In RST, m RST = 46 and RS ST. Find m STR. 30 Tim has a rectangular prism with a length of 10 centimeters, a width of centimeters, and an unknown height. He needs to build another rectangular prism with a length of 5 centimeters and the same height as the original prism. The volume of the two prisms will be the same. Find the width, in centimeters, of the new prism. 33 In the diagram below of ACD, E is a point on AD and B is a point on AC, such that EB DC. If AE = 3, ED = 6, and DC = 15, find the length of EB. 31 In the diagram below of circle C, QR is a diameter, and Q(1, 8) and C(3.5, ) are points on a coordinate plane. Find and state the coordinates of point R. 34 In the diagram below of TEM, medians TB, EC, and MA intersect at D, and TB = 9. Find the length of TD. 3 Using a compass and straightedge, and AB below, construct an equilateral triangle with all sides congruent to AB. [Leave all construction marks.] 35 In KLM, m K = 36 and KM = 5. The transformation D is performed on KLM to form K L M. Find m K. Justify your answer. Find the length of K M. Justify your answer. 7
8 36 Given: JKLM is a parallelogram. JM LN LMN LNM Prove: JKLM is a rhombus. 38 On the set of axes below, solve the following system of equations graphically for all values of x and y. y = (x ) + 4 4x + y = On the grid below, graph the points that are equidistant from both the x and y axes and the points that are 5 units from the origin. Label with an X all points that satisfy both conditions. 8
9 ID: A 0110ge Answer Section 1 ANS: The length of the midsegment of a trapezoid is the average of the lengths of its bases. x + 30 = 44. x + 30 = 88 x = 58 PTS: REF: ge STA: G.G.40 TOP: Trapezoids ANS: 1 x + x + + 3x + 4 = 180 6x + 6 = 180 x = 9 PTS: REF: 01100ge STA: G.G.30 TOP: Interior and Exterior Angles of Triangles 3 ANS: PTS: REF: ge STA: G.G.55 TOP: Properties of Transformations 4 ANS: PTS: REF: ge STA: G.G.17 TOP: Constructions 5 ANS: 1 The closer a chord is to the center of a circle, the longer the chord. PTS: REF: ge STA: G.G.49 TOP: Chords 6 ANS: PTS: REF: ge STA: G.G.56 TOP: Isometries 7 ANS: 3 PTS: REF: ge STA: G.G.31 TOP: Isosceles Triangle Theorem 8 ANS: 4 x = (4 + 5) 4 x = 36 x = 6 PTS: REF: ge STA: G.G.53 TOP: Segments Intercepted by Circle KEY: tangent and secant 9 ANS: 4 PTS: REF: ge STA: G.G.19 TOP: Constructions 10 ANS: 3 PTS: REF: ge STA: G.G.71 TOP: Equations of Circles 11 ANS: PTS: REF: ge STA: G.G. TOP: Locus 1 ANS: 4 PTS: REF: 01101ge STA: G.G.1 TOP: Planes 1
10 ID: A 13 ANS: 1 Opposite sides of a parallelogram are congruent. 4x 3 = x + 3. SV = () + 3 = 5. 3x = 6 x = PTS: REF: ge STA: G.G.38 TOP: Parallelograms 14 ANS: 3 m = A B = 5. m = A B = 10 4 = 5 PTS: REF: ge STA: G.G.63 TOP: Parallel and Perpendicular Lines 15 ANS: = 1 = 61 PTS: REF: ge STA: G.G.51 TOP: Arcs Determined by Angles KEY: inside circle 16 ANS: 1 a + (5 ) = ( 15 ) a + (5 ) = 4 15 a + 50 = 60 a = 10 a = 10 PTS: REF: ge STA: G.G.48 TOP: Pythagorean Theorem 17 ANS: 4 d = ( 3 1) + ( 0) = = 0 = 4 5 = 5 PTS: REF: ge STA: G.G.67 TOP: Distance 18 ANS: 4 The slope of y = 3x + is 3. The perpendicular slope is = 1 3 (3) + b 1 = 1 + b b = PTS: REF: ge STA: G.G.64 TOP: Parallel and Perpendicular Lines 19 ANS: 4 PTS: REF: ge STA: G.G.44 TOP: Similarity Proofs 0 ANS: PTS: REF: 01100ge STA: G.G.74 TOP: Graphing Circles
11 ID: A 1 ANS: 1 3x x 1 = 6x + 5x + 14 = 6x + x = 1 PTS: REF: 01101ge STA: G.G.3 TOP: Exterior Angle Theorem ANS: Because the triangles are similar, m A m D = 1 PTS: REF: 0110ge STA: G.G.45 TOP: Similarity KEY: perimeter and area 3 ANS: 3. The sum of the interior angles of a pentagon is (5 )180 = 540. PTS: REF: 01103ge STA: G.G.36 TOP: Interior and Exterior Angles of Polygons 4 ANS: 1 PTS: REF: 01104ge STA: G.G.3 TOP: Planes 5 ANS: 3 m = A B = 3 4 PTS: REF: 01105ge STA: G.G.6 TOP: Parallel and Perpendicular Lines 6 ANS: 1 A (, 4) PTS: REF: 01103ge STA: G.G.54 TOP: Compositions of Transformations KEY: basic 7 ANS: 3 V = πr h = π 6 7 = 97π PTS: REF: 01107ge STA: G.G.14 TOP: Volume and Lateral Area 8 ANS: 3 PTS: REF: 01108ge STA: G.G.6 TOP: Conditional Statements 3
12 ID: A 9 ANS: = 67 PTS: REF: 01109ge STA: G.G.31 TOP: Isosceles Triangle Theorem 30 ANS: 4. l 1 w 1 h 1 = l w h 10 h = 5 w h 0 = 5w w = 4 PTS: REF: ge STA: G.G.11 TOP: Volume 31 ANS: (6, 4). C x = Q x + R x. C y = Q y + R y. 3.5 = 1 + R x = 8 + R y 7 = 1 + R x 6 = R x 4 = 8 + R y 4 = R y PTS: REF: ge STA: G.G.66 TOP: Midpoint 3 ANS: PTS: REF: 01103ge STA: G.G.0 TOP: Constructions 33 ANS: 5. 3 x = x = 45 x = 5 PTS: REF: ge STA: G.G.46 TOP: Side Splitter Theorem 34 ANS: 6. The centroid divides each median into segments whose lengths are in the ratio : 1. TD = 6 and DB = 3 PTS: REF: ge STA: G.G.43 TOP: Centroid 4
13 ID: A 35 ANS: 36, because a dilation does not affect angle measure. 10, because a dilation does affect distance. PTS: 4 REF: ge STA: G.G.59 TOP: Properties of Transformations 36 ANS: JK LM because opposite sides of a parallelogram are congruent. LM LN because of the Isosceles Triangle Theorem. LM JM because of the transitive property. JKLM is a rhombus because all sides are congruent. PTS: 4 REF: ge STA: G.G.7 TOP: Quadrilateral Proofs 37 ANS: PTS: 4 REF: ge STA: G.G.3 TOP: Locus 38 ANS: PTS: 6 REF: ge STA: G.G.70 TOP: Quadratic-Linear Systems 5
### JEFFERSON MATH PROJECT REGENTS AT RANDOM
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### 0113ge. Geometry Regents Exam In the diagram below, under which transformation is A B C the image of ABC?
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### 0611ge. Geometry Regents Exam Line segment AB is shown in the diagram below.
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### 0610ge. Geometry Regents Exam The diagram below shows a right pentagonal prism.
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### 0112ge. Geometry Regents Exam Line n intersects lines l and m, forming the angles shown in the diagram below.
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### 0114ge. Geometry Regents Exam 0114
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Thought for the day: Patterned lists of equations or expressions can help students shift their focus from procedures and answers to relationships and meaning.
Concepts: Solutions of linear equations (possibly graphs of linear equations, x-intercepts, translations of graphs, solutions of non-linear equations)
Examples of noticing and wondering
I notice that all three equations look the same except for the number subtracted from x.
I notice that the solutions will not be whole numbers.
I notice that all of the equations are linear.
I notice that each solution increases by 1 as you go down the list.
I notice that the graph of each left side moves one unit to the right as you go down the list.
I wonder why the solutions increase by 1 each time.
I wonder If the solutions would still increase by 1 if you changed the 5 (or the 12) to some other number (but still kept it the same in every equation).
I wonder if I could create a list of linear equations whose solutions increase by some other number each time.
I wonder how the patterns I have seen would change (or stay the same) if (1) the parentheses were removed, or (2) the subtraction were replaced by addition, or (3) the x were replaced by x squared.
Creating something new
Students may create and explore their own patterned lists of equations. Many of their creations may flow out of things that they have wondered about. Examples:
Create patterned lists of equations like the ones in this prompt but without the parentheses.
Create patterned lists of equations like the ones in this prompt but using addition instead of subtraction.
Create patterned lists of equations using x squared instead of x.
Create patterned lists of equations whose solutions follow other predetermined patterns.
Create stories or word problems to fit the equations in this prompt or lists of equations of your own.
Notes
The solutions to the equations are 5.4, 6.4, and 7.4, respectively, from top to bottom. Students' first inclination is often to solve each one separately from scratch, rather than to use the relationships between the equations to help them predict a solution or find a more efficient approach.
The pattern in the solutions mirrors the pattern in the numbers that are subtracted from x. Thinking about why this happens can (at least eventually) shift students' focus from the procedure to the meaning of the word "solution." Since, in each case, 5 times some number (which happens to be 2.4) must equal 12. When the amount subtracted from x increases by 1, the value of x must also increase by 1 in order to compensate and leave the value within the parentheses at 2.4.
Students who draw graphs to represent the equations may recognize that n in the expression 5(x – nis always equal to the x-intercept of the graph of the left side and that this intercept must therefore increase by 1 when n increases by 1. This rightward shift in the graph also shifts the point of intersection of the graph with the horizontal line y = 12 one unit to the right, which offers another way to understand (and visualize) the increase in the solutions.
You can easily adapt this prompt to different needs by adjusting the complexity of the equations. For example, the equations will have simpler solutions if you replace 5 by 6 in each one.
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Mathematics
# $\begin{array} { l } { \text { In fig. find the values of } x \text { and } y \text { and then } } \\ { \text { Show that } A B \| C D } \end{array}$
$x=130$ and $y=130$
##### SOLUTION
From given figure
$50^o+x=180^o$.......(Linear pair)
$x=180^o-50^o$
$x=130^o$
Also,
$y=130^o$ .........(vertically opposite angles)
$\therefore x=y$
So, alternate angles are equal
If a transversal intersects two lines such that pair of alternate interior angles are equal, then lines are parallel.
$AB||CD$
You're just one step away
Single Correct Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 86
#### Realted Questions
Q1 Single Correct Medium
$P$ is any point inside the triangle $ABC$. Hence $\angle BPC> \angle BAC$
• A. False
• B. Ambiguous
• C. Data insufficient
• D. True
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1 Verified Answer | Published on 09th 09, 2020
Q2 Subjective Medium
In questions $1$ and $2,$ given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:
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1 Verified Answer | Published on 09th 09, 2020
Q3 Subjective Medium
In the given figure, $\angle ABC$ and $\angle PQR$ are given such that $AB\parallel PQ$ and $BC\parallel QR$. Find the value of $\angle x$ and $\angle y$
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1 Verified Answer | Published on 09th 09, 2020
Q4 One Word Medium
Measure of one angle of linear pair is $108^{\circ}$, then find the measure of another angle.
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1 Verified Answer | Published on 09th 09, 2020
Q5 Subjective Medium
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1 Verified Answer | Published on 09th 09, 2020
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In a data set, absolute deviation of an element is the absolute difference between a given point and that element. Typically the point from which the deviation is measured is a measure of central tendency, most often we consider median or sometimes the mean of the data set. In this page, we shall learn in detail about absolute deviation.
## Definition
In statistical distribution, absolute deviation is an absolute differences between individual number and their median or mean.
Mainly we can find the absolute deviation for the mean and median:
1) Mean Absolute Deviation
2) Median Absolute Deviation
Mean Absolute Deviation
Mean deviation is defined as the average of the absolute deviations taken from an average usually, mean, median or mode. The mean absolute deviation is also calculated by subtracting each score from its median, taking the absolute value and then dividing by N (total number of scores).
Median Absolute Deviation
Median absolute deviation is defined as the median of the absolute deviation taken from median. The formula for median absolute deviation:
$MAD$ = | $y$i - Median of data set|
The mean absolute deviation is somewhat different than other measures of dispersion. This measure could be used with any measure of central tendency.
## Formula
Absolute deviation is the difference between particular number (mean/median) and a given point.
Absolute Deviation Equation:
$D$i = |$y$i - $m(y)$|
where: $D$i = absolute deviation,
$m(y)$ = measure of mean/median
and $y$i = data element
## Examples
Below are solved examples to illustrate absolute deviation:
Example 1: Find the absolute deviation for the data set {1, 2, 5, 6, 7, 8, 10, 15,13, 14, 11, 9, 10}.
Solution:
Step 1: Arrange given data in ascending order.
1, 2, 5, 6, 7, 8, 9, 10, 10, 11, 13, 14, 15
Step 2: Median $m(y)$ = 9
$y$ $y$ - $m(y)$ |$y$ - $m(y)$| 1 -8 8 2 -7 7 5 -4 4 6 -3 3 7 -2 2 8 -1 1 9 0 0 10 1 1 10 1 1 11 2 2 13 4 4 14 5 5 15 6 6 $\sum$ = 111 $\sum$ = 44
Example 2: Calculate absolute deviation about mean for data set: {12, 4, 31, 23, 25}
Solution:
Let y = 12, 4, 31, 23, 25
Step 1 -
Mean of given data = $\frac{Sum\ of\ Observations}{Total\ number\ of\ Observations}$
= $\frac{12+4+ 31+ 23+ 25}{5}$
= $\frac{95}{5}$
= 19
=> Mean ($\bar{y}$) = 19
Step 2:
y y - $\bar{y}$ |y - $\bar{y}$| 12 -7 7 4 -15 15 31 12 12 23 4 4 25 6 6 $\sum$ = 95 $\sum$ = 44
## How to Calculate Absolute Deviation?
When the absolute value of the deviation is calculated, a different distribution emerges:
The following steps are involved in the calculation of absolute deviations:
1) Calculate the median or mean of a given series.
2) Subtract each score from the median or the mean.
3) Take the absolute value.
The absolute values of the scores from the mean, median and mode are shown below:
$y$ $y$ - $\bar{y}$ |$y$ - $\bar{y}$| $y$ - $y$median |$y$ - $y$median| $y$ - $y$mode |$y$ - $y$mode| 6 1.714 1.714 2 2 3 3 5 0.714 0.714 1 1 2 2 7 2.714 2.714 3 3 4 4 3 -1.286 1.286 -1 1 0 0 3 -1.286 1.286 -1 1 0 0 4 -0.286 0.286 0 0 1 1 2 -2.286 2.286 -2 2 -1 1 $\sum$ = 30 $\sum$ = 10.286 $\sum$ = 10 $\sum$ = 11
From the above table we can see that, the total of the absolute values of the median is the smallest value.
## Absolute Standard Deviation
Standard deviation is the most reliable and most commonly used absolute measure of dispersion. The standard deviation method is derived from the mean absolute deviation. The basic difference between standard deviation and the mean absolute deviation is that in the latter each deviation from the mean is squared.
The absolute standard deviation is the square root of the sum of the squared deviance of all values from the absolute mean, divided by the number of values.The equation for ungrouped data from a sample is a variation of that for the mean deviation:
Standard deviation = $\sqrt{\sum|y-\bar{y}|^2}{n}$
Where, $\bar{y}$ = mean of the data and
$n$ = total number of scores.
## Calculate Average Deviation
Average deviation is a condensed statistic of statistical dispersion. We can calculate average deviation by taking the average of the absolute deviations. In general form, average is the result of measure of central tendency.
Average deviation is help to summarize a set of observations, variability or statistic of statistical dispersion. It is the average of the absolute deviations and also called the mean absolute deviation.
The average deviation for the set {$y_1, y_2, y_3,....., y_k$} is given below:
$\frac{1}{k}$ $\sum_{i=1}^k$ |$y$i - $m(y)$|
$n$ = Total score
$m(y)$ = Measure of central tendency
Follow the following steps to find the average absolute deviation:
Step 1: Subtract the mean from each score in order to get the distance of that score form the mean.
Step 2: Consider all these deviations from the mean as positive.
Step 3:
Take the average of these absolute deviation.
Example: Find the average deviation for the data set {2, 4, 6, 2, 3} and show that total of the absolute values of the median is the smallest value.
Solution:
Arrange the data in ascending order.
{2, 2, 3, 4, 6}
Total number of terms (n) = 5
Mean = $\frac{2+2+3+4+6}{5}$ = 3.4
Median = 3
Mode = 2
$y$ |$y$ - $\bar{y}$| |$y$ - $y$median| |$y$ - $y$mode| 2 1.4 1 0 2 1.4 1 0 3 0.4 0 1 4 0.6 1 2 6 2.6 3 4 $\sum$ = 6.4 $\sum$ = 6 $\sum$ = 7
Now
$\frac{1}{n}$$\sum |y - \bar{y}| = \frac{1}{5} \times 6.4 = 1.28 \frac{1}{n} \sum|y - ymedian| = \frac{1}{5} \times 6 = 1.2 (smallest value) \frac{1}{n}$$\sum$ |$y$ - $y$mode| = $\frac{1}{5}$ $\times$ 7 = 1.4
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# $P(\hat{\theta}\neq \theta) \rightarrow 0$ as the sample size increases implies $\hat{\theta}= \theta+o_p(1)$?
While I think it is reasonable, I cannot show this result.
Suppose $$\hat{\theta}$$ is an estimator of $$\theta$$ and $$P(\hat{\theta}\neq \theta) \rightarrow 0$$ as the sample size increases, that is, $$P(\hat{\theta}\neq \theta)=o(1)$$. Then I can write $$\hat{\theta}=\theta+o_p(1)$$.
How can I show this?
*Note the difference between, $$o(1)$$ and $$o_p(1)$$. While "little oh" deal with sequences of numbers, the "little oh in probability" is related with convergence in probability (to zero).
*The only similar result that I know is $$\hat{\theta}=E(\hat{\theta})+(Var(\hat{\theta}))^{1/2} O_p(1)$$. In my case, $$\hat{\theta}=\min_{\theta_1}\{ RSS(\theta_1)+\lambda \theta_1\}$$
• I'm actually not aware of the difference between $o(1)$ and $o_p(1)$. Does it have to do with strict boundedness vs. boundedness in probability? Jun 3, 2019 at 15:50
• @AdamO ok. I Updated my question! Jun 3, 2019 at 15:57
That $$\hat{\theta} = \theta + o_{p}(1)$$ is a restatement of $$\hat{\theta} \to_{p} \theta$$. (Aware that saying a sequence $$(x_{n})$$ converges to some $$x$$ is equivalent to saying $$x_{n} = x + o(1)$$.) The latter statement holds iff $$\mathbb{P}(|\hat{\theta} - \theta| \geq \epsilon) \to 0$$ for every $$\epsilon > 0$$, by definition. Since $$|\hat{\theta} - \theta| \geq \epsilon$$ implies $$\hat{\theta} \neq \theta$$, we have $$\mathbb{P}(|\hat{\theta} - \theta| \geq \epsilon) \leq \mathbb{P}(\hat{\theta} \neq \theta) \to 0$$ by assumption.
• Nice. I believe the key point is the last inequality. For a given $\epsilon$ , in fact, the inverse image $\{\hat{\theta} \neq \theta\}$ contains $\{\mid \hat{\theta}-\theta\mid\}$. Using the monotonicity of the measure, the result follows. But I also need an argument about the equivalence that you claimed in the first sentence. Jun 3, 2019 at 16:36
• @Danmat, The first claim is by convention. The point seems to be the conclusion of a convergence in probability. Unless there is another meaning assigned to the statement pertaining to $o_{p}(1)$, the first claim is not much crucial.
• @GaryMoore Yes, I thought about the first calim, and concluded that the equivalence follows directly by the definition of convergence in probability and the definition of $o_p$. Thanks for the answer! Jun 3, 2019 at 17:08
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Main content
# Optimization problem: extreme normaline to y=x²
A difficult but interesting derivative word problem. Created by Sal Khan.
## Want to join the conversation?
• What is the source of this problem?
(29 votes)
• And we've got solve it in under 3-5 mins for 6 marks , although
(9 votes)
• at how did you decide to factor out 4/x^2. It seems like a wild goose chase to me.
(29 votes)
• If you notice that a polynomial has degrees such as -2, then 0, then 2, and you want to turn it into 0, 2, and 4 to make it easier to factor, you can multiply by the second degree, but must make sure to also divide by the second degree to keep it equivalent. Dividing by the second degree is the same as "factoring out" the reciprocal of the second degree. Example, a = (1/1)a = (n/n)a = (1/n)(n)a = (1/n)(na). I don't know if this helps.
(20 votes)
• At , why does -1 + 1/(2x_0^2) "reach a maximum or a minimum when it = 0"?
(30 votes)
• When the derivative of a function is zero, that means that the slope of the function is zero, so it has reached a local maximum or minimum value. As seen here: http://imgur.com/vw9uW
(21 votes)
• At , why is the derivative 0 at maximum point? Has Sal done videos on it? If not, can someone please explain to me why is the derivative = 0 , at minimum/maximum point? Thanks in advance.
(14 votes)
• You can view it this way: when a function reaches a maximum point, the tangent line goes horizontally, i.e. it's slope equals 0. So, the derivative of a function at a maximum/minimum point = 0.
(28 votes)
• What do you mean by "not" when you say "x not" Sal?
(9 votes)
• He isn't saying "x not". He is saying "x naught". The word "naught" means zero.
The zero (or naught) is a sub-script of x, that is to say a particular fixed value of x, similar to X1, X2, X3, i.e. specific fixed values of the variable x.
(38 votes)
• Im sure this is a very ignorant question, but what is the difference between x and x naught ?
(9 votes)
• x refers to the variable x in general. x-naught refers to a specific value of x. We don't know what that specific value is, so we can't say 3 or 4 or 1.57. So we call it x-naught.
(4 votes)
• at , when Sal rewrites the second quadrant root as a function of x(0) ((x naught)), why doesn't he write -1/2x instead of -1/2x(0)???
(6 votes)
• Sal mistakenly left off the subscript of the x naught in the first term. There should be no singular "x" because the function was in terms of Xo (x naught).
(5 votes)
• Can someone explain please to me why did he take the derivative of the function in 18.50 I understand everything until that part.
Thanks in advance.
(6 votes)
• At that point the work done has been to figure out a function that tells you what the x value at the intersection on the left (Quadrant 2) will be for a given x value at the intersection on the right (Quadrant 1). Now if you were to graph that function then you would be graphing the X naught value on the x axis and the resulting computed x value, or f(x naught) on the y axis. Since you want to find the maximum value for f(x naught) you would take a derivative of the function set to zero to show you where that curve achieves a maximum, at which point it's slope would be zero.
(3 votes)
• at , why the equation of normal line is: y-x0^2=-(1/2x0)*(x-x0)?
(6 votes)
• Well, that is the point-slope form of the normal line,
the point being (x0, y0), and since y=x^2, y0 = x0^2, so the point is (x0, x0^2),
the slope being perpendicular (which is what "normal" means, it would be nice if Sal mentioned that) to the tangent line at (x0, y0); the slope of the tangent line is just y' = 2x, so the line perpendicular to that has the slope -1/2x (the negative reciprocal of the slope of the tangent line), which at the point x0 is m = -1/2x0
So, in summary, the equation is of the form:
y - y0 = m(x - x0)
y0 = -x0^2
m = -1/2x0
(2 votes)
• At , Sal says x0 squared is equal to 1/2.
Surely this impies that x0 is equal to plus OR minus 1/sqrt2?
Obviously that would put it in the wrong quadrant, but is there a way to mathematically justify discarding a solution like this?
(4 votes)
• You just answered your own question. x0 is defined as the interception point in the first quadrant, thus is has to be positive.If you are on your exams and feel like clarifying just in case, you could add in an extra line, ie."Because x0 is defined to be in the first quadrant, x0 > 0. Therefore, x0 = 1/sqrt2"
Hope that helps
(4 votes)
## Video transcript
I just got sent this problem, and it's a pretty meaty problem. A lot harder than what you'd normally find in most textbooks. So I thought it would help us all to work it out. And it's one of those problems that when you first read it, your eyes kind of glaze over, but when you understand what they're talking about, it's reasonably interesting. So they say, the curve in the figure above is the parabola y is equal to x squared. So this curve right there is y is equal to x squared. Let us define a normal line as a line whose first quadrant intersection with the parabola is perpendicular to the parabola. So this is the first quadrant, right here. And they're saying that a normal line is something, when the first quadrant intersection with the parabola is normal to the parabola. So if I were to draw a tangent line right there, this line is normal to that tangent line. That's all that's saying. So this is a normal line, right there. Normal line. Fair enough. 5 normal lines are shown in the figure. 1, 2, 3, 4, 5. Good enough. And these all look perpendicular, or normal to the parabola in the first quadrant intersection, so that makes sense. For a while, the x-coordinate of the second quadrant intersection of the normal line of the parabola gets smaller, as the x-coordinate of the first quadrant intersection gets smaller. So let's see what happens as the x-quadrant of the first intersection gets smaller. So this is where I left off in that dense text. So if I start at this point right here, my x-coordinate right there would look something like this. Let me go down. my x-coordinate is right around there. And then as I move to a smaller x-coordinate to, say, this one right here, what happened to the normal line? Or even more important, what happened to the intersection of the normal line in the second quadrant? This is the second quadrant, right here. So when I had a larger x-value here, my normal line intersected here, in the second quadrant. Then when I brought my x-value in, when I lowered my x-value, my x-value here, because this is the next point, right here, my x-value at the intersection here, went-- actually, their wording is bad. They're saying that the second quadrant intersection gets smaller. But actually, it's not really getting smaller. It's getting less negative. I guess smaller could be just absolute value or magnitude, but it's just getting less negative. It's moving there, but it's actually becoming a larger number, right? It's becoming less negative, but a larger number. But if we think in absolute value, I guess it's getting smaller, right? As we went from that point to that point, as we moved the x in for the intersection of the first quadrant, the second quadrant intersection also moved in a bit, from that line to that line. Fair enough. But eventually, a normal line second quadrant intersection gets as small as it can get. So if we keep lowering our x-value in the first quadrant, so we keep on pulling in the first quadrant, as we get to this point. And then this point intersects the second quadrant, right there. And then, if you go even smaller x-values in the first quadrant then your normal line starts intersecting in the second quadrant, further and further negative numbers. So you can kind of view this as the highest value, or the smallest absolute value, at which the normal line can intersect in the second quadrant Let me make that clear. Up here, you were intersecting when you had a large x in the first quadrant, you had a large negative x in the second quadrant intersection. And then as you lowered your x-value, here, you had a smaller negative value. Up until you got to this point, right here, you got this, which you can view as the smallest negative value could get, and then when you pulled in your x even more, these normal lines started to push out again, out in the second quadrant. That's, I think, what they're talking about. The extreme normal line is shown as a thick line in the figure. Right. This is the extreme normal line, right there. So this is the extreme one, that deep, bold one. Extreme normal line. After this point, when you pull in your x-values even more, the intersection in your second quadrant starts to push out some. And you can think of the extreme case, if you draw the normal line down here, your intersection with the second quadrant is going to be way out here someplace, although it seems like it's kind of asymptoting a little bit. But I don't know. Let's read the rest of the problem. Once the normal line passes the extreme normal line, the x-coordinates of their second quadrant intersections what the parabola start to increase. And they're really, when they say they start to increase, they're actually just becoming more negative. That wording is bad. I should change this to more, more negative. Or they're becoming larger negative numbers. Because once you get below this, then all of a sudden the x-intersections start to push out more in the second quadrant. Fair enough. The figures show 2 pairs of normal lines. Fair enough. The 2 normal lines of a pair have the same second quadrant intersection with the parabola, but 1 is above the extreme normal line, in the first quadrant, the other is below it. Right, fair enough. For example, this guy right here, this is when we had a large x-value. He intersects with the second quadrant there. Then if you lower and lower the x-value, if you lower it enough, you pass the extreme normal line, and then you get to this point, and then this point, he intersects, or actually, you go to this point. So if you pull in your x-value enough, you once again intersect at that same point in the second quadrant. So hopefully I'm making some sense to you, as I try to make some sense of this problem. OK. Now what do they want to know? And I think I only have time for the first part of this. Maybe I'll do the second part in the another video. Find the equation of the extreme normal line. Well, that seems very daunting at first, but I think our toolkit of derivatives, and what we know about equations of a line, should be able to get us there. So what's the slope of the tangent line at any point on this curve? Well, we just take the derivative of y equals x squared, and y prime is just equal to 2x. This is the slope of the tangent at any point x. So if I want to know the slope of the tangent at x0, at some particular x, I would just say, well, let me just say, slope, it would be 2 x0. Or let me just say, f of x0 is equal to 2 x0. This is the slope at any particular x0 of the tangent line. Now, the normal line slope is perpendicular to this. So the perpendicular line, and I won't review it here, but the perpendicular line has a negative inverse slope. So the slope of normal line at x0 will be the negative inverse of this, because this is the slope of the tangent line x0. So it'll be equal to minus 1 over 2 x0. Fair enough. Now, what is the equation of the normal line at x0 let's say that this is my x0 in question. What is the equation of the normal line there? Well, we can just use the point-slope form of our equation. So this point right here will be on the normal line. And that's the point x0 squared. Because this the graph of y equals x0, x squared. So this normal line will also have this point. So we could say that the equation of the normal line, let me write it down, would be equal to, this is just a point-slope definition of a line. You say, y minus the y-point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the x-point that we're at. Minus x, minus x0. This is the equation of the normal line. So let's see. And what we care about is when x0 is greater than 0, right? We care about the normal line when we're in the first quadrant, we're in all of these values right there. So that's my equation of the normal line. And let's solve it explicitly in terms of x. So y is a function of x. Well, if I add x0 squared to both sides, I get y is equal to, actually, let me multiply this guy out. I get minus 1/2 x0 times x, and then I have plus, plus, because I have a minus times a minus, plus 1/2. The x0 and the over the x0, they cancel out. And then I have to add this x0 to both sides. So all I did so far, this is just this part right there. That's this right there. And then I have to add this to both sides of the equation, so then I have plus x0 squared. So this is the equation of the normal line, in mx plus b form. This is its slope, this is the m, and then this is its y-intercept right here. That's kind of the b. Now, what do we care about? We care about where this thing intersects. We care about where it intersects the parabola. And the parabola, that's pretty straightforward, that's just y is equal to x squared. So to figure out where they intersect, we just have to set the 2 y's to be equal to each other. So they intersect, the x-values where they intersect, x squared, this y would have to be equal to that y. Or we could just substitute this in for that y. So you get x squared is equal to minus 1 over 2 x0 times x, plus 1/2 plus x0 squared. Fair enough. And let's put this in a quadratic equation, or try to solve this, so we can apply the quadratic equation. So let's put all of this stuff on the lefthand side. So you get x squared plus 1 over 2 x0 times x minus all of this, 1/2 plus x0 squared is equal to 0. All I did is, I took all of this stuff and I put it on the lefthand side of the equation. Now, this is just a standard quadratic equation, so we can figure out now where the x-values that satisfy this quadratic equation will tell us where our normal line and our parabola intersect. So let's just apply the quadratic equation here. So the potential x-values, where they intersect, x is equal to minus b, I'm just applying the quadratic equation. So minus b is minus 1 over 2 x0, plus or minus the square root of b squared. So that's that squared. So it's one over four x0 squared minus 4ac. So minus 4 times 1 times this minus thing. So I'm going to have a minus times a minus is a plus, so it's just 4 times this, because there was one there. So plus 4 times this, right here. 4 times this is just 2 plus 4 x0 squared. All I did is, this is 4ac right here. Well, minus 4ac. The minus and the minus canceled out, so you got a plus. There's a 1. So 4 times c is just 2 plus 4x squared. I just multiply this by two, and of course all of this should be over 2 times a. a is just 2 there. So let's see if I can simplify this. Remember what we're doing. We're just figuring out where the normal line and the parabola intersect. Now, what do we get here. This looks like a little hairy beast here. Let me see if I can simplify this a little bit. So let us factor out-- let me rewrite this. I can just divide everything by 1/2, so this is minus 1 over 4 x0, I just divided this by 2, plus or minus 1/2, that's just this 1/2 right there, times the square root, let me see what I can simplify out of here. So if I factor out a 4 over x0 squared, then what does my expression become? This term right here will become an x to the fourth, x0 to the fourth, plus, now, what does this term become? This term becomes a 1/2 x0 squared. And just to verify this, multiply 4 times 1/2, you get 2, and then the x0 squares cancel out. So write this term times that, will equal 2, and then you have plus-- now we factored a four out of this and the x0 squared, so plus 1/16. Let me scroll over a little bit. And you can verify that this works out. If you were to multiply this out, you should get this business right here. I see the home stretch here, because this should actually factor out quite neatly. So what does this equal? So the intersection of our normal line and our parabola is equal to this. Minus 1 over 4 x0 plus or minus 1/2 times the square root of this business. And the square root, this thing right here is 4 over x0 squared. Now what's this? This is actually, lucky for us, a perfect square. And I won't go into details, because then the video will get too long, but I think you can recognize that this is x0 squared, plus 1/4. If you don't believe me, square this thing right here. You'll get this expression right there. And luckily enough, this is a perfect square, so we can actually take the square root of it. And so we get, the point at which they intersect, our normal line and our parabola, and this is quite a hairy problem. The points where they intersect is minus 1 over 4 x0, plus or minus 1/2 times the square root of this. The square root of this is the square root of this, which is just 2 over x0 times the square root of this, which is x0 squared plus 1/4. And if I were to rewrite all of this, I'd get minus 1 over 4 x0 plus, let's see, this 1/2 and this 2 cancel out, right? So these cancel out. So plus or minus, now I just have a one over x0 times x0 squared. So I have 1 over x0-- oh sorry, let me, we have to be very careful there-- x0 squared divided by x0 is just x0, let me do that in a yellow color so you know what I'm dealing with. This term multiplied by this term is just x0, and then you have a plus 1/4 x0. And this is all a parentheses here. So these are the two points at which the normal curve and our parabola intersect. Let me just be very clear. Those 2 points are, for if this is my x0 that we're dealing with, right there. It's this point and this point. And we have a plus or minus here, so this is going to be the plus version, and this is going to be the minus version. In fact, the plus version should simplify into x0. Let's see if that's the case. Let's see if the plus version actually simplifies to x0. So these are our two points. If I take the plus version, that should be our first quadrant intersection. So x is equal to minus 1/4 x0 plus x0 plus 1/4 x0. And, good enough, it does actually cancel out. That cancels out. So x0 is one of the points of intersection, which makes complete sense. Because that's how we even defined the problem. But, so this is the first quadrant intersection. So that's the first quadrant intersection. The second quadrant intersection will be where we take the minus sign right there. So x, I'll just call it in the second quadrant intersection, it'd be equal to minus 1/4 x0 minus this stuff over here, minus the stuff there. So minus x0 minus 1 over 4 mine x0. Now what do we have? So let's see. We have a minus 1 over 4 x0, minus 1 over 4 x0. So this is equal to minus x0, minus x0, minus 1 over 2 x0. So if I take minus 1/4 minus 1/4, I get minus 1/2. And so my second quadrant intersection, all this work I did got me this result. My second quadrant intersection, I hope I don't run out of space. My second quadrant intersection, of the normal line and the parabola, is minus x0 minus 1 over 2 x0. Now this by itself is a pretty neat result we just got, but we're unfortunately not done with the problem. Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of that with respect to x0, is equal to, this is pretty straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0. So let's set that equal to 0, and then solve this problem right there. Well, we add one to both sides. We get 1 over 2 x0 squared is equal to 1, or you could just say that that means that 2 x0 squared must be equal to 1, if we just invert both sides of this equation. Or we could say that x0 squared is equal to 1/2, or if we take the square roots of both sides of that equation, we get x0 is equal to 1 over the square root of 2. So we're really, really, really close now. We've just figured out the x0 value that gives us our extreme normal line. This value right here. Let me do it in a nice deeper color. This value right here, that gives us the extreme normal line, that over there is x0 is equal to 1 over the square root of 2. Now, they want us to figure out the equation of the extreme normal line. Well, the equation of the extreme normal line we already figured out right here. It's this. The equation of the normal line is that thing, right there. So if we want the equation of the normal line at this extreme point, right here, the one that creates the extreme normal line, I just substitute 1 over the square root of 2 in for x0. So what do I get? I get, and this is the home stretch, and this is quite a beast of a problem. y minus x0 squared. x0 squared is 1/2, right? 1 over the square root of 2 squared is 1/2. Is equal to minus 1 over 2 x0. So let's be careful here. So minus 1/2 times 1 over x0. One over x0 is the square root of 2, right? All of that times x minus x0. So that's 1 over the square root of 2. x0 is one over square root of 2. So let's simplify this a little bit. So the equation of our normal line, assuming I haven't made any careless mistakes, is equal to, so y minus 1/2 is equal to, let's see. If we multiply this minus square root of 2 over 2x, and then if I multiply these square root of 2 over this, it becomes one. And then I have a minus and a minus, so that I have a plus 1/2. I think that's right. Yeah, plus 1/2, this times this times that is equal to plus 1/2. And then, we're at the home stretch. So we just add 1/2 to both sides of this equation, and we get our extreme normal line equation, which is y is equal to minus square root of 2 over 2x. If you add 1/2 to both sides of this equation, you get plus 1. And there you go. That's the equation of that line there, assuming I haven't made any careless mistakes. But even if I have, I think you get the idea of hopefully how to do this problem, which is quite a beastly one.
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## Linear Programming Problem Using Graphical Method Homework Help
Discover if the unbiased function Z is a function of 2 variables just, then the Linear Programming Problem can be resolved efficiently by the graphical method.Likewise it can be resolved by this method if Z is a function of 3 variables. In this case the graphical option ends up being complex enough. The linear programming issues are fixed in used mathematics designs.Build the unbiased function Action Create the restrictions the goal and restraints should be definable by linear mathematical practical relationships. The problem incorporates an objective, revealed as an unbiased function that the choice maker desires to accomplish. Additive – Terms in the unbiased function and restriction formulas need to be additive.
LP issues are identified by an unbiased function that is to be optimized or lessened, subject to a number of restraints. Both the unbiased function and the restraints should be created in terms of a linear equality or inequality. Normally; the unbiased function will be to optimize revenues (e.g., contribution margin) or to reduce expenses (e.g., variable expenses).Linear programming, or LP, is a method of designating resources in an optimum method. In LP, these resources are understood as choice variables. The requirement for choosing the finest worth’s of the choice variables e.g., to optimize revenues or decrease expenses is understood as the unbiased function.
We can utilize graphical approaches to fix linear optimization issues including 2 variables. The graphical method does not generalize to a big number of variables, the fundamental principles of linear programming can all be shown in the two-variable context. Graphical approaches supply us with a photo to go with the algebra of linear programming, and the photo can anchor our understanding of fundamental meanings and possibilities.The Book Emporium purchases books from 2 providers . How lots of bundles ought to the Emporium order from each provider to please the above requirements at the least possible expense.
If the goal is to optimize contribution from the sale of Products A and B having contribution per system of and respectively, the unbiased function will be Optimum Contribution What this implies is that the goal of resolving the problem is to take full advantage of the overall contribution of the service by offering the maximum mix of items A and B. Given that we just need the slope (gradient) of the unbiased function, we can outline the Goal Function on a chart using any random worth in location of optimum contribution Restriction inequalities, as specified in Action ought to be outlined on a chart. In a linear programming problem, we look for the practical point that optimizes or reduces the unbiased function. If the practical area is unbounded, the unbiased function might not have an optimum or minimum; however if coefficients of the unbiased function are favorable, then the minimum worth of the unbiased function happens at a corner point and there is no optimum worth. Build the unbiased function Action Create the restraints the goal and restrictions should be definable by linear mathematical practical relationships.
The menu is to consist of 2 products and Expect that each ounce of supplies systems of vitamin and systems of iron and each ounce of supplies system of vitamin and systems of iron. Expect the expense of is ounce and the expense of is ounce. If the breakfast menu needs to offer at least systems of vitamin and systems of iron, how numerous ounces of each product ought to be supplied in order to fulfill the iron and vitamin requirements for the least expense what will this breakfast coos.
If the goal is to optimize contribution from the sale of Products A and B having contribution per system of and respectively, the unbiased function will be Optimum Contribution What this implies is that the goal of fixing the problem is to optimize the overall contribution of the service by offering the maximum mix of items A and B. The problem with the above formula is that we can not merely outline it on a chart (as needed in action 5). Considering that we just need the slope (gradient) of the unbiased function, we can outline the Goal Function on a chart using any random worth in location of optimum contribution Restraint inequalities, as specified in Action ought to be outlined on a chart.In a linear programming problem, we look for the possible point that takes full advantage of or reduces the unbiased function. If the practical area is unbounded, the unbiased function might not have an optimum or minimum; however if coefficients of the unbiased function are favorable, then the minimum worth of the unbiased function takes place at a corner point and there is no optimum worth. Utilize these areas to discover optimum and minimum worth’s of each offered unbiased function.
The coordinate system is drawn and each variable is associated to an axis normally is associated to the horizontal axis and to the vertical one as revealed in figure A mathematical scale is marked in axis, proper to the worth’s that variables can take according to the problem restraints. In order to do this, for each variable matching to an axis, all variables are set to absolutely no other than the variable associated to the studied axis in each restraint. Succeeding built tableaux in the Simplex method will offer the worth of the unbiased function at the vertices of the practical area, changing concurrently, the coefficients of slack and preliminary variables.In the preliminary tableau the worth of the unbiased function at the -vertex is computed, the collaborates represent the worth which have the fundamental variables, being the outcome
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## Precalculus (10th Edition)
$1.27\text{ radians}$
Recall that $2\pi\text{ radians}=360^\circ$. Hence, $1^\circ= \frac{2\pi\text{ radians}}{360^\circ}=\frac{\pi\text{ radians}}{180^\circ}$. Therefore: $73^{\circ}=73\cdot1^{\circ}=73^\circ \cdot \frac{\pi\text{ radians}}{180^\circ} \approx1.27\text{ radians}$
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Delightful Designs: A Coloring Book of Magical Patterns
Available from Amazon and Create Space.
From the Introduction:
This coloring book is the result of an unexpected and serendipitous discovery made while I was studying tessellations, attempting to learn how many distinct, tessellating shapes could be derived from a square template when given four identical edges. My results showed that 15 distinct shapes were possible when the edge was asymmetric, only two when the edge had mirror symmetry, and four when the edge had central-point rotation.
What, you may wonder, is center-point rotation? It means that you form an edge by beginning with a line segment (1) that is shaped in some way (2), making sure that the line does not cross itself. A copy of this shaped line (3) is then rotated 180 degrees (4) and one end is connected to an end of the original shaped line (5). This connected line is then used to from a closed figure by rotation and/or flips. Not all lines will work; some will cross when they are arranged to make the four shapes.
The four distinct shapes that can be formed in this way all tessellate, that is, each shape can fit together with copies of itself to fill the plane with no gaps of overlap. So far none of this was too exciting, but then I rotated the shapes 45 degrees so that instead of fitting together edge to edge, they would fit together vertex to vertex, forming a checkerboard-like pattern of tiles and voids. The results amazed me because the voids took the same four shapes that the tiles had. (When I thought about it, I realized that this must happen because the voids share their four edges from the tiles and thus are confined to the same four shapes.)
Three of the four distinct shapes have either mirror or rotational symmetry and one is asymmetric. Two of the symmetric shapes have only two orientations, one has four orientations, and the asymmetric shape has eight orientations, giving a total of 16 possible tiles to use in creating patterns. Using just these four shapes in different orientations results in a huge number of possible patterns that can be altered by simply changing a tile.
I have searched for previous use of the magical possibilities of these sets of tessellating tiles and have found some that are close. David Bailey at http://www.tess-elation.co.uk has three of the four shapes and has done research on the family. "Rep-Tiling the Plane" in the May 2000 issue of Scientific American put all four shapes in a pattern. The article started with a completely different approach to tiling and never mentioned the importance of the edges having central-point rotation. If you know of previous use of patterns of the sort this book uses, I would like to hear about it.
The patterns in this book can all be considered tessellation patterns, though most are not patterns of a single shape. They are two-, three-, or four-tile tessellations, a type of tessellation that previously had not interested me.
In addition to using sets of edges to form tiles based on square templates, I have used the same edges to form tiles based on triangular and hexagonal templates. Some of these patterns that are visually interesting are included at the end of the book.
The story of how this set of tiles was found and some technical details about it is contained in Exploring Tessellations: A Journey through Heesch Types and Beyond (CreateSpace, 2015). If you would like to make your own patterns with these shapes, the typefaces that I used to design this book are available at myfonts.com. Search for FabFours.
(If you do not understand what this Introduction is saying, play with the patterns on the following pages for a while and then come back and read it again.)
I apologize in advance for errors that remain in the book.
Robert Schenk
November 2015
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Quantropy (Part 3)
I’ve been talking a lot about ‘quantropy’. Last time we figured out a trick for how to compute it starting from the partition function of a quantum system. But it’s hard to get a feeling for this concept without some examples.
So, let’s compute the partition function of a free particle on a line, and see what happens…
The partition function of a free particle
Suppose we have a free particle on a line tracing out some path as time goes by:
$q: [0,T] \to \mathbb{R}$
Then its action is just the time integral of its kinetic energy:
$\displaystyle{ A(q) = \int_0^T \frac{mv(t)^2}{2} \; dt }$
where
$\displaystyle{ v(t) = \frac{d q(t)}{d t} }$
is its velocity. The partition function is then
$Z = \displaystyle{\int e^{i A(q) / \hbar} \; Dq }$
where we integrate an exponential involving the action over the space of all paths $q$. Unfortunately, the space of all paths is infinite-dimensional, and the thing we’re integrating oscillates wildly. Integrals like this tend to make mathematicians run from the room screaming. For example, nobody is quite sure what $Dq$ means in this expresson. There is no ‘Lebesgue measure’ on an infinite-dimensional vector space.
Discretizing time
We’ll start by treating time as discrete—a trick Feynman used in his original work. We’ll consider $n$ time intervals of length $\Delta t.$ Say the position of our particle at the ith time step is $q_i \in \mathbb{R}.$ We’ll require that the particle keeps a constant velocity between these time steps. This will reduce the problem of integrating over ‘all’ paths—whatever that means, exactly—to the more manageable problem of integrating over a finite-dimensional space of paths. Later we can study what happens as the time steps get shorter and more numerous.
Let’s call the particle’s velocity between the (i-1)st and ith time steps $v_i.$
$\displaystyle{ v_i = \frac{q_i - q_{i-1}}{\Delta t} }$
The action, defined as an integral, is now equal to a finite sum:
$\displaystyle{ A(q) = \sum_{i = 1}^n \frac{mv_i^2}{2} \; \Delta t }$
We’ll consider histories of the particle where its initial position is
$q_0 = 0$
but its final position $q_n$ is arbitrary. Why? If we don’t ‘nail down’ the particle at some particular time, our path integrals will diverge. So, our space of histories is
$X = \mathbb{R}^n$
and now we’re ready to apply the formulas we developed last time!
We saw last time that the partition function is the key to all wisdom, so let’s start with that. Naively, it’s
$\displaystyle{ Z = \int_X e^{- \beta A(q)} Dq }$
where
$\displaystyle{ \beta = \frac{1}{i \hbar} }$
But there’s a subtlety here. Doing this integral requires a measure on our space of histories. Since the space of histories is just $\mathbb{R}^n$ with coordinates $q_1, \dots, q_n,$ an obvious guess for a measure would be
$Dq = dq_1 \cdots dq_n \qquad \qquad \qquad \qquad \quad \textrm{(obvious first guess)}$
However, the partition function should be dimensionless! You can see why from the discussion of units last time. But the quantity $\beta A(q)$ and thus its exponential is dimensionless, so our mesasure had better be dimensionless too. But $dq_1 \cdots dq_n$ has units of lengthn. To deal with this we can introduce a length scale, which I’ll call $\Delta x,$ and use the measure
$Dq = \displaystyle{ \frac{1}{(\Delta x)^n} \, dq_1 \cdots dq_n } \qquad \qquad \qquad \textrm{(what we'll actually use)}$
I should however emphasize that despite the notation $\Delta x,$ I’m not discretizing space, just time. We could also discretize space, but it would make the calculation a lot harder. I’m only introducing this length scale $\Delta x$ to make our measure on the space of histories dimensionless.
Now let’s compute the partition function. For starters, we have
$\begin{array}{ccl} Z &=& \displaystyle{ \int_X e^{-\beta A(q)} \; Dq } \\ \\ &=& \displaystyle{ \frac{1}{(\Delta x)^n} \int e^{-\beta \sum_{i=1}^n m \, \Delta t \, v_i^2 /2} \; dq_1 \cdots dq_n } \end{array}$
Normally when I see an integral bristling with annoying constants like this, I switch to a system of units where most of them equal 1. But I’m trying to get a physical feel for quantropy, so I’ll leave them all in. That way, we can see how they affect the final answer.
Since
$\displaystyle{ v_i = \frac{q_i - q_{i-1}}{\Delta t} }$
we can show that
$dq_1 \cdots dq_n = (\Delta t)^n \; dv_1 \cdots dv_n$
To show this, we need to work out the Jacobian of the transformation from the $q_i$ coordinates to the $v_i$ coordinates on our space of histories—but this is easy to do, since the determinant of a triangular matrix is the product of its diagonal entries.
We can rewrite the path integral using this change of variables:
$Z = \displaystyle{\left(\frac{\Delta t}{\Delta x}\right)^n \int e^{-\beta \sum_{i=1}^n m \, \Delta t \, v_i^2 /2} \; dv_1 \cdots dv_n }$
But since an exponential of a sum is a product of exponentials, this big fat n-tuple integral is really just a product of n ordinary integrals. And all these integrals are equal, so we just get some integral to the nth power! Let’s call the variable in this integral $v$, since it could be any of the $v_i$:
$Z = \displaystyle{ \left(\frac{\Delta t}{\Delta x} \int_{-\infty}^\infty e^{-\beta \, m \, \Delta t \, v^2 /2} \; dv \right)^n }$
How do we do the integral here? Well, that’s easy…
Integrating Gaussians
We should all know the integral of our favorite Gaussian. As a kid, my favorite was this:
$\displaystyle{ \int_{-\infty}^\infty e^{-x^2} \; d x = \sqrt{\pi} }$
because this looks the simplest. But now, I prefer this:
$\displaystyle{ \int_{-\infty}^\infty e^{-x^2/2} \; d x = \sqrt{2 \pi} }$
They’re both true, so why did my preference change? First, I now like $2\pi$ better than $\pi.$ There’s a whole manifesto about this, and I agree with it. Second, $x^2/2$ is better than $x^2$ for what we’re doing, since kinetic energy is one half the mass times the velocity squared. Originally physicists like Descartes and Leibniz defined kinetic energy to be $m v^2,$ but the factor of 1/2 turns out to make everything work better. Nowadays every Hamiltonian or Lagrangian with a quadratic term in it tends to have a 1/2 in front—basically because the first thing you do with it is differentiate it, and the 1/2 cancels the resulting 2. The factor of 1/2 is just a convention, even in the definition of kinetic energy, but if we didn’t make that convention we’d be punished with lots of factors of 2 all over.
Or course it doesn’t matter much: you just need to remember the integral of some Gaussian, or at least know how to calculate it. And you’ve probably read this quote:
A mathematician is someone to whom
$\displaystyle{ \int_{-\infty}^\infty e^{-x^2/2} \; d x = \sqrt{2 \pi} }$
is as obvious as 2+2=4 is to you and me. – Lord Kelvin
So, you probably learned the trick for doing this integral, so you can call yourself a mathematician.
Stretching the above Gaussian by a factor of $\sqrt{\alpha}$ increases the integral by a factor of $\sqrt{\alpha}$, so we get
$\displaystyle{ \int_{-\infty}^\infty e^{-x^2/2\alpha} \; d x = \sqrt{2 \pi \alpha} }$
This is clear when $\alpha$ is positive, but soon we’ll apply it when $\alpha$ is imaginary! That makes some mathematicians sweaty and nervous. For example, we’re saying that
$\displaystyle{ \int_{-\infty}^\infty e^{i x^2 / 2} \, dx = \sqrt{2 \pi i}}$
But this integral doesn’t converge if you slap absolute values on the function inside: in math jargon, the function inside isn’t ‘Lebesgue integrable’. But we can tame it in various ways. We can impose a ‘cutoff’ and then let it go to infinity:
$\displaystyle{ \lim_{M \to + \infty} \int_{-M}^M e^{i x^2 / 2} \, dx = \sqrt{2 \pi i} }$
or we can damp the oscillations, and then let the amount of damping go to zero:
$\displaystyle{ \lim_{\epsilon \downarrow 0} \int_{-\infty}^\infty e^{(i - \epsilon) x^2 / 2} \, dx = \sqrt{2 \pi i} }$
We get the same answer either way, or indeed using many other methods. Since such tricks work for all the integrals I’ll write down, I won’t engage in further hand-wringing over this issue. We’ve got bigger things to worry about, like: what’s the physical meaning of quantropy?
Computing the partition function
Where were we? We had this formula for the partition function:
$Z = \displaystyle{ \left( \frac{\Delta t}{\Delta x} \int_{-\infty}^\infty e^{-\beta \, m \, \Delta t \, v^2 /2} \; dv \right)^n }$
and now we’re letting ourselves use this formula:
$\displaystyle{ \int_{-\infty}^\infty e^{-x^2/2\alpha} \; d x = \sqrt{2 \pi \alpha} }$
even when $\alpha$ is imaginary, so we get
$Z = \displaystyle{ \left( \frac{\Delta t}{\Delta x} \sqrt{ \frac{2 \pi}{\beta m \, \Delta t}} \right)^n = \left(\frac{2 \pi \Delta t}{\beta m \, (\Delta x)^2}\right)^{n/2} }$
And a nice thing about keeping all these constants floating around is that we can use dimensional analysis to check our work. The partition function should be dimensionless, and it is! To see this, just remember that $\beta = 1/i\hbar$ has dimensions of inverse action, or $T/M L^2$.
Expected action
Now that we’ve got the partition function, what do we do with it? We can compute everything we care about. Remember, in statistical mechanics there’s a famous formula:
free energy = expected energy – temperature × entropy
and last time we saw that similarly, in quantum mechanics we have:
free action = expected action – classicality × quantropy
where the classicality is
$1/\beta = 1/i \hbar$
In other words:
$\displaystyle{ F = \langle A \rangle - \frac{1}{\beta}\, Q }$
Last time I showed you how to compute $F$ and $\langle A \rangle$ starting from the partition function. So, we can use the above formula to work out the quantropy as well:
Expected action $\langle A \rangle = - \frac{d}{d \beta} \ln Z$ Free action $F = -\frac{1}{\beta} \ln Z$ Quantropy $Q = \ln Z - \beta \,\frac{d }{d \beta}\ln Z$
But let’s start with the expected action. The answer will be so amazingly simple, yet strange, that I’ll want to spend the rest of this post discussing it.
Using our hard-won formula
$\displaystyle{ Z = \left(\frac{2 \pi \Delta t}{\beta m \, (\Delta x)^2}\right)^{n/2} }$
we get
$\begin{array}{ccl} \langle A \rangle &=& \displaystyle{ -\frac{d}{d \beta} \ln Z } \\ \\ &=& \displaystyle{ -\frac{n}{2} \frac{d}{d \beta} \ln \left(\frac{2 \pi \Delta t}{\beta m \, (\Delta x)^2}\right) } \\ \\ &=& \displaystyle{ -\frac{n}{2} \frac{d}{d \beta} \left( \ln \left(\frac{2 \pi \Delta t}{m \, (\Delta x)^2}\right) - \ln \beta \right) } \\ \\ &=& \displaystyle{ \frac{n}{2} \; \frac{1}{\beta} } \\ \\ &=& \displaystyle{ n\; \frac{i \hbar}{2} } \end{array}$
Wow! When get an answer this simple, it must mean something! This formula is saying that the expected action of our freely moving quantum particle is proportional to $n,$ the number of time steps. Each time step contributes $i \hbar / 2$ to the expected action. The mass of the particle, the time step $\Delta t,$ and the length scale $\Delta x$ don’t matter at all!
Why don’t they matter? Well, you can see from the above calculation that they just disappear when we take the derivative of the logarithm containing them. That’s not a profound philosophical explanation, but it implies that our action could be any quadratic function like this:
$A : \mathbb{R}^n \to \mathbb{R}$
$\displaystyle{ A(x) = \sum_{i = 1}^n \frac{c_i x_i^2}{2} }$
where $c_i$ are positive numbers, and we’d still get the same expected action:
$\langle A \rangle = \displaystyle{ n\; \frac{i \hbar}{2} }$
The numbers $c_i$ don’t matter!
The quadratic function we’re talking about here is an example of a quadratic form. Because the numbers $c_i$ are positive, it’s a positive definite quadratic form. And since we can diagonalize any positive definite quadratic form, we can state our result in a fancier, more elegant way:
Whenever the action is a positive definite quadratic form on an n-dimensional vector space of histories, the expected action is $n$ times $i \hbar / 2.$
For example, take a free particle in 3d Euclidean space, and discretize time into $n$ steps as we’ve done here. Then the action is a positive definite quadratic form on a 3n-dimensional vector space:
$\displaystyle{ A(q) = \sum_{i = 1}^n \frac{m \vec{v}_i \cdot \vec{v}_i}{2} \; \Delta t }$
since now each velocity $\vec{v}_i$ is a vector with 3 components. So, the expected action is $3n$ times $i \hbar / 2.$
Poetically speaking, $3n$ is the total number of ‘decisions’ our particle makes throughout its history. What do I mean by that? In the path integral approach to quantum mechanics, a system can trace out any history it wants. But takes a bunch of real numbers to determine a specific history. Each number counts as one ‘decision’. And in the situation we’ve described, each decision contributes $i \hbar / 2$ to the expected action.
So here’s a more intuitive way to think about our result:
In the path integral approach to quantum theory, each ‘decision’ made by the system contributes $i \hbar / 2$ to the expected action… as long as the action is given by a positive definite quadratic form on some vector space of histories.
There’s a lot more to say about this. For example, in the harmonic oscillator the action is a quadratic form, but it’s not positive definite. What happens then? But three more immediate questions leap to my mind:
1) Why is the expected action imaginary?
2) Should we worry that it diverges as $n \to \infty$?
3) Is this related to the heat capacity of an ideal gas?
So, let me conclude this post by trying to answer those.
Why is the expected action imaginary?
The action $A$ is real. How in the world can its expected value be imaginary?
The reason is that we’re not taking its expected value with respect to an probability measure, but instead, with respect to a complex-valued measure. Last time we gave this very general definition:
$\langle A \rangle = \displaystyle{ \frac{\int_X A(x) e^{-\beta A(x)} \, dx }{\int_X e^{-\beta A(x)} \, dx }}$
The action $A$ is real, but $\beta = 1 / i \hbar$ is imaginary, so it’s not surprising that this ‘expected value’ is complex-valued.
Later we’ll see a good reason why it has to be purely imaginary.
Why does it diverge as n → ∞?
Consider our particle on a line, with time discretized into $n$ time steps. Its expected action is
$\langle A \rangle = \displaystyle{ n\; \frac{i \hbar}{2} }$
To take the continuum limit we must let $n \to \infty$ while simultaneously letting $\Delta t \to 0$ in such a way that $n \Delta t$ stays constant. Some quantities will converge when we take this limit, but the expected action will not. It will go to infinity!
That’s a bit sad, but not unexpected. It’s a lot like how the expected length of the path of a particle carrying out Brownian motion is infinite. In 3 dimensions, a typical Brownian path looks like this:
In fact the free quantum particle is just a ‘Wick-rotated’ version of Brownian motion, where we replace time by imaginary time, so the analogy is fairly close. The action we’re considering now is not exactly analogous to the arclength of a path:
$\displaystyle{ \int_0^T \left| \frac{d q}{d t} \right| \; dt }$
$\displaystyle{ \int_0^T \left| \frac{d q}{d t} \right|^2 \; dt }$
However, both these quantities diverge when we discretize Brownian motion and then take the continuum limit.
How sad should we be that the expected action is infinite in the continuum limit? Not too sad, I think. Any result that applies to all discretizations of a continuum problem should, I think, say something about that continuum problem. For us the expected action diverges, but the ‘expected action per decision’ is constant, and that’s something we can hope to understand even in the continuum limit!
Is this related to the heat capacity of an ideal gas?
That may seem like a strange question, unless you remember some formulas about the thermodynamics of an ideal gas!
Let’s say we’re in 3d Euclidean space. (Most of us already are, but some of my more spacy friends will need to pretend.) If we have an ideal gas made of $n$ point particles at temperature $T,$ its expected energy is
$\frac{3}{2} n k T$
where $k$ is Boltzmann’s constant. This is a famous fact, which lets people compute the heat capacity of a monatomic ideal gas.
On the other hand, we’ve seen that in quantum mechanics, a single point particle will have an expected action of
$\frac{3}{2} n i \hbar$
after $n$ time steps.
These results look awfully similar. Are they related?
Yes! These are just two special cases of the same result! The energy of the ideal gas is a quadratic form on a $3n$-dimensional vector space; so is the action of our discretized point particle. The ideal gas is a problem in statistical mechanics; the point particle is a problem in quantum mechanics. In statistical mechanics we have
$\displaystyle{ \beta = \frac{1}{k T} }$
while in quantum mechanics we have
$\displaystyle{ \beta = i \hbar }$
Mathematically, they are the exact same problem except that $\beta$ is real in one case, imaginary in the other. This is another example of the analogy between statistical mechanics and quantum mechanics—the analogy that motivated quantropy in the first place!
And this makes it even more obvious that the expected action must be imaginary… at least when the action is a positive definite quadratic form.
30 Responses to Quantropy (Part 3)
1. Arrow says:
Very interesting article.
One thing about the n->inf limit, to me such limit seems unphysical.
In the case of brownian motion the fact that path length is infinite is a consequence of the point particle simplifying assumption, if we consider particles with definite size then a particle cannot change direction arbitrarily many times in any time frame because at some point the local neighborhood of the particle in question will be saturated with particles trying to pass momentum to it and such crowding effects will take over.
Analogously in the quantum case I believe there have to be physical reasons which place limits on how fast a particle can change it’s momentum. There has to be an energy scale which sets the limit, is there some known physics that could set such limit?
Personally I can imagine a scenario where the chaotic motions of particles that are taken into account by path integral formulation are not their intrinsic property but rather a cumulative effect of the background fields associated with all the other particles in either it’s local neighborhood or possibly the whole Universe. In such a picture the energy of the most energetic particle/field in the whole Universe would be a natural cut off scale, sure it could be large but not infinite.
2. Toby Bartels says:
I now like $2\pi$ better than $\pi$.
As well you should, but the really fundamental constant is $2i\pi$. This is the period of the exponential function, which is more fundamental than the cosine function. It shows up all over the place in very basic results, such as the Cauchy integral formula, roots of unity, etc. To be sure, $2\pi$ also appears quite often, especially in applications where everything is (or can be) real, since it’s the magnitude of the fundamental $2i\pi$. In college, we came up with a bunch of barred symbols (in homage to $\hbar$), so we called $2\pi$ ‘pi-bar’ ($\pi$, if that shows up).
• Toby Bartels says:
I forgot some ‘latex’s, and the strikethrough did not show up.
• John Baez says:
I fixed what I could easily fix.
As we’ve seen here, the partition function of the time-discretized free particle is
$\displaystyle{ Z = \left(\frac{2 \pi i \hbar \Delta t}{m \, (\Delta x)^2}\right)^{n/2} }$
so you get to have your $2 \pi i$ or your $h = 2 \pi \hbar$, whichever you want, but not both… which suggests that $2 \pi i \hbar$ could be an interesting constant in its own right.
As you surely know, but others may not, $2 \pi$ is now called $\tau$, because it represents one ‘turn’, and because $\tau$ looks like $\pi$ but twice as simple.
• Toby Bartels says:
I don’t know if this was another mistake of mine or if it came it when John fixed my mistakes, but it’s $2i\pi$ that we called pi-bar.
3. Tim van Beek says:
John wrote:
There is no ‘Lebesgue measure’ on an infinite-dimensional vector space.
I remember the surprised looks of my fellow students when I told them about this, so here is the explanation: The Lesbegue measure on the reals is translationally invariant and finite for a finite interval.
In an infinite dimensional separable Hilbert space a “Lesbegue measure” with these properties would have to assign a finite number to the unit ball (radius 1), and a finite number to the ball with radius, say, 1/10. The problem with this is that you can fit infinite many balls with radius 1/10 inside the unit ball, by positioning one at $0.5 * e_n$ where $e_n$ is the nth element of an orthonormal basis. Since the “Lesbegue measure” should be invariant for translations, it would have to assign each of these balls the same finite number as the one centered around the origin.
• John Baez says:
Thanks for explaining that. This fact played a huge role in my early life, when I was working with Segal on integrals over infinite-dimensional Hilbert spaces. Lebesgue measure is gone… but Gaussian measures (and ‘generalized measures’) still exist!
4. Very nice example. Especially your formula for the expected action. Congratulation for that.
What is next? Things diverge. That does not come unexpected. My initial response (cf. my last comment) to that situation was to restrict the domain for the paths from continuous functions to more regular things like bounded variation, $C^1$ or even $C^\infty$. Since the expected action is somehow the arclength the integrals might converge then. However, such an approach seems to destroy the guiding analogy between statistical mechanics and what you are doing. Does it? After all, Brownian motion is highly irregular and nevertheless connected (by a ‘Wick-rotation’) to heat diffusion which is probably the most regular motion there is. It might even be possible to find some ‘natural’ space of paths.
Such or similar were my initial thoughts. After I read your article, in particular the part your discussion of the ideal gas, it occurred to me that there might be a less technical way to proceed. Roughly speaking, it might not be necessary to consider the $n\rightarrow \infty$ case. What does (can) that mean? In the case of the ideal gas we are only interested in the energy of a huge number of particles, but not for infinitely many of them. Maybe, the analogue of ‘energy of $n$ particles in an ideal gas’ is not the time evolution of something, but ‘expected action of a particle running through $n$ gates’. I am not overly precise what I mean by a gate now. Just that it is in principle possible to have a huge number ($n\geq 1$) of them in series and a particle passing them by ‘making decisions’. While in principle it is possible to let the number of gates run to infinity we are not interested in doing that since ‘reality’ limits the number of gates as it limits the number of particles in an ideal gas. Too weird? OK, I stop here.
In any case you have all tools at hand to introduce a potential and discuss the difficulties that come along.
• John Baez says:
Thanks for your comments! Maybe there’s a nice continuum limit of the expected action of a particle running through $n$ gates, meaning that we don’t take $n \to \infty$, but define a function of the particle’s positions at $n$ chosen times, namely the least possible action for a path going through those points at those times, and then compute its expected value using a real-time path integral. We’d need to introduce a time discretization and then show the limit exists as we remove this discretization. This is definitely true for the free particle, but that’s a specially easy case.
Anyway, I think some creative ideas are needed here…
• some guy on the street says:
Since Uwe said “gate”, I’m inclined to wonder what happens to the expected action of the particle with time in $n$ steps, if we observe the position at step $j$?
• some guy on the street says:
or… that’s what you already said, isn’t it?
• Garett Leskowitz says:
In my mind, the n gates or signposts that are erected in the specification of the particle’s positions are strongly reminiscent of “collapse events.” By increasing n, we increase the finesse with which we specify what the particle can do on its way from point A to point B, and we make our description of the particle’s time evolution “more classical” – more like a trajectory and less like wavefunction evolution.
This increase in information content seems to require an increase in the action, relative to a fundamental unit, i hbar.
The connection to heat capacity is very interesting. Heat capacity is a key quantity in thermal physics, and I would hope to see it make it into the table of analogues!
To a physical chemist, the heat capacity measures how many joules of energy are required to cause a kelvin of temperature change. It is an extensive quantity, and, in my mind anyway, is a measure of “thermally accessible degrees of freedom.” Is this usefully related to the finesse with which we can (or have to) specify microscopic details of a macroscopically defined, thermal system?
The definition of (constant volume) heat capacity involves a derivative of internal energy with respect to temperature. The analogue on the quantum side would seem to involve a derivative of action with respect to i hbar.
This analogue might raise and answer (admittedly vague) questions like “quantitatively, how much more quantum would this system behave if i hbar were bigger?”
One final comment regarding the n “gates”: A study of the harmonic oscillator might be illuminating. In quantum optics it is possible to produce “squeezed states” of a cavity mode by coupling it resonantly to a detector at multiples of a resonance frequency. I wonder if judicious selection of the time schedule of selected gate points changes quantities like expected action, etc., in an understandable way.
5. Jon Rowlands says:
The nice simple expected action looks like it maybe comes from the free particle being classical, so that $A(q)$ doesn’t involve $\beta$ (i.e. $\hbar$). Does anything change if a quantum free particle is used instead?
• John Baez says:
I’m working with the quantum free particle; all the calculations here are quantum-theoretic. In the path integral approach to quantum mechanics, the action never involves Planck’s constant. Planck’s constant shows up in integrals over the space of paths, e.g. the partition function itself:
$\displaystyle{ Z = \int_X e^{i A(q) / \hbar} D q }$
The expected action works out simply in this problem because the action $A(q)$ is a quadratic function of the path $q.$
Linear or quadratic actions give exactly solvable theories and typically serve as the starting-point for further investigations. For example, the free particle and harmonic oscillator have quadratic actions. So, the expected action should also be easy to compute exactly for the harmonic oscillator. Particles in more complicated potentials will get a lot harder.
6. Consider the action
$A(q) = \int_0^T \frac{1}{2}m v^2 - b q dt$
Being just a mathematician I can’t judge whether that is a physically realistic action. I think we have used that in school as a local approximation of gravity. Anyway, using all your assumptions (e.g. $q_0=0$, the Gaussian integral aso.) the expected action can be computed as
$\frac{n}{2 \beta}-\frac{b^2\triangle t}{2 m}\sum_{i=0}^{n-1}i^2$
For $b=0$ this is consistent with your result. The sum over $i^2$ can be evaluated explicitly and since $\triangle t$ is $\frac{t}{n}$ the behaviour of the expected action is like $n^2$.
• John Baez says:
Wow – you’re just a mathematician? You look like you know what you’re doing!
I’ve really been appreciating your comments in this quantropy discussion—you’re the only one brave enough to join in and actually do some calculations.
As you suspect, the action you wrote down describes a particle in a potential
$V(q) = b q$
which describes a constant force
$F = - V'(q) = -b$
So, if you’re doing experiments dropping neutrons and seeing how their wavefunctions spread out, this is the right action to use—at least until the neutron hits the floor and bounces!
• C. Codau, V.V. Nesvizhevsky, M. Fertl, G. Pignol and K.V. Protasov, Transitions between levels of a quantum bouncer induced by a noise-like perturbation.
The quantum problem of a ball bouncing above an ideal mirror was considered a long time ago but it was thought as a mere exercise in elementary quantum mechanics (see, for instance, [1]). Things did change since the quantization of energy levels of Ultra Cold Neutrons (UCN) bouncing above a mirror in the Earth’s gravitational fi eld has been demonstrated in an experiment performed at the Institute Laue Langevin (ILL) [2, 3, 4]. Together with neutron interferometric measurements in the gravity fi eld [5] and studies of bouncing cold atoms [6, 7], bouncing neutrons provide one of the few situations where quantum e ffects in a gravity fi eld can be observed.
But anyway, that’s a digression…
Any Lagrangian that’s a quadratic function of $q(t)$ and $\dot q(t)$ should be manageable, but some are easier than others. I did the very easiest one, and yours looks very nice too. I hadn’t thought of trying it, but it’s important because a constant force field is a simple approximate model for any force. I’d been wanting to do the harmonic oscillator, which should have its own charms.
• A quantum bouncer. Interesting.
On the harmonic oscillator I can offer support albeit it might take a couple of days to find some spare time. So, if someone jumps in I would not be too disappointed.
you’re the only one brave enough to join in and actually do some calculations.
Arghh! Don’t tell everybody! But honestly, so far I have no idea what quantropy actually means. Do you consider it possible that by extremizing quantropy 1) in the end one only gets garbage 2) one gets an alternative way to explain some known physics 3) one gets new effects. What would be the best possible outcome one could hope for ? … or is it too early to pose such questions ? What about a more ‘philosophical’ background article without formulas to get more people into it.
• John Baez says:
Uwe wrote:
But honestly, so far I have no idea what quantropy actually means.
Neither do I. That’s why I’m doing all these @*#$& calculations. Do you consider it possible that by extremizing quantropy 1) in the end one only gets garbage 2) one gets an alternative way to explain some known physics 3) one gets new effects. What would be the best possible outcome one could hope for ? … or is it too early to pose such questions? It’s not too early to pose them. It may be too early to answer them. What about a more ‘philosophical’ background article without formulas to get more people into it? I’ve explained all the philosophical background I know, especially here and here. I don’t really want more people to get into the subject until I figure out the easy fun stuff myself. If I were a normal physicist I’d keep all these ideas secret until I published a paper or two. But I’ll just repeat: There’s a beautiful and precise 2 × 2 analogy square: classical statics classical mechanics thermal statics quantum mechanics In three out of the four entries, we can derive everything starting from a minimum or maximum principle: principle of least potential energy principle of least action principle of maximum entropy ??? So, it seems insane for the fourth not to work the same way. We can figure out what quantum mechanics ‘extremizes’, and it’s quantropy. However, it’s complex-valued so we should really speak of a stationary principle rather than a minimum or maximum principle. And if you think harder you see that’s true in some of the other cases too. So we can say: principle of stationary potential energy principle of stationary action principal of stationary entropy principle of stationary quantropy Now, it could turn out that the principle of stationary quantropy is just a mirage due to the divergences we’re seeing. Or, maybe it’s fine but only if we treat time as discrete: that would be quite interesting. It could be that we can only use it to derive results that can be derived in other ways. That’s also true about the principle of least action to some extent—Feynman tells how as a young student he was very big on doing everything with just $F = ma$, which is pretty ironic given his later work on path integrals. Or, it could be that the principle of stationary quantropy will give really new insights. That’s also true about the principle of least action. But for now I (or we) just gotta think and calculate. • … I’d keep all these ideas secret until I published a paper or two. Maybe you should do this indeed. It would be less fun though. But for now I (or we) just gotta think and calculate. You are not alone :-) For the action of the harmonic oscillator $A(q) = \int_0^T \frac{1}{2}m v^2 - \frac{1}{2}b q^2 dt$ the quantropic partition function is $Z = e^{\frac{n \beta b m \left(\Delta t\right)^2}{\left(m-b \left(\Delta t\right)^2\right)^{2}}}\left(\frac{2\pi\Delta t}{\beta(\Delta x)^2\left(m-b \left(\Delta t\right)^2\right)}\right)^{\frac{n}{2}}$ Observe that for $b=0$ it is the partition function for the free particle. At least. The expected action is still linear in $n$ and does not depend on $\Delta x.$ The calculations are elementary and actually rather boring. In case you are interested in them I could dump them here or in my blog or private mail them. Whatever. • John Baez says: Uwe wrote: John wrote: If I were a normal physicist I’d keep all these ideas secret until I published a paper or two. Maybe you should do this indeed. It would be less fun though. Right. I have to strike a balance between caution and having fun. Since I have tenure, I can afford to have more fun. But I still like to be the first to publish stuff. I figure that as long as most people think quantropy is a weird, stupid or useless idea, it’s safe for me to talk about it publicly. But I don’t want to recruit a lot of people into working on it before I’ve put something on the arXiv. The calculations are elementary and actually rather boring. In case you are interested in them I could dump them here or in my blog or private mail them. Thanks—but I’ll probably do them myself someday. I find I learn a lot from doing these calculations, even if in some sense they’re elementary and boring. They make me think about lots of things, in a way that reading a calculation does not. But being able to check whether my answer matches yours will be very nice, since I make lots of mistakes. So thanks for that! The expected action is still linear in $n$ and does not depend on $\Delta x.$ Good! But what matters is not just that. What matters is the precise coefficient! And from your results we see the expected action is $\displaystyle{ \langle A \rangle = n \frac{i \hbar}{2} }$ And that’s great! Let me explain why: Since the expected action for the free particle is so simple, I’ve decided that the expected action is what I should focus on for now. As I hinted in this blog post, in statistical mechanics, people make an enormous amount of fuss over the fact that a system has expected energy equal to $\frac{1}{2} k T$ times the number of degrees of freedom. Strictly speaking this is only true for classical systems whose energy is a positive definite quadratic form on some finite-dimensional vector space. The ‘number of degrees of freedom’ is the dimension of that vector space. But since every smooth function can be approximated by a quadratic function near a nondegenerate minimum, this is approximately true for a vast number of classical systems. In fact, you’ll often see this formula mistakenly used as a definition of temperature! In my post, I noticed that for systems where the action is a positive definite quadratic form on a finite-dimensional vector space, the system has expected action equal to $\frac{1}{2} i \hbar$ times the dimension of that vector space. This is the precise analogue of what we know about statistical mechanics! For the harmonic oscillator the action is still a quadratic form, but it’s not positive definite, so some new subtleties intrude. Nonetheless I expected the above principle would still hold, and you’re telling me it does. • I figure that as long as most people think quantropy is a weird, stupid or useless idea, it’s safe for me to talk about it publicly. But I don’t want to recruit a lot of people into working on it before I’ve put something on the arXiv. These are some substantial objections against quantropy and once you elevate this from blog to arXiv you maybe want to deal with them. Thus it might be helpful for you to know why I do not think that this is useless (thereby giving some possible uses). A sole analogy possibly won’t convince people. There is just not enough innovation and probably not a lot of physics and certainly no relevant mathematics contained. Such or similar might be a first opinion when dealing with this. Things change, at least for me, with the introduction of quantropy. Why is this the case? Now one can break the symmetry and introduce something new by making the integrals converge. For example, if you consider particles travelling along the paths with the speed of light you will eventually get a cutoff in the integrals. You have mentioned a cutoff in one of your articles as a way to define things. In this case the purely mathematical $n$ will be replaced by a physically hopefully more tractable velocity $c$. Is such a computation new? Most likely not. Feynman an his follow ups could certainly do these computations. However, there is a slim chance that they simply were not interested in it for two reasons: Their integrals converged anyway and the resulting Schrödinger equation would contain an unwanted velocity parameter. The situation in the quantropy business is different. Here, a choice _has_ to be made since otherwise nothing converges. If it is the above choice of a “speed limit” reevaluating your motivating example for quantropy might lead to what you have called a “Feynman’s sum over histories formulation of quantum mechanics” augmented with a velocity parameter. My knowledge of physics is slim enough to consider this to be interesting. With this thoughts I certainly do not want to diminish other choices to make the integrals converge or even robust approaches that do not consider convergence at all. Anything goes. These are just my (naive?) thoughts why one should not immediately discard this idea. • John Baez says: Uwe wrote: A sole analogy possibly won’t convince people. There is just not enough innovation and probably not a lot of physics and certainly no relevant mathematics contained. Such or similar might be a first opinion when dealing with this. Things change, at least for me, with the introduction of quantropy. What are you talking about, exactly, when you say ‘a sole analogy’? At first I thought you meant the analogy between entropy and quantropy, but the last sentence suggest otherwise. The analogy between quantum mechanics and statistical mechanics? This analogy is already very famous, and there’s a huge amount of physics and mathematics in it. Many books have been written based on this analogy: for example, Glimm and Jaffe’s Quantum Physics: A Functional Integral Point of View constructs quantum field theories using Wick rotation to transform path integrals into statistical mechanics problems, and Barry Simon’s Functional Integration and Quantum Physics covers different aspects of the same territory. The subject of conformal field theory is another example: thanks to this analogy it leads a double life, on the one hand allowing people to understand critical points in 2d statistical mechanics problems, and on the quantum side playing a fundamental role in string theory. So the first interesting thing about quantropy is that it’s an obvious yet apparently largely unexplored aspect of this already famous analogy. 7. Steven S. says: The expression for the expected action makes me think of the expected energy of a quantum harmonic oscillator (and hence QFT). So by analogy maybe one could think of 1/2i * h/2pi as the eigenvalue of the ‘action operator’ and when n steps (for the 1D case) are taken it means n quanta of action (=’actons’?) are emitted just like n photons with energy h nu are absorbed/emitted when an electron moves to its nth eigenstate/drops down to the ground state. • John Baez says: Hmm! I’ve been thinking about strange things like this myself recently: the analogies between classical mechanics, thermodynamics, statistical mechanics and quantum mechanics lead to some strange possibilities like quantizing thermodynamics, quantizing quantum mechanics, etc. Of course ‘second quantization’ is a known thing, and known to be useful… but there could be other things, not yet explored, which are logically consistent and (who knows?) perhaps even useful somehow. You’re right: the formula $\hbar \nu/2$ for the ground state energy of the quantum harmonic oscillator looks suspiciously similar to the formula $\hbar / 2i$ for the ‘expected action per decision’ of a free quantum particle. But I don’t see a mathematical relationship underlying this similarity, so I’m reluctant to plunge in and try stuff! On the other hand, there’s a clear mathematical relationship between $\hbar / 2i$ as the ‘expected action per decision’ for a free quantum particle and $k T / 2$ as the ‘expected energy per degree of freedom’ for a classical harmonic oscillator. That’s what I worked out in this post. So, you (or someone) could try to find a precise mathematical relation between the $\hbar \nu /2$ ground state energy of the quantum harmonic oscillator and the $k T / 2$ expected action per degree of freedom of the classical harmonic oscillator. But the first number shows up as an eigenvalue of a differential operator, while the second shows up as the value of an integral (as in this post). Also, the first depends on the frequency of the oscillator, while the second does not! So I don’t see a connection. 8. Hi there I’ve recently come across this series of posts and found them quite intriguing. I’ve often wondered if there might be a way to get to quantum theory by modifying classical (Bayesian) probability theory in some principled way, and this looks like it might be the beginning of a path that could lead in that direction. It’s very interesting stuff, but there’s a slight conceptual issue that’s been bothering me, and I wonder whether thinking about it more might shed some more light on what this analogy really means. I hope I can express it clearly enough. In the thermal case, when you derive the Boltzmann distribution by maximising the entropy, it’s done subject to to a constraint on the expected energy, i.e. you set $\langle E_x \rangle = E$, for some prescribed value $E$. Once you’ve done the whole Lagrange multiplier thing you then have to choose the particular value for $\beta$ such that the constraint $\langle E_x \rangle = E$ is satisfied. $\beta$ thus becomes a function of $E$. But then it turns out that in practice you often know $\beta$ and not $E$, because the system is in contact with a heat bath that fixes the temperature. In this case you still get $\beta$ as a function of $E$, but you plug in the value of $\beta$ to find $E$ and not the other way around. Now, in your quantropy analogy you’re doing the latter thing. That is, you’re not finding a stationary point of the quantropy subject to a constraint $\langle A_x \rangle = iA$ for some particular known value of $A$ and then choosing $\lambda$ such that the constraint is satisfied. Instead it looks like you’re using the Lagrange multiplier technique to find a functional relationship between $A$ and $\lambda$ and then plugging in a known value of $\lambda$ (equal to $1/i\hbar$) to find $A$. So I’m wondering what that really means. I realise you probably don’t have an answer to that yourself, but it seems to me that thinking about it might (or might not) be a useful next step to take. Naively, it looks like it’s saying the universe is in contact with a sort of Wick-rotated heat bath with an imaginary temperature of $1/i\hbar$. But I wonder if there’s some other, more sensible interpretation of this step. • John Baez says: I’m glad you liked these posts, Nathaniel! You wrote: That is, you’re not finding a stationary point of the quantropy subject to a constraint $\langle A_x \rangle = iA$ for some particular known value of $A$ and then choosing $\lambda$ such that the constraint is satisfied. Instead it looks like you’re using the Lagrange multiplier technique to find a functional relationship between $A$ and $\lambda$ and then plugging in a known value of $\lambda$ (equal to $1/i\hbar$) to find $A$. I don’t think of these as significantly different things. The way it always works when you’re looking for a stationary point of a function$f$subject to a constraint on some other function$g$is that you set $\nabla (f - \lambda g) = 0$ and solve the resulting equations. In this process, it’s pretty much always easier to pick a value of $\lambda$ and find out what value the constraint $g$ than takes, rather than the other way around—though in terms of the original problem, you usually want to think of things the other way around. In particular, I don’t see anything about the math of quantropy that differs from the math of entropy in this respect. • I agree, the math is exactly the same – it’s the interpretation I’m trying to get at. It’s kind of a subtle point, and I was afraid I wouldn’t be able to get it across clearly. The point is that, when doing classical MaxEnt (which I do a lot), sometimes you know $\langle E \rangle$ and want to work out $\beta$, and sometimes you know $\beta$ and want to work out $\langle E \rangle$. I was just pointing out that for quantropy it’s always the latter case.$\lambda$isn’t a variable whose value depends on the system’s state like$\beta$is, but rather it is (or appears to be) a universal constant. Beyond that, my point isn’t really anything more than “hey, that’s interesting, I wonder what it means?” To put it another way, if we lived in an isothermal universe (i.e. one where everything everywhere was connected to a heat bath at a constant temperature) then$\beta$would have the same value for all systems, and so we’d think of it as a universal constant. That isn’t the universe we live in, though, and$\beta$has different values for different systems. But for the quantropy case,$\lambda\$ does seem to always have the same value, for every system, unless I’ve misunderstood something. That’s what I meant when I said it looks, naively, like the universe is in contact with a sort of Wick-rotated heat bath (or rather an action bath) with a constant imaginary temperature. I’m not saying that’s a plausible physical picture, just that it’s interesting that that’s what the maths seems to look like.
9. Robin Oswald says:
Interesting series of posts! Just a small typo at the end:
In statistical mechanics we have
$\beta = \frac{1}{kT}$
while in quantum mechanics we have
$\beta = i\hbar$
The last equation should obviously be
$\beta = \frac{1}{i \hbar}$
• John Baez says:
Glad you liked the posts! Actually Planck’s constant is analogous to inverse temperature, not temperature. To see this in an intuitive way, note that increasing Planck’s constant increases quantum fluctuations, while increasing temperature increases thermal fluctuations. Also: we get classical dynamics obeying the principle of least action when Planck’s constant goes to zero, while we get classical statics obeying the principle of least energy when the temperature goes to zero.
10. If you have carefully read all my previous posts on quantropy (Part 1, Part 2 and Part 3), there’s only a little new stuff here. But still, it’s better organized […]
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1. ## Derivative of Arcsecx
find derivative of (arcsecx)^3
What i've done...
= 3(sec^-1(x))^2 *(1/ (x)sqrt(x^2-1))
= 3(arcsecx)^2 / (xsqrt(x^2-1))
2. ## Re: Derivative of Arcsecx
Let's test it out:
\begin{align*}y & = \left(\text{arcsec } x\right) ^3 \\ y^{1/3} & = \text{arcsec } x \\ \sec y^{1/3} & = x \\ \left( \cos y^{1/3} \right)^{-1} & = x \\ \left( \cos y^{1/3} \right)^{-2} \sin y^{1/3} \dfrac{1}{3}y^{-2/3} \dfrac{dy}{dx} & = 1 \\ \dfrac{dy}{dx} & = 3y^{2/3}\cot y^{1/3} \cos y^{1/3}\end{align*}
Now, we have $\sec y^{1/3} = \dfrac{x}{1} = \dfrac{\text{hyp}}{\text{adj}}$, so $\cot y^{1/3} = \dfrac{1}{\sqrt{x^2-1}}$ and $\cos y^{1/3} = \dfrac{1}{x}$. So, we can rewrite:
$\dfrac{dy}{dx} = 3\dfrac{ \left( \text{arcsec } x \right)^2 }{ x\sqrt{x^2-1} }$
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# 2 taps A and B can fill a bucket in 12 and 18 minutes resp. How long will it take both the taps to fill the bucket.
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
With the tap A is running alone the bucket is filled in 12 minutes and when tap B alone is running the bucket is filled in 18 minutes.
The rate at which tap A can fill the bucket is 1/12 per minute. The rate at which tap B can fill the bucket is 1/18 per minute.
When both the taps are running the bucket is filled at the rate 1/12 + 1/18 = 5/36
To fill the bucket it takes 36/5 = 7.2 minutes
The bucket is filled in 7.2 minutes when both the taps are running.
academicsfirst | High School Teacher | (Level 2) Adjunct Educator
Posted on
We can also set up an equation to solve this work problem.
We are given that tap A can fill a bucket by itself in 12 minutes and tap B can fill the bucket in 18 minutes. We are asked how long it will take for the two taps running together to fill the bucket.
=> let x = total job time for both working together
=> let 1/12x = time it takes for tap A to fill the bucket
=> let 1/18x = time it takes for tap B to fill the bucket
=> The entire completion of the job will be equal to 1
=> 1/12 x + 1/18x = 1 (completed job)
=> 36 [1/12 x + 1/18x = 1] (to clear fractions)
=> 3x + 2x = 36
=> 5x = 36
=> x =36/5 or 7.2 minutes
Both taps running together will fill the bucket in 7.2 minutes.
We’ve answered 317,598 questions. We can answer yours, too.
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# Is this mathematically/statistically possible?
Let’s say there are 4 groups of people, with an unknown number of people in each group. Within each group, every person is either a 1 or a 0. The percentages below describe the percentage of people who are 1s.
Given:
Group A: 50%
Group B: 51%
Group C: 50%
Group D: 50%
Is it possible for Groups A, B, C, D aggregated to be less than 50%? Is it still true if Group B is 50% instead of 51%?
I think it depends on whether or not you allow
1. A group to have 0 people.
2. If you define a group with an odd number of people as meaningful for A,B,C; and exactly how you handle it if you can have an odd number.
NO (assuming your percentages are exact and not 49.8% rounded to 50%). Yes, it’s still true.
Doesn’t matter. Your aggregate has to be a convex combination of 50%, 50%, 51% and 50%, so it can’t be any smaller than 50%.
No, it’s not possible. At least half the people in each group are 1s. Adding this up, we get that at least half the total population are 1s.
Indeed, if Group B is exactly 50%, we have that exactly half the total population are 1s.
How would a group have 0 people and still have 50% or 51% 1s?
When dealing with averages of various sized groups like this, and then trying to think about the total aggregate average, you are dealing with weighted averages. Perhaps the best-known situation where we all (most of us) have dealt with weighted averages is in computing our school grade-point average.
Here’s a helpful idea for thinking about this: Every individual contributor to the total average has the effect of pulling the aggregate average toward itself. Whatever grade you get in a 5-unit class, and whatever your GPA was before that, and however many total units you had before that, the new 5-unit grade will pull the overall average toward itself. It will pull the average more toward itself that a 4-unit grade would do, which will also pull the average toward itself.
If you only have a few units already, the new grade may affect the average a large amount. If you already have hundreds of units, the new grade may be just a drop in the bucket. If your average is a C, then either an A or a B will pull you up, but the A will pull you up more. But in ALL cases, the new grade pulls the average toward itself.
ETA: If you allow a group of 0 people, it would have a weight of 0 and not affect the average at all. So even if you had an empty group and (somehow) computed its average to be 51% ( :dubious: ) that still could not draw the aggregate average down below 50%.
You have the same sort of thing in the OP’s example. Each group contributes to the aggregate average according to the (unspecified) size of the group. The size of the group is its weight. A big group affects the average more than a little group. If the average is 50%, then adding a 75% group makes a bigger difference than adding a 51% group, and the bigger the group is, the more difference it makes.
Here, you have three groups with 50% average, for an aggregate average (of those three groups) of 50%. Adding the 51% group can only pull the aggregate average up toward (but not beyond) 51%. Even if the three 50% groups are small and the 51% group has six billion people in it, it can only pull the average up toward 51% and not above, and not below 50% either.
If the groups are large, then dealing with odd-number groups and halves of people can only make a minuscule and insignificant difference. If the groups are small, then fractional people might matter, and the result would depend on how you want to count them. But there’s no way the 51% group, added to the other 50% groups, could draw the average down below 50%.
0 * .5 = 0?
Yeah, the more pertinent question is, how would a group have an odd number of people and still have exactly 50% (or, for that matter, exactly 51%)?
But, at any rate, the answer is unconditionally “No”, as everyone else has said.
bizerta is right. Basically, this is because if you have two non-empty groups, the average of the aggregated group must be between the averages of the two separate groups. (And an empty group can’t have an average, because you can’t divide by zero.)
Here’s the proof:
Let group A have n members, and an average of m/n.
Let group B have q members, and an average of p/q.
Suppose that m/n < p/q. (The other case where m/n > p/q has a similar proof, and if m/n = p/q the aggregated average is the same number.)
Then, since n and q are positive integers, their product nq is positive, and you can multiply by nq to get mq < np.
Add pq to each side to give (p + m)q < (n + q)p
Now q/(n + q) is positive (since n and q are positive), so multiply each side by that:
(p + m)/(n + q) < p/q
So the average of the aggregate [(p + m)/(n + q)] must be less that the larger average.
You can construct a similar argument to prove that the average of the aggregate must be larger than the smaller average.
Maybe it’s a math thing, but I’m pretty sure there’s no 1s in a group with 0 people.
It depends on how you define the problem.
You say "a group of people of an arbitrary number of people are split into four groups under the following conditions:
A such that half are assigned a 1, and half are assigned a 0.
B such that just over half (51%) are assigned a 1, else 0.
C as A
D as A"
Say three people enter the room, how do you handle the problem? The problem is not worded to exclude odd numbers or people, nor does it place a lower bound, so therefore the following scenario is possible
A has one member
B has zero members
C has one member
D has one member
So for the person in A,C,D, you must assign a one or zero to the person. Since there’s only three people, you must either have the case (unordered tuples) (0,0,1), (0,1,1), (0,0,0), or (1,1,1).
I don’t understand what I’m missing here, the problem is not defined in a sufficiently exclusive way, so corner cases have to be handled. Unless the problem specifically defined that either you can’t have less than 4 people, or else specifically predefines a behavior in such cases, the question allows potentially all possible configurations of corner cases to be valid, meaning that there must exist cases where there isn’t an even split.
Okay, then you assume that if B only has one member, it must be a 1.
Then you have the configurations
(1,0,0,0), (1,1,0,0), (1,1,1,0), (1,1,1,1). I just don’t get why corner cases don’t have to be predefined.
In a group with 0 people, it is simultaneously true that the number of 1s is 0% of everyone, 100% of everyone, 50% of everyone, 51% of everyone, and every which other thing.
As I and everyone else read it, the problem is: Is there any way to have four groups of people, such that in each group 50% of the group are 1s and the rest are 0s (or this is the case in three of the four groups, and in the fourth, 51% are 1s), and overall, less than 50% of the total population is 1s?
The answer to that question is no. There’s no way to have that. There’s no ambiguity that needs to be resolved; it’s simply impossible to have that.
It’s also true, while we’re at it, that the answer to the question “Is there any way to have four groups of people, such that in each group exactly 50% of the group are 1s and the rest are 0s (or this is the case in three of the four groups, and in the fourth, exactly 51% are 1s), and also, at least one of the groups has an odd number of people?” is also no, but that’s irrelevant to the first question.
I don’t understand, the OP made me think we were doing assignment, meaning n people walk into a room and then we assign them groups, and then assign group members numbers due to some algorithm, in which case don’t these cases need to be accounted for in case of strange input?
I don’t know why you thought the OP had anything to do with assignment or an algorithm. They straight up ask “Is this mathematically/statistically possible?” and then describe a situation. The situation they describe has nothing to do with assignment or an algorithm. It also happens to be a situation which is not mathematically/statistically possible. Accordingly, the answer is “No”.
It’s analogous to this:
"Is the following situation mathematically/statistically possible?
There is a group, with some number of men and some number of women.
There are more men than women.
There are twice as many women as men."
The answer to that question is “No”, not “What happens if the number of women is odd?”.
See, this is why I leave math to mathematicians, and cats to physicists.
It’s no crazier than the fact that 0 = 0.00 * 0 = 1.00 * 0 = 0.50 * 0 = 0.51 * 0, etc., which I am surely you will readily agree to (anything times zero is zero). Indeed, it’s the exact same fact.
But, yes, this is why people are generally and reasonably loath to discuss proportions or average values among an empty population.
I guess I just thought that multiplication and division was defined for every number in the set of natural numbers (save division by zero), so that meant there was a definition for n/2 if n is odd, even though there is no INVERSE operation (i.e. is not isomorphic). This means that defining a problem such that 2n = x is not the same as defining a problem such that n/2 = x if the domain is the set of natural numbers (which is the only set that really makes sense with people). Which would mean that your example with “twice as many” isn’t analogous to the OP’s “half of them” problem. I don’t know, I just find this really, really confusing that it uses the set of natural numbers, but is somehow in a different space with different operations defined.
Um, what? Perhaps I should rephrase, then.
"Is the following situation mathematically/statistically possible?
There is a group, with some number of men and some number of women.
There are less women than men.
There are half as many men as there are women."
The answer is “No, that’s not possible”, not “What happens if there are an odd number of women?”.
Or, put another way,
"Is the following situation mathematically/statistically possible?
There is a group of people, each of which is one of American, British, or Canadian.
One third of the group is Americans.
There are as many Americans as Brits and Canadians combined".
The answer is “No, that’s not possible”, not “What if the number of people in the group is not divisible by three?”.
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# Significant Figures
Course:
Chemistry
Understanding Significant Figures (also known as “sig figs”) is incredibly important in Science! When you collect data, whether you’re doing Chemistry, Biology, Physics, or Engineering, you need to notice how precise your tools allow you to be. There should be one estimated digit in your data. This is where the idea of significant figures comes from - from measurement! Make a note of how precise your data is before you start calculating. Then you round off to the correct number of significant digits at the end of your calculations. Otherwise, you give people the wrong idea of how precise your initial measurements were. In this video, we’ll go over all the rules for determining significant figures. We also do examples of sig figs in calculations! We show you the rules for adding, subtracting, multiplying, and dividing. We even cover the advanced topics of significant figures in logarithms and when you raise 10 to a power.
Understanding Significant Figures (also known as “sig figs”) is incredibly important in Science. When you collect data, whether you’re doing Chemistry, Biology, Physics, or Engineering, you need to notice how precise your tools allow you to be. There should be one estimated digit in your data.
This is where the idea of significant figures comes from—from measurement. Make a note of how precise your data is before you start calculating. Then you round off to the correct number of significant digits at the end of your calculations. Otherwise, you give people the wrong idea of how precise your initial measurements were.
There are 4 rules for determining significant figures:
1. Non-zero digits are always significant.
2. A zero is only significant if it is at the right end of a number AND after a decimal.
3. If a quantity is known to be exact, it has an unlimited number of significant figures.
4. When a number ends in zeros but contains no decimal point, the zeros may or may not be significant.
This video covers all the rules for determining significant figures, and includes examples of sig figs in calculations. We show you the rules for adding, subtracting, multiplying, and dividing. We even cover the advanced topics of significant figures in logarithms and when you raise 10 to a power.
Video: Significant Figures
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What is the degree of the polynomial? 8x2
Question
Updated 8/12/2014 4:07:36 AM
Flagged by yeswey [8/12/2014 3:05:39 AM], Edited by jeifunk [8/12/2014 4:07:36 AM]
Original conversation
User: What is the degree of the polynomial? 8x2
Question
Updated 8/12/2014 4:07:36 AM
Flagged by yeswey [8/12/2014 3:05:39 AM], Edited by jeifunk [8/12/2014 4:07:36 AM]
Rating
3
The degree of the polynomial 8x^2 is 2.
Confirmed by jeifunk [8/12/2014 4:03:50 AM]
Questions asked by the same visitor
Factor x2 - 10x + 16. A. (x - 4)(x + 4) B. (x - 4)(x + 6) C. (x - 2)(x - 8) D. (x + 2)(x + 8)
Question
Updated 7/6/2014 4:20:08 PM
x^2 - 10x + 16 = (x - 2)(x - 8)
Confirmed by yumdrea [7/6/2014 4:20:55 PM]
Factor x2 - x - 90 = 0. A. (x + 10)(x + 9) = 0 B. (x - 10)(x - 9) = 0 C. (x - 10)(x + 9) = 0 D. (x + 10)(x - 9) = 0
Weegy: D. (x + 10)(x - 9) = 0 (More)
Question
Updated 7/9/2014 12:15:16 PM
x^2 - x - 90 = 0 = (x - 10)(x + 9) = 0
Confirmed by jeifunk [7/9/2014 12:36:05 PM]
Multiply the two binomials: (x – 6)(x – 6) A. x2 – 6x + 36 B. x2 – 12x - 36 C. x2 – 6x - 36 D. x2 – 12x + 36
Question
Updated 6/20/2014 1:38:43 PM
(x – 6)(x – 6)
= x^2 - 6x - 6x + 36
= x^2 - 12x + 36
Multiply the two binomials: (a + 4)(a + 4)
Question
Updated 6/20/2014 1:24:12 PM
(a + 4)(a + 4)
= a^2 + 4a + 4a + 16
= a^2 + 8a + 16
Simplify the following expression by adding and subtracting: 6v9 - 7v20 - 3v45
Weegy: 3-3 = 0 (More)
Question
Updated 3/31/2014 10:00:53 AM
6v9 - 7v20 - 3v45
= 18 - 14v5 - 9v5
= 18 - 23v5
Confirmed by jeifunk [3/31/2014 10:14:10 AM]
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# Evenness of Fourier coefficients
Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$ and let $F(n)$ denote its Fourier coefficients $$F(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$ I want to prove that $f$ is even if and only if $F(n)=F(-n)$ for all $n$.
Suppose $f$ is even. Then I have to prove $\int_{-\pi}^\pi f(x)e^{-inx}dx=\int_{-\pi}^\pi f(x)e^{inx}dx$. This is true because the value on the left-hand side at $a$ is $f(a)e^{-ina}$, while the value of the right-hand side at $-a$ is $f(-a)e^{-ina}=f(a)e^{-ina}$.
What about the converse? Suppose $F(n)=F(-n)$ for all $n$. How can I show that $f(x)=f(-x)$ for all $x$?
-
How about making a change of variables and applying an inverse Fourier transform? – Riemann1337 Oct 26 '13 at 18:34
@Riemann1337 Are there more elementary ways than that? (i.e. without making the inverse Fourier transform) – Paul S. Oct 26 '13 at 18:45
Maybe there is a way to argue it by just subtracting the integrals, but I am not sure; using the inverse Fourier transform seems to be the most straightforward way to solve it. – Riemann1337 Oct 26 '13 at 18:59
Changing variables, we have $$F(n)=F(-n)\rightarrow \int_{-\pi}^\pi f(x) e^{inx}dx=\int_{-\pi}^\pi f(x) e^{-inx}dx =\int_{-\pi}^\pi f(-x) e^{inx}dx,$$ so $$\int_{-\pi}^\pi[f(x)-f(-x)]e^{inx}dx = 0.$$ If $\mathcal{F}$ denotes the Fourier transform, then we have that $$0=\mathcal{F}[f(x)-f(-x)]\rightarrow\mathcal{F}[f(x)] = \mathcal{F}[f(-x)].$$ Applying the inverse Fourier transform should give the result.
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# 2000 AIME I Problems/Problem 15
## Problem
A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$?
## Solution
We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$, meaning that there were $\boxed{927}$ cards are above the one labeled $1999$.
## Solution 2
To simplify matters, we want a power of $2$. Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$. From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$, so $\boxed{927}$ cards are above it. - Spacesam
## Solution 3 (Recursion)
Consider the general problem: with a stack of $n$ cards such that they will be laid out $1, 2, 3, ..., n$ from left to right, how many cards are above the card labeled $n-1$?
Let $a_n$ be the answer to the above problem.
As a base case, consider $n=2$. Clearly, the stack must be (top to bottom) $(1, 2)$, so $a_n=0$.
Next, let's think about how we can construct a stack of $n+1$ cards from a stack of $n$ cards. First, let us renumber the current stack of $n$ by adding $1$ to the label of each of the cards. Then we must add a card labeled "$1$".
Working backwards, we find that we must move the bottom card to the top, then add "$1$" to the top of the deck.
Therefore, if $a_n \ne n-1$ (meaning the card $n-1$ is not at the bottom of the deck and so it won't be moved to the top), then $a_{n+1} = a_n + 2$, since a card from the bottom is moved to be above the $n-1$ card, and the new card "$1$" is added to the top. If $a_n = n-1$ (meaning the card $n-1$ is the bottom card), then $a_{n+1}=1$ because it will move to the top and the card "$1$" will be added on top of it.
With these recursions and the base case we found earlier, we calculate $a_{2000} = \boxed{927}$. To calculate this by hand, a helpful trick is finding that if $a_n=1$, then $a_{2n-1}=1$ as well. Once we find $a_{1537}=1$, the answer is just $1+(2000-1537)\cdot2$. - Frestho
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Find Perimeter
```Name
17.2
?
Geometry and
Measurement—3.7.B
MATHEMATICAL PROCESSES
3.1.C, 3.1.E
Find Perimeter
Essential Question
How can you measure perimeter?
You can estimate and measure perimeter in standard units,
such as inches and centimeters.
Hands
On
Unlock
Unlock the
the Problem
Problem
Find the perimeter of the cover of a notebook.
Activity Materials ■ inch ruler
STEP 1 Estimate the perimeter of a notebook
in inches. Record your estimate.
_ inches
STEP 2 Use an inch ruler to measure the length of each
side of the notebook to the nearest inch.
STEP 3 Record and add the lengths of the sides measured
to the nearest inch.
Math Talk
_+_+_+_=_
So, the perimeter of the notebook cover measured
to the nearest inch is _ inches.
Mathematical Processes
Explain how your estimate
compares with your
measurement.
© Houghton Mifflin Harcourt Publishing Company
Try This! Find the perimeter.
Use an inch ruler to find the length
of each side.
Use a centimeter ruler to find the length
of each side.
Add the lengths of the sides:
Add the lengths of the sides:
_+_+_+_=_
_+_+_+_=_
The perimeter is _ inches.
The perimeter is _ centimeters.
Module 17
551
Hands
On
Share
Share and
and Show
Show
1.
Use an inch ruler to find the perimeter of the triangle.
Math Talk
in.
in.
Mathematical Processes
Explain how many
numbers you add together
to find the perimeter of
a figure.
Think: How long is
each side?
in.
_ inches
Use a centimeter ruler to find the perimeter.
2.
3.
cm
cm
cm
cm
cm
cm
cm
cm
cm
_ centimeters
4.
_ centimeters
Use the grid paper to draw a figure that has a perimeter
of 24 centimeters. Label the length of each side.
552
© Houghton Mifflin Harcourt Publishing Company
1 cm
Name
Problem
Problem Solving
Solving
5 in.
Use the photos for 5–6.
5.
6.
Which of the animal photos has a perimeter
of 26 inches?
7 in.
8 in.
Multi-Step Analyze How much greater
is the perimeter of the bird photo than the
perimeter of the cat photo?
8 in.
4 in.
4 in.
7 in.
5 in.
Write Math
© Houghton Mifflin Harcourt Publishing Company • Image Credits: (tl) ©Comstock Images/Getty Images; (tr) ©Alan Carey/Corbis
7.
Multi-Step Erin is putting a fence
around her square garden. Each side of her
garden is 3 meters long. The fence costs
\$5 for each meter. How much will the fence cost?
Show Your Work
Write Math
8.
Gary’s garden is shaped like
a rectangle with two pairs of sides of equal length, and
it has a perimeter of 28 feet. Explain how to find the
lengths of the other sides if one side measures 10 feet.
9.
Evaluate Jill says that finding the perimeter of a figure
with all sides of equal length is easier than finding the
perimeter of other figures. Do you agree? Explain.
Module 17 • Lesson 2
553
Mathematical Processes
M
Model ¥ Reason ¥ Communicate
M
Daily
Daily Assessment
Assessment Task
Task
Fill in the bubble completely to show your answer.
10.
11.
12.
Sally is putting frosting around the edges of the
roof of a gingerbread house. What is the perimeter
of the roof?
A
18 cm
C
8 cm
B
16 cm
D
20 cm
Kyle is adding a border to his triangular flag.
What is the perimeter of the flag?
A
2 inches
C
3 inches
B
6 inches
D
1 inch
Multi-Step Pete glues a rope around his
rectangular rodeo sign. His sign has side lengths
of 2 feet and 3 feet. The rope costs \$4 for each foot.
How much does Pete pay for rope?
A
\$24
C
\$10
B
\$20
D
\$40
TEXAS Test Prep
9 ft
Austin’s class is making a poster for
Earth Day. What is the perimeter of
the poster?
A
24 feet
B
21 feet
C
15 feet
D
30 feet
6 ft
6 ft
9 ft
554
© Houghton Mifflin Harcourt Publishing Company
13.
Geometry and Measurement—3.7.B
MATHEMATICAL PROCESSES 3.1.C, 3.1.E
H o mewo rk
and Practice
17.2
Name
Find Perimeter
Use a centimeter ruler to find the perimeter.
cm
1.
2.
cm
cm
cm
cm
cm
cm
cm
cm
_ centimeters
_ centimeters
Problem
Problem Solving
Solving
Use the drawings for 3–4.
8 in.
6 in.
8 in.
6 in.
A
4 in.
8 in.
© Houghton Mifflin Harcourt Publishing Company
3.
Carly drew quadrilaterals A and B.
Which quadrilateral has a perimeter
of 28 inches?
B
4 in.
7 in.
4.
How much greater is the perimeter of
quadrilateral A than the perimeter of
quadrilateral B?
Module 17 • Lesson 2 555
TEXAS Test Prep
Lesson
Lesson Check
Check
Fill in the bubble completely to show your answer.
7.
Benjamin builds a fence in the
shape of a triangle. Each side of
the fence is the same length. If the
perimeter is 36 feet, how long is
each side of the fence?
6.
Anton puts a rail around his patio.
The patio is in the shape of a
rectangle with side lengths of 7 feet
and 9 feet. What is the perimeter of
Anton’s patio?
A
6 feet
A
16 feet
B
12 feet
B
63 feet
C
9 feet
C
22 feet
D
18 feet
D
32 feet
9 cm
C
3 cm
D
6 cm
Alexander
B
18 cm
FP
A
Alexander makes this name plate
from wood in art class. What is the
perimeter of the name plate?
556
Multi-Step Iris sews a border
around a blanket. The blanket has
side lengths that are 4 feet and
6 feet. The border material costs \$2
for each foot. How much does Iris
pay for the border?
8.
FP
9.
Multi-Step An artist paints two
pictures. Each picture has side
lengths of 2 feet and 4 feet. Framing
costs \$3 for each foot. How much
will the artist pay to put a frame
around both paintings?
A
\$20
A
\$18
B
\$12
B
\$72
C
\$40
C
\$36
D
\$16
D
\$24
© Houghton Mifflin Harcourt Publishing Company
5.
```
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Science Forums
# phillip1882
Members
419
15
1. ## What makes humans different from all the rest
i think there are two things that separate humans from the rest of the animals. one is the ability to trade an item for something completely different, and the other is to be on friendly terms with an animal before eating it.
2. ## golf payout problem
so my dad was playing with three other people in golf for 9 holes. The bet is each player puts in \$5 dollars. There are 2 games being played. \$3 dollars of the \$5 goes to who has the most points and \$2 of the \$5 goes to the trash pot. At he conclusion of the 9 holes my dad was the only winner of the points pot and Larry was the only winner of the trash pot. The points pot should have \$12 dollars and the trash pot should have \$8. However my dad didn't have correct change nor did Larry so the pot was left with \$10 dollars. Dad took \$6 of the \$10 and Larry took \$4. The next time they
3. ## fast prime test?
i don't really understand why this is a fast test.firstly, you would need a number of terms about equal to the prime value. that means a lot of multiplication second you would need to divide by all those terms, also expensive. i see no improvement to trail divide up to square root algorithm.
4. ## Want to learn basic maths
i can't think of anything more basic than khan academy. never the less here is my vain attempt numbers here i show all numbers between 1 and 9. if i want to add one more, i need one more symbol for the empty case, 0. when you add 1 to 9, the basket on the left, empty, gets a dot, or 1, and the basket with 9 gets to be 0. so 10 is equal to ten ones. so if i want to add two numbers, here's several examples. 17 +14 = so starting on the right, 7 +4 = 11, that is, 1 ten, and 1 one, the 1 ten gets carried over, and i have 1+1+1, = 30 so 3 for the
5. ## planet questions
matter is made up of atoms. i agree, not everything is made up of the same atoms. matter is not motion. yes it's in motion, but it's not motion itself. what does that even mean? nor did you specify why they are different. you don't understand what i'm asking. why do planets have moons? i agree things do tend to go to nothing, or space as you say, but as it does so, it generally doesn't leave behind matter that interacts with each other.
6. ## planet questions
okay that doesn't make sense at all. firstly the planets are of a wide variety of sizes. secondly most have moons. thirdly how would you get solid rock from hydrogen, what the sun is made of? fourthly how would the matter get distant enough to combine into it's own separate object?
7. ## planet questions
can you elaborate on that?
umm how?
9. ## planet questions
i'm not as curious as to where our water came from, i'm really curious as to where our magma core came from. our atmosphere too. rocks can't really carry air.
11. ## Speed of Light (again...)
in my understanding, only atoms have mass. light is made up of quarks, which are mass-less. light bends from a black hole because it is made up of particles. light should be constant even near a black hole. i'm no expert though so take this with a grain of salt.
12. ## The Crank Obsession With Einstein Being Wrong?
that diagram of objects orbiting is interesting, but don't forget it also has to be warped toward the earth not just around it.
13. ## I Just Wrote A Book Challenging Nihilism And The Theory Of Gravity
i don't have kindle but i would like to read your thoughts. can you put it someplace more accessible?
14. ## I Just Wrote A Book Challenging Nihilism And The Theory Of Gravity
I'm not convinced unlimited energy is impossible. for example: you can take an iron bar, and a iron bar magnet, rub the standard bar with the magnet bar, and get a second magnet though i must admit, i have no idea how to achieve it.
15. ## The Crank Obsession With Einstein Being Wrong?
my problem with warped space as gravity is it would have to be warped on every point on earth and in every dimension. what that would even look like, let alone how to allow orbits, i have no idea.
16. ## I Know Why Primes Are Primes
so on the pyramid prime, every prime has to be of the form 6n+1 or 6n+5. that's nothing new, not quite sure what you mean by complex primes. do you mean guass primes? for example 2 is (1+i)*(1-i); 5 is (2+i)*(2-i) again i don't know what you mean by complex primes. sort of, there's a distinct difference between rational numbers and irrational numbers. umm can you clarify that? do you mean Turing machines have state? The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. got by a simple google s
17. ## The Crank Obsession With Einstein Being Wrong?
mm, i can only speak for myself, but i more commonly say warp space doesn't make sense. i focus on Einstein when talking about that because he's the person generally associated with it.
18. ## Dot Division Method
http://thescienceexplorer.com/universe/use-quick-trick-speed-long-division can someone show me what 975 / 15 would look like?
19. ## Faster Than Light?
well for example if i played a trillion notes per second,my fingers would need to move at that speed. there must be some aspect of the camera that's moving at that speed.
20. ## Faster Than Light?
this claims a trillion frames per second. isn't that faster than light?
21. ## New Discovery? Prime Pattern
i don't understand. E = 6 5 11 between 5 and 16 7 13 between 7 and 20 11 17 between 11 and 28 13 19 between 13 and 32 23 29 between 23 and 52 E = 4 3 7 between 3 and 10 7 11 between 7 and 18 13 17 between 13 30 19 23 between 19 42 37 41 E = 2 (between 1 and 4) 3 5 between 3 and 8 5 7 between 5 12 11 13 (only the first number P needs to be in range) 11 24 17 19 between 17 36 29 31
22. ## New Discovery? Prime Pattern
Let E be an even number. between 1 and 2*E there will be a number such that it is prime, and it plus E is prime. lets call this number P. after that, between P and 2*P+E, there will be another number such that it and it plus E is prime. repeat indefinitely. for example, let E = 8. 3,11 the next between 3 and 14 5,13 the next between 5 and 18 11,19 the next between 11 and 30 23,31 the next between 23 and 54 29,37 the next between 29 and 66 for E = 14 3, 17 5, 19 17, 31 23, 37 29, 43
23. ## The Philosophy Of Money
hmm. i think a rich person can have a greater impact, but doesn't necessarily mean for good. as was once said by a comedian, Micheal Jordan is rich. the person who pays Micheal Jordan's salary is wealthy. if i had to choose between wealth and love though, i would choose love.
24. ## Warped Space-Time
i never fully understood the concept of warped space-time. i would like to issue a challenge to the scientific community. show me how warped space time would look like uniformly distributed around the earth, left to right, top to bottom, back to front, before and after. and then explain in layman's terms how you achieve a orbit with this.
25. ## Ai Learning Site
are there any sites that you provide the training data and it puts it through a learning algorithm for you?
×
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CC-MAIN-2021-25
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https://www.electricalvolt.com/coulomb-unit-symbol-definition/
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# Coulomb-Unit, Symbol, Definition
Coulomb is the SI unit of electric charge. One coulomb is equal to the amount of electric charge transferred by a current of one ampere in one second. Coulomb is the SI unit of electric charge.
## What is Coulomb?
The coulomb is defined as the quantity of electric charge transported in one second when the current flow is one ampere. Coulomb is named for French physicist Charles -Augustin de Coulomb. One Coulomb is approximately equivalent to 6.24 × 1018 electrons. The charge of one electron is equal to 1.602176634 × 10−19 C. Number of charges in one coulomb is;
One coulomb is the charge of approximately 6241509074460762607.776 elementary charges,
## Formula for Finding charges in Coulombs
The following formula shows the relationship between coulomb and ampere.
For example, if 3 amperes current flow in a circuit for 3 seconds, then the charges transported in the circuit are;
## Symbol of Coulomb?
Coulomb is denoted by the letter C. It is the SI unit of electric charge.
Mathematically, coulomb is;
1 Coulomb = 1 ampere X 1 second
## Other Units of Coulomb
We can get the other units of coulomb by adding a prefix a multiplies it by a power of 10.
## Definition of One Coulomb on the basis of coulomb’s law
Coulomb’s law states that the electrical force between two charged particles is directly proportional to the product of the charge quantity and inversely proportional to the square of the separation distance between the charged particles.
Now, to find the coulomb charge, we consider the followings.
Distance between the charges = 1 Meter
The force between charges = 9 X 109 N
Value of constant = 9 X 109
Q1=Q2 =Q ( Two equal charges)
Putting the above values in the coulomb’s law formula, we get
From the above expression, it is clear that the one coulomb is the quantity of charge that experiences a repulsive force F=9×109 N when placed at a distance of 1 meter in vacuum or air from an equal charge.
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You are here:
# Astronomy/Mars / earth oppostions and mars 686 day earth year
Question
I am trying to understand how to determine Mars' earth year of 686 days. through observation of 33 oppositions between mars and earth I have confirmed the average opposition time to be 780 days. From this I am having difficulty taking this information and determining an outcome of 686 days for Mars' year cycle once around the sun. thank for any help you can give me.
By a lucky chance, I just finished writing and posting a page about how the synodic period (which is the same as the average time between oppositions or any other specific aspect, such as Eastern or Western quadrature) can be determined from the orbital period of the planet. The equation involved can be worked in reverse to calculate the orbital period from the synodic period. (To refer to the web page in question, see http://cseligman.com/text/sky/synodicperiods.htm -- the page includes a link to a page about Planetary Aspects, where it mentions the fact that the average time from one aspect to the next recurrence of the same aspect is the same as the synodic period.)
For outer planets, the equation involved can be expressed as
1/S = 1/E - 1/P,
where E is the orbital period of the Earth, P is the orbital period of the other planet, and S is the synodic period. E is 365.256 days, S (as determined from your data) is 780 days (which is an exceptionally good average, as the exact value is 779.935 days), so moving the 1/P and 1/S to the opposite sides of the equation,
1/P = 1/E - 1/S, = 0.0027377 - 0.0012821 = 0.0014556
The reciprocal then yields almost exactly P = 687 days, which is very close to the correct orbital period for Mars, which is 686.98 days. So you are to be congratulated for obtaining an excellent result.
Astronomy
Volunteer
#### Courtney Seligman
##### Expertise
I can answer almost any question about astronomy and related sciences, such as physics and geology. I will not answer questions about astrology and similar pseudo-scientific rubbish.
##### Experience
I have been a professor of astronomy for over 40 years, and am working on an online text/encyclopedia of astronomy, and an online catalog of NGC/IC objects.
Publications
Astronomical Journal, Publications of the Astronomical Society of the Pacific (too long ago to be really relevant, but you could search for Courtney Seligman on Google Scholar)
Education/Credentials
I received a BA in astronomy and physics and a MA in astronomy, both from UCLA. I was working on my doctoral dissertation when I started teaching, and discovered that I preferred teaching to research.
Awards and Honors
(too long ago to be relevant, but Phi Beta Kappa and Sigma Xi still keep trying to get me to become a paying member)
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## Directed Acyclic Graphs | DAGs | Examples
Directed Acyclic Graph-
Directed Acyclic Graph (DAG) is a special kind of Abstract Syntax Tree.
• Each node of it contains a unique value.
• It does not contain any cycles in it, hence called Acyclic.
## Optimization Of Basic Blocks-
DAG is a very useful data structure for implementing transformations on Basic Blocks.
• A DAG is constructed for optimizing the basic block.
• A DAG is usually constructed using Three Address Code.
• Transformations such as dead code elimination and common sub expression elimination are then applied.
## Properties-
• Reachability relation forms a partial order in DAGs.
• Both transitive closure & transitive reduction are uniquely defined for DAGs.
• Topological Orderings are defined for DAGs.
## Applications-
DAGs are used for the following purposes-
• To determine the expressions which have been computed more than once (called common sub-expressions).
• To determine the names whose computation has been done outside the block but used inside the block.
• To determine the statements of the block whose computed value can be made available outside the block.
• To simplify the list of Quadruples by not executing the assignment instructions x:=y unless they are necessary and eliminating the common sub-expressions.
## Construction of DAGs-
Following rules are used for the construction of DAGs-
### Rule-01:
In a DAG,
• Interior nodes always represent the operators.
• Exterior nodes (leaf nodes) always represent the names, identifiers or constants.
### Rule-02:
While constructing a DAG,
• A check is made to find if there exists any node with the same value.
• A new node is created only when there does not exist any node with the same value.
• This action helps in detecting the common sub-expressions and avoiding the re-computation of the same.
### Rule-03:
The assignment instructions of the form x:=y are not performed unless they are necessary.
## Problem-01:
Consider the following expression and construct a DAG for it-
( a + b ) x ( a + b + c )
## Solution-
Three Address Code for the given expression is-
T1 = a + b
T2 = T1 + c
T3 = T1 x T2
Now, Directed Acyclic Graph is-
## NOTE
From the constructed DAG, we observe-
• The common sub-expression (a+b) has been expressed into a single node in the DAG.
• The computation is carried out only once and stored in the identifier T1 and reused later.
This illustrates how the construction scheme of a DAG identifies the common sub-expression and helps in eliminating its re-computation later.
## Problem-02:
Consider the following expression and construct a DAG for it-
( ( ( a + a ) + ( a + a ) ) + ( ( a + a ) + ( a + a ) ) )
## Solution-
Directed Acyclic Graph for the given expression is-
## Problem-03:
Consider the following block and construct a DAG for it-
(1) a = b x c
(2) d = b
(3) e = d x c
(4) b = e
(5) f = b + c
(6) g = f + d
## Solution-
Directed Acyclic Graph for the given block is-
## Problem-04:
Optimize the block in the Problem-03.
## Solution-
### Step-01:
Firstly, construct a DAG for the given block (already done above).
### Step-02:
Now, the optimized block can be generated by traversing the DAG.
• The common sub-expression e = d x c which is actually b x c (since d = b) is eliminated.
• The dead code b = e is eliminated.
The optimized block is-
(1) a = b x c
(2) d = b
(3) f = a + c
(4) g = f + d
## Problem-05:
Consider the following basic block-
B10:
S1 = 4 x I
S2 = addr(A) – 4
S3 = S2[S1]
S4 = 4 x I
S5 = addr(B) – 4
S6 = S5[S4]
S7 = S3 x S6
S8 = PROD + S7
PROD = S8
S9 = I + 1
I = S9
If I <= 20 goto L10
1. Draw a directed acyclic graph and identify local common sub-expressions.
2. After eliminating the common sub-expressions, re-write the basic block.
## Solution-
Directed Acyclic Graph for the given basic block is-
In this code fragment,
• 4 x I is a common sub-expression. Hence, we can eliminate because S1 = S4.
• We can optimize S8 = PROD + S7 and PROD = S8 as PROD = PROD + S7.
• We can optimize S9 = I + 1 and I = S9 as I = I + 1.
After eliminating S4, S8 and S9, we get the following basic block-
B10:
S1 = 4 x I
S2 = addr(A) – 4
S3 = S2[S1]
S5 = addr(B) – 4
S6 = S5[S1]
S7 = S3 x S6
PROD = PROD + S7
I = I + 1
If I <= 20 goto L10
To gain better understanding about Directed Acyclic Graphs,
Watch this Video Lecture
Next Article-Misc Problems On Directed Acyclic Graphs
Get more notes and other study material of Compiler Design.
Watch video lectures by visiting our YouTube channel LearnVidFun.
## Basic Blocks-
Basic block is a set of statements that always executes in a sequence one after the other.
The characteristics of basic blocks are-
• They do not contain any kind of jump statements in them.
• There is no possibility of branching or getting halt in the middle.
• All the statements execute in the same order they appear.
• They do not lose lose the flow control of the program.
### Example Of Basic Block-
Three Address Code for the expression a = b + c + d is-
Here,
• All the statements execute in a sequence one after the other.
• Thus, they form a basic block.
### Example Of Not A Basic Block-
Three Address Code for the expression If A<B then 1 else 0 is-
Here,
• The statements do not execute in a sequence one after the other.
• Thus, they do not form a basic block.
## Partitioning Intermediate Code Into Basic Blocks-
Any given code can be partitioned into basic blocks using the following rules-
### Rule-01: Determining Leaders-
Following statements of the code are called as Leaders
• First statement of the code.
• Statement that is a target of the conditional or unconditional goto statement.
• Statement that appears immediately after a goto statement.
### Rule-02: Determining Basic Blocks-
• All the statements that follow the leader (including the leader) till the next leader appears form one basic block.
• The first statement of the code is called as the first leader.
• The block containing the first leader is called as Initial block.
## Flow Graphs-
A flow graph is a directed graph with flow control information added to the basic blocks.
• The basic blocks serve as nodes of the flow graph.
• There is a directed edge from block B1 to block B2 if B2 appears immediately after B1 in the code.
## Problem-01:
Compute the basic blocks for the given three address statements-
(1) PROD = 0
(2) I = 1
(3) T2 = addr(A) – 4
(4) T4 = addr(B) – 4
(5) T1 = 4 x I
(6) T3 = T2[T1]
(7) T5 = T4[T1]
(8) T6 = T3 x T5
(9) PROD = PROD + T6
(10) I = I + 1
(11) IF I <=20 GOTO (5)
### Solution-
We have-
• PROD = 0 is a leader since first statement of the code is a leader.
• T1 = 4 x I is a leader since target of the conditional goto statement is a leader.
Now, the given code can be partitioned into two basic blocks as-
## Problem-02:
Draw a flow graph for the three address statements given in problem-01.
## Solution-
• Firstly, we compute the basic blocks (already done above).
• Secondly, we assign the flow control information.
The required flow graph is-
To gain better understanding about Basic Blocks and Flow Graphs,
Watch this Video Lecture
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# Search by Topic
#### Resources tagged with Binomial Theorem similar to Giants:
Filter by: Content type:
Age range:
Challenge level:
### There are 11 results
Broad Topics > Algebraic expressions, equations and formulae > Binomial Theorem
### Remainder Hunt
##### Age 16 to 18 Challenge Level:
What are the possible remainders when the 100-th power of an integer is divided by 125?
### Elevens
##### Age 16 to 18 Challenge Level:
Add powers of 3 and powers of 7 and get multiples of 11.
### Telescoping Series
##### Age 16 to 18 Challenge Level:
Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.
### Growing
##### Age 16 to 18 Challenge Level:
Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)
### Tens
##### Age 16 to 18 Challenge Level:
When is $7^n + 3^n$ a multiple of 10? Can you prove the result by two different methods?
### Discrete Trends
##### Age 16 to 18 Challenge Level:
Find the maximum value of n to the power 1/n and prove that it is a maximum.
### Binomial
##### Age 16 to 18 Challenge Level:
By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn
### Bina-ring
##### Age 16 to 18 Challenge Level:
Investigate powers of numbers of the form (1 + sqrt 2).
### Summit
##### Age 16 to 18 Challenge Level:
Prove that the sum from t=0 to m of (-1)^t/t!(m-t)! is zero.
### The Harmonic Triangle and Pascal's Triangle
##### Age 16 to 18
The harmonic triangle is built from fractions with unit numerators using a rule very similar to Pascal's triangle.
### The Kth Sum of N Numbers
##### Age 16 to 18
Yatir from Israel describes his method for summing a series of triangle numbers.
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https://www.physicsforums.com/threads/force-exerted-by-m1-on-m2-problem.788286/
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# Force exerted by M1 on M2 problem
• JSmithDawg
In summary, the conversation discusses the forces exerted by two masses, M1 and M2, on each other. According to Newton's 3rd law, the forces come in pairs and are equal in magnitude but opposite in direction. To solve the problem, it is suggested to consider the two masses as a system and use F=ma to find the acceleration. Then, separate free body diagrams for M1 and M2 can be drawn to determine the magnitude of the force between the two masses.
## Homework Statement
The illustration is not drawn to scale. M2 = 4 kg and M1=20kg. They're both on a frictionless surface. A 36N constant force is applied to M2. What's the force exerted by m1 on m2?
F=ma
## The Attempt at a Solution
According to Newton's 3rd law, forces come in pairs. If there's 36N of force exerted on M2, the block should move and transfer the force onto M1. Since the surface is frictionless, I don't have to worry about any force being lost due to friction. Thus, 36N of force should be exerted onto M2 by M1. However, I know my logic is wrong, since the online quiz told me that the correct answer is 30N. How do I correctly approach this problem?
JSmithDawg said:
According to Newton's 3rd law, forces come in pairs.
Yes, but as action and reaction. You can use that to determine the relationship between the forces the two masses exert on each other.
Consider the acceleration of the system. What force therefore acts on m1?
Since M1 and M2 move together, if a force of 36 N is applied to them, what is their acceleration?
Chet
Hi JSmithDawg.:)
Consider m1 and m2 as a system and find the acceleration of the system by using F=ma ,as Chestermiller said. Then draw free body diagrams of m1 and m2 separately. Can you find the magnitude of force acting on m1 due to m2? What is the relation between the acting on m1 by m2 and force acting on m2 by m1?
Your logic is partially correct. According to Newton's 3rd law, the force exerted by M1 on M2 will be equal in magnitude but opposite in direction to the force exerted by M2 on M1. This means that the force exerted by M1 on M2 will also be 36N, but in the opposite direction.
To understand why the correct answer is 30N, we need to consider the forces acting on each block separately. Since M2 is being pushed with a constant force of 36N, it will have an acceleration of 9 m/s^2 (F=ma, 36N=4kg*9m/s^2). This means that M2 will have a net force of 36N acting on it in the direction of the applied force.
On the other hand, M1 is not being pushed or pulled by any external forces, so its net force is 0N. However, it is in contact with M2 and thus experiences a reaction force from M2 of 36N (equal in magnitude but opposite in direction). This means that M1 has a force of 36N acting on it in the direction opposite to the applied force on M2.
Now, since M1 has a mass of 20kg, its acceleration will be 1.8 m/s^2 (F=ma, 36N=20kg*1.8m/s^2). This means that M1 will be pushing on M2 with a force of 36N in the opposite direction of the applied force, but due to its lower acceleration, this force will have a magnitude of 1.8 m/s^2. Therefore, the force exerted by M1 on M2 is 30N (36N-1.8N).
In summary, the force exerted by M1 on M2 is 30N because M1 is pushing on M2 with a force of 36N, but due to its lower acceleration, the force has a magnitude of 1.8N less than the applied force on M2.
## 1. What is the formula for calculating the force exerted by M1 on M2 in this problem?
The formula for calculating the force exerted by M1 on M2 in this problem is F = G * (M1 * M2) / r^2, where G is the gravitational constant, M1 and M2 are the masses of the objects, and r is the distance between them.
## 2. How do I determine the direction of the force in this problem?
The force exerted by M1 on M2 will always be in the direction of the line connecting the two objects. This means that if M1 and M2 are both positive masses, the force will be attractive (towards each other), while if one of the masses is negative, the force will be repulsive (away from each other).
## 3. Can I use this formula for calculating the force between any two objects?
No, this formula specifically applies to the force exerted by two point masses on each other due to gravity. It cannot be used for other types of forces or for objects with non-negligible sizes.
## 4. How does the force exerted by M1 on M2 change if the distance between them is doubled?
If the distance between M1 and M2 is doubled, the force exerted by M1 on M2 will decrease by a factor of 4. This is because the force is inversely proportional to the square of the distance between the objects (F ∝ 1/r^2).
## 5. Is the force exerted by M1 on M2 affected by the masses of other nearby objects?
No, the force exerted by M1 on M2 is only affected by the masses of M1 and M2 themselves. The presence of other objects nearby will not have an impact on this force.
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https://www.jiskha.com/questions/580502/3L-to-the-power-0-5-Over-L-to-the-power-2-all-in-brackets-to-the-power-2-outside
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# statistics-varsity level
(3L to the power 0.5. Over L to the power 2)all in brackets to the power 2 outside brackets?
1. (((3L).5)/22 )2
square if first.
(3L)^1 / L^4
3L^(1-4)
3/L^3
posted by bobpursley
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Express each power as an equivalent radical. K. I don't know how to write this, but I'll explain. There is a fraction. 1 over 9. And 9 has a power up to it which is 5 and a power down which is 3. I don't get how to express this
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Thursday
May 5, 2016
# Homework Help: Math
Posted by mary on Sunday, March 18, 2012 at 7:25pm.
Consider a game where Ann and Bob play against each other in a dice game, Ann wins if her score is higher than Bob's, Bob wins if his score is higher or if the scores are the same. What is the probability of Ann winning? I calculated and got 1 - summation from 1 to 6 of n over 36 which gave me 15/36.
Now if Ann can roll a second time what is the probability that she will win?I calculated and got 1 - summation from 1 to 6 of n^2/216 which got me 125/216
Finally if Ann can roll twice and Bob rolls twice what is the probability that Ann will win? Again I calculated and got 1 - summation from 1 to 6 of (n^2)(2n-1) over 1296 which got me the answer 505/1296. Now my question how do I calculate the probability in further cases where Ann and Bob roll n times, by finding a general formula?
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https://stats.stackexchange.com/questions/282501/prove-that-ols-residuals-have-0-expectation-even-when-the-linear-model-is-false
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# Prove that OLS residuals have 0 expectation even when the linear model is false
For the sake of this question, I will consider a linear model as a statistical model such that $\mathbb{E}[y|\mathbf{x}]=\mathbf{x}\boldsymbol{\beta}$. I know that usually some more assumptions go in the definition of the linear model. However, the other properties are mostly needed for OLS estimation of the linear model, and all of them can be relaxed (see the excellent post by Glen_b here).
In my answer to the question
Estimating error - residuals vs. fitted values plot
I stated that, even when $\mathbb{E}[y|\mathbf{x}]\neq\mathbf{x}\boldsymbol{\beta}$, the OLS residuals have expectation 0 because of the Law of Large Numbers. However, this might be wrong: if we have a random sample $S=\{(y_i,\mathbf{x}_i)\}_{i=1}^n$, then we know that $$\sum_{i=1}^ne_i=\sum_{i=1}^n(y_i-\hat{\boldsymbol{\beta}}\mathbf{x}_i)=0 \tag{1}\label{1}$$
whatever the sample size (in other words, residuals are not independent). Now, Let $e$ denote the random variable whose realizations are the $e_i$. The Law of Large Numbers would tell you that
$$\frac{\sum_{i=1}^ne_i}{n}=0 \ \ \forall n\rightarrow\mathbb{E}[e]=0$$
if the residuals were independent, but they aren't. However, property $\ref{1}$ is true because of the definitions of the OLS estimator, thus the sample mean of the residuals is always zero, even when the linear model is "false", i.e., when $\mathbb{E}[y|\mathbf{x}]\neq\mathbf{x}\boldsymbol{\beta}$. My question is: is $\mathbb{E}[e]=0$ true, even when $\mathbb{E}[y|\mathbf{x}]\neq\mathbf{x}\boldsymbol{\beta}$? Simulations seem to indicate that this is indeed the case.
I tried to prove it rigorously by linear algebra and rules of expectation, but I got stuck pretty soon:
$$e=y-(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^Ty\Rightarrow \mathbb{E}[e|X]=\mathbb{E}[y|X]-(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbb{E}[y|X]$$
If I knew that $\mathbb{E}[y|\mathbf{x}]=\mathbf{x}\boldsymbol{\beta}$, then from the above I would immediately get $\mathbb{E}[e|X]=0$, which implies $\mathbb{E}[e]=0$ by the law of iterated expectations. But I can't assume that. Suggestions?
PS yes, I know that using lower-case for both random variables and their realizations is horrible, but if I use upper-case for the RV $\mathbf{X}$, then I don't know what symbol I to use for the design matrix.
• You define the $e_i$ as the residuals with respect to the OLS solution. Obviously their sum is always zero: that's a mathematical property of the solution. But if the model is not what you suppose, then isn't it equally obvious that the expectation of any individual $e_i$ could be any number whatsoever? For instance, let $x_1$ be the mean of the regressors of all the other data points. Suppose the model is correct for all the other $x_i,$ $i\ne 1$, but $E(y_1)=x_1\beta+A$ for some constant $A\ne 0$. You can easily compute that this makes $E(e_1)=A/n\ne 0$. What, then, are you trying to ask? – whuber May 30 '17 at 13:54
• @whuber not sure what $x_1$ is. Is it the mean of the regressors vector? i.e., $x_1=\mathbb{E}[\mathbf{x}]=(\mu_{x_1},\dots,\mu_p)$ where $p$ is the number of regressors. – DeltaIV May 30 '17 at 15:05
• No, it's the mean of all the other regressors: they aren't (or don't have to be) considered as stochastic. Why not do a calculation with a very simple example? Try the model $(0,\beta_0+\epsilon_1),(-1,\beta_0-\beta_1+\epsilon_2),(1,\beta_0+\beta_1+\epsilon_3)$ fit via OLS, for instance. Compare two situations, both with independent errors $\epsilon_i$ and homogeneous variances. In the usual situation they all have zero means. In the "wrong model" alternative, suppose $\epsilon_1$ has a mean of $A$ but you are still computing the OLS estimator. It's easy to figure out $\mathbb{E}(e_1)$. – whuber May 30 '17 at 16:04
• @whuber ok, I think I understood your setting. I will try the exercise and let you know. Thanks! – DeltaIV May 30 '17 at 18:00
• I fixed a typographical error in my previous comment, in order to make all three of the $\epsilon_i$ distinct. It's intended to be the usual OLS setting with three data points, of which the first has its x-value equal to the mean of the other two x-values. By making that common mean zero, the matrix $X^\prime X=\pmatrix{3&0\\0&2}$ is easily inverted, making all your calculations simple. – whuber May 30 '17 at 18:08
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## Class 9 (Assamese)
### Course: Class 9 (Assamese)>Unit 6
Lesson 1: Pairs of angles
# Complementary & supplementary angles
Learn about complementary and supplementary angles, as well as the definitions of adjacent and straight angles. Created by Sal Khan.
## Want to join the conversation?
• Do complementary and supplementary angles have to be adjacent?
• They don't have to be adjacent but if they are they form a straight line on the sides that they don't share.
• Does anyone have any hacks for remembering this I have a test on Friday
• Try this:
Complementary angles add up to 90°
- example: 15° & 75° are complementary
(added together, they form a right angle)
-and-
Supplementary angles add up to 180°
- example: 50° & 130° are supplementary
(added together, they form a straight line)
Two facts:
(1) 90° comes before 180° on the number line
(2) "C" comes before "S" in the alphabet
90° goes with "C" for complementary
so complementary angles add up to 90°
180° goes with "S" for supplementary
so supplementary angles add up to 180°
Hope this helps!
• Definition of linear pair?
• Two angles are linear if they are adjacent angles formed by two intersecting lines. A straight angle is 180 degrees so 2 linear pairs of angles must always add up to 180 degrees.
Hope that this is helpful to u
• can you tell me a trick for remembering suplementory and complimentory angles.
• The easiest way to remember the difference between supplementary and complementary angles is to use the starting letters of the words themselves. Think of the "C" as a corner, which reminds you of a right angle, which is 90 degrees. Think of the "S" as straight, which reminds you of a straight angle, which is 180 degrees.
Hope this helps :)
• So how do you know if its complementary
• If two angles add up to 90 degrees then they are complementary, if they add up to 180 degrees they they are supplementary
• Do we have names for angles that add up to 270 and 360 degrees?
• There are currently no name for 2 angles that add up to 270 degrees but Explementary Angles are the name for a pair of angles that adds up to 360 degrees, even if an angle is more than 180 degrees
• Is there a video about the angle pairs, specifically? Because I don't remember seeing anything about it.
• I think this is the only one. A pair is a group of two. So a pair of angles = two angles.
• How do I determine if it's complementary or supplementary?
• complementary angles add up to 90 degrees. a supplementary angle adds up to 180 degrees.
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# Search by Topic
#### Resources tagged with Resourceful similar to Factor Track:
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Broad Topics > Habits of Mind > Resourceful
### Gabriel's Problem
##### Age 11 to 14 Challenge Level:
Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
### Dozens
##### Age 7 to 14 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Funny Factorisation
##### Age 11 to 14 Challenge Level:
Using the digits 1 to 9, the number 4396 can be written as the product of two numbers. Can you find the factors?
### Shifting Times Tables
##### Age 11 to 14 Challenge Level:
Can you find a way to identify times tables after they have been shifted up or down?
### Special Numbers
##### Age 11 to 14 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### The Remainders Game
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Play this game and see if you can figure out the computer's chosen number.
### Missing Multipliers
##### Age 7 to 14 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Cuboids
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Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
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The clues for this Sudoku are the product of the numbers in adjacent squares.
### A Chance to Win?
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Imagine you were given the chance to win some money... and imagine you had nothing to lose...
### Stars
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Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
### Cryptarithms
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Can you crack these cryptarithms?
### On the Edge
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If you move the tiles around, can you make squares with different coloured edges?
### Reversals
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Where should you start, if you want to finish back where you started?
### Isosceles Triangles
##### Age 11 to 14 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
##### Age 11 to 14 Challenge Level:
A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .
### Wipeout
##### Age 11 to 14 Challenge Level:
Can you do a little mathematical detective work to figure out which number has been wiped out?
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Triangles to Tetrahedra
##### Age 11 to 14 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### Marbles in a Box
##### Age 11 to 16 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### Square It
##### Age 11 to 16 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Dicey Operations
##### Age 11 to 14 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
##### Age 11 to 14 Challenge Level:
What happens when you add a three digit number to its reverse?
### Tower of Hanoi
##### Age 11 to 14 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### How Much Can We Spend?
##### Age 11 to 14 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Two and Two
##### Age 11 to 14 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Where Can We Visit?
##### Age 11 to 14 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Charlie's Delightful Machine
##### Age 11 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Cosy Corner
##### Age 11 to 14 Challenge Level:
Six balls of various colours are randomly shaken into a trianglular arrangement. What is the probability of having at least one red in the corner?
### Two's Company
##### Age 11 to 14 Challenge Level:
7 balls are shaken in a container. You win if the two blue balls touch. What is the probability of winning?
### Odds and Evens
##### Age 11 to 14 Challenge Level:
Is this a fair game? How many ways are there of creating a fair game by adding odd and even numbers?
### Sociable Cards
##### Age 11 to 14 Challenge Level:
Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
### Nice or Nasty
##### Age 7 to 14 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Unequal Averages
##### Age 11 to 14 Challenge Level:
Play around with sets of five numbers and see what you can discover about different types of average...
### M, M and M
##### Age 11 to 14 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Which Solids Can We Make?
##### Age 11 to 14 Challenge Level:
Interior angles can help us to work out which polygons will tessellate. Can we use similar ideas to predict which polygons combine to create semi-regular solids?
### Shapely Pairs
##### Age 11 to 14 Challenge Level:
A game in which players take it in turns to turn up two cards. If they can draw a triangle which satisfies both properties they win the pair of cards. And a few challenging questions to follow...
### Semi-regular Tessellations
##### Age 11 to 16 Challenge Level:
Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?
### Constructing Triangles
##### Age 11 to 14 Challenge Level:
Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
### Treasure Hunt
##### Age 7 to 14 Challenge Level:
Can you find a reliable strategy for choosing coordinates that will locate the treasure in the minimum number of guesses?
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### Property Chart
##### Age 11 to 14 Challenge Level:
A game in which players take it in turns to try to draw quadrilaterals (or triangles) with particular properties. Is it possible to fill the game grid?
### Substitution Cipher
##### Age 11 to 14 Challenge Level:
Find the frequency distribution for ordinary English, and use it to help you crack the code.
### What Does Random Look Like?
##### Age 11 to 14 Challenge Level:
Engage in a little mathematical detective work to see if you can spot the fakes.
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×
Get Full Access to Probability And Statistical Inference - 9 Edition - Chapter 7.6 - Problem 11e
Get Full Access to Probability And Statistical Inference - 9 Edition - Chapter 7.6 - Problem 11e
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# (The information presented in this exercise comes from the
ISBN: 9780321923271 41
## Solution for problem 11E Chapter 7.6
Probability and Statistical Inference | 9th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
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4
Problem 11E
PROBLEM 11E
(The information presented in this exercise comes from the Westview Blueberry Farm and National Oceanic and Atmospheric Administration Reports [NOAA].) For the following paired data, (x, y), x gives the Holland, Michigan, rainfall for June, and y gives the blueberry production in thousands of pounds from the Westview Blueberry Farm:
The data are from 1971 to 1989 for those years in which the last frost occurred May 10 or earlier.
(a) Find the correlation coefficient for these data.
(b) Find the least squares regression line.
(c) Make a scatter plot of the data with the least squares regression line on the plot.
(d) Calculate and plot the residuals. Does linear regression seem to be appropriate?
(e) Find the least squares quadratic regression curve.
(f) Calculate and plot the residuals. Does quadratic regression seem to be appropriate?
(g) Give a short interpretation of your results.
Step-by-Step Solution:
Step 1 of 3
Demographic Data 1. Population data collection 2.Coverage error 3. Content error 4. Defacto population 5. De Jure population 6. Published info. on population data. 1. Population Data Collection: ● How many people are within a specific country’s borders ● How are they distributed geographically ● How many people are being born ● How many people are dying ● How many people are getting married ● Internal and external migrants ● At town/city level, census taking done every five years. ● At National level, census taking done every 10 years. Decennial census taking. ● Formal census taking began in 1790. ● More progressive and inclusive. ● Census questionnaires administered to permanent households/ residential homes/ apartments. Nursing homes 2. Coverage e
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321923271
This textbook survival guide was created for the textbook: Probability and Statistical Inference , edition: 9. The answer to “(The information presented in this exercise comes from the Westview Blueberry Farm and National Oceanic and Atmospheric Administration Reports [NOAA].) For the following paired data, (x, y), x gives the Holland, Michigan, rainfall for June, and y gives the blueberry production in thousands of pounds from the Westview Blueberry Farm: The data are from 1971 to 1989 for those years in which the last frost occurred May 10 or earlier.(a) Find the correlation coefficient for these data.(b) Find the least squares regression line.(c) Make a scatter plot of the data with the least squares regression line on the plot.(d) Calculate and plot the residuals. Does linear regression seem to be appropriate?(e) Find the least squares quadratic regression curve.(f) Calculate and plot the residuals. Does quadratic regression seem to be appropriate?(g) Give a short interpretation of your results.” is broken down into a number of easy to follow steps, and 137 words. This full solution covers the following key subjects: regression, plot, data, squares, least. This expansive textbook survival guide covers 59 chapters, and 1476 solutions. Probability and Statistical Inference was written by and is associated to the ISBN: 9780321923271. Since the solution to 11E from 7.6 chapter was answered, more than 306 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 11E from chapter: 7.6 was answered by , our top Statistics solution expert on 07/05/17, 04:50AM.
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# Thread: investment question
1. ## investment question
Tammy has joined a 401K plan with her company. Her goal is a total value of $2,000,000 when she retires in 45 years. With an expectant growth rate of 10% and being paid bi-weekly (26 times a year). What amount should she have deducted from her paycheck in order to meet her goal? 2. Hello, Linda! Tammy has joined a 401K plan with her company. Her goal is a total value of$2,000,000 when she retires in 45 years.
With an expectant growth rate of 10% and being paid bi-weekly (26 times a year),
what amount should she have deducted from her paycheck in order to meet her goal?
This is an Annuity . . .
. . I will assume that the interest is computed bi-weekly.
The formula is: . $D \;=\;A\cdot\frac{i}{(1+i)^n-1}$
. . where: . $\begin{Bmatrix}D &=&\text{periodic deposit} \\ A &=& \text{final amount} \\ i &=& \text{periodic interest rate} \\ n &=& \text{no. of periods} \end{Bmatrix}$
We have: . $\begin{array}{ccc}A &=& 2,000,000 \\ i &=& \frac{0.1}{26} \\ n &=& 1170 \end{array}$
Hence: . $D \;=\;2,000,000\cdot\frac{\frac{0.1}{26}}{\left(1 + \frac{0.1}{26}\right)^{1170} - 1} \;=\;87.17140754$
Therefore, Tammy should authorize an $87.17 bi-weekly deduction. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Actually, she should authorize an$87.18 deduction.
Rounding down, she will be about \$32 short of her two million dollars.
. . (But being a bi-millionaire at the time, she probably won't care.)
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http://www.afunzone.com/May27TeaserSolution.html
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May 27th Brain Teaser Solution
Brain Teasers: A brain teaser is a form of puzzle that requires thought to solve. It often requires thinking in unconventional ways with given constraints in mind; sometimes it also involves lateral thinking. Logic puzzles and riddles are specific types of brain teasers. Today's can be found below....
The Teaser
1. Tom is younger than Rose, but older than Will and Jack, in that order. Rose is younger than Susie, but older than Jack. Jack is younger than Jim. Susie is older than Rose, but younger than Jim. Jim is older than Tom. Who is the oldest?
2. Allie takes fruit, cake, and cookies for her picnic. She has three boxes for them. One is labeled FRUIT. One is labeled COOKIES. One is labeled CAKE. But she knows her Mom likes to fool her and has put every single thing in the wrong box. The only other thing she knows for sure is that the fruit is not in the CAKE box. Where is the cake?
3. A block of wood in the form of a cube 3" x 10" x 13" has all its six faces painted pink. If the wooden block is cut into 390 cubes of 1" x 1" x 1", how many of these would have pink paint on them?
The Solution
1. Jim
2. The cake is not in the CAKE BOX, and it cannot be in the COOKIE BOX, because that is the only place for the fruit; so it must be in the box labeled FRUIT.
3. The 1 inch x 1 inch x 1 inch cubes that do not have any pink paint on them will be at the core of the wooden block. This core will be 1 inch x 8 inch x 11 inch, and will contain 88 cubes. Out of a total of 390 cubes, there are 88 cubes without any pink paint on them. Therefore, the remaining 302 cubes will have one, two or three sides with pink paint on them [depending on whether they were at the face, edge or corner of the wooden block].
Today's Puzzles:
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# Corner Detection
## Changes from 2124387a52aca1e909aca1e26a6fcb9d6c143099 to ab5d487401cb98a8a74dbe4ec69d852ad416a0b7
---
title: Corner Detection
categories: session
---
**Reading** Ma 4.3 and 4.A and Szelisky Chapter 3.2
**Warning** Ma starts Chapter 4 by discussing *tracking*,
which means that motion as a function of time is considered as well
as the image as a function of spatial co-ordinates.
This is a lot of concepts and quantities to process at the same time.
We will instead start by discussing features in a still image.
When we have a good idea of what features are and how they behave,
we shall introduce motion.
**Briefing** [Corner Lecture]()
# Exercises
## Learning Objectives
1. What makes a feature in visual terms?
2. What makes a feature in mathematical terms?
3. How do we differentiate a sampled signal?
4. How does the Harris corner detector work?
We use the same setup as in [Image Filters]().
## Exercise 1
Learning goal: 1D derivatives and 1D convolutions
**Part 1**<br>
Extract one row from the grayscale image, and visualize it (e.g. with matplotlib.pyplot).<br>
Does the values correspond to what you would expect from the row?
<details>
<summary>Hint (Click to expand)</summary>
</br>
python
# Load image im as usual and convert to greyscale.
row = im[50,:] # Row number 50
n = row.size
import matplotlib.pyplot as plt
plt.plot(range(n),row,color="blue",label="Pixel Row")
plt.show()
</details>
**Part 2**<br>
Convolve the row with a $[1/2,-1/2]$ kernel, using numpy, and visualize.<br>
Does the values make sense?
**Part 2**
Convolve the row with a $[1/2,-1/2]$ kernel and visualize.<br>
Do the values make sense?
<details>
<summary>Hint 1 (Click to expand)</summary>
<summary>Manual with Numpy (Click to expand)</summary>
</br>
Create a new 1D array with np.zeros(<shape>) and iterate over the row and the kernel.
To implement convolution manually using a loop over a numpy array, you can
create a new 1D array with np.zeros(<shape>) and fill it in by
iterating over the row and the kernel.
Remember that the resulting array should be smaller than the original.
</details>
<details>
<summary>Hint 2 (Click to expand)</summary>
<summary>Using the API (Click to expand)</summary>
</br>
If you are not able to get a result using numpy, use scipy.signal (can also be used to compare your result):
If you are not able to get a result using numpy, use scipy.signal
(can also be used to compare your result):
python
from scipy import signal
row_d = signal.convolve(row, kernel)
# or
# row_d = signal.correlate(row, cv.flip(kernel, -1))
</details>
<details>
<summary>Hint 3 (Click to expand)</summary>
<summary>Cross-correlation (Click to expand)</summary>
</br>
If you use cross-correlation instead of convolution, flip the kernel.
</details>
## Exercise 2
Learning goals: 2D Derivatives
**Part 1**<br>
Apply the sobel operator
$$G_x = 1/8 \begin{bmatrix} 1 & 0 & -1 \\ 2 & 0 & -2 \\ 1 & 0 & -1 \\ \end{bmatrix}$$
to the entire grayscale image
using scipy.signal.convolve2d, scipy.signal.correlate2d or cv.filter2d.
Either show the image or write it to file with cv.imwrite.
This should give you the derivative $I_x$ of the image $I$ with
respect to $x$.
<details>
<summary>Hint</summary>
</br>
1. Look at the exercises from last Friday. We apply the Sobel filter just like a blurring
filter.
2. You can define the matrix $G_x$ as given above using the numpy.array() function.
</details>
+ How does this relate to the 1D derivative you did in exercise 1.2?
+ What are the minimum and maximum values of the $I_x$ matrix?
**Part 2**<br>
Show $I_x$ as an image.
You probably have negative pixel values, so you may have to
scale the image.
+ Try to take the absolute values of the luminence values.
+ Try to scale the luminences into the $0\ldots255$ range,
e.g. by adding $255$ and dividing by two.
+ What does the different visualisations tell you?
+ You may scale further to use the full $0\ldots255$ range and thus
increas contrast.
**Part 3**<br>
Repeat Parts 1 and 2 with vertical derivation, i.e. use $G_y$ instead
of $G_x$.
$$G_y = 1/8 \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ -1 & -2 & -1 \\ \end{bmatrix}$$
+ Compare the images. What differences can you make out?
## Exercise 3
Learning Goal: find rotation invariant heuristics for edges
In Exercise 2, we calculated $I_x$ and $I_y$ which give a lot of
edge information. Now we want to aggregate this information over
a Window.
Note that $I_x$ and $I_y$ are matrices with the same dimensions as
the original image. The index of an entry in these matrices will
be denoted $\mathbf{x}$ below, and we are going to make more
*pseudo-images* with the same dimensions.
### 3.1
For every point $\mathbf{x}$ we calculate the matrix
$$G(\mathbf{x}) = \begin{bmatrix} \sum I_x^2 & \sum I_xI_y \\ \sum I_xI_y & \sum I_y^2 \end{bmatrix},$$
where the summations are made over a window, say a $5\times5$ window,
around $\mathbf{x}$.
Note that this is not a *pseudo-image*. For each $\mathbf{x}$ we
have $2\times2$ matrix and not just a scalar.
### 3.2
For each pixel position $\mathbf{x}$ calculate the eigenvectors/-values
of $G(\mathbf{x})$.
We know that large eigenvalues indicate features, and we want
to visualise this information.
Several variants are possible, and you may need only one or two
to get the picture.
You can make matrices containing, for each $\mathbf{x}$
+ The maximum of the eigenvalues of $G(\mathbf{x})$
+ The minimum of the eigenvalues of $G(\mathbf{x})$
+ The sum of the eigenvalues of $G(\mathbf{x})$
+ The product of the eigenvalues of $G(\mathbf{x})$
1. Scale these matrices so that they can be interpreted as
grey scale images, and visualise them.
2. Compare these visulisations to the edge plots from Exercise 2.
3. What do you see?
## Exercise 4
Learning goals: Introduction to Harris corner detector
If you have completed Exercise 3, you have done 90% of the
implementation of the Harris detector.
The missing part is the heuristics
$$C(G)=\lambda_{1}\lambda_{2} - k\cdot (\lambda _{1}+\lambda _{2})^{2} =\det(G)-k\cdot \mathrm {tr} (G)^{2}$$
and the threshold used.
You can choose if you want to complete the implementation
of your very own Harris detector or use OpenCV's implementation.
**Part 1**<br>
Consider the grayscale image we have been working with, and the gradient magnitude from 3.3.<br>
Where do you expect to find corners?
Apply opencv's built in harris-detector.<br>
E.g. cv.cornerHarris(img_gray, block_size, kernel_size, k) with block_size = 2, kernel_size = 5 and k = 0.06.<br>
Here, block_size is the size of neighbourhood considered for corner detection, kernel_size is the size of the sobel derivative kernel, while k is the harris free parameter.
Make a copy of the original image (with colors) and make circles around any corners found by the harris-detector.<br>
Example code for drawing circles is added below.
<details>
<summary>Code</summary>
</br>
python
cx = cv.cornerHarris(img_gray, bsize, ksize, k)
T = 0.1 # Threshold
c_image = img
for i in range(c_image.shape[0]):
for j in range(c_image.shape[1]):
if c_x[i, j] > T:
cv.circle(c_image, (j, i), 2, (0, 0, 255), 2)
</details>
Save/visualize the result, how does it compare with your expectations?
**Part 2**<br>
Adjust the threshold T when drawing circles, what does this do?
**Part 3**<br>
Adjust the kernel_size (must be positive and odd),
block_size and/or $k$, and observe how they change the result.
## Optional Exercises
### Laplacian of Gaussian
Test different variations of LoG filters on your test images.
You can construct Gaussians as we did in the last session on
[Image Filters]() and the Laplacian as suggested in the
[Corner Lecture]() today.
How does LoG compare to the techniques you ghave tested above?
### Building on Exercise 1
Repeat part 2 with an image column (instead of row) and the transpose of the kernel.
### Building on Exercise 2
Using the same method as in exercise 1.2 and 1.3, compute the gradient of all
rows and columns, and compute the magnitude (as in exercise 3.3), compare with the
magnitude from 3.3.
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Sei sulla pagina 1di 3
TD5 Statistical Physics (M1)
Sublimation of a Solid and Surface Adsorption
Exercise 1: Sublimation.
In this exercise, we will study the sublimation of a crystalline solid, with a simple
microscopic model.
In a box with volume V, and temperature T, coexist a crystalline solid with its
monoatomic gas. The mass of the atoms is m. They have no spin. The volume of
the solid is negligible in comparison with the volume occupied by the gas. The
vapor is considered as a classical perfect gas.
The total number of atoms is N, with Ng atoms in the gas phase, and Ns=N-Ng
atoms in the solid phase.
1) Gas. Give the expression of the partition function Zg(Ng) of the gas, its
free energy, its pressure, its chemical potential, its entropy.
2) Solid. In the solid, the atoms are positioned at the nodes of a crystalline
lattice. The atoms are supposed to be independent from each other and
behave as a 3D harmonic oscillator, as in the Einsteins model. The energy
of each atom is thus given by three quantum numbers n x, ny and nz:
3
n x ,n y ,n z o n x n y n z
2
2-a) What do represent o and ? Give an order of magnitude for these
constants.
2-b) Give the expression of the partition function Zs(Ns) of the solid, its
free energy and its chemical potential.
3) Equilibrium condition between solid and vapor.
The number of atoms in the gas phase, Ng, can take a priori every value
between 0 and N. In this question, we will determine its value at a given
temperature.
3-a) Give the partition function of the total system (solid + gas) as a
function of Zg and Zs.
3-b) Give the probability P(Ng=Ng o), where Ngo is a given value of Ng, as a
function of Zg and Zs.
3-c) Check that the most probable value Ng m of Ng is given by the equality
of the chemical potentials of the gas phase and of the solid phase.
3-d) At a given temperature, how does Ng m depend on the volume V? At a
given N and T, prove that the solid-vapor equilibrium is possible only if
V<Vo. Determine the volume Vo. What happens if V>Vo?
We will now consider only the case where V<Vo.
4) Equilibrium pressure.
4-a) Compute the pressure for a given volume V<Vo and at a given
temperature T. Prove that it is proportional to the average number <Ng>
of atoms in the gas phase.
4-b) Considering that for a large system <Ng>Ng m, prove that the
pressure P depends only on T, and on parameters characteristics of the
solid phase, but is independent on V.
Exercise 2: Adsorption of a gas on a solid surface.
We consider in this exercise a gas in contact with a solid surface containing M
sites that can get gas molecules. The gas quantity is sufficiently large for the gas
not to be disturbed by adsorption of some molecules on the solid surface.
I. We consider first that each of the M sites can get only one
molecule at maximum.
The energy of each site is =0 when it is empty, and =-1 (1>0) when it gets one
molecule.
The distance between the sites is supposed to be sufficiently large for the
adsorbed molecules not to interact. When n molecules are adsorbed, the energy
1) Chemical potential of the gas. The gas is a classical perfect gas. The
mass of the molecules in the gas, is m. The molecules have no internal
degree of freedom. They are confined in a volume V.
1-a) Compute the grand-canonical partition function, and the grand-
potential of the gas.
1-b) Find the equation of state PV=<N>kBT
1-c) Prove that the chemical potential of a molecule in the gas can be
written:
o T k B T ln P / Po
P is the pressure of the gas, P o is a reference pressure, for example
Po=1 bar, o(T) is called standard chemical potential.
o T k B T. ln f T
with f(T) a function of temperature T.
2) Solid. What imposes the equilibrium condition for the chemical
potential of the molecules adsorbed at the solid surface?
3) Adsorption sites. Write first the grand-canonical partition function for
a given site on the solid surface, and then the grand-canonical function
for the M adsorption sites. What is the corresponding grand-potential?
4) Coverage ratio. Compute the average number <n> of adsorbed
molecules and the coverage ratio =<n>/M as a function of 1, T, .
Compute and draw (T,P) as a function of the temperature T and of the
pressure P in the gas. These curves are called Langmuirs isotherms.
II. Lets now consider the case where each site can get many
molecules.
This case corresponds to the case of successive adsorption layers.
On each site, the energy of adsorption for the first molecule is still equal to 1.
The energy of adsorption for the following molecules are equal to -2 (0<2<1).
We suppose now that the number of molecules that can be adsorbed on a given
site is unlimited. As previously, the chemical potential of the adsorbed molecules
is given by the equilibrium condition with the surrounding gas.
sites.
2) Compute the coverage ratio =<n>/M as a function of the temperature
T and of the pressure P in the gas. Prove that
cx
(1 x cx )(1 x )
3) When c>>1, (T,P) is shown in Figure 1. What does represent the
elbow in the beginning of the figure?
(T,P) is called the Brunauer-Emmett-Teller adsorptions isotherms
(B.E.T. isotherms). It is used to measure the effective surface of divided
solids (like powders).
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The Golden Ratio For Designers
What is the Golden ratio?
The ratio of two numbers that equals roughly 1.618 is called the golden ratio, sometimes referred to as the divine proportion, the golden number, or the golden proportion. Often represented by the Greek letter phi, it has a close relationship to the Fibonacci sequence, which is a set of numbers where each successive number is added to the previous one. The ratio of each Fibonacci number to the preceding number progressively approaches 1.618, or phi. The Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, and so forth.
History of the Golden Ratio
The golden ratio was originally mentioned in Euclid’s Elements, a classical Greek treatise on geometry and mathematics, which dates to approximately 300 BCE. The proportion was known to Euclid and other ancient mathematicians such as Pythagoras, but they did not refer to it as the golden ratio. It would be many years before the proportion became mysterious. The ratio was hailed as symbolizing divinely inspired simplicity and orderliness in the book De divina proportione, written by Italian mathematician Luca Pacioli and including paintings by Leonardo da Vinci, which was published in 1509.
Mathematicians and artists began to recognize the golden ratio as a result of Pacioli’s book and Leonardo’s drawings. Many aficionados have asserted over the years since Pacioli’s book that the number is inherently beautiful, that it is a mathematical representation of beauty, and that golden triangles, golden rectangle side lengths, and golden ratio line segments are shown in works of art throughout history.
Since the golden ratio appears frequently in the natural world, proponents of the ratio contend that it is visually beautiful. Natural examples of the golden ratio are the proportions of nautilus shells and human bodies; however, these can differ significantly between individuals. Not all seashells expand in a pattern called a golden spiral that corresponds to the golden ratio. While it is true that nautiluses retain the same shell dimensions throughout their lives, the ratio of their shells is typically not an expression of phi but rather a logarithmic spiral.
There are other facets of nature where phi appears. Sunflower spirals and other seeds tend to hew closely to phi, while tree leaves and pine cone seeds typically grow in patterns that resemble the golden ratio. Because phi facilitates effective packing or dispersion, leaves that grow by the golden ratio will not shade one another and will instead rest at the golden angle.
Although there’s no proof that using the golden ratio is superior to using other proportions, designers, and artists are constantly striving to give their creations an intriguing composition, balance, and order.
The golden ratio in art and graphic design
A few designers and artists have purposefully centered their creations around the golden ratio. The golden ratio served as the foundation for most of the architectural system developed by renowned mid-century modern architect Le Corbusier. When Salvador Dali, the surrealist painter, painted The Sacrament of the Last Supper, he purposefully utilized a canvas that resembled a golden rectangle. American prog-metal band Tool released “Lateralus,” a song featuring time signatures inspired by Fibonacci, in 2001.
Additional instances of the golden ratio have been discovered by art historians in the Great Pyramid of Giza, the Mona Lisa, and the Parthenon in classical Athens. But most of the time, there isn’t any concrete proof that artists deliberately employed ratios in the manner that Le Corbusier, Dali, or Tool did. We are unable to determine whether phi was intentionally used by ancient engineers in the pyramids’ design notes or specifications.
How to use the golden ratio in your work
Mathematical rules do not apply to aesthetics or design. Even if you can make a bad design that still follows the golden ratio, you can utilize the ratio to guide your composition in order to reduce clutter and produce a balanced, ordered design. According to graphic designer Jacob Obermiller, “placement is everything on a graphic that might be pretty busy.” The golden ratio is a useful tool for guidance.
The rule of thirds and the golden ratio can function similarly. It can serve as a compositional convention or guidance, but it is not a strict guideline for organizing your work. In the end, guidelines of any type are beneficial and space is crucial. According to Sara Berndt, a student of human factors engineering, “if everything is important, then nothing is important.” You could alienate your reader, viewer, or user if you just center every image or arrange text in an arbitrary block. Make sure everything in your work is well-composed and appropriately spaced out by using the golden ratio as a guide.
You can generate diversity and white space that appeals to the eye and facilitates comprehension of text by using a convention like the golden ratio or the rule of thirds. According to Berndt, “the relationship between blank space and the ‘pay attention’ space is the essence of the golden ratio.” People can only process so much information visually. This is a guiding concept that will assist you in comprehending the span of human attention and producing visually appealing content.
If you choose to base your artwork or design on the golden ratio, it can make your work appear more balanced, even, and visually appealing. However, as long as you create thoughtfully and imaginatively, your ratios don’t have to be precisely 1.618. Adobe Illustrator can assist you in crafting your work so that everything is balanced to your own golden standards, regardless of the ratios and proportions you choose to use.
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# Thread: How to check if an object is moving in direction of a target
1. ## How to check if an object is moving in direction of a target
Hi,
I need a sanity check on a problem I’m trying to solve. I know where a number of moving objects are positioned at t=6 and t=7. I know where the destination is at t=7. So my goal is that I need to pick the “best” object i.e. the object moving most in the direction of the destination.
So for every object I’m doing the following:
I know position of object at t=7 is (x1,y1)
I know position of destination at t=7 is (x2,y2)
dot_product = (x1*x2)+(y1*y2)
length_a = sqrt(x1*x1 + y1*y1)
length_b = sqrt(x2*x2 + y2*y2)
result_degrees = result_radians * 180 / PI
So my first question is do you think this is correct? It doesn’t seem to give me the right answer e.g. if an object A is at (2,5) at t=6 and (2,2) at t=7 and the destination is at (4,2) at t=7, I seem to get an angle of approx 57 degrees when I would expect an angle of 90 degrees.
Also am I approaching the problem correctly? After I determine the angle of every object with the destination, is it true to say that to pick the “best” object i.e. the one moving towards the destination, is by picking the one with the smallest angle?
2. Originally Posted by kerrymaid
Hi,
I need a sanity check on a problem I’m trying to solve. I know where a number of moving objects are positioned at t=6 and t=7. I know where the destination is at t=7. So my goal is that I need to pick the “best” object i.e. the object moving most in the direction of the destination.
So for every object I’m doing the following:
I know position of object at t=7 is (x1,y1)
I know position of destination at t=7 is (x2,y2)
dot_product = (x1*x2)+(y1*y2)
length_a = sqrt(x1*x1 + y1*y1)
length_b = sqrt(x2*x2 + y2*y2)
result_degrees = result_radians * 180 / PI
So my first question is do you think this is correct? It doesn’t seem to give me the right answer e.g. if an object A is at (2,5) at t=6 and (2,2) at t=7 and the destination is at (4,2) at t=7, I seem to get an angle of approx 57 degrees when I would expect an angle of 90 degrees.
Also am I approaching the problem correctly? After I determine the angle of every object with the destination, is it true to say that to pick the “best” object i.e. the one moving towards the destination, is by picking the one with the smallest angle?
1. To answer your question you have to state for each moving object not only the position but the direction of the movement too (vulgo course).
2. You calculated the value of the angle between the two position vectors of a moving object and the target.
3. Ok, Well if I know an object is at position (2,5) at t=6 and (2,2) at t=7 I can assume by working out its speed (assuming no change) that it will be at (2,-1) at t=8. So can this information be used to infer a direction (if so how?) and how is then compared against the destination to pick the best object? Am I even looking at the right thing with my above calculations?
Many thanks.
4. Originally Posted by kerrymaid
Ok, Well if I know an object is at position (2,5) at t=6 and (2,2) at t=7 I can assume by working out its speed (assuming no change) that it will be at (2,-1) at t=8. So can this information be used to infer a direction (if so how?) and how is then compared against the destination to pick the best object? Am I even looking at the right thing with my above calculations?
Many thanks.
1. Draw a sketch (see attachment)
2. I assume that you are allowed to use vectors (?). If so:
3. The object moves along a line which passes through the point (2, 5) (at t = 6) and the point (2, 2) (at t = 7). That means the direction is determined by the vector $\displaystyle \vec d = (2,2)-(2,5)=(0,-3)$.Therefore the equation of the line is
$\displaystyle l: (x,y)=\underbrace{(2,5)}_{\rm start point}+(t-6) \cdot \underbrace{(0, -3)}_{\rm direction}$
4. The direction from the start point to the target is determined by the vector $\displaystyle \vec a = (4,2)-(2,5) = (2,-3)$
5. To judge which objects is moving best in the direction to the target you only have to calculate the angle (I've labeled it aberration) between the direction of the move and the direction to the target. In your case the angle $\displaystyle \alpha$ has a value of:
$\displaystyle \cos(\alpha)=\dfrac{(0,-3) \cdot (2, -3)}{\sqrt{9} \cdot \sqrt{13}}=\dfrac 3{\sqrt{13}}~\implies~\alpha \approx 33.69^\circ$
5. Hi Earboth,
That was a really helpful answer. Thanks so much for taking the time to do that.
I have what is hopefully my final query (apologies for the length). The moving objects that I’m referring to are actually cars moving around a city. The reason that I will simply “know” my current and previous position is that I’m assuming that in reality a car will be able to access its current (x,y) position from GPS which in the case of my program will be a file where I can simply read in the cars current and past position (this is a valid enough assumption to make). From this I can work out the distance travelled by the car in a given time interval and thus the cars speed at that time. Also, from your previous answer, I can now infer a direction. This is all great so far.
Finally, I want to ensure that I’m implementing everything correctly. According to the algorithm that I’m implementing, it says the following.
“The following notations are used in the subsequent discussion:
v(I,T): motion vector of node I at time T and p(I,T): position vector of node I at time T”. So it assumes that I know these.
Firstly can I confirm the following: From the above definition, I assume if the current time t=7 that the position vector, , for node I (based on our previous examples) is its position at time t=6 and t=7 i.e. (2,5) and (2,2). I also assume the motion vector is its direction and speed. So the motion vector is direction (0,-3) and speed of 3m/s. is this correct in your opinion?
Secondly: Part of the algorithm expects me to work out the “future” position of a car at time Tc + Δt where Tc is the current time and Δt is a small time slot such as 1 second. I can work this out by assuming the vehicle will move in a straight line. E.g. I know the position of the car at t=6 is (2,5) and at t=7 is (2,2) so calculating the direction vector as (0,-3) (based on your previous example) I can then add the direction vector to (2,2) to get (2, -1). All good.
However obviously a car can make a turn at an intersection. The paper seems to address this by saying “we rely on the accuracy of the motion vector of each vehicle. To handle sudden changes in movement, such as turning at a junction, a vehicle cannot directly report its motion vector accessed from GPS but should refer to past moving behaviour. Let m denote the motion vector access from GPS. Mold and Mnew are the reported motion vectors at the last time and this time respectively.
Mnew = α*Mold +(1- α)*m, 0<= α<=1
Where α=0.3 if the car is located in a junction and 0.7 otherwise”.
So I assume from the above they’re trying to somehow add a weight to the “predictability” of estimating the future position based on linear motion. I can assume that I simply know when a vehicle is located in a junction (this again is a valid assumption).
Now, I’m getting a bit confused about what to sub into the above formula.
So lets assume it’s currently t=7. Do I assume Mold is my old motion vector from t=6 to t=7 i.e. in the example above it is (0,-3) with speed 3m/s or does it represent the motion vector before that from t=5 to t=6. And what do you think lower case ‘m’ actually is? I would have thought ‘m’ and Mnew are the same thing. I would really appreciate if you could show me an example (based on our previous figures) of what you think this formula should be.
Really appreciate the help.
6. Originally Posted by kerrymaid
Hi Earboth,
That was a really helpful answer. Thanks so much for taking the time to do that.
You're welcome!
I have what is hopefully my final query (apologies for the length). The moving objects that I’m referring to are actually cars moving around a city. The reason that I will simply “know” my current and previous position is that I’m assuming that in reality a car will be able to access its current (x,y) position from GPS which in the case of my program will be a file where I can simply read in the cars current and past position (this is a valid enough assumption to make). From this I can work out the distance travelled by the car in a given time interval and thus the cars speed at that time. Also, from your previous answer, I can now infer a direction. This is all great so far.
Finally, I want to ensure that I’m implementing everything correctly. According to the algorithm that I’m implementing, it says the following.
“The following notations are used in the subsequent discussion:
v(I,T): motion vector of node I at time T and p(I,T): position vector of node I at time T”. So it assumes that I know these.
Firstly can I confirm the following: From the above definition, I assume if the current time t=7 that the position vector, , for node I (based on our previous examples) is its position at time t=6 and t=7 i.e. (2,5) and (2,2). I also assume the motion vector is its direction and speed. So the motion vector is direction (0,-3) and speed of 3m/s. is this correct in your opinion?
As far as I understand the situation you have a position vector p(I, T) and a motion vector v(I, T). (This motion vector contains the direction vector and the speed = absolute value of the direction vector). Then the next place can roughly predicted by:
$\displaystyle p(I+1, T+\Delta t) = p(I, t)+\Delta t \cdot v(I,T)$
where T means the actual time and Δt menas the elapsed time, a time interval.
Secondly: Part of the algorithm expects me to work out the “future” position of a car at time Tc + Δt where Tc is the current time and Δt is a small time slot such as 1 second. I can work this out by assuming the vehicle will move in a straight line. E.g. I know the position of the car at t=6 is (2,5) and at t=7 is (2,2) so calculating the direction vector as (0,-3) (based on your previous example) I can then add the direction vector to (2,2) to get (2, -1). All good.
To be exact you add the direction vector multiplied by the elapsed time. With the given example of the previous post the elapsed time was Δt = 1. If you want to go directly from (2, 5) to (2, -1) you have to multiply the direction vector by Δt = 2.
However obviously a car can make a turn at an intersection. The paper seems to address this by saying “we rely on the accuracy of the motion vector of each vehicle. To handle sudden changes in movement, such as turning at a junction, a vehicle cannot directly report its motion vector accessed from GPS but should refer to past moving behaviour. Let m denote the motion vector access from GPS. Mold and Mnew are the reported motion vectors at the last time and this time respectively.
Mnew = α*Mold +(1- α)*m, 0<= α<=1
Where α=0.3 if the car is located in a junction and 0.7 otherwise”.
So I assume from the above they’re trying to somehow add a weight to the “predictability” of estimating the future position based on linear motion. I can assume that I simply know when a vehicle is located in a junction (this again is a valid assumption).
Now, I’m getting a bit confused about what to sub into the above formula.
So lets assume it’s currently t=7. Do I assume Mold is my old motion vector from t=6 to t=7 i.e. in the example above it is (0,-3) with speed 3m/s or does it represent the motion vector before that from t=5 to t=6. And what do you think lower case ‘m’ actually is? I would have thought ‘m’ and Mnew are the same thing. I would really appreciate if you could show me an example (based on our previous figures) of what you think this formula should be.
Really appreciate the help.
The last part of your question is a bit foggy to me: Where these values of 0.3 and 0.7 come from?
In my opinion $\displaystyle \alpha = 1$ if there isn't any change in direction and speed:
$\displaystyle M_{new} = 1 \cdot M_{old}+(1-1) \cdot m$
where m is derived from previous GPS data. So in my opinion this m could be calculated by:
$\displaystyle m = M_{oldold}-M_{old}$
so that are always three positions are involved: The current position and the positions one or two steps (one or two Δt) before the momentary position.
Let us assume the car is in a junction. Then $\displaystyle \alpha = 0.3$ according to the text of your question. That means:
$\displaystyle M_{new} = 0.3 \cdot M_{old}+\underbrace{(1-0.3)}_{\rm{= 0.7}} \cdot m$
And this result looks somehow funny to me.
Let's assume the car is at (2, 5) and is moving with $\displaystyle \vec v = (0, -3)$. The time intervall between two measures is $\displaystyle \Delta t = 1$. After 1 s the car makes a right turn, the speed doesn't change. Then the new direction vector is $\displaystyle \vec v = (-3,0)$. The car passes through the points: (2, 5) --> (2, 2) --> (-1, 2)
According to the text of the question (as I understand it!) the car would be at the position (-0.1 , 1.1) (see attachment)
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Martingale System
The Math Behind the Martingale System. Let’s keep things simple. Say that you’re starting with a bankroll of \$11,000 which is subsequently divided into 100 units of \$110 each. We’ll apply the martingale system to an NFL bet to cover the spread, with a starting wager of one unit (\$110) on the Cleveland Browns. As is standard with NFL. The Martingale System was designed as a way to recoup losses and progressively build a bankroll. However, it is an incredibly risky strategy, as it requires you to place progressively larger bets.
See All Guides
Whether it’s a lengthy stretch of victories, a dismal run of consecutive outright losses, or regular bad luck at your sportsbook, everyone eventually finds themselves on a streak. And like all good things, every streak is bound to end.
While lengthy streaks can leave loyal fans brimming with confidence or reeling in despair, many sports bettors see an opportunity to profit off the inevitable end to the streak.
We’ll introduce you to the martingale system, which is widely-used by strategic sharp bettors looking to find an edge over bookmakers.
## What Is the Martingale System?
### Martingale System Sports Betting
Using the martingale system, a sports bettor places a bet on a specific team and an expected outcome. If the first bet proves to be a loser, the sports bettor increases their stake on each subsequent bet on the same outcome. This continues until a win is achieved.
With each subsequent losing bet, bettors wager double the stake laid on their previous losing bet – and so on – until the desired expected outcome pays out and turns a profit.
## The Math Behind the Martingale System
Let’s keep things simple. Say that you’re starting with a bankroll of \$11,000 which is subsequently divided into 100 units of \$110 each.
We’ll apply the martingale system to an NFL bet to cover the spread, with a starting wager of one unit (\$110) on the Cleveland Browns. As is standard with NFL spread bets, let’s say that the Brown’s odds against the spread are -110.
In this case, your initial Week 1 wager of \$110 (which includes the vig typically paid to the sportsbook) has potential winnings of \$100.
Win draw win bet explained. In the event your Week 1 bet is a loser, the martingale system dictates that you’ll need to bet again on Cleveland in Week 2. You’ll need to increase the to provide a return equal to both the amount you wagered in Week 1 (plus the amount you would have won if your Week 1 bet had been a winner).
### Breakdown of Martingale System Applied to a Bet on the Cleveland Browns
If your week 1 bet is a loser, your week 2 wager is determined by the size your week 1 wager, plus what you could have potentially won in week 1.
Further, you’ve got to account for the vig, which in this case is \$21. As such, your wager would be calculated by adding \$110 + \$100 + \$21, giving you a total of \$231.
If your week 2 bet is a loser, the size of your week 3 wager is determined by your week 2 wager plus the what you could have potentially won in week 2. Additionally, you’ve got to account for the vig. As such, your bet would be calculated as follows: \$231 + \$210 + \$43.10 = \$474.10. During the NFL season all these betting lines are available to wager on in real time, from week 1 to the Super Bowl.
You can continue to use the martingale method until you eventually produce a winner.
## Limitations of the Martingale System
Faithfully adhering to the martingale system will inevitably produce a winning bet. However, the method does have some limitations.
There’s a real possibility of your bankroll running dry before the end of a streak that produces a winning bet. If you managed to pick a winner in Week 3, you’d have wagered a total of \$805.
If the streak had dragged on beyond Week 3, you’d quickly find that it would get harder and harder to diversify your betting strategy. Your ability to make wagers on other events would be severely restricted, and eventually, your bankroll would be depleted all together if the streak dragged into Week 7.
While the NFL’s short schedule may make this a reasonable risk when using the martingale system to bet against the spread, lengthy streaks are much more common in sports with long schedules including the NHL, NBA, and MLB.
Since the start of the 2008-2009 season, there have been over 60 outright winning streaks that lasted 10 or more games in the NBA. In addition, the Atlanta Hawks’ epic ATS win streak in January 2015 lasted 13 games and would have required you to bump up your wager on the streak-ending 14th contest to a prohibitive \$1,685,987.23 – just to win your desired profit of \$100.
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# Search by Topic
#### Resources tagged with Reflections similar to Mirror, Mirror...:
Filter by: Content type:
Stage:
Challenge level:
### There are 35 results
Broad Topics > Transformations and their Properties > Reflections
### Mirror, Mirror...
##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### ...on the Wall
##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Matching Frieze Patterns
##### Stage: 3 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Decoding Transformations
##### Stage: 3 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### Combining Transformations
##### Stage: 3 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Simplifying Transformations
##### Stage: 3 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Shaping up with Tessellations
##### Stage: 2 and 3
This article describes the scope for practical exploration of tessellations both in and out of the classroom. It seems a golden opportunity to link art with maths, allowing the creative side of your. . . .
### The Frieze Tree
##### Stage: 3 and 4
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Friezes
##### Stage: 3
Some local pupils lost a geometric opportunity recently as they surveyed the cars in the car park. Did you know that car tyres, and the wheels that they on, are a rich source of geometry?
### Orbiting Billiard Balls
##### Stage: 4 Challenge Level:
What angle is needed for a ball to do a circuit of the billiard table and then pass through its original position?
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Screen Shot
##### Stage: 4 Challenge Level:
A moveable screen slides along a mirrored corridor towards a centrally placed light source. A ray of light from that source is directed towards a wall of the corridor, which it strikes at 45 degrees. . . .
### Frieze Patterns in Cast Iron
##### Stage: 3 and 4
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### A Resource to Support Work on Transformations
##### Stage: 4 Challenge Level:
This resources contains a series of interactivities designed to support work on transformations at Key Stage 4.
### It's Times Again
##### Stage: 2 and 3 Challenge Level:
Numbers arranged in a square but some exceptional spatial awareness probably needed.
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
### Transformation Game
##### Stage: 3 Challenge Level:
Why not challenge a friend to play this transformation game?
### Can You Explain Why?
##### Stage: 3 Challenge Level:
Can you explain why it is impossible to construct this triangle?
### Reflecting Squarely
##### Stage: 3 Challenge Level:
In how many ways can you fit all three pieces together to make shapes with line symmetry?
### Snookered
##### Stage: 4 and 5 Challenge Level:
In a snooker game the brown ball was on the lip of the pocket but it could not be hit directly as the black ball was in the way. How could it be potted by playing the white ball off a cushion?
### A Problem of Time
##### Stage: 4 Challenge Level:
Consider a watch face which has identical hands and identical marks for the hours. It is opposite to a mirror. When is the time as read direct and in the mirror exactly the same between 6 and 7?
### Surprising Transformations
##### Stage: 4 Challenge Level:
I took the graph y=4x+7 and performed four transformations. Can you find the order in which I could have carried out the transformations?
### Reflecting Lines
##### Stage: 3 Challenge Level:
Investigate what happens to the equations of different lines when you reflect them in one of the axes. Try to predict what will happen. Explain your findings.
### A Roll of Patterned Paper
##### Stage: 4 Challenge Level:
A design is repeated endlessly along a line - rather like a stream of paper coming off a roll. Make a strip that matches itself after rotation, or after reflection
### The Fire-fighter's Car Keys
##### Stage: 4 Challenge Level:
A fire-fighter needs to fill a bucket of water from the river and take it to a fire. What is the best point on the river bank for the fire-fighter to fill the bucket ?.
### The Eyeball Theorem
##### Stage: 4 and 5 Challenge Level:
Two tangents are drawn to the other circle from the centres of a pair of circles. What can you say about the chords cut off by these tangents. Be patient - this problem may be slow to load.
### Paint Rollers for Frieze Patterns.
##### Stage: 3 and 4
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Plex
##### Stage: 2, 3 and 4 Challenge Level:
Plex lets you specify a mapping between points and their images. Then you can draw and see the transformed image.
### So It's Times!
##### Stage: 2 and 3 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### One Reflection Implies Another
##### Stage: 4 Challenge Level:
When a strip has vertical symmetry there always seems to be a second place where a mirror line could go. Perhaps you can find a design that has only one mirror line across it. Or, if you thought that. . . .
### Two Triangles in a Square
##### Stage: 4 Challenge Level:
Given that ABCD is a square, M is the mid point of AD and CP is perpendicular to MB with P on MB, prove DP = DC.
### Rotations Are Not Single Round Here
##### Stage: 4 Challenge Level:
I noticed this about streamers that have rotation symmetry : if there was one centre of rotation there always seems to be a second centre that also worked. Can you find a design that has only. . . .
### Tricircle
##### Stage: 4 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
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| 3.640625 | 4 |
CC-MAIN-2017-30
|
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| 0.868578 |
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