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1000 | https://brilliant.org/wiki/system-of-linear-equations/ | System of Linear Equations (Simultaneous Equations)
Niranjan Khanderia,
Mei Li,
Arindam Kumar Paul,
and
Hemang Agarwal
Andrew Ellinor
Thaddeus Abiy
Yuxuan Seah
Daniel Lim
Arron Kau
Mahindra Jain
Hobart Pao
Aditya Virani
James Erb
Chung Kevin
A Former Brilliant Member
Abhishu Brahmecha
A Former Brilliant Member
Lii Gang Hah
Calvin Lin
Jimin Khim
contributed
Contents
Types
Substitution Method
Elimination Method
System of Linear Equations - More Variables
Elementary Row Operation or Gaussian Elimination Method
System of Linear Equations Word Problems - Basic
System of Linear Equations Word Problems - Intermediate
System of Linear Equations - Problem Solving
The system of linear equations are shown in the figure bellow:
Inconsistent: If a system of linear equations has no solution, then it is called inconsistent.
Consistent: If a system of linear equations has at least one solution, then it is called consistent.
Homogeneous system of equations: If the constant term of a system of linear equations is zero, i.e. the value after the "=" sign is zero, then it is called the homogeneous system of equations.
Trivial and non-trivial solutions: Every homogeneous system of equations has a common solution and that is zero because they have a common solution for all variables, which is 0. This solution is called the trivial solution. If there are other solutions, then they are called non-trivial solutions.
We will explore how to solve a system of linear equations using the substitution and elimination methods.
Substitution Method
In this method, we
find a relation that isolates one of the variables by changing the subject;
substitute the relation into the other equation(s) to reduce the number of variables by 1;
repeat until we are left with a single variable, and solve for it;
substitute the solved value back into the relations;
state the complete solution.
Let's follow the above steps to solve the following system of equations:
x+yx−y=5=3.
Step 1: Isolate variable x using the second equation: x=3+y.
Step 2. Substitute the relation into the other equation: (3+y)+y=5⇒3+2y=5.
Step 3. Repeat, and then solve. Since we now have only 1 variable, solve for it: 3+2y=5⇒y=25−3=1.
Step 4. Substitute back into the relations: x=3+y=3+1=4.
Step 5. State the complete solution: (x,y)=(4,1).
Solve the system of equations
2x+y2y=10=8.
In this example, because we are already given 2y=8 in the second equation, we know that y=4. Substituting this into the first equation, we get 2x+4=10, and thus x=3. □
Note: It matters which equation we use and which variable we isolate out. In this example, even if we obtained x=210−y from the first equation, we find that we are unable to substitute it into the second equation as there is no x term.
Solve the system of equations
2x+y3x+2y=4=7.
Solving the first equation for y gives y=4−2x. Substituting this into the second equation gives
3x+2(−2x+4)3x−4x+8−xx=7=7=−1=1.
Then substituting this value into the first equation, we have y=−2x+4=−2(1)+4=−2+4=2. Therefore, the solution to the system of equations is x=1,y=2. □
Using the process of substitution may not be the quickest nor the easiest approach for a given system of linear equations. However, we are always guaranteed to find the solution, if we work through the entire process. The word "system" indicates that the equations are to be considered collectively, rather than individually. Thus the solution must not lose validity for any of the equations. Select your options so that your calculations are simple and use any method that suits you.
Elimination Method
The elimination method multiplies the given n equations with suitable constants so that when the modified equations are added, one of the variables is eliminated. Once this is done, the system will have effectively been reduced by one variable and one equation. This process is repeated until one variable and one equation remain (namely, the value of the variable). From there, the obtained value is substituted into the equation with 2 variables, allowing a solution to be found for the second variable. The process is repeated until the values of all n variables are found.
Method
Find two equations that have the same variable. Multiply each equation by a number such that their coefficients are equal.
Subtract the two equations.
Repeat until we are left with a single variable, and solve for it.
Substitute the solved value back into the original equations to solve for the remaining variables.
Let's follow the above steps to solve the following system of equations:
3x−4y9x−8y=0=12.
Step 1: Multiply each equation by a number such that the coefficients of a variable are the same.
Let's say that we want to eliminate the variable x.
Multiply the first equation by 3 and the second equation by 1, and we obtain
9x−12y=0,9x−8y=12.
Step 2: Subtract the two equations:
9x−12y−(9x−8y−4y=0=12)=−12.
Step 3: Repeat and solve.
We are already down to one variable. Solving it gives us y=−4−12=3.
Step 4: Substitute the solved value back into the relations.
Substituting into the first equation, we get 3x−4×3=0⇒x=312=4.
System of Linear Equations - More Variables
When we have more variables to work with, we just have to remember to stick to a particular method, and keep on reducing the number of equations or variables.
We will solve the following system of equations using both approaches:
x+3y−z2x−y+2z3x+2y−z=6=1=2.
Substitution method
We begin with the first of the above three equations:
Step 1: The first equation gives x+3y−z=6⇒x=6−3y+z.
Step 2: Substituting for x in the second equation, we obtain
2(6−3y+z)−y+2z12−6y+2z−y+2z−7y+4z=1=1=−11.(1)
Substituting for x in the third equation, we obtain
3(6−3y+z)+2y−z18−9y+3z+2y−z2z−7y=2=2=−16.(2)
Step 3. We need to repeat until we have only one equation.
Now, we begin with (1) of the two equations (1) and (2) above:
Step 1: The equation (1) gives us y=74z+11.
Step 2: Substituting for y in (2) gives
2z−774z+112z−(4z+11)−2z=−16=−16=−5.
Step 3: We are now down to one equation. Solving it gives us z=−2−5=25.
Step 4: Substitute z=25 into y=74z+11 to obtain
y=74z+11=74×25+11=721=3.
Now, substitute z=25 and y=3 into x=6−3y+z to obtain
x=6−3×3+25=−21.
Step 5: Hence, the solution is
(x,y,z)=(−21,3,25). □
Elimination method
We are given the same system of linear equations as above:
x+3y−z2x−y+2z3x+2y−z=6=1=2.
Step 1: Let's eliminate x from the equations.
Step 2. Twice of the first equation minus the second is
[ \begin{align}
2(x + 3y - z &= 6)\
-(2x - y + 2z &= 1)\
\hline
7y - 4z & = 11. \qquad (4) \\end{align} ]
Thrice of the first equation minus the third is
3(x+3y−z−(3x+2y−z7y−2z=6)=2)=16.(5)
We need to repeat till we just have one variable.
Step 1. Let's eliminate y from the equations.
Step 2. The fourth equation minus the fifth is
7y−4z−(7y−2z−2z=11=16)=−5.(6)
Step 3. We are now down to one variable. Solving it gives us z=−2−5=25.
Step 4. Substitute z=25 into the fourth equation 7y−4z=11 to obtain
y=74z+11=74×25+11=721=3.
Now, substitute z=25,y=3 into the first equation x=6−3y+z to obtain
x=6−3×3+25=−21.
Step 5. Hence, the solution is
(x,y,z)=(−21,3,25). □
Solve the following system of equations:
x+2y−3z2x−5y+4z5x+4y−z=−3=13=5.
Let us start with the last equation. Solving for z, we obtain
z=5x+4y−5.
Substituting this into the second equation gives
2x−5y+4(5x+4y−5)22x+11y−332x+y−3=13=0=0.
Substituting this into the first equation gives
x+2y−3(5x+4y−5)x+2y−15x−12y+15+3−14x−10y+18−7x−5y+9=−3=0=0=0.
Thus, we now have reduced our system into a pair of equations with two variables:
2x+y−3−7x−5y+9=0=0.
Solving for y in the first equation, we get
y=3−2x.
Substituting this into the second equation gives
−7x−5(3−2x)+9−7x−15+10x+93x−6x=0=0=0=2.
Hence,
y=3−2⋅2=−1,z=5⋅2+4⋅(−1)−5=1.
Thus, the values of x,y, and z which satisfy the given system of equations are (2,−1,1). □
Elementary Row Operation or Gaussian Elimination Method
Main Article: Solving Linear Systems Using Matrices
Elementary row operation or Gaussian elimination is a popular method for solving system of linear equations. By this method, everyone can solve system of linear equations only by matrix row operations that you already know. The terms we need to know for this section are explained below.
Augmented Matrix:
An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation and each column represents all the coefficients for a single variable.
Suppose the system of linear equation is
x1+2x2+5x33x1+x2+4x32x1−7x2+x3=2=1=5.
Then the augmented matrix of this system of linear equations is 13221−7541215.
What are row echelon form, reduced row echelon form, and leading 1?
The matrix is in reduced row echelon form
100010001122.
To be of this form, a matrix must have the following properties:
If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1. We call this a leading 1.
If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix.
In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row.
Each column that contains a leading 1 has zeros everywhere else in that column.
A matrix that has the first three properties is said to be in row echelon form. If that matrix has the 4th property, then it's called reduced row echelon form.
Changing a matrix into row echelon form by elementary row operation:
A matrix can be changed to its reduced row echelon form, or row reduced to its reduced row echelon form using the elementary row operations, which are as follows:
Interchange one row of the matrix with another of the matrix.
Multiply one row of the matrix by a nonzero scalar constant.
Replace the one row with the one row plus a constant times another row of the matrix.
Suppose that a system of linear equation is
x1+2x2+2x3x1+3x2+3x32x1+6x2+5x3=4=5=6,
and its augmented matrix is 112236235456.
Now, we'll apply elementary row operation on this matrix and convert it to reduced row echelon form:
⇒102216215416−1R1+R2⇒10021221141−2−2R1+R3⇒10021021−141−4−2R2+R3⇒100210211414−1R3.
Finally the matrix is in the row echelon form: 100210211414.
So, at last we get the reduced form of the system of linear equation, which is
x1+2x2+2x3x2+x3x3=4=1=4.
By solving this, we get
x1x2x3=2=−3=4.
Md Nur Uddin is a student of Mathematics Discipline Of Khulna University. Now, you have to find his student ID by solving the following system of linear equations:
−3x+2y−6z5x+7y−5zx+4y−2z=6=6=8.
If the solution satisfies x+y=z, what is the value of x+y, or equivalently z?
The correct answer is: 1
System of Linear Equations Word Problems - Basic
What weight will the fourth scale display?
The correct answer is: 27
Let R, C, and D be the weights of the rabbit, cat, and dog, respectively. Each picture lends itself to an equation, giving us the following system:
C+R=10R+D=20C+D=24
If we add these equations together, we get 2C+2R+2D=54. If we divide by 2, we are left with the sum of the weights of each animal C+R+D=27.
A bottle that fully contains honey weighs 1500 grams.
The bottle with half of the honey weighs 900 grams.
What is the weight of the empty bottle in grams?
The correct answer is: 300
Today, in a 10-member committee, an old member was replaced by a young member. As such, the average age is the same today as it was 4 years ago.
What is the (positive) difference in ages between the new member and the replaced old member?
The correct answer is: 40
System of Linear Equations Word Problems - Intermediate
The brothers Luiz and Lucio bought a land surrounded by a wall of 340 meters. They built an inner wall to divide the land into two parts. Now, the part of Luiz is surrounded by a wall of 260 meters, and the part of Lucio by a wall of 240 meters. What is the length of the inner wall?
The correct answer is: 80
Today, there are 4 times as many days to the end of my exams, as there are to the start of my exams.
Tomorrow, there are 5 times as many days to the end of my exams, as there are to the start of my exams.
How many days do my exams last for?
The correct answer is: 12
There was a thief who went to a wine shop to steal wine. He went in at 10 pm, stole 15 L of wine and added 15 L of water to top up the barrel. He came back at 1 am, stole 15 L of the mixture from the same barrel and added 15 L of water to top it up. He came back at 4 am, stole 15 L of the mixture from the same barrel and added 15 L of water to top it up. In the morning, the ratio of wine to water in the barrel was 343:169. Find the initial amount of wine in the barrel.
The correct answer is: 120
System of Linear Equations - Problem Solving
The solutions of the system of equations
3x−y=a,x+y=5
are the same as those of the system of equations
2x+y=24,x−3y=b.
What is the value of a+b?
Since the two systems of equations have the same solutions, we first use substitution to find the solutions to the simultaneous equations
x+y=5,2x+y=24.
Solving the first equation for y gives y=5−x and plugging into the second equation gives 2x+(5−x)=24, or x=19. Plugging this into the first equation gives 19+y=5, or y=−14. Thus, x=19 and y=−14.
Substituting these values into 3x−y=a gives
a=3x−y=3×19−(−14)=71.
Similarly, substituting into x−3y=b gives
b=x−3y=19−3×(−14)=61.
Therefore,
a+b=71+61=132. □
Cite as:
System of Linear Equations (Simultaneous Equations).
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P. polycephalum is capable of distinguishing between “self” and “nonself” (Masui et al., 2018). This process is termed allorecognition. It is likely that allorecognition evolved independently multiple times during the history of life (Grosberg and Plachetzki, 2010). Thus it represents a fundamental example of convergent evolution of a behavioral phenotype. The idea that complex behaviors are also subject to recurrent evolution, both within closely related and across distant taxa, is relatively new (York and Fernald, 2017). It is very likely that natural selection acts not only on several biological pathways that are unique to behavioral traits, for example, the form and function of nervous systems, but also sensory–motor coupling.
Phagocytosis, the incorporation of foreign material into the cell, is another universal mechanism that is used by many primordial life-forms such as protists to ingest food. Phagocytosis is also involved in the immune response of higher organisms. However, there are great differences between the sophisticated components of the human immune system (e.g., the presence of lymphocytes and immunoglobulin) and that of lower invertebrates (Bayne, 1990), where such specialized cells and structures are often absent. Instead, there is such a diversity of different phagocytotic mechanisms, which are often unique to a particular taxon, that convergent evolution seems plausible (Stuart and Ezekowitz, 2008). This observation is not only limited to phagocytosis and allorecognition but also extends to other areas of physiology.
Therefore it is conceivable that cognition has evolved in the same way. Convergent evolution implies that the same end result—cognition—occurred independently several times throughout the history of life. Different mechanisms are employed by neural and nonneural organisms to achieve the same goal. van Duijn et al. (2006) proposed that basal cognition is immanent in sensorimotor systems (i.e., the integration of the sensory system and the motile machinery). This function is quintessentially realized in nervous systems, but analogous mechanisms exist in unicellular eukaryotes and even in bacteria.
Thus cognition is not restricted to nervous systems but may be a universal feature of life (Keijzer, 2017). Even the evolution of complex brains is considered convergent: The brains of highly intelligent octopi and primates have evolved independently, although from a common ancestor (Hirth and Reichert, 2007; Roth, 2015). Regarding habituation, a very ancient form of learning, available molecular evidence suggests that it evolved independently in quite different species, such as protists, plants, and neural organisms (van Duijn, 2017). P. polycephalum is also capable of habituation (Boisseau et al., 2016; Vogel and Dussutour, 2016). Different life-forms seem to have arrived at the same adaptive solution because of the benefits of possessing such a mechanism.
However, certain parallels and similarities can be detected between the cognition in neural and nonneural organisms. Sufficiently complex information-processing structures need to be present. In the brain, anatomical and functional connectivity plays a big role. In the myxomycete, interacting networks of cellular signaling and the motile machinery fulfill a roughly analog function. However, the overall cognitive complexity of P. polycephalum is by no means equivalent to that of a brain. On the spectrum of basal cognition, myxomycetes and other protists are probably at its lower end. Still, there are some similarities between myxomycetes and synaptic connections in the brain (Boussard et al., 2021), see also earlier sections in this chapter). The interconnected macroplasmodial transport network, for example, exhibits distinct oscillation patterns that are intimately linked to behavior (Fleig et al., 2020). The oscillation patterns show great variability in frequency, amplitude, and response to stimuli and could be involved in memory and basal cognition. It needs to be investigated whether they are truly analogous to information-processing mechanisms.
The concept of convergent evolution puts cognition into perspective: The fact that similar features evolved independently shows that the evolution of the mind was probably not a historical accident. Instead, cognitive traits are a replicable feature within the evolutionary process (Powell et al., 2017).
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2022, Trends in Ecology & EvolutionPeter Buston, Tim Clutton-Brock
Convergent evolution of strategic growth beyond vertebrates
Convergent evolution of strategic growth in response to similar socioecological conditions occurs across social vertebrates. Likely, there has been convergent evolution of strategic growth in response to these same conditions in social invertebrates; for example, ants [98,99] cnidarians [100,101], shrimp [102,103], and termites [104,105]. It is also plausible that there has been convergent evolution of strategic growth in some plants [106,107]: saplings of some forest trees; for example, sugar maple (Acer saccharum) and beech (Fagus grandifolia), increase growth and become taller in response to the opening up of canopy gaps, but decrease growth and remain smaller in the presence of closed canopies [108,109]. This can lead to the formation of size hierarchies, analogous to those observed in clown anemonefish , with individuals going through multiple periods of fast growth and slow growth as they progress through the understory before obtaining a dominant canopy position [110,111]. In light of this synthesis, this convergence makes sense, because these trees share similar socioecological conditions to social vertebrates: high reproductive skew among individuals within a neighbourhood; long-lived individuals and stable neighbourhood membership, with little (no) movement between neighbourhoods (Figure 5).
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Journal2022, Trends in Ecology & EvolutionPeter Buston, Tim Clutton-Brock
Chapter
Functional Groups
2001, Encyclopedia of BiodiversityRobert S. Steneck
IV. Functional Convergences
Convergent evolution results in distantly related organisms converging on the same body plan or ecological function. Charles Darwin noted in his Origin of Species (1859) that
for animals belonging to two most distant lines may have become adapted to similar conditions, and thus have assumed a close external resemblance; but such resemblances will not reveal—will rather tend to conceal—their blood relationship (page 463).
Relatively few convergently evolved anatomical, morphological, or physiological characteristics are the cornerstones of functional groupings.
IV.A. Groupings among Mobile Organisms
Convergent evolution is well-known and documented in the terrestrial realm. Marsupial and placental mammals have converged to similar morphologies and ecological function (Fig. 5). The limited variations on the mammalian body plan are evident in the wolf and catlike carnivores, the arboreal gliders, fossorial herbivores, anteaters, and subterranean insectivores that evolved independently in Australia for the marsupials and on the other continents for the placentals.
Convergent functions can be found among dissimilar-looking organisms. Distantly related marine molluskan herbivores, chitons and limpets, provide an excellent example. Chitons (class Polyplacophora) and snails (class Gastropoda) evolved in the Cambrian at the beginning of the molluskan diversification. The true limpets, Patellogastropoda, evolved from the Archaeogastropoda, much later in the Triassic. However, the radula (the teeth) of these two groups are functionally similar (Fig. 6). Both groups have relatively few teeth that contact the substrate, and those that do are hardened by mineralization of iron or silica compounds (note the black teeth in Fig. 6). Both groups have strong buccal musculature for applying downward forces. Within the molluskan body plan, only chitons and limpets have such a large foot area-to-mass relationship and an excavating-type radula. These morphological and anatomical characteristics, along with their size and mobility, allow species of these groups to specialize on large and expansive macrophytes such as sea grasses, kelp, and encrusting coralline algae. Although species diversity is much higher in other groups of mollusks such as nonlimpet gastropods, it is the functional characteristics of these two groups that make them capable of consuming and even trophically specializing on the tough or limestone-imbedded cells of kelp or coralline algae, respectively.
Convergent functions are numerous among terrestrial organisms. For example, a diversity of flowering plants and their pollinators possess similar morphological and anatomical characteristics despite significant phyletic separation among the plants and the pollinators. The geometry of the flowers, such as the length and width of the floral tube, as well as the placement of nectaries are common features among the plants. The flying characteristics and mouthparts are convergent among the pollinating insects, butterflies, moths, bats, and birds. It is surprisingly easy to find functional similarities among distantly related organisms.
IV.B. Groupings among Sessile Organisms
Convergent morphology and anatomy is common among sessile organisms on land and in the sea. I briefly consider terrestrial plant communities that have many of the same convergences described previously for animals. Families of cactus-like plants in the Old World (Euphdrbiacea) and New World (Cactaceae) dominate arid environments but are unrelated. Much has been written about the remarkably convergent forest and biomes that develop under similar environmental conditions (Huston, 1994). There appear to be relatively few adaptive solutions to common environmental conditions. Often, the suite of characteristics that improve ecological function under specific conditions define functional groupings.
Functional groupings among sessile organisms were studied by terrestrial ecologists, who grouped plants according to their size and the location of their growth and regenerative structures. Raunkiaer (1934) showed that particular functional groups (called “life-forms”) persisted and dominated the vegetation under similar environmental conditions. Using this scheme, Arctic regions were clearly distinguished from arid or tropical regions. Similar floras, defined at the functional group level, dominated under similar conditions in different regions even though there were often no species in common. This early focus on the size and placement of growth and regenerative structures was based on the realization that gradients in environmental stress and productivity relative to disturbances are critical to the success of specific life-forms. Plants with protected meristems and perenniating structures persisted in stressful or highly disturbed habitats. Where environmental limitations were relaxed, plant size and competitive ability were more important to the flora.
Recently, the field has become popular among marine ecologists and paleontologists who study sessile marine organisms. Again, morphological and functional convergences are viewed more as the rule than the exception. Several growth forms have evolved conver-gently among widely disparate groups, such as marine algae, Cnidaria, and Bryozoa (Fig. 7). Most extant and extinct species within those higher taxa can be placed within the groups illustrated in Fig. 7. The growth forms represent a morphological progression. The simplest growth form is solitary and can be a single cell as is the case of a diatom or a solitary organism. This group is often composed of small organisms. The next step in a morphological progression of growth forms simply involves a linear series of the single cells or modules, creating simple filaments. These can grow prostrate along the substrate as recumbent algal filaments or as runners or vines among sessile invertebrates. Similar morphologies growing vertically escape the benthic boundary layer. They experience greater water flow, and most of the cells of these groups are in contact with their aqueous environment. The more massive multiserial or multiseriate organisms allow three-dimensional space to be exploited. Several ecological processes, such as productivity and competition, vary predictably as sessile organisms become larger and more massive. With increasing mass, fewer cells directly contact the marine environment. Fewer still are in contact with the environment among the encrusting forms because half of their surface area is affixed to primary substrate. The growth forms in Fig. 7 exhibit progressively increasing size and generally decreasing surface area to volume ratio.
For benthic marine algae of all three major divisions (= phyla), the progression from single-celled microalgae to filamentous, foliose, larger erect forms and finally to crustose algae (Fig. 7) corresponds to a reduction in productivity per unit biomass (Fig. 8A; Steneck and Dethier, 1994). In general, microalgae and filamentous forms with a high surface area to volume ratio are capable of the highest mass-specific productivity. This decreases progressively from filamentous forms to the more massive erect forms including kelp and is lowest among the crustose forms. Thallus longevity (Fig. 8B) increases along the same continuum of growth forms. Canopy height, however, shows somewhat of a bell-shaped curve, with the tallest algal forms such as kelp and fucoid algae having intermediate levels of productivity and longevity (Fig. 8C). These emergent ecological properties were shown for cold-water floras of the Atlantic and Pacific as well as for tropical floras of the Caribbean (Fig. 8).
Morphologically and functionally similar organisms have evolved similar ways to overcome similar problems. For example, as organisms become more massive, fewer cells are in direct contact with the marine environment. That is, they have a plumbing problem. Sessile organisms such as plants, algae, and invertebrates evolved physiological integration or a degree of interdependence and cooperation among cells within an organism or modules within a colony. Often this involves sharing cytoplasm or photoassimilates within the organisms and among modules of a colony.
Physiologically integrated colonial organisms have an advantage over physiologically isolated aclonal organisms by sharing cytoplasm among members of the colony. Sessile organisms can respond from a stimulus by growing at a place distant from the stimulus. Wound healing can be facilitated and mortality of the genet reduced by physiological integration. Physiological integration evolved iteratively among sessile organisms on land and in the sea. This allows land plants such as vines to colonize the inhospitable habitats of sand dunes or tree canopies by being physiologically integrated. The most abundant marine algae, the red encrusting coralline algae, are integrated and this is thought to have contributed to their ecological and evolutionary success over unintegrated or poorly integrated algae. Arguably, physiological integration in algae is essential for their morphological diversity. The least physiologically integrated group, Chlorophyta or green algae, have the lowest morphological diversity despite having the same long period of evolutionary history dating back to the Precambrian.
Physiological integration may be a prerequisite for the development of coral reef ecosystems. It evolved independently among cnidarian reef builders: the tabulate, rugose, and finally the scleractinian corals. In each case, the proportion of species with a high degree of physiological integration increased during times of significant reef-building activity. Coral reef ecosystems require high rates of coral growth to keep up or catch up with rising sea level. Reef-building corals today are physiologically plants, with endosymbiotic algae producing carbohydrates necessary for growth. Transport of photosynthetic products throughout the coral colony requires a high degree of physiological integration. Without high rates of coral growth translating to high rates of reef growth, coral reef ecosystems would stand little chance of remaining in the shallow turbulent zone in which they thrive. In fact, it is likely that many coral reef ecosystems that have dominated shallow seas for much of the past 600 million years require the framework-building corals to be physiologically integrated.
Physiological integration must be functionally important to sessile organisms because it evolved independently many times. It is necessary for the similar functional groups found among the corals, algae, and bryozoans (Fig. 7). Colonial integration, or the degree of interdependence and cooperation among modules within a colony, is important for all these groups because it increases colonial function. It also allows the development of nonfeeding defensive or reproductive functions within the colony or clone, and it allows greater structural integrity and morphological diversity. The important functional advantages of physiological integration may account for why it has persisted and become ubiquitous among so many unrelated sessile, colonial, and clonal organisms.
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Reference work2001, Encyclopedia of BiodiversityRobert S. Steneck
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Gases and liquids
Chemistry
Structure of matter
Atomic models
Chemical bonds
Material science
Structure of metals
Ductility of metals
Solidification of metals
Alloys
Steelmaking
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Heat treatment of steels
Material testing
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Basics
Gear types
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HomeMechanicsGases and liquidsHow does a hot air balloon work: Buoyancy in gases
Mechanics
Gases and liquids
How does a hot air balloon work: Buoyancy in gases
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tec-science
08/17/2019
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Buoyant forces act not only in liquids but also in gases. This will be illustrated in the following by the example of a hot air balloon.
Buoyancy
In the article Buoyancy, the physical cause of the buoyant force were explained in detail. For the sake of clarity, liquids in which the objects were submerged were considered. The magnitude of the buoyant force was determined by the volume of liquid displaced by the immersed object Δ V Δ V on the one hand and the density of the liquid ρ l ρ l on the other hand (with g g as gravitational acceleration):
F b=Δ V⋅ρ l⋅g buoyant force(1)(1)F b=Δ V⋅ρ l⋅g buoyant force
Figure: Buoyancy of a balloon in water
The Archimedes’ principle was also derived from this equation, which states that the buoyancy of a body is equal to the weight of the displaced liquid. For this purpose, in the equation above, the product of the displaced liquid volume and the liquid density can be interpreted as the displaced liquid mass Δ m Δ m. The product of mass and gravitational acceleration then shows up as the weight of the displaced liquid F g,d i s F g,d i s:
F b=Δ V⋅ρ lΔ m⋅g F b=Δ m⋅gF g,d i s F b=F g,d i s Archimedes’ principle(2)(3)(4)(2)F b=Δ V⋅ρ l⏟Δ m⋅g(3)F b=Δ m⋅g⏟F g,d i s(4)F b=F g,d i s Archimedes’ principle
From liquids to gases
In principle, one can now also imagine a liquid whose density becomes smaller and smaller in thought. At some point one will have finally approached the density of gases. So there is no reason why buoyancy should not also occur in gases. And indeed, practice shows that buoyant forces also act in gases. These are calculated with the same equations as for liquids. The density ρ l ρ l therefore generally refers to the density of the surrounding fluid (whether liquid or gas).
Animation: From buoyancy in liquids to buoyancy in gases
The calculation of the buoyant force in gases is usually easier compared to liquids, since a body is usually completely submerged in the gas. The displaced gas volume thus corresponds to the volume of the body. With liquids, on the other hand, it must be noted that they can only be partially submerged in the liquid. The displaced liquid volume then corresponds to the actually submerged body volume.
Since gases have relatively low densities compared to liquids, the displaced gas mass into which a solid object is immersed is often negligibly small compared to its own mass. In such a case, the buoyancy can usually be neglected compared to the weight of the body. For example, a person displaces about 80 litres of air by his own body volume. With an air density of approximately 1.25 gram per litre, the displaced air mass is thus 100 g. A person with a mass of 80 kg thus appears to be 100 g lighter due to the buoyancy caused by the surrounding air (this corresponds to only about 0.1% of the body weight).
However, in cases where the weight of a body is relatively small compared to its volume, the buoyancy also plays an important role in gases. This will be especially the case when two gaseous substances are considered. Party balloons filled with helium are a typical everyday example of this, in which the buoyancy of the surrounding air causes the balloon to rise.
Figure: Helium filled balloons in different states
In this case, the helium mass inside the balloon is lighter than the displaced air mass, so that in accordance with the Archimedes’ principle, the resulting buoyancy is greater than the weight of the helium. This buoyant force is so great that it lifts not only the weight of the helium but also the weight of the balloon and the cord. Over time, however, the helium will leak from the balloon and the balloon will shrink. At some point the buoyancy will only be sufficient to keep the balloon floating but no longer make it rise. In this case, the buoyancy is equal to the weight of helium plus that of the balloon (and the cord). If even more helium leaks, the volume of the balloon will eventually be so small that significantly less air mass will be displaced. The buoyancy decreases according to the Archimedes’ principle. Finally, the buoyancy can no longer balance the weight of the helium and that of the balloon and the balloon sinks to the ground.
Hot air balloon
Hot air balloons use the same principle of buoyancy. Instead of helium, heated air is used. A hot air balloon consists of a airtight bag (called envelope) consisting of several segments sewn together. The basket is attached to the lower end of the balloon with suspension ropes. The burners, which are fed by gas cylinders, are mounted on a frame.
Figure: Hot air balloon
Typical specifications of a hot air balloon are given in the figure below, which illustrates how it works. The envelope of the balloon has a volume of about 4000 m³. This means that the balloon displaces a volume of 4000 m³ of cold atmospheric air. At an ambient temperature of about 24 °C, the air has a density of about 1.17 kg/m³. With a volume of 4000 m³, the balloon thus displaces a cold air mass of about 4700 kg. According to the Archimedes’ principle, this results in a buoyant force of 47 kN. If the same air were inside the balloon as outside, then of course there would be an air mass of 4700 kg inside the envelope. The buoyancy would then correspond to the weight and the air would practically float inside the balloon, but would not generate any effective upward force.
Figure: Typical figures of a hot air balloon
Therefore, the air inside the balloon is heated to over 100 °C with burner flames. As a result, the air density drops and with it the air mass inside the balloon. The decrease of the air mass inside the balloon can be explained by the fact that the heated air expands and partly flows out of the balloon. Unlike a party balloon, a hot air balloon is not a closed system, but open at the bottom, where the gas burner heats the air (and also open at the top, more about this later). Note that the volume of the balloon practically does not change during heating, so the displaced air mass or the associated buoyant force does not change either.
At an internal air temperature of 104 °C, for example, the density of the air has dropped to about 0.92 kg/m³, so that inside the balloon volume of 4000 m³ there is only an air mass of approx. 3700 kg. The buoyancy of 47 kN is now only countered by a weight of 37 kN. Thus the heated air generates an upward force of 10 kN. This is sufficient to lift a total mass of 1000 kg! The envelope with about 150 kg and the burner as well as the gas bottles with a total of 250 kg have to be taken into account. After deducting a basket mass of 100 kg, this leaves 500 kg for transporting the passengers.
Since the buoyancy of a hot air balloon is determined by the displaced air mass (i.e. by the balloon volume) and can practically not be changed during the flight, the air mass in the balloon must be increased in order to descent the balloon. This is achieved by a hole on the top of the balloon, which is closed during ascent and can be opened with a rope to allow the balloon to sink. This causes the hot light air to escape upwards and cold heavy air to flow in from below. The air mass inside the balloon increases again and the weight is greater than the buoyancy, so that the hot air balloon now sinks downwards.
Outlook
In the article Buoyancy, the cause of buoyant forces in liquids was explained by the different hydrostatic pressures on the bottom or top of the submerged object. The fact that buoyant forces also act in gases suggests that there must also be something like “hydrostatic” pressures in gases. In the same way as the pressure in liquids increases with increasing depth, the pressure in gases should also increase with increasing depth or decrease with increasing height. And in fact this is exactly the reason why the atmospheric air pressure decreases with increasing altitude – for example on mountains. More information can be found in the article Barometric Formula.
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What are type 1 and type 2 errors?
Type 1 & Type 2 Errors Explained - Differences & Examples
Understanding type 1 and type 2 errors is essential. Knowing what and how to manage them can help improve your testing and minimize future mistakes.
Table of Contents
Many teams use statistical methods to test the quality and performance of software products and websites, but these methods aren’t foolproof.
Type 1 and type 2 errors are crucial to the testing process, helping you understand its accuracy and reliability.
Encountering these errors is not a setback but a guidepost. They highlight areas for improvement in your testing and reassure you of its effectiveness.
Understanding the scenarios in which each type occurs is essential. It's not just about knowing the differences but about using that knowledge to improve your testing, address errors, and minimize future mistakes.
Let’s examine type 1 and type 2 errors in depth so you can better understand their meaning and how to manage them.
Browse this guide
Types of errors in statistics
Type 1 Error
Type 2 Error
Probability in error types
Type 1 Error Probability
Type 2 Error Probability
Balancing Type 1 and Type 2 Errors
Type 1 error examples
Mistaken A/B test result
Usability testing false positive
Inaccurate performance issue detection
Type 2 error examples
Missed bug detection
Overlooked performance issues
Undetected security vulnerability
How to manage and minimize type 1 and 2 errors
Adjust significance levels
Increase sample size
Implement more thorough testing methodologies
Use multiple testing techniques
Continuously monitor and feedback
Conduct root cause analysis
Foster a culture of quality
Using Amplitude to reduce errors
Types of errors in statistics
In product and web testing, we generally categorize statistical errors into two main types—type 1 and type 2 errors. These are closely related to the ideas of hypothesis testing and significance levels.
Researchers often develop a null (H0) and an alternate hypothesis (H1) when conducting experiments or analyzing data. The null hypothesis usually represents the status quo or the baseline assumption, while the alternative hypothesis represents the claim or effect being investigated.
The goal is to determine whether the observed data provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
With this in mind, let’s explore each type and the main differences between type 1 errors vs type 2 errors.
Type 1 Error
A type 1 error occurs when you reject the null hypothesis when it is actually true. In other words, you conclude there is a notable effect or difference when there isn’t one—such as a problem or bug that doesn’t exist.
This error is also known as a “false positive” because you’re falsely detecting something insignificant. Say your testing flags an issue with a feature that’s working correctly—this is a type 1 error.
The problem has not resulted from a bug in your code or product but has come about purely by chance or through unrelated factors. This doesn’t mean your testing was completely incorrect, but there isn’t enough weighting to confidently say the flag is genuine and significant enough to make changes.
Type 1 errors can lead to unnecessary reworks, wasted resources, and delays in your development cycle. You might alter something or add new features that don’t benefit the application.
Type 2 Error
A type 2 error, or “false negative,” happens when you fail to reject the null hypothesis when the alternative hypothesis is actually true. In this case, you’re failing to detect an effect or difference (like a problem or bug) that does exist.
It’s called a “false negative,” as you’re falsely concluding there’s no effect when there is one. For example, if your test suite gives the green light to a broken feature or one not functioning as intended, it’s a type 2 error.
Type 2 errors don’t mean you fully accept the null hypothesis—the testing only indicates whether to reject it. In fact, your testing might not have enough statistical power to detect an effect.
A type 2 error can result in you launching faulty products or features. This can massively harm your user experience and damage your brand’s reputation, ultimately impacting sales and revenue.
Probability in error types
Understanding and managing type 1 and type 2 errors means understanding some math, specifically probability and statistics.
Let’s unpack the probabilities associated with each type of error and how they relate to statistical significance and power.
Type 1 Error Probability
The probability of getting a type 1 error is represented by alpha (α).
In testing, researchers typically set a desired significance level (α) to control the risk of type 1 errors. This is the statistical probability of getting those results (p value). You get the p value by doing a t-test, comparing the means of two groups.
Common significance levels (α) are 0.05 (5%) or 0.01 (1%)—this means there’s a 5% or 1% chance of incorrectly rejecting the null hypothesis when it’s true.
If the p value is lower than α, it suggests your results are unlikely to have occurred by chance alone. Therefore, you can reject the null hypothesis and conclude that the alternative hypothesis is supported by your data.
However, the results are not statistically significant if the p value is higher than α. As they could have occurred by chance, you fail to reject the null hypothesis, and there isn’t enough evidence to support the alternative hypothesis.
You can set a lower significance level to reduce the probability of a type 1 error. For example, reducing the level from 0.05 to 0.01 effectively means you’re willing to accept a 1% chance of a type 1 error instead of 5%.
Type 2 Error Probability
The probability of having a type 2 error is denoted by beta (β). It’s inversely related to the statistical power of the test—this is the extent to which a test can correctly detect a real effect when there is one.
Statistical power is calculated as 1 - β. For example, if your risk of committing a type 2 error is 20%, your power level is 80% (1.0 - 0.02 = 0.8). A higher power indicates a lower probability of a type 2 error, meaning you’re less likely to have a false negative. Levels of 80% or more are generally considered acceptable.
Several factors can influence statistical power, including the sample size, effect size, and the chosen significance level. Increasing the sample size and significance level increases the test's power, indirectly reducing the probability of a type 2 error.
Balancing Type 1 and Type 2 Errors
There’s often a trade-off between type 1 and type 2 errors. For instance, lowering the significance level (a) reduces the probability of a type 1 error but increases the likelihood of a Type 2 error (and vice versa).
Researchers and product teams must carefully consider the relative consequences of each type of error in their specific context.
Take medical testing—a type 1 error (false positive) in this field might lead to unnecessary treatment, while a type 2 error (false negative) could result in a missed diagnosis.
It all depends on your product and context. If the cost of a false positive is high, you might want to set a lower significance level (to lower the probability of type 1 error). However, if the impact of missing a genuine issue is more severe (type 2 error), you might choose a higher level to increase the statistical power of your tests.
Knowing the probabilities associated with type 1 and type 2 errors helps teams make better decisions about their testing processes, balance each type's risks, and ensure their products meet proper quality standards.
Type 1 error examples
To help you better understand type 1 errors or false positives in product software and web testing, here are some examples.
In each case, the Type 1 error could lead to unnecessary actions or investigations based on inaccurate or false positive results despite the absence of an actual issue or effect.
Mistaken A/B test result
Your team runs an A/B test to see if a new feature improves user engagement metrics, such as time spent on the platform or click-through rates.
The results show a statistically significant difference between the control and experiment groups, leading you to conclude the new feature is successful and should be rolled out to all users.
However, after further investigation and analysis, you realize the observed difference was not due to the feature itself but an unrelated factor, such as a marketing campaign or a seasonal trend.
You committed a Type 1 error by incorrectly rejecting the null hypothesis (no difference between the groups) when the new feature had no real effect.
Usability testing false positive
Imagine you’re testing that same new feature for usability. Your testing finds that people are struggling to use it—your team puts this down to a design flaw and decides to redesign the element.
However, after getting the same results, you realize that the users’ difficulty using the feature isn’t due to its design but rather their unfamiliarity with it.
After more exposure, they’re able to navigate the feature more easily. Your misattribution led to unnecessary design efforts and a prolonged launch.
This is a classic example of a Type 1 error, where the usability test incorrectly rejected the null hypothesis (the feature is usable).
Inaccurate performance issue detection
Your team uses performance testing to spot your app’s bottlenecks, slowdowns, or other performance issues.
A routine test reports a performance issue with a specific component, such as slow response times or high resource utilization. You allocate resources and efforts to investigate and confront the problem.
However, after in-depth profiling, load testing, and analysis, you find the issue was a false positive, and the component is working normally.
This is another example of a Type 1 error: testing incorrectly flagged a non-existent performance problem, leading to pointless troubleshooting efforts and potential resource waste.
Type 2 error examples
In these examples, the type 2 error resulted in missed opportunities for improvement, the sending out of faulty products or features, or the failure to tackle existing issues or problems.
Missed bug detection
Your team has implemented a new feature in your web application, and you have designed test cases to catch each bug.
However, one of the tests fails to detect a critical bug, leading to the release of a faulty feature with unexpected behavior and functionality issues.
This is a type 2 error—your testing failed to reject the null hypothesis (no bug) when the alternative (bug present) was true.
Overlooked performance issues
Your product relies on a third-party API for data retrieval, and you regularly conduct performance testing to ensure optimal response times.
However, during a particular testing cycle, your team didn’t identify a significant slowdown in the API response times. This results in performance issues and a poor user experience for your customers, with slow page loads or delayed data updates.
As your performance testing failed to spot an existing performance problem, this is a type 2 error.
Undetected security vulnerability
Your security team carries out frequent penetration testing, code reviews, and security audits to highlight potential vulnerabilities in your web application.
However, a critical cross-site scripting (XSS) vulnerability goes undetected, enabling malicious actors to inject client-side scripts and potentially gain access to sensitive data or perform unauthorized actions. This puts your users’ data and security at risk.
It’s also another type 2 error, as your testing didn’t reject the null hypothesis (no vulnerability) when the alternative hypothesis (vulnerability present) was true.
How to manage and minimize type 1 and 2 errors
Using Amplitude to reduce errors
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Spleen removal
URL of this page: //medlineplus.gov/ency/article/002944.htm
Spleen removal
Spleen removal is surgery to remove a diseased or damaged spleen. This surgery is called splenectomy.
The spleen is in the upper part of the belly, on the left side underneath the ribcage. The spleen helps the body fight germs and infections. It also helps filter the blood.
Description
The spleen is removed while you are under general anesthesia (asleep and pain-free). Your surgeon may do either an open splenectomy, a laparoscopic splenectomy, or a robotic splenectomy.
During open spleen removal:
Your surgeon makes a cut (incision) in the middle of the belly or on the left side of the belly just below the ribs.
The spleen is located and removed.
If you are also being treated for cancer, lymph nodes in the belly are examined. They may also be removed.
The incision is closed using stitches or staples.
During laparoscopic or robotic spleen removal:
Your surgeon makes 3 or 4 small cuts in the belly.
Your surgeon inserts an instrument called a laparoscope through one of the cuts. The scope has a tiny camera and light on the end, which allows the surgeon to see inside the belly. Other instruments are inserted through the other cuts.
A harmless gas is pumped into the belly to expand it. This makes it easier for your surgeon to work.
Your surgeon uses the scope and other instruments to remove the spleen.
The scope and other instruments are removed. The incisions are closed using stitches or staples.
With laparoscopic or robotic surgery, recovery is often faster and less painful than with open surgery. Talk to your surgeon about which type of surgery is right for you or your child.
Why the Procedure is Performed
Conditions that may require spleen removal include:
Abscess or cyst in the spleen.
Blood clot (thrombosis) in the blood vessels of the spleen.
Cirrhosis of the liver.
Diseases or disorders of blood cells, such as idiopathic thrombocytopenia purpura (ITP), hereditary spherocytosis, thalassemia, hemolytic anemia, and hereditary elliptocytosis. These are all rare conditions.
Hypersplenism (overactive spleen).
Cancer of the lymph system such as Hodgkin disease.
Some types of leukemia.
Other tumors or cancers that affect the spleen.
Sickle cell anemia.
Splenic artery aneurysm (rare).
Trauma to the spleen.
Risks
Risks of having anesthesia and surgery in general are:
Reactions to medicines
Breathing problems
Bleeding, blood clots, infection
Risks associated with this surgery include:
Blood clot in the portal vein (an important vein that carries blood to the liver)
Collapsed lung
Hernia at the surgical cut site
Increased risk for infection after splenectomy (children are at higher risk than adults for infection)
Injury to nearby organs, such as the pancreas, stomach, and colon
Pus collection under the diaphragm
Risks are about the same for both open, robotic and laparoscopic spleen removal.
Before the Procedure
You or your child will have many visits with your health care providers and several tests before surgery. You may have:
A complete physical exam
Immunizations, such as the pneumococcal, meningococcal, Haemophilus influenzae, flu, and COVID-19 vaccines
Blood tests, special imaging tests, and other tests to make sure you are healthy enough to have surgery
Transfusions to receive extra red blood cells and platelets, if you need them
Tell the surgeon or nurse if:
You are, or could be pregnant.
You are taking any medicines, including medicines, drugs, supplements, or herbs you bought without a prescription.
You have been drinking a lot of alcohol, more than 1 or 2 drinks a day.
Planning for your surgery:
If you have diabetes, heart disease, or other medical conditions, your surgeon may ask you to see the provider who treats you for these conditions.
If you smoke, it's important to cut back or quit. Smoking can slow healing and increase the risk for blood clots. Ask your provider for help quitting smoking.
If needed, prepare your home to make it easier to recover after surgery.
Ask your surgeon if you need to arrange to have someone drive you home after your surgery
During the week before your or your child’s surgery:
You or your child may be asked to temporarily stop taking medicines that keep your blood from clotting. These medicines are called blood thinners. This includes over-the-counter medicines and supplements such as aspirin, ibuprofen (Advil, Motrin), naproxen (Aleve, Naprosyn), and vitamin E. Many prescription medicines are also blood thinners.).
Ask your surgeon which medicines you or your child should still take on the day of surgery.
Let your surgeon know about any illness you may have before your surgery. This includes COVID-19, cold, flu, fever, herpes breakout, or other illness. If you do get sick, your surgery may need to be postponed
On the day of surgery:
Follow instructions about when you or your child should stop eating or drinking.
Take the medicines the surgeon told you or your child to take with a small sip of water.
Arrive at the hospital on time.
After the Procedure
You or your child will spend less than a week in the hospital. The hospital stay may be only 1 or 2 days after a laparoscopic splenectomy. Healing will likely take 4 to 6 weeks.
After going home, follow instructions on taking care of yourself or your child.
Outlook (Prognosis)
The outcome of this surgery depends on what disease or injuries you or your child has. People who do not have other severe injuries or medical problems often recover after this surgery.
After the spleen is removed, a person is more likely to develop infections. Talk with your provider about getting needed vaccinations, particularly the flu and COVID-19 vaccines. Children may need to take antibiotics to prevent infections. Most adults do not need antibiotics long-term.
Alternative Names
Splenectomy; Laparoscopic splenectomy; Spleen removal - laparoscopic
Patient Instructions
Laparoscopic spleen removal in adults - discharge
Open spleen removal in adults - discharge
Spleen removal - child - discharge
Surgical wound care - open
When you have nausea and vomiting
Images
Red blood cells, target cells
Spleen removal - series
References
Cameron J. Spleen. In: Cameron J, ed. Current Surgical Therapy. 14th ed. Philadelphia, PA: Elsevier; 2023:651-662.
Nassar AK, Hawn M. The spleen. In: Townsend CM, Jr, Beauchamp RD, Evers BM, Mattox KL, eds. Sabiston Textbook of Surgery. 21st ed. Philadelphia, PA: Elsevier; 2022:chap 57.
Remiker AS, Brandow AM. Hyposplenism, splenic trauma, and splenectomy. In: Kliegman RM, St. Geme JW, Blum NJ, et al, eds. Nelson Textbook of Pediatrics. 22nd ed. Philadelphia, PA: Elsevier; 2025:chap 536.
Review Date 1/21/2025
Updated by: Jonas DeMuro, MD, Diplomate of the American Board of Surgery with added Qualifications in Surgical Critical Care, Assistant Professor of Surgery, Renaissance School of Medicine, Stony Brook, NY. Review provided by VeriMed Healthcare Network. Also reviewed by David C. Dugdale, MD, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team.
Related MedlinePlus Health Topics
Spleen Diseases |
1005 | https://teksguide.org/resource/82-conservation-momentum | 8.2 Conservation of Momentum | TEKS Guide
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8.2 Conservation of Momentum
8.2 Conservation of Momentum
Resource ID: zY5-70ge@7
Grade Range: HS - 12
Sections
Learning ObjectivesConservation of MomentumCheck Your Understanding
Learning Objectives
Learning Objectives
By the end of this section, you will be able to do the following:
Describe the law of conservation of momentum verbally and mathematically
Section Key Terms angular momentum isolated system law of conservation of momentum
Conservation of Momentum
Conservation of Momentum
It is important we realize that momentum is conserved during collisions, explosions, and other events involving objects in motion. To say that a quantity is conserved means that it is constant throughout the event. In the case of conservation of momentum, the total momentum in the system remains the same before and after the collision.
You may have noticed that momentum was not conserved in some of the examples previously presented in this chapter. where forces acting on the objects produced large changes in momentum. Why is this? The systems of interest considered in those problems were not inclusive enough. If the systems were expanded to include more objects, then momentum would in fact be conserved in those sample problems. It is always possible to find a larger system where momentum is conserved, even though momentum changes for individual objects within the system.
For example, if a football player runs into the goalpost in the end zone, a force will cause him to bounce backward. His momentum is obviously greatly changed, and considering only the football player, we would find that momentum is not conserved. However, the system can be expanded to contain the entire Earth. Surprisingly, Earth also recoils—conserving momentum—because of the force applied to it through the goalpost. The effect on Earth is not noticeable because it is so much more massive than the player, but the effect is real.
Next, consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—in the example shown in Figure 8.4 of one car bumping into another. Both cars are coasting in the same direction when the lead car, labeled m 2, is bumped by the trailing car, labeled m 1. The only unbalanced force on each car is the force of the collision, assuming that the effects due to friction are negligible. Car m 1 slows down as a result of the collision, losing some momentum, while car m 2 speeds up and gains some momentum. If we choose the system to include both cars and assume that friction is negligible, then the momentum of the two-car system should remain constant. Now we will prove that the total momentum of the two-car system does in fact remain constant, and is therefore conserved.
Figure 8.4 Car of mass m 1 moving with a velocity of v1 bumps into another car of mass m 2 and velocity v2. As a result, the first car slows down to a velocity of v′1 and the second speeds up to a velocity of v′2. The momentum of each car is changed, but the total momentum ptot of the two cars is the same before and after the collision if you assume friction is negligible.
Using the impulse-momentum theorem, the change in momentum of car 1 is given by
Δ p 1=F 1 Δ t,Δ p 1=F 1 Δ t,Δ p 1=F 1 Δ t,
where F1 is the force on car 1 due to car 2, and Δ t Δ t Δ t is the time the force acts, or the duration of the collision.
Similarly, the change in momentum of car 2 is Δ p 2=F 2 Δ t Δ p 2=F 2 Δ t Δ p 2=F 2 Δ t where F2 is the force on car 2 due to car 1, and we assume the duration of the collision Δ t Δ t Δ t is the same for both cars. We know from Newton’s third law of motion that F2 = –F1, and so Δ p 2=−F 1 Δ t=−Δ p 1 Δ p 2=−F 1 Δ t=−Δ p 1 Δ p 2=−F 1 Δ t=−Δ p 1.
Therefore, the changes in momentum are equal and opposite, and Δ p 1+Δ p 2=0 Δ p 1+Δ p 2=0 Δ p 1+Δ p 2=0.
Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is,
p 1+p 2=constant p 1+p 2=constant p 1+p 2=constant
p 1+p 2=p′1+p′2,p 1+p 2=p′1+p′2,p 1+p 2=p′1+p′2,
where p′1 and p′2 are the momenta of cars 1 and 2 after the collision.
This result that momentum is conserved is true not only for this example involving the two cars, but for any system where the net external force is zero, which is known as an isolated system. The law of conservation of momentum states that for an isolated system with any number of objects in it, the total momentum is conserved. In equation form, the law of conservation of momentum for an isolated system is written as
p tot=constant p tot=constant p tot=constant
or
p tot=p′tot,p tot=p′tot,p tot=p′tot,
where ptot is the total momentum, or the sum of the momenta of the individual objects in the system at a given time, and p′tot is the total momentum some time later.
The conservation of momentum principle can be applied to systems as diverse as a comet striking the Earth or a gas containing huge numbers of atoms and molecules. Conservation of momentum appears to be violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not.
Tips For Success
Momenta is the plural form of the word momentum. One object is said to have momentum, but two or more objects are said to have momenta.
Fun In Physics
Angular Momentum in Figure Skating
So far we have covered linear momentum, which describes the inertia of objects traveling in a straight line. But we know that many objects in nature have a curved or circular path. Just as linear motion has linear momentum to describe its tendency to move forward, circular motion has the equivalent angular momentum to describe how rotational motion is carried forward.
This is similar to how torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr 2 is analogous to mass or inertia. You may recall learning that the quantity mr 2 is called the rotational inertia or moment of inertia of a point mass m at a distance r from the center of rotation.
We already know the equation for linear momentum, p = mv. Since angular momentum is analogous to linear momentum, the moment of inertia (I) is analogous to mass, and angular velocity (ω)(ω)(ω) is analogous to linear velocity, it makes sense that angular momentum (L) is defined as
L=I ω L=I ω L=I ω
Angular momentum is conserved when the net external torque (τ τ τ) is zero, just as linear momentum is conserved when the net external force is zero.
Figure skaters take advantage of the conservation of angular momentum, likely without even realizing it. In Figure 8.5, a figure skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice, and because the friction is exerted very close to the pivot point. Both F and r are small, and so τ τ τ is negligibly small.
Figure 8.5(a) An ice skater is spinning on the tip of her skate with her arms extended. In the next image, (b), her rate of spin increases greatly when she pulls in her arms.
Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that L = L′.
Expressing this equation in terms of the moment of inertia,
I ω=I′ω′,I ω=I′ω′,I ω=I′ω′,
where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I′ is smaller, the angular velocity ω′ω′ω′ must increase to keep the angular momentum constant. This allows her to spin much faster without exerting any extra torque.
A video is also available that shows a real figure skater executing a spin. It discusses the physics of spins in figure skating.
Grasp Check
Based on the equation L = Iω, how would you expect the moment of inertia of an object to affect angular momentum? How would angular velocity affect angular momentum?
Large moment of inertia implies large angular momentum, and large angular velocity implies large angular momentum.
Large moment of inertia implies small angular momentum, and large angular velocity implies small angular momentum.
Large moment of inertia implies large angular momentum, and large angular velocity implies small angular momentum.
Large moment of inertia implies small angular momentum, and large angular velocity implies large angular momentum.
Check Your Understanding
Check Your Understanding
Exercise 5
When is momentum said to be conserved?
When momentum is changing during an event
When momentum is increasing during an event
When momentum is decreasing during an event
When momentum is constant throughout an event
Exercise 6
A ball is hit by a racket and its momentum changes. How is momentum conserved in this case?
Momentum of the system can never be conserved in this case.
Momentum of the system is conserved if the momentum of the racket is not considered.
Momentum of the system is conserved if the momentum of the racket is also considered.
Momentum of the system is conserved if the momenta of the racket and the player are also considered.
Exercise 7
State the law of conservation of momentum.
Momentum is conserved for an isolated system with any number of objects in it.
Momentum is conserved for an isolated system with an even number of objects in it.
Momentum is conserved for an interacting system with any number of objects in it.
Momentum is conserved for an interacting system with an even number of objects in it.
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1006 | https://www.accountingtools.com/articles/what-is-the-accounting-entry-for-depreciation.html | The accounting entry for depreciation — AccountingTools
The accounting entry for depreciation
/
Steven Bragg
What is the Accounting Entry for Depreciation?
The accounting for depreciation requires an ongoing series of entries to charge a fixed asset to expense, and eventually to derecognize it. These entries are designed to reflect the ongoing usage of fixed assets over time.
Depreciation is the gradual charging to expense of an asset's cost over its expected useful life. The reason for using depreciation to gradually reduce the recorded cost of a fixed asset is to recognize a portion of the asset's expense at the same time that the company records the revenue that was generated by the fixed asset. Thus, if you charged the cost of an entire fixed asset to expense in a single accounting period, but it kept generating revenues for years into the future, this would be an improper accounting transaction under the matching principle, because revenues are not being matched with related expenses.
In reality, revenues cannot always be directly associated with a specific fixed asset. Instead, they can more easily be associated with an entire system of production or group of assets, such as a production line.
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Fixed Asset Accounting
How to Audit Fixed Assets
The journal entry for depreciation can be a simple entry designed to accommodate all types of fixed assets, or it may be subdivided into separate entries for each type of fixed asset. The basic journal entry for depreciation is to debit the Depreciation Expense account (which appears in the income statement) and credit the Accumulated Depreciation account (which appears in the balance sheet as a contra account that reduces the amount of fixed assets). Over time, the accumulated depreciation balance will continue to increase as more depreciation is added to it, until such time as it equals the original cost of the asset. At that time, stop recording any depreciation expense, since the cost of the asset has now been reduced to zero.
Example of the Depreciation Entry
ABC Company calculates that it should have $25,000 of depreciation expense in the current month. The entry is:
In the following month, ABC's controller decides to show a higher level of precision at the expense account level, and instead elects to apportion the $25,000 of depreciation among different expense accounts, so that each class of asset has a separate depreciation charge. The entry is:
Cash Impact of Depreciation
Depreciation is considered an expense, but unlike most expenses, there is no related cash outflow. This is because a company has a net cash outflow in the entire amount of the asset when the asset was originally purchased, so there is no further cash-related activity. The one exception is a capital lease, where the company records it as an asset when acquired but pays for the asset over time, under the terms of the associated lease agreement.
The Capitalization Limit
Depreciation and a number of other accounting tasks make it inefficient for the accounting department to properly track and account for fixed assets. They reduce this labor by using a capitalization limit to restrict the number of expenditures that are classified as fixed assets. Any expenditure for which the cost is equal to or more than the capitalization limit, and which has a useful life spanning more than one accounting period (usually at least a year) is classified as a fixed asset, and is then depreciated.
The Difference Between Carrying Cost and Market Value
Finally, depreciation is not intended to reduce the cost of a fixed asset to its market value. Market value may be substantially different, and may even increase over time. Instead, depreciation is merely intended to gradually charge the cost of a fixed asset to expense over its useful life.
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1007 | https://www.jimpryor.net/teaching/courses/phil735/notes/sets.html | Sets and Tuples
Phil 735: Sets and Tuples
There are a variety of “collections” that a mathematician or logician might work with. What are sometimes called strings or sequences or lists pay attention to the order of their elements; they also pay attention to the multiplicity of their elements. The string "a" is different from the string "aa". Other collections make other choices. Sets, for example, don’t pay attention to either the order or the multiplicity of their elements. The set {"a", "a", "b"} is the same as the set {"b", "a"}. We will be working with sets a lot this term.
Another collection, less often discussed, is a multiset. This pays attention to the multiplicity of its elements but not their order. So a multiset containing the elements "a", "a", and "b" will be the same as a multiset containing the elements "a", "b", and "a", but different from a multiset containing only "b" and "a" (once).
Graphs and trees are collections with more complex structures.
I’ll try to stick to the convention of naming sets with capital greek letters, like Δ and Γ. But I may sometimes slip; and there is no universal practice. (And sometimes ☺ I will use capital Greek letters like Α, Β, …) As a general convention, though, you should try to use capital letters of some sort to refer to sets (and expect others to be doing so too). Sometimes people use capital letters for other roles, too.
Sometimes these notes will be composed in the first-person singular, other times in the first-person plural. Understand them in every case as being offered by Matt and Jim jointly --- though it may happen that one or the other of us had a larger role in writing a given passage, and we may not always fully agree about all details.
As we said, sets are collections that may have elements or members. Their members may, but need not, also be sets. When the members aren’t sets, they’re called urelements or individuals or atoms. When d is a member of set Δ, we write d ∈ Δ. (Note the difference between the symbol ∈ for set membership, and the symbol ɛ sometimes used to designate the empty string.)
The empty set, written as ∅ or {}, has no members.
Boxes like this will contain various arcane comments that could be useful to some readers, but you'll probably want to ignore if you're learning these materials for the first time.
For a given set of things Δ, there is exactly one empty set of Δs, which contains no Δs. Where Δs and Γs are sets of things that are fundamentally different in certain ways, it’s less clear and more contentious whether the empty set of Δs and the empty set of Γs should be the same entity. But we will just ignore that issue.
Sets can have lots of members — even infinitely many — and they can contain sets which have lots of members. But on the standard (post-Russellian) picture, no set can have itself as a member. Nor can there be an infinite descending chain of sets, each of whom is a member of the next outermost set. That is, no sets like this:
{0, {1, {2, {3, ...}}}}
These claims aren’t indisputable, and alternative set theories have been developed which don’t respect them. But those kinds of controversies aren’t going to be our topic here. We’re just going to assume we’re working with well-behaved sets, we’re not going to worry about Russell’s paradox, and so on.
Some notable sets are:
ℕ, the set of natural numbers {0, 1, 2, ...} (a few authors exclude zero from what they call “the natural numbers,” but most include it)
ℤ, the set of integers {..., -2, -1, 0, 1, 2, ...}
ℤ₆, the set of integers modulo 6 {0, 1, 2, 3, 4, 5} (and similarly for other ℤ n)
ℝ, the set of real numbers (also called “the continuum”)
𝟚, the set of of truth-values {false, true} (This notation is less common than the preceding ones. Sometimes this set is identified with ℤ₂.)
Sometimes mathematicians or logicians will specify a set by listing all its members:
{0, 16, 128}
Other times, they’ll specify the set using set-builder notation, which can look like this:
```
{x² | x ∈ ℕ}
{y | y ∈ ℕ and ∃x ∈ ℕ (y = x²)}
{y ∈ ℕ | ∃x ∈ ℕ (y = x²)}
```
All three of these expressions specify the same set. In the first two variants, the part to the right of the vertical bar | introduces some variables that range over a domain (in these examples, always ℕ) and may also be constrained to satisfy a guard condition; the part to the left of the vertical bar is then a term. The third variant can be understood as just a different style of writing the second.
Sometimes instead of a vertical bar, a colon is used: {x² : x ∈ ℕ}.
The language of these webpages, and most of your proofs, and most math textbooks and articles, will mix English and symbols in a way that may be precise, but nonetheless won’t count as a formal language like those of a logical system, with strictly regimented syntax. It’ll be an informal language, albeit one that contains lots of math notation. Similarly, most of the proofs you’ll be reading and constructing won’t be formal proofs in a strictly regimented system like first-order logic. That is not to impugn them, or say they can’t be rigorous and convincing.
I draw attention to this now because the symbolism we use in our informal vs our formal languages can differ somewhat.
There are formal languages (and proof systems) for working with sets, but we’re not aiming to master them. Our references to sets will always be informal.
Although our informal language will use quantifiers like first-order logic does, authors will often specify what domain they’re drawing from when they use a quantifier. Thus I may write ∃x ∈ ℕ and not just ∃x. In the formal logical languages you first encountered in your studies, you’ll only have had the second, “unrestricted” quantifier. But some formal languages also introduce a “restricted” kind of quantifier too. The reasons for doing that are a topic for another discussion.
I’ll assume you’re familiar with basic set-theoretic operations and relations:
What does {2,3} ∈ {2,3,4} mean, and is it true?
What does {2,3} ⊆ {2,3,4} mean, and is it true? (See below.)
What does Γ ∩ Δ mean? What is {1,2,3} ∩ {2,3,4}?
What does Γ ∪ Δ mean? What is {1,2,3} ∪ {2,3,4}?
If Γ₀, Γ₁, ... Γ n are sets, then another way to write Γ₀ ∪ Γ₁ ∪ ... ∪ Γ n is . Similarly for .
If Δ is a set of sets, then is the union of all the members of Δ. Or you could write . Similarly for .
The “difference” between Γ and Δ — that is, the set of all members of Γ that aren’t also members of Δ — is sometimes written Γ ∖ Δ, but I prefer to write it just as Γ – Δ.
Sometimes people write or (or Δ′, (Δ ∪ Γ)′). Here they are assuming some specific larger set Ω that Δ and Γ are subsets of, and these are just alternative ways of specifying Ω – Δ and Ω – (Δ ∪ Γ).
Δ is a subset of Ω when everything that’s a member of Δ is also a member of Ω. We write this relation as Δ ⊆ Ω. Ω may have additional members besides the things that are also in Δ. Or it may not. If it doesn’t, then Ω has exactly the same members as Δ. And as we understand sets, that means that they are exactly the same set. So among the sets that Δ is a subset of is itself.
If you want to say that Δ is a subset of and not identical to some other set Γ, then you say that Δ is a proper subset of Γ. Often that is written like this: Δ ⊂ Γ. But some authors use the symbol ⊂ the way we’ve explained ⊆. Sorry! If you want to avoid ambiguity, you could express the claim that Γ is a proper subset of, that is, is a subset of and is not equal to, Δ, like this: Δ ⊊ Γ. But I’ve tried doing that and people found it confusing or hard to read. (And some misinterpreted it as saying that Δ ⊈ Γ, that is, that Δ is not a subset of Γ.)
As we said, sets are the same when they have exactly the same members. For collections like multisets, which care about the multiplicity of their elements, then the collections will be the same when they have all the same elements to the same multiplicity. For collections like strings, which care about the order of their elements, then the collections will be the same when they have all the same elements in the same order. For collections with other sorts of structure, the rules will be analogous. There’s a general idea here, that the identity of a collection depends upon the identity of its elements and on any structure by which the collection organizes them, and on nothing more. You can’t have two collections with all the same elements organized (in the way that is inherent to that kind of collection) in the same way, yet those collections be numerically distinct. This general idea is called extensionality. (The root term “extensional” also has other meanings you may be familiar with. These meanings are conceptually related, but don’t expect that you could by unguided first principles derive specifically the one meaning from the other.)
When mathematicians or logicians work with a collection, they’ll almost always be understanding the collection to be one that’s extensional in the way I described. But in computer science contexts, this can’t be taken for granted. There it can be useful to introduce notions of structures that aren’t extensional, where we can have two numerically distinct but indiscernible collections, with the same elements organized in the same way. (For example, in Python, frozensets are extensional but what they call sets are not.) We’re not going to get into those contexts here. I’m just flagging that although the idea of an extensional collection is very natural and useful to work with, it’s not conceptually mandatory, and in some formal settings we can make good sense of, and it’s useful to work with, collections that aren’t extensional.
Here are some basic facts about unions, intersections, and subsets:
∅ ⊆ Α ⊆ Α (that means that ∅ ⊆ Α and Α ⊆ Α)
∪ and ∩ are each commutative: this means, for example, that Α ∪ Β = Β ∪ Α
∪ and ∩ are each associative: this means, for example, that Α ∪ (Β ∪ Γ) = (Α ∪ Β) ∪ Γ
Α ∩ Β ⊆ Α ⊆ Α ∪ Β
Α ⊆ Β iff Α ∩ Β = Α iff Α ∪ Β = Β (iff means “if and only if” — you’ve encountered this by now in your philosophy studies, right?)
Α ∩ Α = Α ∪ Α = Α ∪ ∅ = Α - ∅ = Α
Α ∩ ∅ = Α - Α = ∅
Α ∪ (Β ∩ Γ) = (Α ∪ Β) ∩ (Α ∪ Γ)
Α ∩ (Β ∪ Γ) = (Α ∩ Β) ∪ (Α ∩ Γ)
Α ∪ (Β - Α) = Α ∪ Β
Α ∩ (Β - Α) = ∅
Α ∪ (Α ∩ Β) = Α
Α ∩ (Α ∪ Β) = Α
Β - (Β - Α) = Β ∩ Α
The power set of a set Ω contains all and only the sets that are subsets of Ω. This includes Ω itself, and it includes the empty set ∅. That’s because all of the empty set’s members (all none of them) are also members of Ω. Sometimes the power set of Ω is written ℘(Ω), but I will write it as 𝟚 Ω. (At some point, this notational choice may make sense. Until then, just take it as an arbitrary symbol.)
Thus the powerset of {1,2,3} is {∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}.
When two sets have members in common, we say that the sets overlap. When the sets fail to have any members in common, we say that they’re disjoint. (The empty set is an edge case: should we define “disjointness” in a way to count it as disjoint from other sets, because their intersection is empty? Or should we count it as non-disjoint, since after all, it’s a subset of those other sets?) When two sets fail to be numerically identical, then — regardless of whether they overlap or are disjoint — we say that the sets are distinct.
In mereology, the formal theory meant to model our intuitive understanding of part-whole relations, they also use the term “overlapping.” But there instead of “disjoint,” they talk about things with no parts in common as being wholly “discrete.” Don’t confuse these vocabularies.
Partitions and Covers
A partition of a set is a division of the set into one or more (non-empty) “cells,” where everything from the original set gets to be in one of the cells, and none of the cells overlap. So, for example, here is one partition of the set {1,2,3}:
{1} {2,3}
The partition is itself understood to be a set — the set containing the two sets just listed. But for the moment I’ll just write out the elements of the partition separated with space, to make the intuitive idea more obvious. Another partition of the set {1,2,3} is:
{1,3} {2}
A cover is similar to a partition, except that now we’re allowed to have some overlap. It’s allowed (but not required) that some of the members of the original set get to be in multiple cells. So both of the above partitions of {1,2,3} also count as covers of that set. But so too does this:
{1,3} {2,3}
Covers come up sometimes in linguistics. We won’t need to work with them in this course.
Tuples (Ordered pairs, triples, …)
Sometimes we want to talk about several items taken from several sets. We do this using the notion of an ordered pair (or triple, or …) An ordered pair (a,b) has a “first” element a and a “second” element b. It’s permitted that these be the same object. (Instead of (a,b), some authors will write ⟨a,b⟩.) Two ordered pairs count as the same just in case they have the same first elements and the same second elements. The two elements might both be the same type of thing (they might both be members of the set of natural numbers ℕ, for example). Or they might be different types of things (maybe a is a number but b is a person). When Α and Β are two, not-necessarily-distinct, sets, we describe the set of all ordered pairs whose first element comes from Α and whose second element comes from Β as Α ⨯ Β. We can write this as:
Α ⨯ Β = {(a,b) | a ∈ Α and b ∈ Β}
This is called the Cartesian product of Α and Β. As I said, it’s permitted that Α and Β be the same set. We’ll see another notation for Α ⨯ Α below.
The ordered triple of Α, Β, and Γ we write as:
Α ⨯ Β ⨯ Γ = {(a,b,c) | a ∈ Α and b ∈ Β and c ∈ Γ}
In many mathematical and logical texts, you will see authors proposing reductions or equations between various of the notions we’re discussing. For example, it’s common to see the ordered triple (a,b,c) defined as the pair (a,(b,c)) whose second element is itself a pair. But there might be settings where I need to differentiate between pairs that contain pairs, on the one hand, and triples on the other. In many computer science settings, for example, where we pay close attention to the types of things we’re working with, conflating these can be disastrous.
Another sort of reduction you will often encounter in math and logic texts is one that identifies ordered pairs with certain kinds of sets. There are different ways to do this. The most widespread proposal is that:
(a,b) =def {{a},{a,b}}
Other proposals reduce numbers to sets. (There are different ways to do this, too.)
It makes no mathematical difference for this course whether any of these equations or reductions are accepted. (Sometimes it can make a mathematical difference, if a proposed reduction is carried to other contexts, where it breaks.) The proposals may also matter in the philosophy of math, but for that very reason they are substantive and can be intelligibly debated. In any case, those are not our topics in this course. So I’m going to downplay these reductions in our discussion. Even if we decided to accept some or all of them, I don’t think it’s right to offer them as introductory definitions of these notions.
I will myself tend to refrain from reductions like (a,b,c) = (a,(b,c)) that you see other authors going in for. The other authors may say that’s what a triple is, or how it’s defined. I’ll tend to say instead that the author is mathematically modelling or representing some notion (a notion we have some intuitive, somewhat pre-theoretical grasp on) in terms of (a,(b,c)). In some settings it’s worth keeping track of the difference. That’s why I’m flagging these things for you. I also think you’ll have better conceptual hygiene if you’re more aware of the choices to be made here. But for the material we’ll be focusing on, it won’t matter mathematically.
I talked about the “first” and “second” element of an ordered pair. But the suggestion of an “order” to these elements can mislead. Really the important thing is just that we keep track of which element comes from which set, or which element comes from a single set playing the role of one side of the Cartesian product rather than the other. Instead of the “first” element and the “second” element, we could instead talk about the “west” element and the “east” element. And for ordered quadruples, about the “north,” “south,” “east” and “west” elements. If someone then took it in mind to ask whether the west element comes before or after the east element, or whether the north element came in between them, these questions wouldn’t have any established sense.
Similarly, if you take the ordered quadruple (a,b,c,d), we might call it “increasing” iff a ≤ b ≤ c ≤ d. But I could just as easily define another notion, call it “ascending,” which holds iff a ≤ d ≤ b ≤ c. There’s no sense in which one of these two notion is more intrinsically natural or less gruesome than the other.
With strings, on the other hand, the collection does have a more natural intrinsic ordering. It’s genuinely more natural to count the letter "b" as coming “between” the letters "a" and "c" in "abc" than it is to count "c" as coming “between” "a" and "b", because of way the string "abc" is inherently structured.
Sometimes ordered pairs, triples, and so on — the general class of things I will call n-tuples or just tuples — are referred to as “ordered sets.” Avoid this usage; it will be too confusing when we look at a different notion of ordered set, in a few classes.
The set Α ⨯ Α of pairs of Αs is sometimes written as A², and the set of triples Α ⨯ Α ⨯ Α written as Α³. But then that raises the question whether:
Α ⨯ (Α ⨯ Α), that is, Α ⨯ Α², should = Α³ as well?
Those who identify (a₁,a₂,a₃) with (a₁,(a₂,a₃)) will say sure, because that’s just the same thing as Α ⨯ Α ⨯ Α. But then should it also be the case that:
(Α ⨯ Α) ⨯ Α = Α³
and so that ((a₁,a₂),a₃) always = (a₁,(a₂,a₃))? The Α³ notation papers over this issue. Nonetheless, I will sometimes go along with the notation because it’s very prevalant.
Don’t confuse Α² and 𝟚 Α! The first is the set of all pairs both of whose elements come from Α. The second is the set of all subsets of Α (which as we said, some authors write as ℘(Α)).
Some More Vocabulary about Sets
Singletons are sets with just one member. For example, {"a"} is the singleton set containing the string "a" as its sole member. {{}} is the singleton set containing just the one member {}, that is, the empty set. Since another notation for {} is ∅, we could also write {{}} as {∅}.
If Γ is a subset of Δ (that is, Γ ⊆ Δ), then we can also say that Δ is a superset of Γ.
Pure sets are sets like {}, {{}}, {{},{{}}}, and so on, where no matter how deeply nested, none of the sets involved have any members that aren’t themselves sets.
We will discuss issues about the size or cardinality of sets later.
You may encounter talk of the rank of a set. The way this talk is used is this. At rank 0, we have all the sets whose members are only urelements. This includes at least the empty set. Potentially it could also include the set containing you and me, but standardly the rank-talk is used in theoretical settings where we assume everything being talked about is set-like — which you and I are not. So in that setting, there is only one set of rank 0, the empty set. A set belongs to rank k+1 when all of its members are (either urelements or) sets of rank ≤ k, and at least some of its members are from rank k.
You may encounter talk of the cumulative hierarchy of sets. This is another way of talking about rank — though sometimes the sets of rank k are counted as level k+1 of the cumulative hierarchy. On that way of talking, urelements belong to level 0, the empty set ∅ has rank 0 and belongs to level 1, the set {∅} has rank 1 and belongs to level 2, and so on.
You may encounter talk of classes. One can begin by thinking of these as collections like sets — and indeed every set is a class. However there are also thought to be some classes that are too “big” or “complex” to be sets. If there is some rank such that all the members of the class are ≤ that rank, then the class is allowed to be a set. If not, it’s not a set but a proper class. A distinguishing property of these is that sets can be members of other classes (sets or proper classes). However, proper classes can’t be members of anything.
The reason for making this distinction is that it enables us to avoid some paradoxes, like Russell’s set of all sets that aren’t members of themselves. In today’s mainstream set theory, every set fails to be a member of itself — but there is no set that contains them. (There is a class that contains them, but it’s not a set. It’s a proper class.) Neither is there any set that contains all the singleton sets (of arbitrary rank). In both cases, this is because there is no bound on the maximum rank of the things we’re talking about.
You might wonder: well, if there can be classes of such things but not sets, why don’t we just repeat Russell’s paradox at the level of classes? That is, what about the class of all classes that aren’t members of themselves? Well, among the classes that aren’t members of themselves are some proper classes, and these can’t be the members of anything. So there can be no class whose members are all classes that aren’t members of themselves, any more than there can be a set of all sets that aren’t members of themselves.
It may be helpful to think of all the class-talk as really just a covert way of talking about predicates. When we say that some classes are sets, we mean that some predicates define a set. When we say that other classes aren’t sets, we mean that other predicates don’t. In the latter cases, the classes can’t be members of anything because really there is no object that contains exactly the objects satisfying the predicate. Some of our talk about classes makes it look like they are a special kind of object, that somehow magically resists being a member of anything. But perhaps its best to think of that talk as superficial and misleading.
If you study set theory, you will learn more about this. We are not going to discuss it further here. I mention the notions of rank and class just to give you some orientation, so that if you encounter talk about these, you know where it should go in your mental map.
Created by jimpryor@unc.edu
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2023-09-16 10:55 EDT |
1008 | https://artofproblemsolving.com/wiki/index.php/Mathematical_notation?srsltid=AfmBOooBJPSFn6fHA4bUlDDKbZpzREGKH7ZqQPoFv1SnnK6lbIXpPH_G | Art of Problem Solving
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Mathematical notation
This is a list of symbols and conventions in mathematical notation.
Sets
: the integers (a unique factorization domain).
: the natural numbers. Unfortunately, this notation is ambiguous -- some authors use it for the positive integers, some for the nonnegative integers.
: Also an ambiguous notation, use for the positive primes or the positive integers.
: the rationals (a field).
: the reals (a field).
: the complex numbers (an algebraically closed and complete field).
: the -adic numbers (a complete field); also and are used sometimes.
: the residues (a ring; a field for prime).
When is one of the sets from above, then denotes the numbers (when defined), analogous for . The meaning of will depend on : for most cases it denotes the invertible elements, but for it means the nonzero integers (note that these definitions coincide in most cases). A zero in the index, like in , tells us that is also included.
Definitions
For a set , denotes the number of elements of .
divides (both integers) is written as , or sometimes as . Then for , or is their greatest common divisor, the greatest with and ( is defined as ) and or denotes their least common multiple, the smallest non-negative integer such that and
When , one often says that are called "coprime".
For to be squarefree means that there is no integer with . Equivalently, this means that no prime factor occurs more than once in the decomposition.
Factorial of :
Binomial Coefficients:
For two functions the Dirichlet convolution is defined as . A (weak) multiplicative function is one such that for all with .
Some special types of such functions:
Euler's totient function: .
Möbius' function: .
Sum of powers of divisors: ; often is used for , the number of divisors, and simply for .
For any it denotes the number of representations of as sum of squares.
Let be coprime integers. Then , the "order of " is the smallest with .
For and , the -adic valuation can be defined as the multiplicity of in the factorisation of , and can be extended for by . Additionally often is used.
For any function we define as the (upper) finite difference of . Then we set and then iteratively for all integers .
Legendre symbol: for and odd we define
Then the Jacobi symbol for and odd (prime factorization of ) is defined as:
Hilbert symbol: let and . Then is the "Hilbert symbol of in respect to " (nontrivial means here that not all numbers are ).
When , then we can define a counting function. One special case of a counting function is the one that belongs to the primes , which is often called . With counting functions, some types of densities can be defined:
Lower asymptotic density:
Upper asymptotic density:
Asymptotic density (does not always exist):
Shnirelman's density:
Dirichlet's density(does not always exist):
and are equal iff the asymptotic density exists and all three are equal then and equal to Dirichlet's density.
Often, density is meant in relation to some other set (often the primes). Then we need with counting functions and simply change into and into :
Lower asymptotic density:
Upper asymptotic density:
Asymptotic density (does not always exist):
Shnirelman's density:
Dirichlet's density(does not always exist):
Again, the same relations as above hold.
See Also
Mathematical convention
This article is a stub. Help us out by expanding it.
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1009 | http://aris.gusc.lv/ChemFiles/CarnegieMellonUChem/Programs/Courses/03740/FOD/ABC97PS0FOD.html | First Order Decay
Advanced Biochemistry Fall Term, 1998 First Order Decay Kinetics
This page expands on some of the topics in PS #0, prob 2. Recall that we derived equations that described the following first order reaction, A --> B. The fraction A remaining after time, t, is [A]/[Ao] = exp(-kt). One way to evaluate k, the rate constant, is to take logarithms, yielding, ln([A]/[Ao]) = - kt; alternatively, 2.303 x log([A]/[Ao]) = -kt. A graph of the LHS vst will give us k as the -slope.
First order decay simply means that for a population of atoms (e.g. radioactive), molecules (our example of A --> B), or anything else, a constant fraction/unit time is converted to something else. The actual fraction/unit time is expressed as k (the rate constant, in units of time).
In the above example, if k = 0.10/min, then 10% of A at any time will be converted to B in the next one minute. The following four graphs show different ways of displaying first-order decay data. The values for [A]/[Ao] at the times indicated were calculated with k = 0.10 min. All four of these display formats are in common use in the biochemical literature. Which do you prefer?
A. Plot of [A]/[Ao] vs time.
B. Plot of [A]/[Ao] vs time, with the Y-axis logarithmic.
C. Plot of ln([A]/[Ao]) vs time.
D. Plot of log([A]/[Ao]) vs time.
E. Plot of [A]/[Ao] vs time, for a short time interval.
Here we plot only the first portion of the reaction, with linear fits to the data. This approach is termed the "initial rate" method. Let's evaluate k from the slopes of these linear fits. a) The "black data" correspond to the first 5% of the reaction. A linear fit yields k = 0.098 min, about 2% lower than the true value of 0.10 min. b) The "blue data" correspond to the first 10% of the reaction. A linear fit yields k = 0.095 min, about 5% lower than the true value of 0.10 min. c) The "red data" (first 20% of the reaction) give us k = 0.091 min, about 9% too low. This example shows that an exponential curve (or any other curve) can be approximated by a straight line over small regions. However, it is usually best to confine linear fits of initial rate data to the first 5% or less of a first order decay curve. This limitation can be an advantage if we know that a very slow process is characterized by first order decay; then we can determine k without following the reaction for large fraction reactions, i.e. excessive times.
A given rate constant can be converted to any other units of timeby multiplication. For example,
k = 0.1minx 1/60 = 0.00167sec.
k = 0.1minx 60 = 6hr.
etc.
The biochemistry and chemistry literature usually reports first-order rate constants in units of sec. In addition, two other terms are used frequently:
Half-life (or half-time):is the time required for A to decay to 50% of Ao. This measure is a standard term in radioactive decay, e.g.has a half-life of 14.3 days. As shown in the problem set answers,= ln2/k.
Average lifetime (or relaxation time):is the time required for A to decay to 1/e of Ao. The fraction, 1/e, is about 37%; the time required for this fraction reaction is 1/k (i.e.= 1/k). [Can you show that 1/k is, in fact, the average lifetime of the entire population of A molecules? Hint: what is the definition of an average?]
Note that both of the above terms are expressed in units of time. If you have the idea of first order decay under control, you should be able to solve the following problems (ideally, without recourse to graphs or equations) with a calculator.
1. A radioactive sample containingregistered 1000 cpm at noon today.
How many cpm will be registered four weeks from now? How many cpm will be registered at noon tomorrow?
What is the average lifetime of a-labeled compound?
2. For our reaction A --> B (k = 0.1min), what fraction of A will be converted to B after 2 relaxation times? 3 relaxation times? 4 relaxation times?
3. An average American car depreciates at about 20% per year. How much will my new $10,000 Geo be worth in two years? How much will it be worth in four years? When will it be worthless?
Answers to the above problems are available.
Final comment: e, the base of natural logarithms, appears in all of the equations for first order decay (and in many other equations in science). A history of this curious number (2.718...) and a lucid description of simple calculus can be found in
Manor, Eli (1994). e The Story of a Number Princeton Univ. Press, Princeton, NJ.
Return to thePS #0 Graphs page.
Back to ABC98 Home Page |
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Use of inequality $1 - \cos (x) \leq x^2 /2$
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$\begingroup$
In this answer the value of $1 - \cos(x)$ has to be evaluated in order to find its upper limit, if it exists.
In particular, $x = 2 \pi / n$. The answer is related to the length of a side of a regular $n$-gon inscribed into a unit-radius circumference; because the perimeter of the $n$-gon is always less than $2 \pi$, the single side must always be less than $2 \pi / n$.
The inequality
$$1 - \cos (x) \leq \displaystyle \frac{x^2}{2}$$ (1)
is used and the proof is completed with
$$2(1 - \cos(x)) \leq (2 \pi / n)^2$$
$$\sqrt{2(1 - \cos(x))} \leq 2 \pi / n$$
But it is well known that the cosine is a function $f(x) \in [-1;1]$, so $1 - \cos (x) \in [0,2]$. By using this information, we would obtain
$$1 - \cos (x) \leq 2$$ (2)
The proof would provide
$$2(1 - \cos(x)) \leq 4$$
$$\sqrt{2(1 - \cos(x))} \leq 2$$
which is a completely different result.
Why in that case it is preferable to use (1) instead of (2)?
How to choose when it is convenient to use (1) and when to use (2) in a proof?
calculus
real-analysis
trigonometry
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edited Apr 13, 2017 at 12:20
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asked Mar 16, 2016 at 18:52
BowParkBowPark
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1
$\begingroup$ Strange question. The inequality $(2)$ is trivial and doesn't help addressing the $n$-gon problem. $\endgroup$
user65203
– user65203
2017-12-09 16:16:41 +00:00
Commented Dec 9, 2017 at 16:16
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3 Answers 3
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$\begingroup$
The bound $-1\le \cos(x)\le 1$ is correct, but rather a crude one.
We can easily obtain tighter bounds using the inequality from elementary geometry
$$|\sin(\theta)|\le |\theta| \tag 1$$
for all $\theta$, and the half-angle formula for the sine function
$$1-\cos(\theta)=2\sin^2(\theta/2) \tag 2$$
Squaring both sides of $1$, substituting $\theta =x/2$, and using $(2)$ reveals
$$\begin{align} \sin^2(x/2)&=\frac{1-\cos(x)}{2}\\ &\le x^2/4 \tag 3 \end{align}$$
whereupon we find for all $x$
$$\bbox[5px,border:2px solid #C0A000]{1-\cos(x)\le \frac12 x^2} \tag 4$$
For values of $x<2$, $\frac12 x^2<2$ and $(4)$ provides a tighter bound than $1-\cos(x)\le 2$. For $x\ge 2$, it is still correct that $1-\cos(x)\le \frac12 x^2$, but the inequality $1-\cos(x)\le 2$ is obviously tighter. Therefore, we can write
$$1-\cos(x)\le \begin{cases}\frac12 x^2&,x<2\\2&,x\ge 2\tag 5\end{cases}$$
So, $(5)$ provides a guideline for the appropriate use of the bounds for $1-\cos(x)$.
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edited Jun 26, 2021 at 15:25
answered Mar 16, 2016 at 19:37
Mark ViolaMark Viola
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4
$\begingroup$ Yesterday I open this question and when I realize result was irrelevant, I did not even read the whole. Clearly, I made a mistake and I am struggling to understand how. But in any case I would like to see 'Would the downwoter care to comment?'. Cowardly? Please be respectful. $\endgroup$
Atbey
– Atbey
2017-12-09 02:24:01 +00:00
Commented Dec 9, 2017 at 2:24
$\begingroup$ Unfortunately, I am unable to redo downvote. Please make an unnoticeable edit if possible. $\endgroup$
Atbey
– Atbey
2017-12-09 02:24:54 +00:00
Commented Dec 9, 2017 at 2:24
$\begingroup$ @atbey I've edited and removed the comment. $\endgroup$
Mark Viola
– Mark Viola
2017-12-09 16:13:02 +00:00
Commented Dec 9, 2017 at 16:13
$\begingroup$ @atbey Much appreciative! $\endgroup$
Mark Viola
– Mark Viola
2017-12-09 20:02:29 +00:00
Commented Dec 9, 2017 at 20:02
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Let $x\gt 0$. You know that $\sin(x) < x$ for such $x$. Integrating $$ 1 - \cos(x) =\int_0^x \sin(t)\, dt < \int_0^x t\,dt = {t^2\over 2}$$
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answered Mar 16, 2016 at 19:40
ncmathsadistncmathsadist
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ncmathsadist's technique can be extended for a tighter bound as described below. It also extends the region over which the bound is better than $1 - \cos (x) \leq 2$ per Mark Viola's answer.
Cutting off the Taylor series for $\sin(t)$ before a negative term (the terms alternate sign), one gets a polynomial bound on $\sin(t)$ that can be integrated.
For example $\sin (t) \leq t-t^3/3!+t^5/5!$, therefore
$$1-\cos (x)=\int_0^x \sin (t)\ dt \leq \int_0^x t-\frac{t^3}{3!}+\frac{t^5}{5!}\ dt = \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} $$
The cut-off approximation is an upper bound. The cut-off residual will always be negative. This can be shown by summing the terms pairwise. Since $\sin (t)$ is absolutely convergent, there isn't an issue with ordering of terms.
A bound which is better than $1 - \cos (x) \leq 2$ from $[0,2\,\pi]$ is: $$1-\cos (x) \leq \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \frac{x^{8}}{8!} + \frac{x^{10}}{10!} - \frac{x^{12}}{12!} + \frac{x^{14}}{14!} \leq 2$$
You can then see that all this really says is: $$\cos (x) \geq 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^{8}}{8!} - \frac{x^{10}}{10!} + \frac{x^{12}}{12!} - \frac{x^{14}}{14!}$$ ...or $\cos (x)$ is greater than its Taylor expansion with the cut-off starting at a positive term.
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edited Feb 13, 2024 at 22:08
answered Oct 22, 2020 at 22:40
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1011 | https://www.youtube.com/watch?v=J-JxOoEABeI | Geometric Construction of Tangents to a Circle through an External Point Lesson
Lindsay Bowden
3020 subscribers
2 likes
Description
522 views
Posted: 5 Dec 2024
Welcome to this comprehensive lesson on Geometric Constructions: Tangents to a Circle Through an External Point! In this high school geometry video, we’ll explore the essential construction technique of drawing tangents to a circle from an external point. Using clear, step-by-step instructions, I’ll guide you through this important geometric skill, perfect for building a strong foundation in geometry.
🔑 What you'll learn:
✔ How to use a compass and straightedge to construct tangents from an external point to a circle accurately
➡️ Find these notes here:
💡 For more lessons and resources, check out my educational store:
🔗 Other helpful videos:
✔ Constructing Incircle of a Triangle:
✔ Constructing Circumcircle of a Triangle:
✔ Constructing Hexagon Inscribed in a Circle:
🎓 About Lindsay Bowden:
With a Bachelors of Science in Education, M.Ed in Curriculum and Instruction, and Ed.S in Curriculum and Instruction, I specialize in creating engaging secondary math resources tailored for on-level and special education classrooms. Check out my full curriculum bundles and additional lessons that will make your math class a breeze!
🌐 Connect with me:
● Website:
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Transcript:
welcome to this video on constructing tangents to a circle through an external point so in this video you'll be given a circle and you'll be given a point the first thing that you want to do is connect the center of the circle and the point then you want to draw the perpendicular bisector of that segment that you just drew so remember you go about three fourths away up down up down and then connect then without adjusting we're going to put our center or our compass on the center and we're going to draw an arc and an arc and then we will connect all right so let's try that so i'm going to connect these i'm going to draw the perpendicular bisector of that segment so about three-fourths of the way above below above below all right i'm going to label this a and this b like i have all right next we're going to put our compass on the center of the circle and extend our compass out to point b then we're going to put our compass on point b and draw arcs on the circle above and below like so and then we will connect those arcs to point a all right if you want to pause the video now and try this one you can all right so the first thing we do is we draw a line segment so let's make sure it's really straight by the way you definitely want to use a pencil when constructing all right so perpendicular bisector of that line segment let's go over halfway above below above below and as you can see here i didn't make the arc long enough that's okay just go back and make it longer that's why you want to use a pencil so you can make adjustments all right and then i'm going to connect those all right then i'm going to measure from b to a then i'm going to draw an arc up and down on the circle let's see let's use a out here like i have in the example and then we're going to connect a to both arcs on the circle to be tangent and remember a tangent just touches once it touches the circle at one spot so both of these are tangents and we also know because they are both tangents they go to the same point these should be equal congruent segments |
1012 | https://www.youtube.com/watch?v=4yo8pv1NXUs | Limite de resistencia a la fatiga parte 2
Consultores en Ingeniería Mecánica
694 subscribers
78 likes
Description
3360 views
Posted: 26 Mar 2020
En este video se define la ecuación de Marín, exlicando los factores de superficie, tamaño, carga, temperatura y confiabilidad
4 comments
Transcript:
[Música] ahora ya calculamos cuál va a ser la resistencia a la fatiga del material con el que está hecho la probeta pero cuando vayamos a diseñar una pieza seguramente no va a tener ni la forma que tiene la probeta tampoco va a tener el tamaño que tiene la probeta tampoco va a tener el acabado superficial que tiene la probeta tampoco va a tener la temperatura de operación con el que se fracturó la probeta tampoco va a tener el tipo de carga que es flexión en la probeta entonces esas condiciones que se tuvo por la fractura de la probeta cuando vayamos a diseñar una pieza algunas si se van a aparecer otras no por lo tanto todo lo que se explicó de forma verbal las diapositivas pasadas aquí lo tenemos con una ecuación esta ecuación se le conoce como la ecuación de marín y nos dice la resistencia a la fatiga real de la pieza que vamos a diseñar va a depender de cinco factores esos factores van a modificar la resistencia y la fatiga de la probeta que se conoce como ese prima y entonces nosotros tenemos que calcular cada uno de esos factores el factor k se refiere el factor de modificación de la condición superficial recuerden que la probeta es un acabado tipo espejo las piezas que vamos a diseñar no todos van a tener ese acabado superficial tenemos un factor cabe que es el de tamaño el tamaño de la probeta ya está normalizado pero nosotros seguramente vamos a diseñar piezas mucho más grandes o más pequeñas el factor de modificación de carga casi la probeta se llevó la fractura por flexión a lo mejor nosotros vamos a diseñar piezas que se sometidas a flexiones o atención o torsión etcétera factor de modificación de la temperatura que es el factor care el factor de confiabilidad el factor k en el s prima que es el límite de resistencia la fatiga de la probeta y este es el valor que queremos encontrar que es el límite de resistencia a la fatiga en la ubicación crítica en una parte de la máquina en la geometría y condición de humus factores superficie y como lo calculamos esto en la probeta teníamos una probeta pulida con un pulimento muy fino en dirección axial en la pieza real que vayamos a diseñar se va a tener un acabado superficial bajo esto va a provocar el límite de resistencia a la fatiga de la probeta disminuye porque porque hay una mayor rugosidad en la superficie que produce un fenómeno de concentración de esfuerzos que es lo que ya dijimos anteriormente y tenemos esta actuación el factor de superficie ocupamos conocer su esfuerzo último a la atención y dependiendo el acabado superficial van a hacer los valores de a y debe que vamos a utilizar en esta ecuación muchas de las piezas que vamos a diseñar por darles un ejemplo van a tener que ser maquinadas para alcanzar su tamaño definitivo si nosotros estamos trabajando en el sistema internacional con unidades en mega pascal es primero calculamos el s ute del material y el valor a para esta condición maquinal va a ser 4.51 y el exponente b va a ser menos punto 265 por otra parte si nosotros estamos manejando el sistema inglés entonces vamos a utilizar el factor al de 2.7 y el exponente de va a ser exactamente el mismo factor de tamaño la probeta su sección transversal es circular y tenemos un diámetro normalizado con estas dimensiones en la vida real o en la pieza real puede que tengamos una sección circular así como la probeta pero con un diámetro diferente si la pieza que vamos a diseñar tiene esas condiciones de sección circular pero con diámetro diferente y la carga que provoca fatiga en el material es una carga axial entonces el cable es igual a 1 independientemente si la pieza está muy grande o muy pequeña ese factor de tamaño es uno sin embargo si la pieza que vamos a diseñar es sección circular y tiene un diámetro diferente y está sometida a flexión o torsión entonces tenemos que identificar cuál es el diámetro y si estamos nosotros en el sistema internacional tenemos que checar este diamante diferente en qué rango cae si por ejemplo acá en el rango de 2.79 a 51 milímetros la ecuación para encontrar el factor de tamaño va a ser cabe igual a 1.24 veces el diámetro elevado a la menos punto 107 por otra parte si el diámetro que vamos a diseñar la pieza cae entre 51 y 254 milímetros ahora la ecuación que vamos a usar es cabe es igual a 1.51 veces el diámetro de diseño elevado a la menos punto 157 ahora si la pieza que vamos a diseñar es circular y tiene un diámetro diferente pero la pieza no es rotativa vamos a usar este diámetro equivalente antes de seguir a qué se refiere esto la probeta cuando la montamos a la máquina la probeta gira y la carga es constante hacia abajo en la máquina entonces la variación de esfuerzos es provocado porque la pieza está girando habrá casos en el que nosotros diseñamos una pieza donde la pieza no vaya a girar y quien esté provocando la variación de esfuerzos va a ser flexión alternante o torsión alternante si tenemos ese caso donde la pieza que se va a diseñar es estática y quien provoca la variación de esfuerzos es flexión relativa y torsión rotativa entonces el diámetro de la pieza que vamos a calcular lo multiplicamos por lo multiplicamos por punto 37 y este diámetro equivalente lo vamos a buscar en cuál de estos dos rangos cae si cae en éste metemos el diámetro equivalente y hacemos esta operación si la pieza que vamos a enseñar es una sección no circular entonces vamos a usar esta ecuación dice aquí el diámetro de una probeta de sección circular sometida a flexión rotativa que tenga un área del 95% del esfuerzo igual al área del 95% del esfuerzo de la sección considerada si tenemos una sección rectangular que va a ser sometida a flexión a torsión yo tengo aquí esta es la base esta es la altura si yo conozco las dimensiones de la pieza que voy a enseñar respecto a su base y esa altura multiplicamos estos dos valores le sacamos la raíz y esto lo multiplicamos por 808 este diámetro equivalente que nos arrojó va a ser el diámetro que se puede ligar a una sección circular con esta ecuación ya que tengamos este auto equivalente lo vamos a introducir en cualquiera de estas dos relaciones para obtener el talle correspondiente tenemos aquí un factor de modificación de tanna si la pieza que vamos a diseñar está sujeta flexión pues es uno porque porque la probeta se llevó la fractura con flexión entonces la pieza que voy a diseñar también la forma en que vamos a fatigar la flexión entonces el tc es el factor de modificación la tarde pues es igual a 1 si la pieza la vamos a llevar la fractura con una fuerza axial el factor que hace es 0.85 si vamos a aplicar torsión para fracturar la pieza de 0.59 factor de temperatura la probeta cuando se llevó la fractura se se hizo bajo una temperatura de 20 grados centígrados si la pieza que vamos a diseñar va a operar a una temperatura de 20 grados centígrados entonces el factor es igual a 1 sin embargo si nosotros esta pieza que vamos a diseñar va a ser sometida a una temperatura de 400 grados centígrados entonces ese factor de temperatura su valor va a ser cero un problema y así sucesivamente factor de confiabilidad en esta tabla del lado izquierdo tenemos los factores de confiabilidad más bien los porcentajes de confiabilidad y la tercera columna tenemos el factor de confiabilidad si nosotros queremos que los datos de la resistencia y la fatiga tengo una confiabilidad del 50 por ciento entonces vamos a utilizar un factor de confiabilidad de pero como es lógico nosotros queremos que los resultados sean mucho más confiables entonces entre más alto sea el valor de confiabilidad vean que pasan con los factores de confiabilidad empiezan a reducirse entonces si yo necesito que los datos de mi diseño tengan una confiabilidad del 99.9 por ciento entonces yo voy a utilizar un factor de confiabilidad de 0.750 y tres bien pues con esto terminamos la parte teórica del diseño bajo carga dinámica o fatiga y en los próximos vídeos vamos a ver algunos ejemplos de cómo hacer el cálculo de el límite de resistencia y la fatiga de la pieza real vamos a calcular los esfuerzos alternantes los esfuerzos medios dependiendo el problema pues vamos a calcular el factor de seguridad o definir tamaños y formas de las piezas para que cumplan con las condiciones del problema muchas gracias y nos vemos en la próxima sesión |
1013 | https://battlingblades.com/blogs/news/legendary-blades-10-famous-sword-names-from-mythology?srsltid=AfmBOoqTsT0rp0wkmNOLplvk8Vut4RDquP9Ra27IPKQ66d1OAIdGlaWm | Fall Deals: Up To 80% Off
Legendary Blades: 10 Famous Sword Names From Mythology
Famous swords and mythological weapons have played a crucial role in human history and culture, especially in warfare and symbolism.
Throughout, legendary blades have been revered as mythical weapons wielded by heroes and villains, embodying humanity's virtues and vices.
From holy swords blessed by gods to demonic swords in mythology, these legendary weapons have captured our imagination for centuries.
As fellow medieval weaponry enthusiasts, we've researched and compiled a list of the top 10 famous sword names from mythology.
Whether you're a fan of famous swords in Greek mythology, Norse sagas, or Arthurian weaponry, you'll find fascinating insights here!
And if you're looking to own a piece of history yourself, we also recommend swords from Battling Blades.
So grab your scabbard, and let's journey through the realm of famous weapons in mythology!
Famous Sword #1: Excalibur:
One of the most famous sword names in mythology is Excalibur, the holy sword name of King Arthur’s weapon.
Excalibur was a sword best known for its origins as the “sword in the stone”.
This meant only the man who was prophesied to be the ‘true’ monarch of Avalon (England) could pull it out.
Young Arthur, a lowly squire, pulled out the sword and claimed his rightful place as king.
With this claim, the Lady of the Lake gave him the ownership of the sword.
Excalibur thus was more than just a famous sword; it represented Arthur's legitimacy, power, and destiny.
It was also said that the sword was indestructible and magical, as all real legendary swords tend to be.
These included the ability to glow brightly, inflict mortal wounds, and return to its owner's hand when thrown.
Excalibur also had a sheath, which granted its bearer invincibility in battle but was eventually lost or stolen.
More importantly, however, Excalibur's significance in Arthurian legend extends beyond its physical attributes.
The famous sword represented the ideals of chivalry, the Knight’s Code, and the moral values of justice, loyalty, and honor.
It was also the key to many of Arthur's quests, including the search for the Holy Grail.
If you're looking for a sword that captures the spirit of legendary blades like Excalibur, you’re in the right place.
Battling Blades offers a range of high-quality medieval swords that are similar in design and craftsmanship.
One such range of products are the Claymore swords, two-handed swords with long, straight blades and cross-shaped hilts.
Made of carbon steel and featuring intricate details on the handle and pommel, these swords are both functional and decorative.
Famous Sword #2: Durandal
Durandal is a legendary sword with its roots in French mythology.
According to the legends, Durandal was originally wielded by the great knight Roland, one of Charlemagne's twelve peers.
The story is closely tied to the Battle of Roncevaux Pass, where Roland fought against an army of invading Saracens.
Note: Learn more about the battle here!
As the legend goes, Roland used Durandal to strike a rock, splitting it in half.
This created a pass through the mountains that allowed the retreating Christian army to escape.
Moreover, Durandal was known for its incredible sharpness, strength, and divine origins.
Durandal was a longsword with a single-edged blade and a unique gold hilt that was adorned with precious gems.
More importantly, it was ‘indestructible’, which means it even sliced through iron without being destroyed.
At Battling Blades, we offer a range of longswords, including the Dragon Longsword.
It features a high-carbon Damascus steel blade and is ornately crafted.
While it belongs to a far later time period, this blade follows the long history of longswords.
And as a recipient of that history, the replica swords by Battling Blades are expertly crafted.
Famous Sword #3: Kusanagi-no-Tsurugi
Kusanagi-no-Tsurugi, or the Grass-Cutting Sword, is a legendary sword from Japanese mythology.
According to legend, the sword was discovered by storm god Susanoo in the body of an eight-headed serpent he killed.
The famous sword is said to have been given to the sun goddess Amaterasu.
Ameterasu then passed it down to her grandson, the first emperor of Japan.
Kusanagi-no-Tsurugi is a long, curved sword with a double-edged blade.
It is often depicted with a jewel or mirror on its hilt.
The sword is considered one of the three Imperial Regalia of Japan and is a symbol of the emperor's authority.
Battling Blades offers a wide range of Japanese swords in memory of the coolest swords in history.
This refers to their range of replica katana swords, which may well be some of the most popular real legendary swords.
Their "Ninja Sword Katana" features a straight, single-edged blade inspired by traditional Japanese designs. And ninjas, of course!
Famous Sword #4: Gram
Gram is a legendary blade in mythology owned and used by Sigurd, the hero of Norse mythology.
According to the legend, the dwarves forged Gram.
It was thus said to be a powerful weapon capable of cutting through iron as if it were cloth.
The famous sword was given to Sigurd by his mentor, Regin.
Regin then convinced his student to slay the dragon Fafnir and take the treasure it had been guarding.
With Gram, Sigurd went on to accomplish this feat and became one of the greatest heroes of Norse mythology.
And indeed, he went on to become a very famous swordsman in mythology.
In any case, Gram was described as a longsword.
This meant it could be wielded with one hand or two, as with all ‘Bastard’ swords.
Furthermore, it claimed a keen edge and was remarkably enduring, thus allowing it to remain unmarked even past the toughest battles.
And, of course (because famous swords are opulent), the sword's hilt was made of gold and decorated with intricate designs.
Most importantly, its blade claimed to be sharp enough to cut through an anvil.
At Battling Blades, we offer a variety of swords that would be similar to Gram, especially our range of longswords.
In fact, our Longsword/Bastard Sword is the perfect example!
As a hand-and-a-half sword, it’s a perfect representation of Gram, especially since it’s forged with High Carbon Damascus Steel.
With its sharp, sturdy blade and elegant design, this longsword is the perfect addition to any collection of historical weaponry.
Of course, if you’re more interesting in checking out viking swords, Battling Blades offers a whole collection of those, too!
Famous Sword #5: Joyeuse
Joyeuse is a famous sword name that most associate with Charlemagne, the first Holy Roman Emperor.
According to the legend, Joyeuse was a holy sword forged by a famed smith, Galas.
This famous sword had the power to protect its owner and give victory in battle.
In short, though, Joyeuse is defined as a longsword.
It has a blade that is 32 inches long and 1.8 inches wide.
It was also said to have three different colors on the blade - gold, red, and blue.
These represented the virtues of Faith, Hope, and Charity, thus also cementing Joyeuse’s position as one of the religious swords.
And, of course, as a sword, Joyeuse was known for its enduring, long-lasting power.
Moreover, it's supposed 'magical properties' could apparently help its wielder in battle.
Battling Blades offers a variety of longswords that are similar in style and function to Joyeuse.
One such sword is the Norman Knightly Longsword.
This particular blade is based on the design of medieval European swords.
Furthermore, it features a blade made of high-carbon Damascus steel and is just slightly longer at 38 inches.
Like Joyeuse, the Norman Knightly Longsword is an incredible blade, amazing as a replica sword for a collector or re-enactor.
Famous Sword #6: Zulfiqar
Zulfiqar is a legendary blade that is pretty important in Islamic mythology and history, especially Shia history.
According to the story, the angel Jibrael gifted it to the Prophet Muhammad.
The sword was then used by Ali ibn Abi Talib, the son-in-law of the Prophet, in the Battle of Uhud.
Moreover, it is believed to have been used by other leaders and is considered a symbol of strength and power.
In terms of design, the sword has a distinctive bifurcated structure with two points.
This is in some ways believed to represent the dual nature of Islam: the temporal and the spiritual.
Moreover, the blade is evidently made of steel and is adorned with decorative inscriptions and religious symbols.
And as one of the most popular religious swords, it also holds a place in the hall of mythological weapons.
In any case, in popular Shia depictions, this famous holy sword is seen as a scimitar.
Battling Blades offers a similar selection of replica legendary blades in its collection of scimitars.
One of these is the Arabian Scimitar made of of Damascus Steel and reaching 37 inches.
This sword features a curved blade and a double edged design.
Most importantly, as a replica, it’s a powerful representation of a legendary blade wielded by a famous swordsman in mythology.
Famous Sword #7: Tyrfing
Tyrfing is a famous sword name from Norse mythology that plays a significant role in many legendary tales.
According to the legends, the sword was created by dwarves for Svafrlami, the king of Gardariki.
However, the sword was cursed (associated with demonic sword names), bringing great misfortune to those who wielded it.
The sword is said to be exceptionally sharp and deadly, capable of cutting through any armor or shield.
It’s also said never to miss its target, making it a deadly weapon in the hands of a skilled warrior.
Note: Learn more about the story behind Tyrfing here.
Keeping such demonic swords in mythology in mind, let’s talk about replicas.
Battling Blades offers a range of swords similar in memory and association to Tyrfing.
There’s a great divide in terms of the functionality of the sword, but it’s association to Norse mythology is clear.
So why not check out Viking swords due to their association with Norse history?
Moreover, these viking swords are crafted from high-quality materials which makes them both durable and functional.
You can specifically check out the Carolingian Viking Sword! It’s 38 inches long and made of high carbon Damascus steel.
With their sleek and stylish designs, these swords are perfect for collectors, enthusiasts, and anyone else who’s interested!
Famous Sword #8: Hrunting
In Anglo-Saxon mythology, Hrunting is one of the famous weapons names wielded by Beowulf in his battle against Grendel.
According to the tale, Hrunting was given to Beowulf by the blacksmith Unferth as a symbol of support and admiration.
The sword was said to be enchanted and was believed to have magical powers to invincible the wielder in battle.
Hrunting was a relatively famous sword - it had a keen edge and a finely crafted hilt according to the story.
It was believed to be a longsword or broadsword; in other words, a double-edged blade with a pointed tip.
The legendary blade was also said to be made from the finest materials.
Additionally, the edge was razor-sharp and nigh-unbreakable.
Some legends also claim that the sword had magic powers!
These powers, of course, ensured the blade would glow for the user in the presence of danger.
Of course, magic isn’t real… but blades can still be replicated.
Battling Blades offers a variety of longswords that are similar in construction and design, especially in terms of edges and lengths.
Choose a replica sword today!
Famous Sword #9: Harpe
Harpe is probably one of the most famous swords in Greek mythology, and its story is steeped in significance.
However, Harpe was first wielded by Kronos, father of Zeus, when he fought against the Olympian gods with other titans.
Perseus later used it to slay Medusa, who had snakes as hair and could petrify people with a look.
If you're interested in owning a sword similar to Harpe, Battling Blades offers a wide range of Greek swords.
Their collections include swords with curved blades, hooks, and even knives, helmets, and armor!
One of their products is the Kopis sword, made of high carbon steel and 19 inches long.
It is the perfect start for collecting replicas of the coolest swords in history.
Famous Sword #10: Ascalon
Ascalon was one of mythology’s legendary holy swords with a significant history in Christian mythology.
According to the legends, the famous sword was given by Saint Catherine of Alexandria to Saint George.
Saint George was a Christian martyr and a protector of the Kingdom of Lydda in Palestine.
Thus, Saint George used Ascalon to slay the dragon terrorizing the people of Lydda, making him a revered figure.
Even beyond his persona, though, Ascalon itself remains one of the most venerable religious swords in mythology.
It is often referred to as a long, broadsword capable of piercing through any armor.
Moreover, it is said to have been made from the same metal as the nails used to crucify Jesus Christ.
Additionally, the sword also has a cross engraved in it, symbolizing its connection to Christianity as a holy sword.
Keeping note of the religious perspective, Battling Blades recalls another conneciton to the battle for Christianity.
The Crusader Bundle, in memory of Christianity’s battle for domination, offers multiple products.
One of these is a longsword representing a high quality replica of a famous sword.
Alone, it might not represent Ascalon, but with the other items, you can portray a true Christian Crusader.
The Bottom Line
Swords have played a significant role in shaping the mythologies and histories of various cultures worldwide.
From the Norse and Irish mythologies to Christian and Islamic beliefs, swords have been revered as symbols of power, heroism, and justice.
Even today, swords captivate people's imaginations and inspire awe and respect.
Of course, there are many more mythological weapons and legendary blades throughout history that we haven’t covered.
Find a more exhaustive list of those mythical weapons (specifically swords) here!
On the other hand, if you’d like a replica or a real life sword for your own collection, Battling Blade offers great products!
These products are meticulously crafted to match the original swords in appearance and performance, making them an absolute must-have!
In summary, the timeless appeal of swords continues to endure, and Battling Blades is keeping that legacy alive by providing authentic replicas of the most famous swords from myth and history.
With their commitment to quality and craftsmanship, Battling Blades is the go-to destination for sword enthusiasts and collectors.
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Chapter 2: Graphing Trigonometric Functions
Math 175: Plane Trigonometry
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Thu, 13 May 2021 16:12:38 GMT
2.1: Radian Measure
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Contents
1. Home
2. Campus Bookshelves
3. Rio Hondo
4. Math 175: Plane Trigonometry
5. Chapter 2: Graphing Trigonometric Functions
6. 2.1: Radian Measure
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Math 175: Plane Trigonometry
Front Matter
Chapter 1: Right Triangles and an Introduction to Trigonometry
Chapter 2: Graphing Trigonometric Functions
Chapter 3: Trigonometric Identities and Equations
Chapter 4: Inverse Trigonometric Functions
Chapter 5: Triangles and Vectors
Chapter 6: The Polar System
Back Matter
2.1: Radian Measure
Last updated May 13, 2021
Save as PDF
Chapter 2: Graphing Trigonometric Functions
2.2: Applications of Radian Measure
Page ID 61242
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Contributors and Attributions
Learning Objectives
Define radian measure.
Convert angle measure from degrees to radians and radians to degrees.
Calculate the values of the 6 trigonometric functions for special angles in terms of radians or degrees.
Figure 2.1.1 Angle θ and intercepted arc AB⏜ on circle of circumference C=2πr
In Figure 2.1.1 we see that a central angle of 90∘ cuts off an arc of length π 2 r, a central angle of 180∘ cuts off an arc of length π r, and a central angle of 360∘ cuts off an arc of length 2π r, which is the same as the circumference of the circle. So associating the central angle with its intercepted arc, we could say, for example, that
360∘"equals”2π r(or 2π’radiuses’).
The radius r was arbitrary, but the 2π in front of it stays the same. So instead of using the awkward "radiuses'' or "radii'', we use the term radians:
(2.1.1)360∘=2π radians
The above relation gives us any easy way to convert between degrees and radians:
(2.1.2)Degrees to radians:x degrees=(π 180⋅x)radians(2.1.3)Radians to degrees:x radians=(180 π⋅x)degrees
Equation 2.1.2 follows by dividing both sides of Equation 2.1.1 by 360, so that 1∘=2π 360=π 180 radians, then multiplying both sides by x. Equation 2.1.3 is similarly derived by dividing both sides of Equation 2.1.1 by 2π then multiplying both sides by x.
The statement θ=2π radians is usually abbreviated as θ=2π rad, or just θ=2π when it is clear that we are using radians. When an angle is given as some multiple of π, you can assume that the units being used are radians.
Example 2.1.1
Convert 18∘ to radians.
Solution
Using the conversion Equation 2.1.2 for degrees to radians, we get
18∘=π 180⋅18=π 10 rad.
Example 2.1.2
Convert π 9 radians to degrees.
Solution
Using the conversion Equation 2.1.3 for radians to degrees, we get
π 9 rad=180 π⋅π 9=20∘.
Table 4.1 Commonly used angles in radians
Table 4.1 shows the conversion between degrees and radians for some common angles. Using the conversion Equation 2.1.3 for radians to degrees, we see that
1 radian=180 π degrees≈57.3∘.
Figure 4.1.2
Formally, a radian is defined as the central angle in a circle of radius r which intercepts an arc of length r, as in Figure 4.1.2. This definition does not depend on the choice of r (imagine resizing Figure 4.1.2).
One reason why radians are used is that the scale is smaller than for degrees. One revolution in radians is 2π≈6.283185307, which is much smaller than 360, the number of degrees in one revolution. The smaller scale makes the graphs of trigonometric functions (which we will discuss in Chapter 5) have similar scales for the horizontal and vertical axes. Another reason is that often in physical applications the variables being used are in terms of arc length, which makes radians a natural choice.
The default mode in most scientific calculators is to use degrees for entering angles. On many calculators there is a button labeled DRG for switching between degree mode (D), radian mode (R), and gradian mode (G). On some graphing calculators, such as the the TI-83, there is a MODE button for changing between degrees and radians. Make sure that your calculator is in the correct angle mode before entering angles, or your answers will likely be way off. For example,
sin4∘=−0.0698,sin(4 rad)=−0.7568,
so the values are not only off in magnitude, but do not even have the same sign. Using your calculator's sin−1, cos−1, and tan−1 buttons in radian mode will of course give you the angle as a decimal, not an expression in terms of π.
You should also be aware that the math functions in many computer programming languages use radians, so you would have to write your own angle conversions.
Contributors and Attributions
Michael Corral (Schoolcraft College).The content of this page is distributed under the terms of theGNU Free Documentation License, Version 1.2.
This page titled 2.1: Radian Measure is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation.
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2.2: Applications of Radian Measure
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1015 | https://math.stackexchange.com/questions/4875338/constructing-a-circle-tangent-to-another-circle-and-two-sides-of-a-triangle | Skip to main content
Constructing a circle tangent to another circle and two sides of a triangle
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Given the circle tangent to the sides AB and BC, I want to construct another circle that is tangent to this circle and also tangent to the sides AB and AC.
The center of such circle lies on the angle bisector of ∠A. Furthermore, the locus of the points that are equidistant from both AB and the circle with center O is a parabola with its focus at O and the directrix parallel to AB with the distance equal to the radius of the circle as mentioned here. Thus, the center of such circle can be found by intersecting the parabola and the angle bisector of ∠A But I don't think it's possible to do this construction by using just a compass and a straight-edge.
Another thing that I've noticed is that when two circles are tangent to each other, their centers and their tangent point are placed on the same line. So, in order to do this construction it is enough to find the point of tangency and connect it to the center O and extend it so that it intersects the angle bisector A which gives the center of this circle. But I'm not sure how to find the point of tangency.
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asked Mar 5, 2024 at 7:05
SoheilSoheil
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Can we do co-ordinate bashing, then make some observations maybe?
– D S
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Mar 5, 2024 at 13:26
@DS Thanks for the idea! It is just strange as the construction doesn't "look" hard to be doable with only ruler and compass.
– Soheil
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Mar 5, 2024 at 23:16
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This is one of the Problem of Apollonius cases. Is the two lines and one circle case. Look here: en.wikipedia.org/wiki/Problem_of_Apollonius
– MEEL
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@MEEL Thanks! I think you should post it as an answer.
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As noted in the comments and answers, this is a special case of the Problem of Apollonius, which dates back to antiquity. This special case is examined at cut-the-knot.org, from which this answer is adapted.
OP has already noted that the solution boils down to finding the point of tangency between the given and sought circles. What makes this possible is the observation that the point S at which the circles touch would be a center of a homothety that takes either circle (and its tangents) to the other (and its tangents).
The construction is simple. We construct tangents to the given circle that are parallel to the sides AC and AB. These tangents meet at H, and the line AH consists of centers of homotheties that take tangents AB and AC of the sought circle to the constructed tangents of the given circle (and vice versa). There are two intersections of AH with the given circle. One intersection is on BC and leads to a trivial solution. But the other intersection S gives us the point of contact between the two circles, and the intersection of OS with the angle bisector of ∠A gives us K, the center of the sought circle.
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edited Mar 7, 2024 at 23:48
answered Mar 7, 2024 at 23:41
brainjambrainjam
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Hint.
Considering the segment s=[A,B] and calling α=12∠BAC^, the unknown radius r and x the distance over s from A to the vertical projection of the unknown circle's center, and d0 the distance from A over s to the known circle's vertical projection, we have:
(r+r0)2=(d0−x)2+(r−r0)2
with x=rtanα and r0 the known circle's radius, thus we can calculate r.
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answered Mar 7, 2024 at 21:58
CesareoCesareo
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I am posting this as an answer because I was asked to by the inquirer
This is one of the Problem of Apollonius cases. Is the two lines and one circle case. Look here:
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answered Mar 7, 2024 at 22:29
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Your start and instinct were right, here is how to complete your construct (clockwise from upper left):
Draw the directrix d, bisect the angle of interest, from O draw the perpendicular to the angle bisector with intersection K, mark off KO′=KO, and mark the intersection with the directrix as J. With diameter JK create a circle centered on the line JK (at the unmarked midpoint), and with diameter OO′ create a circle centered on line OO′ (at K), mark their intersection at L. With radius JL and center J draw a circle and mark the intersection with the directrix at M. Draw a perpendicular at M and mark its intersection with the angle bisector at N.
By construction JM=JL, JLK is a circle with diameter JK so ∠JLK is right and therefore JL is tangent to circle OLO′ which has center K. Since JL is a tangent and JOO′ a secant to circle OLO′, JL2=JO×JO′. But JM=JL so JM2=JO×JO′. Now, by symmetry NO=NO′ so the circle with center N and radius NO has the secant JOO′, and since JM2=JO×JO′ it has tangent JM. Thus NO=NM, and N is the intersection of the angle bisector and the parabola defined by focus O and directrix d.
As others have commented, Apollonius tackled constructs like this in his Conics and some of his lost manuscripts. But the above demonstration depends only on Books 1-3 of Elements, and mainly on the proofs on tangents toward the end of Book 3.
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edited Mar 7, 2024 at 23:57
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Double Integrals in Polar Coordinates
To this point we’ve seen quite a few double integrals.
However, in every case we've seen to this point the region
could be easily described in terms of simple functions in Cartesian coordinates. In this section we want to look at some regions that are much easier to describe in terms of polar coordinates. For instance, we might have a region that is a disk, ring, or a portion of a disk or ring. In these cases using Cartesian coordinates could be somewhat cumbersome.
For instance let’s
suppose we wanted to do the following integral, To this we would have to determine a set of inequalities for
x
and
y
that describe this region. These would be,
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With these limits the integral would become, Due to the limits on the inner integral this is liable to be an unpleasant integral to compute. However, a disk of radius 2 can be defined in polar coordinates by the following inequalities, These are very simple limits and, in fact, are constant limits of integration which almost always makes integrals somewhat easier. So, if we could convert our double integral formula into one involving polar coordinates we would be in pretty good shape.
The problem is that we can’t just
convert the
dx
and the
dy
into a
dr
and a . In computing double integrals to this point we have been using the fact that and this really does require Cartesian coordinates to use.
Once we’ve moved into
polar coordinates
and so we’re going to
need to determine just what
dA
is under polar coordinates.
So, let’s step back a little bit and start off with a general region in terms of polar
coordinates and see what we can do with that. Here is a sketch of some region using polar coordinates.
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So, our general region will be defined by inequalities, Now, to find
dA
let’s redo the figure above as follows,
As shown, we’ll break up the region into a mesh of radial lines and arcs.
Now, if we pull one of the pieces of the mesh out as shown we have something that is almost, but not quite a rectangle. The area of this piece is . The two sides of this piece both have length where
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is the radius of the outer arc and is the radius of the inner arc. Basic geometry then tells us that the length of the inner edge is while the length of the out edge is where is the angle between the two radial lines that form the sides of this piece.
Now, let’s assume that we’ve taken the mesh so small that we can assume
that and with this assumption we can also assume that our piece is close enough to a rectangle that we can also then assume that, Also, if we assume that the mesh is small enough then we can also assume that, With these assumptions we then get . In order to arrive at this we had to make the assumption that the mesh was very small. This is not an unreasonable assumption.
Recall
that the definition of a double integral is in terms of two limits and as limits go to infinity the mesh size of the region will get smaller and smaller. In fact, as the mesh size gets smaller and smaller the formula above becomes more and more accurate and so we can say that,
We’ll see another way of deriving this once we reach the
Change of
Variables
section later in this chapter. This second way will not involve any assumptions either and so it maybe a little better way of deriving this. Before moving on it is again important to note that . The actual formula for
dA
has an
r
in it. It will be easy to forget this
r
on occasion, but as you’ll see without it some integrals will not be possible to
do.
Now, if we’re going to be converting an in
tegral in Cartesian coordinates into an
integral in polar coordinates we are going to have to make sure that we’ve also
converted all the
x
’s and
y
’s into polar coordinates as well.
To do this we’ll need to
remember the following conversion formulas,
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1017 | https://maximerobeyns.com/fragments/matdist | Pairwise Distances for Kernel Methods
Maxime Robeyns
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Pairwise Distances for Kernel Methods
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Contents
Squared Euclidean Distance (from Features)
Computing the Full Matrix
Squared Euclidean Distance (from a Gram Matrix)
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Machine Learning
Pairwise Distances for Kernel Methods
An esoteric note on computing the pairwise distance between the rows of two matrices (with PyTorch examples).
November 26,2021
London, UK
When working with kernel methods, particularly with translation invariant kernels k(x,x′)=k(x−x′), it is common to compute a matrix of distances between pairs of inputs R i j=‖x i−x j′‖ℓ. 0: The notation ‖⋅‖ℓ implies that the distance metric is norm based (e.g. a Euclidean norm). While we only consider this type of distance metric in this article, this needn’t be the case—depending on the type of the input points X. 0
While the idea of computing such a matrix is straightforward, the implementation (for instance, in your favourite Python linear algebra package) is not necessarily so obvious. In this Fragment, I want to step through the implementation of some common distance matrices R to try to demystify it all.
Squared Euclidean Distance (from Features)
Perhaps the most common such matrix of pairwise distances is the squared Euclidean distance between the rows:
R m n=‖x m−x n′‖2 2,
where X∈R M×D and X′∈R N×D can be thought of as design matrices, and the notation x i∈R 1×D denotes a single row of X.
This distance matrix is used in popular kernels such as the Squared Exponential (RBF) and rational quadratic. It should be clear that R is an M×N matrix 1: When M=N, the square matrix R is also symmetric positive semidefinite; since a distance metric is a symmetric, non-negative real-valued mapping d:X×X→R. 1:
Each element of R is a scalar:
R m n=‖x m−x n′‖2 2=(x m−x n′)(x m−x n′)⊤=x m x m⊤−2 x m x′n⊤+x n′x′n⊤,
where in the second line we have written the squared Euclidean norm as: ‖a‖2 2=∑d=1 D(a d)2=a a⊤∈R, for a∈R 1×D a row vector.
Implementing this in PyTorch 2: you could just as easily use Numpy, but we’ll be using PyTorch for these examples. 2 for two row vectors is simple enough:
def r(xm: Tensor, xn: Tensor) -> Tensor:
return (xm - xn)@(xm - xn).T
The returned value, although defined as a Tensor, is just a singleton list:
r(x1, x2).shape
torch.Size([1, 1])
Computing the Full Matrix
While the above is a good start, we’re interested in finding a function R(X1, X2) that will compute the entire R matrix of pairwise squared Euclidean distances between each row of X and X′.
We can use PyTorch’s broadcasting semantics to do this efficiently.
Consider the three parts of the solution above for an individual point:
R m n=x m x m⊤−2 x m x′n⊤+x n′x′n⊤.
The first term is the sum of the squared elements of the first argument x m. For the full R matrix, if the first argument X is an M×D matrix, then this x x⊤ term for each element of the i th row of R is x i x i⊤=∑d=1 D x i d 2, which we can compute as Xa.pow(2).sum(dim=1)[i]3: For those unfamiliar with PyTorch syntax, A.pow(2) raises each element of the matrix A to the power of 2. A.sum(1) sums along the 1st (column) dimension of a matrix A with at least 2 dimensions. 3 in Python.
The second term is −2 multiplied by the inner product of the two arguments (remember, x is a row vector). Computing this for each element of the full R matrix is can be done with a single outer product, giving us an M×N matrix: -2 Xa @ Xb.T4: In PyTorch, @ is the matrix multiplication operator. 4
The last term can be found in the same way as the first. The x′x′⊤ term for each element of the j th column of R is computed as Xb.pow(2).sum(1)[j].
Computing the sum of squared elements for all rows of X leaves us with a tensor of shape (M,)—the i th element of this tensor contains the first term for all solutions in the i th row of R. Similarly doing this for X′ results in a tensor of size (N,); the j th element of this row vector corresponds to the x′x′⊤ term in all solutions for the j th column of R.
Making one of these (e.g. the first) into a column vector 5: For a tensor T of size (A,) (i.e. a row vector), the PyTorch syntax T[:, None] or equivalently T.reshape(-1, 1) will result in a tensor of size (A,1) (i.e. a column vector). 5 while the other remains a row vector will give us tensors of size (M,1) and (N,). Due to PyTorch’s broadcasting semantics, adding these two will result in a tensor of size (M,N)6: The first tensor has 2 dimensions while the second has 1, so PyTorch will first make them equal length by prepending 1 to the dimension of the shorter tensor: giving (M,1) and (1,N). Now each dimension of the result is given by the max of the two arguments: (max(M,1), max(1,N)) = (M,N). 6, which contains the first and last terms of all the pairwise squared Euclidean distances R: that is, x m x m⊤+x n′x n′⊤ for all m∈[1,M],n∈[1,N].
We now have all the pieces we need to write the PyTorch method for computing the matrix of pairwise squared Euclidean distances:
def R(Xa: Tensor, Xb: Tensor) -> Tensor:
assert Xa.size(1) == Xb.size(1) # ensure D matches
return Xa.pow(2).sum(1)[:,None] + Xb.pow(2).sum(1) - 2Xa @ Xb.T
Squared Euclidean Distance (from a Gram Matrix)
Recall that a Gram matrix is the following matrix of inner products between all pairs of points, which in our context are the rows of a design matrix X:
G=[⟨x 1,x 1⟩⟨x 1,x 2⟩⋯⟨x 1,x N⟩⟨x 2,x 1⟩⟨x 2,x 2⟩⋯⟨x 2,x N⟩⋮⋮⋱⋮⟨x N,x 1⟩⟨x N,x 2⟩⋯⟨x N,x N⟩]
That is, given G, we can easily find x i x j⊤ by simply reading off the corresponding element of the Gram matrix: G i j.
As we did above, consider finding the squared distance between just two vectors, for i,j∈[1,N]:
R i j=‖x i−x j‖2 2=x i x i⊤−2 x i x j⊤+x j x j⊤=⟨x i,x i⟩−2⟨x i,x j⟩+⟨x j,x j⟩=G i i−2 G i j+G j j.
In PyTorch, we can implement this as:
def r(G: Tensor, i: int, j: int) -> Tensor:
return G[i, i] - 2G[i, j] + G[i, j]
Now considering the full R matrix, observe that the squared Euclidean norm of each row of X lies on the diagonal of the Gram matrix; ‖x i‖2 2=G[i,i] = G.diag()[i].
We can make use of PyTorch’s broadcasting semantics once more to compute the N×N matrix of all G i i+G j j terms as G.diag()[:, None] + G.diag().
The matrix of squared Euclidean distances is thus computed from the Gram matrix as:
def R(G: Tensor) -> Tensor:
g = G.diag()
return g[:, None] + g - 2 G
BibTeX Citation
@article{robeyns2021matdist
author={Maxime Robeyns},
title={Pairwise Distances for Kernel Methods},
year={2021},
url={
}
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Footnotes
The notation ‖⋅‖ℓ implies that the distance metric is norm based (e.g. a Euclidean norm). While we only consider this type of distance metric in this article, this needn’t be the case—depending on the type of the input points X.
↩︎
When M=N, the square matrix R is also symmetric positive semidefinite; since a distance metric is a symmetric, non-negative real-valued mapping d:X×X→R.
↩︎
you could just as easily use Numpy, but we’ll be using PyTorch for these examples.
↩︎
For those unfamiliar with PyTorch syntax, A.pow(2) raises each element of the matrix A to the power of 2. A.sum(1) sums along the 1st (column) dimension of a matrix A with at least 2 dimensions.
↩︎
In PyTorch, @ is the matrix multiplication operator.
↩︎
For a tensor T of size (A,) (i.e. a row vector), the PyTorch syntax T[:, None] or equivalently T.reshape(-1, 1) will result in a tensor of size (A,1) (i.e. a column vector).
↩︎
The first tensor has 2 dimensions while the second has 1, so PyTorch will first make them equal length by prepending 1 to the dimension of the shorter tensor: giving (M,1) and (1,N). Now each dimension of the result is given by the max of the two arguments: (max(M,1), max(1,N)) = (M,N).
↩︎ |
1018 | https://ciliajournal.biomedcentral.com/articles/10.1186/s13630-014-0010-9 | Situs inversus and ciliary abnormalities: 20 years later, what is the connection? | Cilia | Full Text
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Cilia
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Situs inversus and ciliary abnormalities: 20 years later, what is the connection?
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Published: 14 January 2015
Situs inversus and ciliary abnormalities: 20 years later, what is the connection?
Petra Pennekamp1,
Tabea Menchen1,
Bernd Dworniczak2&
…
Hiroshi Hamada3
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Ciliavolume 4, Article number:1 (2015) Cite this article
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Abstract
Heterotaxy (also known as situs ambiguous) and situs inversus totalis describe disorders of laterality in which internal organs do not display their typical pattern of asymmetry. First described around 1600 by Girolamo Fabrizio, numerous case reports about laterality disorders in humans were published without any idea about the underlying cause. Then, in 1976, immotile cilia were described as the cause of a human syndrome that was previously clinically described, both in 1904 by AK Siewert and in 1933 by Manes Kartagener, as an association of situs inversus with chronic sinusitis and bronchiectasis, now commonly known as Kartagener’s syndrome. Despite intense research, the underlying defect of laterality disorders remained unclear. Nearly 20 years later in 1995, Björn Afzelius discussed five hypotheses to explain the connection between ciliary defects and loss of laterality control in a paper published in the International Journal of Developmental Biology asking: ‘Situs inversus and ciliary abnormalities: What is the connection?’. Here, nearly 20 research years later, we revisit some of the key findings that led to the current knowledge about the connection between situs inversus and ciliary abnormalities.
Review
Laterality defects in humans
Left-right asymmetry is common in nature. Disorders of left-right asymmetry can cause randomization (heterotaxy/situs ambiguous) or complete reversal of organ position (situs inversus totalis) (reviewed in [1-3]). The first reports describing laterality defects came from Girolamo Fabrizio (Hieronymus Fabricius; 1537 to 1619; Italian anatomist and surgeon) around 1600, followed by Marco Aurelio Severino (1580 to 1656, Italian anatomist and surgeon) who first documented the finding of a human heart being on the right hand side of the body in 1643 , and also Matthew Baillie (1761 to 1823, Scottish physician and pathologist) who described the complete mirror image reversal of the thoracic and abdominal organs more than a century later in 1788 .
In the following centuries numerous case reports about laterality disorders were published. In a frequently overlooked case study published in 1904, Dr. AK Siewert from Kiev (Ukraine) described a patient ‘who since birth had the unusual combination of symptoms of bronchiectasis and situs inversus totalis’ . This was the first description of what has since become known as ‘Kartagener’s syndrome (KS)’ based on the description by Manes Kartagener in 1933 , sometimes still referred to as Siewert’s (Zivert’s) syndrome, Siewert(Zivert)-Kartagener syndrome, or ‘immotile cilia syndrome’. It was later renamed ‘primary ciliary dyskinesia (PCD)’ after dysmotile cilia had been observed in individuals diagnosed clinically as having KS [8,9].
Although several decades of research were required to understand the connection behind laterality disorders and the role of cilia, Kartagener was already thinking in the right direction. He noted in one simple sentence that ‘cystic kidneys, which are often compared with congenital bronchiectasis, have been observed in combination with situs inversus’ [7,10]. At that time nobody was aware that two types of cilia would be involved in laterality development and that genes causing cystic kidney disease would also play a role .
Structural defects of cilia as the cause for Kartagener’s syndrome
Numerous case reports about laterality defects in humans were published in the following years, but still lacked any direct correlation between laterality disorders and cilia function. During that time, Björn Afzelius had improved sample preparation for electron microscopy, leading to a better preservation and analysis of ciliary ultrastructure . Years later in 1974, Afzelius met Henning Pedersen, who showed Afzelius his unpublished electron micrograph of an immotile human sperm tail lacking dynein arms. Afzelius suggested that Pedersen publish these findings soon. After meeting Pedersen, Afzelius analyzed - together with Rune Eliasson - sperm samples from male patients with stiff, immotile sperm tails. As expected, they also found lack of dynein arms as the cause for immotile human spermatozoa in these patients. Pedersen and Afzelius subsequently published their findings nearly at the same time in 1975 [13-15].
Following these landmark studies, the hypothesis was raised that KS is caused by a genetic lack of dynein arms in cilia . Both Afzelius and Pedersen proved independently that dynein arms were missing in cilia of nasal and bronchial mucosa of affected patients, confirming this hypothesis [17,18]. Afzelius sent a letter to Kartagener to notify him about these findings, but Kartagener’s daughter reported to Afzelius that he had died in August 1975. Evidently this information was never brought to Kartagener’s attention [15,19].
Special type of cilia at the embryonic organizer
Although the ultrastructural defect causing ciliary immotility and bronchiectasis in patients with KS was identified, the connection between ciliary immotility and body axis development remained elusive. Afzelius early on raised the hypothesis that motile cilia defects can explain a variety of phenotypes/defects observed in patients with KS, but not all. He also discussed the existence of ‘so-called sensory hairs protruding from the cell surface into the extracellular space’ to explain the poor sense of smell and decreased hearing ability in these patients . To explain the connection between cilia and the situs inversus observed in patients with KS, Afzelius also mentioned that ‘a further category of cells which carry cilia is the differentiating cells of vertebrate embryos’, referring to a publication analyzing cilia during cardiac development of the chicken . He hypothesized that ‘it is not unreasonable to assume that a malrotation may occur when the ciliary movement causing rotation is lacking’ and that ‘chance alone will determine whether the viscera will take up the normal or the reversed position during embryogenesis, when normal dynein arms are missing’ . In 1976 he noted that the cause of situs inversus remains elusive despite many theories, but he also presciently suggested that ‘cilia on the embryonic epithelia have a certain position and fixed beat direction (in normal embryos) and that their beating somehow is instrumental in determining the visceral situs’ .
It is unclear whether Afzelius in the early years of his research was aware of experiments performed by numerous embryologists that identified the ‘organizer’ , a group of cells necessary and sufficient to initiate a complex program of spatial organization in competent embryonic tissue (reviewed in ). For the groundbreaking research leading to the identification of the organizer performed by Hans Spemann (1869 to 1941) and Hilde Mangold (1898 to 1924), the Nobel Prize in Physiology or Medicine was awarded to Hans Spemann in 1935 . Organizer activity was subsequently also demonstrated in embryos of higher vertebrates such as birds and mammals [23-25] in a structure that Victor Hensen (1835 to 1924) had described in 1876 in an extensive paper and named ‘the node’ , also known as ‘Hensen’s node’ .
Nearly 20 years after Afzelius expressed his hypothesis , scientists were just starting to put pieces of this puzzle together. It was only in 1994 that Sulik et al. published an extensive study on the development of the node and formation of the notochordal plate in embryonic day (E) 7 to 9 mice (Theiler stages 10 to 14). It is of note that Sulik et al. still found it important to define several terms such as ‘the rostral end of the murine primitive streak will be referred to as the node (analogous to Hensen’s node in avian species)’ because the structure was not properly assigned in previous studies [28,29]. It is important to note here that the ‘node’ in mouse was renamed several times based on morphological and functional studies. First named ‘archenteron’ by Theiler in 1972 the term ‘node’ was introduced by Beddington in 1991: ‘This (the archenteron) is a misnomer as it is not equivalent to the archenteron in amphibians but, as far as we can tell, corresponds to the dorsal blastopore lip of Xenopus or Hensen’s node of the chick. Therefore, I would suggest that we call it the ‘node” . Later analyses of ciliation and gene expression demonstrated in 2007 that this definition unites two entities, the node and the posterior notochord (PNC), the latter characterized by bi-lateral Nodal expression, motile monocilia and cilia-driven leftward flow and functioning as left-right (LR)-organizer. Using this functional characteristic, the PNC, still commonly named ‘node’ in mouse embryos, corresponds to the gastrocoel roof plate (GRP) and not to the dorsal blastopore lip of amphibians . Nevertheless, in the study of Sulik et al. the authors described ‘the presence of groupings of cells in the area of the rostral midline that had small ventral surface areas relative to adjacent cells’ , each with ‘a prominent single, central cilium-like structure on their ventral side’ at the ventral layer of the node . They also performed video microscopy of the node showing that these monocilia cilia were motile, although they failed to find ‘evidence of synchronized activity in time or direction’ .
A year later Afzelius, who now interpreted KS as a disease caused by defective or absent cilia, presented five hypotheses to explain the ‘curious’ connection between ciliary defects and loss of laterality control. These hypotheses included combined loss of function of two closely linked genes, one responsible for the asymmetry of viscera and the second responsible for synthesis or assembly of ciliary structures, cytoskeletal defects , lack of structural coordination and defective ciliation-or-division switch . As the most likely hypothesis he suggested that ‘there are cilia that have determined positions and a fixed beat direction, much the same as they have on the epidermis of amphibian embryos’ and that ‘ciliary beating in normal embryos is assumed to be instrumental in pushing the heart to the left side, whereas chance alone will determine whether viscera will take up the normal or the reversed position during embryogenesis, when there is no regular ciliary motility’ [17,34]. We can only speculate whether Afzelius had been aware of the study on the development of the node and formation of the notochordal plate presented by Sulik et al. .
The nodal signaling cascade and the nodal flow
It still took several years to gain insight into the function of cilia during the process of LR-axis development. Initially, the discovery of molecular networks acting in and around the node during patterning of the body axes constituted the major scientific breakthrough in the analysis of axis development. Although we acknowledge outstanding scientific contributions stemming from research on other model organisms, such as chicken and Xenopus, this review will focus on findings obtained in mouse studies unless otherwise stated.
In 1993, Nodal, one of the key factors of LR-axis development and a member of the transforming growth factor-beta (TGF-beta) superfamily, was identified in mice. Nodal expression was detected in a symmetric fashion exclusively at the node or ‘associated with the node’ dependent on the developmental stage analyzed (approximately E7 (primitive streak embryos) to E8.5 (‘coinciding with the disappearance of the node as distinct structure’ )) and named Nodal due to its localized expression at the node . Surprisingly, three years later, the correlation between asymmetric gene expression and LR-asymmetry in mice became obvious in two independent studies published in Nature [37,38]. In these studies, it was demonstrated that expression of Nodal was dependent on the developmental stage and in contrast to the previously published study not only in the mouse node but also in the left lateral plate mesoderm (LPM) [37,38]. In the same studies, Nodal expression was examined in mouse mutants with perturbed LR-development, especially the situs inversus viscerum (iv) displaying random LR-asymmetry and the inversion of embryonic turning (inv) developing situs inversus . Both mouse models were extensively used for the analysis of LR-development. It was anticipated that these mutant mice would provide important insights into the understanding of LR-development, although the genetic basis of either mutation was still unknown at that time. The authors demonstrated that Nodal expression in these mutants was either normal, reversed or bilateral depending on the morphological LR-asymmetry. This was similar to previous findings in chicken and Xenopus , thus demonstrating the evolutionary conservation of Nodal expression and suggesting that Nodal is one of the key regulators of LR-development [37,42]. In the same volume of Nature it was shown that another gene named Lefty (left right determination factor) was transiently and asymmetrically expressed in the left LPM and the left half of the prospective floorplate during LR-axis development. Similar to Nodal expression the site of Lefty expression correlated with the morphological asymmetry observed in iv/iv and inv/inv mutants . Extensive studies on Lefty function surprisingly demonstrated that the original expression pattern attributed to a single Lefty gene in fact derived from two highly-related and chromosomally linked genes, Lefty-1 and Lefty-2, which were both asymmetrically expressed but with distinct expression domains functioning downstream of iv and inv function . In 1998, the paired-like homeodomain transcription factor 2 (PITX2) was identified in two independent studies as an additional evolutionarily conserved downstream effector of the signaling cascade that establishes asymmetries along the entire LR-axis, the Nodal signaling cascade (Figure1) [45,46].
Figure 1
Timeline for laterality development (A), node structure (B) and genetic cascade of LR patterning (C) in mouse. LR patterning can be divided into three steps: breakage of symmetry at the node, patterning of the LPM and asymmetric morphogenesis (A). The initial breakage of bilateral symmetry in the mouse occurs in the vicinity of the node (asterisk in A and B) at the headfold stage corresponding to embryonic day (E)7.5 to E7.8. The node is a transient structure composed of two epithelial layers: the dorsal node and the ventral node [27,123,124]. The ventral node can be separated into the pit region (red in B and C) and the crown region (marked by Cerl2 expression (black) in B). Pit cells carry mainly motile monocilia on their apical surface, whereas crown cells mainly carry immotile monocilia [11,118]. Motile monocilia of the node rotate in a clockwise orientation generating a leftward fluid flow over the node cavity (nodal flow; direction is marked by arrows in C). Nodal flow is sensed by cilia of the crown cells and converted into asymmetric signaling involving Nodal-Lefty1-Lefty2 regulatory loops in the LPM and induction of Pitx2 expression in the left LPM (C). Key players of this asymmetric signaling cascade are shown with Nodal expression in crown cells (purple) and the left LPM (green), Lefty-2 and Pitx2 expression in the left LPM (green) and Lefty-1 expression in the left side of the midline (yellow) [120,123,125-130]. I and II: scanning electron micrographs of wild type mouse embryos (I: 0 somite; II: 5 somites). III: wild type mouse embryo at E9.0 viewed from the left; B: cryo section of a mouse node after in situ hybridization using a Cerl2 probe photographed using DIC optics. c: crown cells; p: pit cells; R: right; L: left. Scale bars: 50 μm.
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Figure 2
Types of cilia found at the left-right-organizer of vertebrates. In mouse, two types of ciliated cells at the node have been demonstrated (A). Pit cells possess motile cilia visualized by a GFP-fusion protein of DNAH11 (LRD-GFP; arrowhead). Most crown cells possess immotile cilia containing polycystin-2 but lacking LRD-GFP (arrow). It is suggested that immotile cilia of crown cells act as flow sensors. Primary cilia containing polycystin-2 are found on endodermal cells adjacent to the node (asterix). [11,101,118]. Classical motile cilia (B) possess nine outer microtubule doublets forming a ring around a central pair (CP) of single microtubules, called 9 + 2 axoneme. Outer microtubule doublets possess dynein arms that drive rhythmic movement of the axonemes. Radial spoke complexes project from each outer doublet towards the CP and are thought to be important in regulating motion of the axoneme (B). Motile monocilia (B) are found in cells of the LR-organizer in various vertebrates, such as ventral node in mouse, notochordal plate in rabbit or Kupffer’s vesicle in zebrafish and medaka (C). In mouse and medaka, cilia of the LR-organizer usually display the 9 + 0 configuration lacking the CP. In other vertebrates, such as zebrafish, they display 9 + 2 or in rabbit 9 + 0, 9 + 2 and 9 + 4 configuration. Irrespective of the structure, these cilia move in a rotational manner, establishing a leftward-directed fluid flow within the cavity of the LR-organizer. Immotile cilia (B) lack motility components such as dynein arms, CP and radial spokes and act as specialized sensors of the cell transducing signals from extracellular stimuli to a cellular response [125,131]. A: Immunofluorescent staining detecting polycystin-2 (red) in node monocilia of a 2 somite stage LRD-GFP (green) mouse embryo. A: anterior; l: left; p: posterior; r: right. Scale bar: 20 μm [132-142].
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At this time it also emerged that cilia function played a role during LR-axis development, and both the iv and the inv mouse mutants were important in this respect. First described in 1956 , iv was mapped to mouse chromosome 12 in 1989 and identified by a positional cloning approach in 1997 as an axonemal dynein heavy-chain gene named left/right-dynein (Lrd, currently known as dynein, axonemal, heavy chain 11; Dnah11) . Lrd was shown to be expressed in the node of the embryo at E7.5, consistent with having a role in LR-development . Based on the observation that the asymmetric expression patterns of Nodal and Lefty were randomized in iv/iv embryos, it was suggested that iv functions early in the genetic hierarchy of LR-specification. Nevertheless, the connection between Lrd and cilia at the node was considered unlikely; at that time, it had been supposed that cilia at the node were immotile monocilia lacking dynein arms even though ciliary motility at the node, despite no evidence of synchronized activity in time or direction, had previously been observed . The gene defect underlying the LR-axis defects observed in inv mutants was identified a year later in 1998 in two independent studies. These were published five years after the first description of the inv phenotype (inversion of embryonic turning and cystic kidneys), and the gene was appropriately named Inversin [40,50,51]. Despite these findings, the function of Inversin and its connection to axis development remained unknown.
The first tangible evidence in mice that cilia of the node play an important role during LR-development came from the characterization of a mouse mutant lacking the kinesin family member 3b (KIF3B) microtubule-dependent motor protein although it was not the first mouse mutant in which impaired ciliogenesis was reported . Loss of Kif3b function resulted in prenatal lethality, neural tube disorganization and randomized LR-asymmetry. Lefty-2 expression was either bilateral or absent, suggesting - comparable to iv and inv mutants - that Kif3b has a function upstream of the Nodal cascade. It was further shown that KIF3B was localized within the axoneme of node monocilia and that Kif3b mutant nodes lacked monocilia, suggesting that monocilia of the node play an important role during LR-development. To further elucidate the function of these cilia, video microscopy of the node was performed. In contrast to the then accepted understanding that these cilia lack dynein arms and appear to be immotile [48,49] it was convincingly demonstrated that node monocilia were motile and capable of generating a leftward flow. Based on these data, the authors suggested that motile monocilia at the node generate a directed leftward flow leading to the concentration of a secreted factor to the left side of the node, which then triggers the downstream signaling cascade of left-defining genes, the Nodal signaling cascade, still known as the ‘morphogen hypothesis’ [48,52,53].
Aware of the findings of Nonaka et al., Afzelius hypothesized in 1999 that exposure of embryos to highly viscous methyl cellulose during the critical stage would increase the percentage of situs inversus . In fact, this hypothesis was confirmed years later using indeed methyl cellulose to influence flow in Xenopus and mouse embryos in 2007 and 2012, respectively [55,56].
The finding of nodal flow was a major scientific breakthrough with regard to the connection between cilia function and LR-axis development. Although additional mouse mutants, such as those lacking the KIF3A subunit of kinesin-II, provided supportive evidence for the relevance of cilia at the node during LR-axis development [57,58], it was still necessary to eliminate the last serious doubts.
The first supporting evidence came from an extensive study showing that abnormal nodal flow precedes situs inversus in iv and inv mutant mice, demonstrating immotile cilia in iv mutants lacking LRD and a slower net leftward flow in inv mutants, due to a more ‘turbulent flow’ despite the fact that cilia lacking Inversin rotate more rapidly . These data suggested that abnormal flow is not the consequence of the abnormal LR-body axis determination but may be its source . The hypothesis that the directed mechanical fluid flow across the node is implicated in the symmetry breaking event was further confirmed by elegant experiments in which wild type embryos and iv mutants lacking LRD were cultured under artificial fluid flow conditions. Reversal of laterality was checked by using genetic markers at the LPM, demonstrating randomized or reversed expression previous to organ laterality development, and further confirmed by analysis of organ situs demonstrating reversal of heart looping and embryonic turning . This study convincingly demonstrated that a strong rightward artificial flow could reverse fluid flow at the node and thereby the laterality of wild type embryos, proving that the laterality of treated embryos was successfully controlled by the direction of the artificial flow. Similar experiments were then performed with homozygous iv/iv mutant embryos that lack nodal flow due to immotile cilia at the node and exhibit situs inversus in half of the mutant embryos. Here, it was demonstrated that iv/iv embryos responded even to slow artificial flow, manifesting normal or reversed laterality dependent on the direction of the artificial flow. Although the exact mechanism leading to the initial breakage of symmetry at the node was still unknown, these experiments unequivocally showed that efficient and directed nodal flow generated by motile monocilia at the node was necessary to establish left-right laterality in the embryo (reviewed in [61-63]).
The laterality-kidney connection
By the year 2000 more than 24 genes involved in axis development had been published, not only by analyses of existing spontaneous mouse mutations but also due to the increasing ability to genetically manipulate the mouse to generate targeted mutations (reviewed in ). Nevertheless, knowledge about cilia structure and function at the node was still limited.
Further insight came from a completely unexpected source, namely research performed on autosomal recessive and autosomal dominant polycystic kidney diseases (ARPKD and ADPKD, respectively). Until 1994, several spontaneous mutations in mouse lines had been described resembling ARPKD in humans, namely cpk (congenital polycystic kidneys) mapped in 1991 to mouse chromosome 12 , bpk (BALB/c polycystic kidneys) , pcy (polycystic kidneys) mapped in 1991 to mouse chromosome 9 and jck (juvenile cystic kidneys) and others (reviewed in ), but none of the disease-causing genes had been identified at that time, making it impossible to study the pathogenesis of PKD in more detail.
Then in 1994, a candidate gene was described to be associated with a mutation causing recessive polycystic kidney disease in mice. This gene was mutated during a large scale mutagenesis program and the line was called TgN(Imorpk)-737Rpw (Imorpk: insertional mutation, Oak Ridge polycystic kidneys), abbreviated TgN737Rpw and later on Oak Ridge Polycystic Kidney (orpk) insertional mutation [72,73]. The targeted allele of this mouse Tg737 gene, Tg737 D 2-3 b Gal, was published in 2000 . Unlike the original orpk allele where all homozygotes survived to birth, embryos homozygous for the Tg737 D 2-3 b Gal mutation arrested in development at mid-gestation and exhibited neural tube defects, enlargement of the pericardial sac and, most notably, LR-asymmetry defects with bilateral expression of Nodal and Lefty-2. It was shown that nodal monocilia were lacking, similar to the previously described Kif3b and Kif3a mutants [52,57,58], providing further evidence that motile monocilia at the node are important for proper LR-axis development.
Most interesting with regard to the analysis of both Tg737 alleles and future analysis of ciliary function during development were the different phenotypes observed in these two Tg737 alleles: these included cystic kidneys in the hypomorphic orpk allele and a more severe phenotype involving motile monocilia at the node in the complete loss-of-function Tg737 D 2-3 b Gal allele. It was suggested that the inability to develop and maintain polarity of renal and node cells as demonstrated by lack of monocilia that were found on virtually every cell of the body [75-77] might be the underlying cause of cyst formation in the kidneys and laterality defects. This observation led to the name ‘Polaris’ for the gene product of Tg737 . Further analysis demonstrated that Polaris localized just below the apical membrane in the region of the basal bodies and within cilia or flagellar axonemes and that it was important for ciliogenesis of both motile and non-motile primary cilia in different model organisms [79-81], leading to the name ‘Intraflagellar transport 88’ (Ift88). As a result of these studies, a common key function of cilia in the development of both cystic kidneys and laterality disorders became more apparent.
At that time, based on studies performed in Caenorhabditis elegans [82,83], an association between structural and/or functional defects in primary cilia of vertebrate epithelia and another cystic kidney disease caused by mutations in PKD1 (ADPKD1) encoding polycystin-1 [84-87] and PKD2 (ADPKD2) encoding polycystin-2 , was suggested.
Indeed, it was possible to show that polycystin-2, next to additional polycystic kidney disease proteins, such as polycystin-1, Polaris and Cystin, localized to primary cilia of the kidney [89,90] supporting this hypothesis. With regard to Pkd2 mouse mutants, it was previously shown that somatic inactivation of Pkd2 expectedly resulted in polycystic kidney disease but also prenatal lethality and cardiac defects . Unexpectedly, it was also shown that polycystin-2 was required for LR-axis development and that loss of Pkd2 function resulted in heterotaxy. Loss of Nodal, Lefty-1 and Lefty-2 expression and bilateral Pitx2 expression in Pkd2 mutants suggested that polycystin-2 is active early during axis development and upstream of the Nodal signaling cascade . The findings of ciliary localization of polycystin-2 in primary cilia of the kidney, LR-axis defects of Pkd2 mutant mouse embryos and ion channel activity of polycystin-2 [94-96] suggested that polycystin-2 might have a function in cilia of the node, perhaps as an ion channel necessary to sense and translate the leftward flow similar to the proposed function in the kidney.
The two-cilia model
Just a few months before the involvement of Pkd2 in laterality development was published, mutations in DNAH5 (dynein, axonemal, heavy chain 5) leading to non-functional DNAH5, loss of outer dynein arms and immotile cilia, were published in individuals with PCD and KS and it became increasingly likely that two different types of ciliary defects can cause laterality defects, defects in cilia motility and defects in sensory function.
The next hint came again from the kidney research field showing that polycystin-1 and polycystin-2 mediated mechanosensation in primary cilia of the kidney resulting in calcium influx into the cell . The hypothesis was thus raised that the polycystins might have a similar function at the node.
Only a few months later it was possible to show that two populations of node monocilia initiated LR-asymmetry in the mouse. These populations could be distinguished by LRD, which localized to a motile subset of nodal monocilia, and polycystin-2, which localized to all nodal monocilia although it appeared to be enriched in non-LRD containing cilia (Figure 2) . The same study showed that an asymmetric calcium signaling appeared at the left margin of the node coincident with leftward nodal flow. This signal was absent in both mutant mouse embryos lacking Lrd or Pkd2. These data suggested that LR-asymmetry is established by an entirely ciliary mechanism consisting of motile, LRD-containing monocilia generating the directed nodal flow and non-motile polycystin-2-containing cilia sensing the flow and initiating an asymmetric calcium signal at the left border of the node, appropriately named the ‘two-cilia model’ [11,99] (and reviewed in [100-103]).
The polarization of nodal cells
Nevertheless, several questions still remained with regard to ciliary function at the node: 1) how can rotational movement of node monocilia generate the unidirectional flow; 2) how can this unidirectional nodal flow be sensed by the embryo; and 3) what mechanism specifies the differentiation of the two types of cilia at the node?
A theoretical analysis of fluid dynamics at the node proposed a model in which a productive linear flow could result if the rotation axis of the cilia has a posterior tilt . Careful analyses of cilia orientation and cilia movement indeed demonstrated that cilia did not stand perpendicular to the node surface but were tilted posteriorly, confirming the hypothesis postulated by the theoretical analysis [105,106]. That this posterior tilt was necessary to generate a directed flow and that the flow depends on the tilt angle of the cilia was further confirmed using a mechanical model simulating different scenarios .
However, the question remained how this posterior tilt of cilia at the node is generated. Establishment of the three body axes, anterior–posterior (AP)-, dorso–ventral (DV)- and LR-axis, is central to the vertebrate body plan. Since the LR-axis is the last axis to be determined during development, LR-polarity must be generated by utilizing the pre-existing positional cues from the AP- and DV-axes.
Previously, it had been shown that complete loss of function of Biccaudal C (BicC), the gene which is mutated in both the bpk mouse model leading to an ARPKD phenotype and the jcpk mouse model leading to an ADPKD phenotype [107,108], resulted in LR-axis defects by disrupting the planar alignment of motile cilia required for cilia-driven fluid flow. Furthermore, it had been shown that BICC uncoupled Dishevelled 2 (DVL2) signaling from the canonical Wnt pathway, which has been implicated in antagonizing planar cell polarity (PCP), the orientation of specialized structures within a plane of the epithelial sheet . Thus, it was suggested that establishment of PCP is also involved in the orientation of nodal cilia.
Careful analysis of the basal body orientation demonstrated that the basal bodies of nodal cilia were initially positioned centrally. They then gradually shifted towards the posterior side of the node cells until the majority was located at the posterior side of the ciliated node cells at the two- to three-somite stage, when the velocity of the fluid flow is maximal . Further analysis demonstrated that Dishevelled (Dvl), a key participant of both canonical and non-canonical Wnt signaling pathways and in mice, represented by three widely expressed and functionally redundant Dvl genes (Dvl1, Dvl2 and Dvl3), was relevant for the positioning of basal bodies, confirming this hypothesis .
The involvement of the canonical Wnt signaling was further excluded based on analyses performed with embryos deficient in Wnt3a, the only ligand that activates the canonical Wnt pathway in the node. These analyses demonstrated that neither the directional flow nor the rotational axis of nodal cilia and position of basal bodies were affected by loss of WNT3A, although Wnt3a mutants showed laterality defects [110,111]. On the other hand, blocking RAC1, a small G protein and effector molecule of the non-canonical Wnt pathway, also known as the non-canonical PCP pathway, led to defects in positioning of the basal bodies and vertical nodal flow suggesting that the non-canonical Wnt/PCP pathway is involved . This hypothesis was confirmed by further analyses demonstrating that in the absence of Vangl1 and Vangl2, the two mouse homologues of the Drosophila core PCP gene Van Gogh (Vang) as well as in Cofilin1;Vangl2 double mutants , failure to properly polarize nodal cilia led to randomization of LR-asymmetry (reviewed in [114-116]).
The sensor of nodal flow
Although mechanistic inside of how leftward nodal flow is generated progressed, exactly how this left side-specific signal gets sensed and transduced remained enigmatic. At that time, only the Nodal inhibitor Cerl2 had been identified to act as a critical target of flow suggesting that symmetry is broken by flow-mediated left-asymmetric release of Nodal repression at the midline . Based on the mechanosensory function of polycystin-1 and polycystin-2 in kidney epithelial cells and the elevated left-side specific calcium signal at the node observed in mouse embryos, it was assumed that sensing of this flow occurs through cilia. Pkd2 was a good candidate since mice lacking polycstin-2 exhibited LR-patterning defects and lost the left-sided expression of Nodal, suggesting that Pkd2 is relevant for cilia function at the node. Interestingly, neither the structure nor the motility of nodal cilia was compromised, suggesting that loss of polycystin-2 in these cilia results in the inability to sense flow . This hypothesis was confirmed by using Ca 2+ signaling blockers that interfere with polycystin-2 signaling. Rescue experiments demonstrated that although endogenous polycystin-2 localized to cilia of both crown cells and pit cells of the node, the latter located at the central region of the node and mainly possessing motile cilia generating the nodal flow, polycystin-2 was required only in crown cells for the correct establishment of LR-asymmetry. In addition, it was shown that the Nodal inhibitor Cerl2 was not only a critical target of flow but also the major target of Pkd2 mediated signaling during LR pattern formation. Surprisingly, it was also possible to demonstrate that restoring cilium formation in crown cells of Kif3a mutant embryos, which usually completely lack cilia, was also sufficient to induce LR-asymmetry. These data indicated that cilia of the crown cells of the node are the sensors for the leftward fluid generated at the node (and reviewed in [119,120]).
Motor or sensor?
It was now widely accepted that laterality is initiated at the embryonic LR-organizer, where motile cilia generate leftward flow that is detected by immotile sensory cilia, which then transduce flow into downstream asymmetric signals.
In 2011, GALNT11 (N-acetylgalactosamine-type O-glycosylation enzyme) was identified as a candidate disease gene in a patient with heterotaxy . Functional analyses performed in Xenopus tropicalis demonstrated that galnt11 activated Notch signaling. Live imaging of the cilia of the Xenopus organizer was also performed. These analyses demonstrated that either galnt11 or notch1 depletion increased the ratio of motile cilia at the expense of immotile cilia (producing a laterality defect reminiscent of loss of the ciliary sensor polycystin-2) and that Notch overexpression decreased this ratio (mimicking the ciliopathy PCD). These data demonstrated that galnt11-mediated notch1 signaling modulates the spatial distribution and ratio of motile and immotile cilia, deciding who is motor and who is sensor at the LR-organizer, a decision which is important for the determination of laterality .
Conclusions
This review attempted an historical overview of key publications and experiments that influenced the direction of research and led to our current knowledge connecting the curious link between situs inversus and ciliary abnormalities (Figure3). Of course numerous excellent additional studies exist, which added even more details to this knowledge regarding ciliary structure and function necessary for proper axis development. In addition, a large number of genes influencing node and ciliary structure and function have been identified. Among these, the largest group of genes influences nodal ciliogenesis (ranging from complete absence to short or abnormal cilia) (Additional file 1, Table S1). Another large group of genes influences nodal morphology and shape including orientation of cilia (PCP) necessary to generate the directed leftward flow (Additional file 1, Table S2). A third subset of genes has been shown to cause axonemal defects resulting in dyskinetic cilia and PCD (with or without heterotaxy) (Additional file 1, Table S3). Interestingly, several PCD-causing genes, to the best of our knowledge, do not cause laterality defects in either humans or mouse models but their analyses nonetheless aid to understand the structure of nodal cilia (Additional file 1, Table S4).
Figure 3
Milestones of research towards understanding of laterality disorders. Since the first description of laterality defects around 1600 by Fabrizio several milestones can be identified (marked with red arrows on the time scale) which led to a sharp increase in knowledge with regard to the origin of laterality defects. This included the first description of bronchiectasis in combination with situs inversus totalis by Siewert in 1904 and Kartagener in 1933 followed by the demonstration that ciliary defects are the underlying cause of Kartagener’s syndrome in 1976 [17,18]. The demonstration of motile monocilia at the mouse node in 1994 followed by the identification of the first asymmetrically expressed genes in 1996 [37,38], later known as members of the Nodal signaling cascade, opened the wide field of research on the genetic and molecular levels which led to our current knowledge about the connection of situs inversus and ciliary abnormalities. It is worth mentioning here that around the same periods, Antony van Leeuwenhoek already in 1675 described a living protozoan ‘provided with diverse incredibly thin feet, or little legs, which were moved very nimbly’ and that the term ‘cilia’ was probably first used by OF Muller in 1786 (reviewed in [143,144]).
Full size image
We can only speculate how Bjorn Afzelius, who died April 27 2008, would view the tremendous advances in molecular and genetic research that have convincingly linked ciliary function at the node to LR-axis development and, amazingly, confirmed his hypothesis from twenty years before.
We can anticipate that combined efforts by clinicians and basic researchers as well as the brisk pace of advancements in genetic analyses and modification of animal models will bring even greater understanding of how ciliary function influences LR-axis development and we are looking forward to the advancements in this field.
Abbreviations
ADPKD:
autosomal dominant polycystic kidney disease
AP:
anterior-posterior
ARPKD:
autosomal recessive polycystic kidney disease
BicC :
Biccaudal C
Bpk :
BALB/c polycystic kidneys
Cpk :
Congenital polycystic kidneys
Dnah11 :
Dynein, axonemal, heavy chain 11
DNAH5 :
Dynein, axonemal, heavy chain 5
DV:
dorso-ventral
Dvl :
Disheveled
E:
embryonic day
GALNT11:
N-acetylgalactosamine-type O-glycosylation enzyme
GFP:
green fluorescent protein
GRP:
gastrocoel roof plate
Ift88 :
Intraflagellar transport 88
Inv :
Inversion of embryonic turning
Iv :
Situs inversus viscerum
Jck :
Juvenile cystic kidneys
KIF3A:
Kinesin family member 3A
KIF3B:
Kinesin family member 3B
KS:
Kartagener’s syndrome
LPM:
lateral plate mesoderm
LR:
left-right
Lrd :
Left/right-dynein
Orpk :
Oak Ridge Polycystic Kidney
PCD:
primary ciliary dyskinesia
PCP:
planar cell polarity
pcy :
Polycystic kidneys
Pitx2 :
Paired-like homeodomain transcription factor 2
PKD:
polycystic kidney disease
PKD1 :
Polycystic kidney disease gene 1
PKD2 :
Polycystic kidney disease gene 2
PNC:
posterior notched
RAC1:
RAS-related C3 botulinum substrate 1
TGF:
transforming growth factor
Vangl :
Van Gogh like
Wnt:
Wingless-type MMTV integration site family
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Acknowledgements
All studies presented here and numerous excellent studies not cited within this review for which we apologize helped us to understand a little part of laterality development, cilia structure and function and associated diseases. This gain of knowledge is due not least to the combined efforts of clinicians and basic researchers, the still improving methods to genetically modify and analyze model organisms for functional analysis and the brisk pace of advancements in genetic analysis. The authors thank Martina Brueckner who kindly provided lrd-gfp mice. The authors thank colleagues for continuous discussions, especially Heymut Omran, Gerard W. Dougherty and Niki T. Loges. PP is funded by a research grant of the Deutsche Forschungsgemeinschaft (DFG OM 6/5), TB is funded by a research grant of the Deutsche Forschungsgemeinschaft (DFG OM 6/4), research in the lab of HH is funded by Core Research for Evolutional Science and Technology (CREST) of the Japan Science and Technology Corporation (JST).
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Authors and Affiliations
Department of General Pediatrics, University Children’s Hospital Muenster, 48149, Muenster, Germany
Petra Pennekamp&Tabea Menchen
Department of Human Genetics, University Hospital Muenster, 48149, Muenster, Germany
Bernd Dworniczak
Graduate School of Frontier Biosciences, Osaka University, Osaka, Japan
Hiroshi Hamada
Authors
1. Petra PennekampView author publications Search author on:PubMedGoogle Scholar
2. Tabea MenchenView author publications Search author on:PubMedGoogle Scholar
3. Bernd DworniczakView author publications Search author on:PubMedGoogle Scholar
4. Hiroshi HamadaView author publications Search author on:PubMedGoogle Scholar
Corresponding author
Correspondence to Petra Pennekamp.
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Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
PP and TM wrote the manuscript and designed the figures. PP, BD and HH revised the manuscript. All authors read and approved the final manuscript.
Additional file
Additional file 1: Table S1–S4.
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Pennekamp, P., Menchen, T., Dworniczak, B. et al. Situs inversus and ciliary abnormalities: 20 years later, what is the connection?. Cilia4, 1 (2015).
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Received: 18 August 2014
Accepted: 26 November 2014
Published: 14 January 2015
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Keywords
Situs inversus
Cilia
Left-right organizer
Node
Nodal
Nodal flow
Two cilia model
Planar cell polarity
Motile cilia
Sensory cilia
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Ventilatory Response to Hypoxia
In subject area:Medicine and Dentistry
The ventilatory response to hypoxia is defined as the immediate and rapid increase in ventilation that occurs when oxygen levels in the environment decrease, typically reaching high levels within 5 to 10 minutes. This response is influenced by various factors, including individual differences and hormonal changes.
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Review article
Thoracic/physiology
2008, Anaesthesia & Intensive Care MedicineJane E. Risdall, David P. Gradwell
Ventilatory response to hypoxia: in most individuals hypoxia induces an increase in ventilation. Although this increases the respiratory work, the increased ventilation outweighs this resulting in a fall in the arterial partial pressure of carbon dioxide Paco2 and a rise in arterial Pao2. This hypoxia-induced ventilatory response is mediated through the aortic and carotid bodies (see also pages 437–40, in the October issue). The associated fall in arterial carbon dioxide pressure (PaCO2), and the consequent respiratory alkalosis, acts centrally as a respiratory depressant, setting up a conflict in response, the outcome reflecting the balance between the two influences. Respiratory control commonly is more sensitive to changes in PaCO2 and an acute reduction in arterial oxygen tension (PaO2) does not induce a respiratory response until it falls to 6–6.7 kPa (45–50 mm Hg). Below this level hypoxia induces a brisk increase in ventilation, exerting a stronger effect than the fall in PaCO2 or hydrogen ion concentration.
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Journal2008, Anaesthesia & Intensive Care MedicineJane E. Risdall, David P. Gradwell
Review article
Sleep-Disordered Breathing: Beyond Obstructive Sleep Apnea
2014, Clinics in Chest MedicineKelly Newton MD, ... Teofilo Lee-Chiong MD
Hypoxia
Ventilatory response to hypoxia is blunted during sleep.9–12 A sex difference in hypoxic drive has been described. Hypoxic ventilatory response during wakefulness is much higher in men than women.12 In addition, although hypoxic ventilatory response is lower during non–rapid eye movement (NREM) sleep than wakefulness among men, it is similar during NREM sleep and waking in women.9,10 Whether this contributes to the decreased prevalence of sleep-disordered breathing among premenopausal women is unclear.11–14 The hypoxic ventilatory response declines further during REM sleep.9–12 Isocapnic hypoxia is not a major stimulus for arousal in normal adults, who can remain asleep even if arterial saturation decreases to as low as 70%.9–11
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Journal2014, Clinics in Chest MedicineKelly Newton MD, ... Teofilo Lee-Chiong MD
Chapter
NEUROPHYSIOLOGY | Overview
2006, Encyclopedia of Respiratory MedicineK.M. Spyer, ... A.V. Gourine
Hypoxia
The ventilatory response to hypoxia in neonates and adults is characterized by an initial increase in central respiratory drive, due to an increase in the activity of the peripheral chemoreceptors. This transient increase in breathing is followed by a decline in central respiratory output. Several brain regions – including the hypothalamus, mesencephalon, and brainstem – have been shown to be involved in this secondary respiratory decline. In neonatal animals brainstem transection or lesions in the lateral upper pons help to maintain breathing activity that is normally abolished during acute sustained hypoxia. It has been proposed that several brain sites within the thalamus, hypothalamus, pons, and medulla may function as central oxygen sensors. Their physiological role as yet remains largely unclear.
A number of neurotransmitters and neuromodulators have been put forward as putative mediators of hypoxic ventilatory decline. GABA, serotonin, adenosine, opioids, nitric oxide, and others have all been implicated. A temporal sequence in the release of these factors in the brainstem (within the VRC) during hypoxia suggests a specific role for each of them in the sequential development of the hypoxic ventilatory response. In addition, there is pharmacological evidence showing that blockade of GABA, serotonin, or adenosine receptors in the lower brainstem reduces the degree of respiratory decline evoked by hypoxia.
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Reference work2006, Encyclopedia of Respiratory MedicineK.M. Spyer, ... A.V. Gourine
Chapter
VENTILATION | Control
2006, Encyclopedia of Respiratory MedicineF.L. Powell
Ventilatory Response to Arterial
The ventilatory response to is called the hypoxic ventilatory response, and is notable for its nonlinearity. Figure 4 shows that increases in have relatively little effect on ventilation, whereas decreases in below about 60 mmHg cause large increases in ventilation. This contrasts with the effects of CO2 on ventilation, which are maximal when is near its normal value (Figure 3). The hypoxic ventilatory response is not large until falls to a level at which O2–hemoglobin saturation starts decreasing significantly. Ventilation is a linear function of arterial O2 saturation but this is a coincidence and not a mechanistic explanation. Arterial chemoreceptors respond only to O2 partial pressure, and not O2 content or saturation. They are necessary for the hypoxic ventilatory response because patients without functional carotid bodies lack the response.
Figure 4 illustrates the interactions between and as ventilatory stimuli. The lower response poikilocapnic (i.e., changing CO2) curve shows a normal ventilatory response to hypoxia where decreases as a result of hypoxic ventilatory stimulation. This decreases ventilatory drive from central chemoreceptors. The net effect on ventilation is the result of stimulation by hypoxia and inhibition from decreased . In contrast, if is held constant during hypoxia, for example, by increasing inspired , then the hypoxic ventilatory response is greater as shown on the normocapnic curve in Figure 4. The increase in ventilation between the poikilocapnic and normocapnic curves is explained by the removal of hypocapnic inhibition. Increasing above normal reveals a synergistic or multiplicative interaction between and , that is, the combined effects of hypoxia and hypercapnia exceed the sum of the individual effects.
The time course of the ventilatory response to changes in begins within a single breath of breathing a low O2 mixture, as expected, given the rapid response of arterial chemoreceptors. However, the ventilatory response changes if hypoxemia persists, as described in the section on ‘Plasticity in ventilatory control’.
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Reference work2006, Encyclopedia of Respiratory MedicineF.L. Powell
Chapter
HYPOXIA | Respiratory Responses to Hypoxia in Fishes
2011, Encyclopedia of Fish PhysiologyS.F. Perry
What Is the Benefit of the Hypoxic Ventilatory Response?
The ventilatory responses to hypoxia, namely hyperventilation of the gill, ASR, and/or air breathing, represent attempts to maintain O2 uptake in the face of declining aquatic O2 levels. The rate of branchial O2 transfer is governed by diffusivity, convection (ventilation and perfusion), and the blood-to-water PO2 gradient (the driving force for O2 diffusion). During hyperventilation, increased water flow across the gill serves to raise the mean blood-to-water PO2 gradient. Effectively, fewer O2 molecules can be removed from the water during the shortened duration of transit over the gill, which raises the average PO2 of the ventilatory water and results in an elevation of arterial PO2. Hyperventilation during hypoxia therefore serves to minimize the extent of the reduction in arterial PO2 (and hence arterial O2 content) that is the inevitable consequence of lowering water PO2.
Although theory readily predicts a beneficial effect of hyperventilation on raising arterial PO2, it is more challenging to provide direct empirical data in support especially considering that many protocols or environmental conditions known to elicit hyperventilation may simultaneously influence other factors (e.g., circulating catecholamine levels or blood flow) regulating the rate of O2 diffusion across the gill. Hence, most experimental protocols cannot easily distinguish between the specific effects of ventilatory adjustments from other factors such as changes in blood-flow patterns and acute changes in functional lamellar surface area. Recently, investigators were able to record continuously in real time, arterial PO2 and ventilation from goldfish (Carassius auratus) experiencing injections of the ventilatory stimulant NaCN into the buccal cavity. These data (Figure 3) clearly demonstrate a cyclical rise and fall of PaO2 that is in phase with the increasing and decreasing .
The benefit of hyperventilation becomes particularly important as arterial PO2 approaches the P50 value of hemoglobin (the PO2 at which hemoglobin is 50% saturated with O2), where the steep slope of the O2 equilibrium curve means that even small differences in arterial PO2 can have a relatively large impact on arterial O2 content. In addition to defending arterial PO2, hyperventilation during hypoxia produces a respiratory alkalosis as arterial PCO2 is lowered by equilibration of the arterial blood with ventilatory water of lower mean PCO2. Elevation of red blood cell pH following from the respiratory alkalosis can, in turn, increase the affinity of hemoglobin for O2 via the Bohr effect, thereby aiding O2 uptake (see also TRANSPORT AND EXCHANGE OF RESPIRATORY GASES IN THE BLOOD | Gas Transport and Exchange: Interaction Between O2 and CO2 Exchange and TRANSPORT AND EXCHANGE OF RESPIRATORY GASES IN THE BLOOD | Root Effect: Root Effect Definition, Functional Role in Oxygen Delivery to the Eye and Swimbladder). The benefit of the hypoxic hyperventilatory response, then, is increased branchial O2 transfer which contributes to the maintenance of metabolic rate under hypoxic conditions in oxyregulators (fish that maintain a constant metabolic rate during hypoxia).
The trade-off is that more and more water is moved across the gills as less O2 is removed from the water per unit volume moved. Thus, the downside of hypoxic hyperventilatory response is an increasing energy expenditure on ventilation, the cost of which is high even at rest in water-breathing fish. The maintenance of metabolic rate under these circumstances eventually becomes futile when the cost of increasing ventilation to maintain O2 uptake from an environment of reduced O2 availability eventually exceeds the benefits of the O2 so obtained. The switch to oxygen conforming (the strategy of lowering metabolic rate with reduced ambient PO2), often with a concomitant reduction in , will lower the energetic expenditure on ventilation (see also HYPOXIA | Metabolic Rate Suppression as a Mechanism for Surviving Hypoxia).
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Reference work2011, Encyclopedia of Fish PhysiologyS.F. Perry
Chapter
Respiratory Physiology: Control of Ventilation
2005, Principles and Practice of Sleep Medicine (Fourth Edition)Neil J. Douglas
Hypoxic Ventilatory Response during Sleep
In adult humans, the ventilatory response to hypoxia falls during sleep1234 (Fig. 18-1). The hypoxic ventilatory response was lower during non-rapid eye movement (NREM) sleep than during wakefulness when the study subjects were exclusively or mostly men,12 but the responses were similar in wakefulness and in NREM sleep in the studies in which women predominated.345 It is not clear why there is such a difference between the sexes in the effect of sleep on the hypoxic ventilatory response. Comparison of the results in men and women (Fig. 18-2) shows that the major sex difference is in the levels of ventilatory drives during wakefulness, which are much higher in men than in women.4
The hypoxic ventilatory response during rapid eye movement (REM) sleep is lower than in NREM sleep in both men and women.1234 The hypoxic ventilatory response during REM sleep was remarkably similar in all three directly comparable studies124: about 0.4 L/min per oxygen saturation (Sao2) percentage (see Fig. 18-2).
In studies on adult animals, the isocapnic hypoxic ventilatory response has been found to be unchanged6 or decreased7 during sleep in tracheostomized dogs and decreased8 during sleep in goats.
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Book2005, Principles and Practice of Sleep Medicine (Fourth Edition)Neil J. Douglas
Chapter
Control of Breathing
2017, Nunn's Applied Respiratory Physiology (Eighth Edition)Andrew B Lumb MB BS FRCA
Acute Hypoxic Response
This is the first immediate and rapid increase in ventilation. Sudden imposition of hypoxia results in stimulation of ventilation within the lung-to-carotid body circulation time (about 6 s), but in most studies the response appears slower due to the delay between reducing inspired oxygen and the reduction in alveolar and then arterial Po2. Ventilation continues to increase for between 5 and 10 minutes, rapidly reaching high levels.
Many factors affect the acute ventilatory response. There are wide variations between individuals, within an individual on different days, between male and female subjects and with the hormonal changes of the menstrual cycle. A small number of otherwise normal subjects lack a measurable ventilatory response to hypoxia when studied at normal Pco2. This is of little importance under normal circumstances, because the Pco2 drive from the central chemoreceptors will normally ensure a safe level of Po2. However, in certain therapeutic and abnormal environmental circumstances, such as at high altitude, it could be dangerous.
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Book2017, Nunn's Applied Respiratory Physiology (Eighth Edition)Andrew B Lumb MB BS FRCA
Review article
Special Issue “Development of the Carotid Body”
2013, Respiratory Physiology & NeurobiologyMachiko Shirahata, ... Luis E. Pichard
4Development of the hypoxic ventilatory response
As described above, breathing measurement in newborn mice requires the careful control of many factors. Therefore, to discuss the hypoxic ventilatory response we have selected data in which measurement conditions were clearly described and they were maintained in optimal conditions. Fig. 3 shows the hypoxic ventilatory response during the early phase of life (<24 h). Most data were obtained from Gallego and Gautier's laboratory (Dauger et al., 2001; Renolleau et al., 2001a,b; Aizenfisz et al., 2002). In two experiments (Swiss-IOPS and C57BL/6–129SvEv) the hypoxic challenge was given using a gas mixture of 10% O2/3% CO2/87%N2 for 90 s. We can observe a clear difference in TTOT (total breathing cycle time) at 1 h of age between the two experiments. The reason for this difference is not apparent, but genetic differences (strain differences) may be a possible explanation. A striking observation here is the presence of the hypoxic ventilatory response at very early age (1 h) due to mostly increased VT (tidal volume). 3% CO2 added to the inspiratory gas may contribute to the increase in ventilation, but at 3 h of age, severe hypoxia (5% O2) without CO2 clearly increased VT and decreased TTOT resulting increased VE. These data suggest that mice show the hypoxic ventilatory response at a very early age. Within Swiss IOPS (Renolleau et al., 2001a), the hypoxic ventilatory response is weaker at 1 and 6 h of age compared to 12 h of age. The response did not significantly change among 12, 24 and 48 h of age. Two studies examined the later phase of postnatal development of the hypoxic ventilatory response (Robinson et al., 2000; Bissonnette and Knopp, 2001). Both used head out plethysmography. When we focus on the early time periods in hypoxia, relative changes in ventilation decrease from P1 and the response plateaued at ∼P20 in both Swiss CD1 outbred and C57BL/6 inbred mice. In summary, available data indicate that the ventilatory response to hypoxia is present even 1 h after birth. The responses of frequency and tidal volume to hypoxia develop differently with age, but they appear to reach steady state level around 3 weeks of age.
An important consideration is whether we can use breathing parameters from newborn mice as an indication of carotid body function. Since we do not have any information about carotid body neural output of newborn mice, here we examined the data from rats, hoping that new insights may be obtained in the relationship between carotid body function and hypoxic ventilatory response during postnatal development. Liu et al. (2006) have found dynamic changes in hypoxic ventilatory responses in rats from P0 to P21. Increases in breathing parameters in response to early time periods (∼30 s) of 10% O2 challenge was minimal at P0, but started to be evident from P1 and reached at higher steady state level between P6 and P8. During this time, both breathing frequency and tidal volume increased. There was a big dip in the hypoxic response at P13, but subsequently the response recovered and gradually reached a steady-state toward P21 with a predominant increase in tidal volume. Usually, the hypoxic ventilatory response during early time periods of the hypoxic challenge is considered to be mediated by the carotid body. However, carotid body chemoreceptor neural activity minimally responded to anoxic stimuli until ∼P7 and after P10 the response became significantly larger than the younger ages (Kholwadwala and Donnelly, 1992). A similar response pattern was seen in the calcium response of chemoreceptor glomus cells (Bamford et al., 1999). These in vitro studies used very severe hypoxia (1% O2) or anoxia, and it is difficult to interpolate the data to intact animals. Nonetheless, it appears that the carotid body in rats did not show a vigorous response to hypoxia until ∼P10, while VE response to hypoxia between P6 and P8 was almost comparable to that of P19 and P21. In mice, the hypoxic ventilatory response is seen earlier than in rats. The dichotomy between the breathing response and the carotid body response to hypoxia at early postnatal ages in the rat suggests that the breathing responses are not appropriate to be used as a surrogate of carotid body function in neonatal rodents. It is possible that the hypoxic ventilatory response in these animals is mediated by chemosensitive organs other than the carotid body such as the pulmonary neuroepithelial body or adrenal chromaffin cells (Nurse et al., 2006). To confirm these points, more comprehensive and comparative studies are required in measuring carotid body neural output and breathing in response to similar stimuli at the same developmental stages.
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Journal2013, Respiratory Physiology & NeurobiologyMachiko Shirahata, ... Luis E. Pichard
Review article
Respiratory Disorders and Sleep
2008, Sleep Medicine ClinicsHiren Muzumdar MD, MSc, Raanan Arens MD
Altered ventilatory responses
Ventilatory responses to hypoxia are absent or significantly reduced in subjects with PWS and are independent of degree of obesity.59 In contrast, hypercapnic ventilatory responses were shown to be normal in nonobese PWS and blunted in obese PWS subjects compared with controls.59 These findings are supported by studies showing absence of peripheral chemosensitivity in PWS60 and studies showing poor arousal and cardiorespiratory responses to hypoxia and hypercapnia from slow wave sleep in these subjects.57,61 Recent reports document sudden death during sleep in some subjects when treated with growth hormone to improve growth velocity and improve body composition.62–68 Although mechanisms leading to sudden death have not been elucidated, abnormalities in ventilatory control and respiratory infections have been noted. Consequently, polysomnography before treatment with growth hormone and careful monitoring during respiratory infections is recommended.66–68
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Journal2008, Sleep Medicine ClinicsHiren Muzumdar MD, MSc, Raanan Arens MD
Review article
Clinical Challenges to Ventilatory Control
2013, Respiratory Physiology & NeurobiologyMelissa L. Bates, ... Marlowe W. Eldridge
3The ventilatory response to hypoxia
We begin with a discussion of studies examining the ventilatory response to hypoxia as a result of impaired chemoreceptor function. The increase in ventilation in response to hypoxia is largely the result of carotid body stimulation by low arterial PO2. Birth is a critical transitional period in the development of normal cardiopulmonary control and, specifically, the carotid body (Teppema and Dahan, 2010). In the fetus, hypoxia inhibits normal fetal breathing movements and, although the carotid bodies are active when PaO2 < 25 mm Hg, their ability to increase ventilation is overridden by suprapontine input (Dawes et al., 1983).
3.1Post-natal development of normal carotid body function
In utero and at birth, the carotid bodies are less sensitive to hypoxia and neuronal output from the carotid body is not essential for the initiation of normal breathing (Blanco et al., 1984; Forster et al., 2000). Three key events occur post-natally in the development of the carotid body: (1) the activation threshold of the carotid body to PaO2 increases to a level similar to that observed in the adult (from <25 mm Hg PaO2 to ∼55 mm Hg), (2) the ability of CO2 to modulate responsiveness develops and (3) the carotid bodies increase in volume and mature in receptor and neurotransmitter expression. These changes allow the carotid bodies to assume an essential role in the generation of a normal ventilatory response (Blanco et al., 1984).
3.2Alterations in normal carotid body function in the infant with BPD
The carotid chemoreceptors of preterm infants are physiologically unprepared to undergo the changes necessary for the transition to the ex utero environment, making these infants more prone to apneas, periodic breathing and respiratory decompensation. These infants may require continuous positive airway pressure (CPAP), supplemental oxygen, and unplanned intubation and re-intubation until ∼36 weeks post-menstrual age (Clark et al., 2013).
The development of normal carotid body morphometry, and a normal ventilatory response, is critically dependent on the perinatal environment. Calder et al. first described the ventilatory response to hypoxia in a group of three month old infants born at ∼27 weeks, with and without BPD, using alternating breaths of 21% and 16% O2 (Calder et al., 1994). In contrast to the premature infants that did not need mechanical ventilation or supplemental oxygen, the infants with BPD failed to increase ventilation in response to a hypoxic challenge. Katz-Salamon and colleagues followed this work by examining how short hyperoxic exposures, which inhibit carotid sinus nerve output and depress ventilation, affect the ventilatory response (Katz-Salamon et al., 1995, 1996). They found a lack of hyperoxic ventilatory depression in infants with BPD. Taken together, this suggests that BPD is associated with impairments in carotid body function such that both the contribution of carotid body signaling to baseline ventilation and to the response to hypoxia is diminished.
Katz-Salamon et al. also found an association between the severity of BPD and alterations in ventilatory drive, such that the infants with the most severe disease experienced the smallest change in ventilation in response to hyperoxia (Katz-Salamon et al., 1995). BPD severity and the degree of hyperoxia-induced ventilatory depression were both related to the time spent mechanically ventilated, suggesting that the perinatal ventilatory support links the BPD severity with impairments in ventilatory drive.
3.3Rodent models of perinatal hyperoxia and hypoxia and their relationship to BPD
Subsequent research in rodents has improved our understanding of the mechanisms by which perinatal hyperoxia impacts carotid body function, demonstrating that supplemental oxygen at birth profoundly blunts future carotid body development. An excellent, detailed review of this work was recently published (Bavis et al., 2013) for those interested in an in-depth discussion of this field. Briefly, hyperoxia at birth blunts the ventilatory response to hypoxia in feline and rodent models, similarly to that seen in infants with BPD (Hanson et al., 1989; Ling et al., 1996). Deficits in normal carotid body function that result from perinatal hyperoxia are caused by a combination of impaired O2 sensitivity (Hanson et al., 1989; Donnelly et al., 2005, 2009; Bavis et al., 2011), carotid body hypoplasia and decreased total afferent neuron number (Bisgard et al., 2003; Chavez-Valdez et al., 2012; Bavis et al., 2013). Perinatal hyperoxia also reduces neurotrophin expression and increases caspase-3 levels in the nucleus tractus solitarii (nTS), the critical site for the integration of chemosensory input. Neurons in the nTS exhibit a proapoptotic phenotype, potentially impairing the normal formation of this neural circuit (Chavez-Valdez et al., 2012). The ventilatory response to CO2 is unchanged in this model (Ling et al., 1996).
Given our limited ability to access and manipulate the carotid bodies in humans for the purpose of research, it is unknown whether these physiological alterations in carotid body morphometry and signaling are present in humans with BPD. It is, however, of important clinical consequence that these ventilatory control impairments appear to be permanent in adult rats with BPD (Ling et al., 1997). Katz-Salamon studied the ventilatory response to hyperoxia in a cohort of preterm infants with BPD (Katz-Salamon et al., 1996). These authors found that while hyperoxia failed to depress ventilation soon after birth, the response recovered over the subsequent several months, suggesting recovery of chemoreceptor function. It is important to note that the magnitude of the response was not compared to term infants, nor were these infants followed beyond infancy. Indeed, our recent preliminary data from 20 to 22 year old adults with BPD, exposed to 5-min periods of eucapnic hypoxia and hyperoxia, suggest that these impairments may also be permanent in humans (Beshish et al., 2012) (see Fig. 1).
Although rodent models of BPD provide a physiologically plausible explanation for observations in humans, the physiologic contributors to the disrupted ventilatory response in the human are probably more complex. Infants with BPD rarely experience sustained normal or above normal PaO2, but are more likely to experience cycles of high, normal, and low PaO2 as they experience apneas and adjustments to their ventilatory support, overlaid with their lung disease. In contrast to the effects of hyperoxic exposure, postnatal intermittent hypoxia sensitizes the carotid chemoreceptor to hypoxia, leading to hyperventilation, a fall in PaCO2, and an increase in apnea number and breathing instability (Khan et al., 2005; Pawar et al., 2008; MacFarlane et al., 2013). In this way, chronic, intermittent hypoxia may be self-perpetuating in the BPD infant.
3.4Clinical consequences of an impaired hypoxic ventilatory drive
Chemoreceptor hypersensitivity may manifest itself clinically in the BPD patient, contributing to feeding difficulties that are not observed in preterm infants without BPD. Craig et al. found that babies with severe BPD had a lower tolerance for breathing pause during sucking and swallowing than term infants. The shorter than normal breath pauses in the BPD infants were followed by a period of irregular hyperventilation (Craig et al., 1999). In some preterm infants, these periods of hyperventilation are followed by an apnea (Gryboski, 1969). While some conclude that this hyperventilation is the result of lower SpO2 during the swallowing phase (Mizuno et al., 2007), others speculate that ventilatory control dysfunction is an important contributor (Gewolb et al., 2001). Indeed, ventilatory control dysfunction, occurring in combination with gas exchange inefficiency, muscle immaturity and an increased work of breathing, may pose an important barrier to the transition from gavage to nipple feeding in the parenteral nutrition-dependent BPD infant (McCain et al., 2012).
The inability to generate a hypoxic ventilatory response is also associated with an increased risk of acute mountain sickness after ascent to altitude (Nespoulet et al., 2012). In a 2001 consensus statement on children at altitude, the International Society for Mountain Medicine noted cases of mountain sickness and high altitude pulmonary edema in children previously diagnosed with BPD (Pollard et al., 2001). Whether a poor ventilatory response contributes to mountain sickness in individuals with BPD remains an area for future exploration.
3.5The response to hypercapnia
In comparison to the hypoxic ventilatory drive, the ventilatory response to hypercapnia is less well understood. Infants with BPD may retain CO2, transiently suppressing central chemoreceptor function. This is discussed in more detail in Section 6. In rodent models, perinatal hyperoxia does not impact the long-term ventilatory response to hypercapnia (Ling et al., 1996). The ventilatory response to CO2 has not been systematically studied in children and adults with BPD but, given the known interactions between carotid body function and the ventilatory response to CO2 (Chenuel et al., 2004) and the fact that carotid body function is altered in individuals with BPD, this is an important future area of study.
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Journal2013, Respiratory Physiology & NeurobiologyMelissa L. Bates, ... Marlowe W. Eldridge
Related terms:
Rapid Eye Movement Sleep
Apnea
Hypercapnia
Carotid Body
Obstructive Sleep Apnea
Hypoxic Ventilatory Response
Opiate
Carbon Dioxide
Hypoxia
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To convert atomic mass into mega-electronvolts (MeV), you can use the following relationship between atomic mass units (amu) and energy:
Conversion Factor: 1 atomic mass unit (1 amu) is equivalent to approximately 931.5 MeV.
Conversion Formula:
Energy (MeV)=Mass (amu)×931.5 MeV/amu Energy (MeV)=Mass (amu)×931.5 MeV/amu
Steps to Convert:
Determine the Atomic Mass: Find the atomic mass of the element or isotope you are interested in (in amu).
Apply the Conversion:
Multiply the atomic mass by 931.5 MeV/amu.
Example:
If you have an atomic mass of 12 amu (like carbon-12), the conversion would be:
\text{\text{
Continue Reading
To convert atomic mass into mega-electronvolts (MeV), you can use the following relationship between atomic mass units (amu) and energy:
Conversion Factor: 1 atomic mass unit (1 amu) is equivalent to approximately 931.5 MeV.
Conversion Formula:
Energy (MeV)=Mass (amu)×931.5 MeV/amu Energy (MeV)=Mass (amu)×931.5 MeV/amu
Steps to Convert:
Determine the Atomic Mass: Find the atomic mass of the element or isotope you are interested in (in amu).
Apply the Conversion:
Multiply the atomic mass by 931.5 MeV/amu.
Example:
If you have an atomic mass of 12 amu (like carbon-12), the conversion would be:
Energy (MeV)=12 amu×931.5 MeV/amu≈11178 MeV Energy (MeV)=12 amu×931.5 MeV/amu≈11178 MeV
This means that the energy equivalent of 12 amu is approximately 11178 MeV.
If you have a specific atomic mass in mind, let me know, and I can help with the calculation!
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Related questions
More answers below
How do you convert MeV to Joules?
How do I express Atomic mass (MA) in atomic mass unit (u)?
What is a gram atomic mass?
How do you convert atomic mass in kilograms into amu?
How do you convert mega electron volts to watts?
Howard Ludwig
Ph.D. in Physics, Northwestern University (Graduated 1982) · Author has 3K answers and 10.3M answer views
·4y
There are a few issues with the question:
The symbol for the megaelectronvolt is MeV (M for the SI scaling prefix mega, which is required to be capitalized as you did, so as not to be taken as m which is required to represent milli; e for electron or, in more modern terminology, the elementary charge, which is never capitalized even though surrounded by capital letters in this case; V for volt, which symbol, but not the unit name, is to be capitalized since it is named after somebody).
Atomic mass is a mass, which is a kind of quantity, not a unit of measurement, whereas MeV is the symbol for a
Continue Reading
There are a few issues with the question:
The symbol for the megaelectronvolt is MeV (M for the SI scaling prefix mega, which is required to be capitalized as you did, so as not to be taken as m which is required to represent milli; e for electron or, in more modern terminology, the elementary charge, which is never capitalized even though surrounded by capital letters in this case; V for volt, which symbol, but not the unit name, is to be capitalized since it is named after somebody).
Atomic mass is a mass, which is a kind of quantity, not a unit of measurement, whereas MeV is the symbol for a particular unit of measurement. One can convert a value from one unit of measurement to another unit of measurement for the same kind of quantity; one can sometimes convert one kind of quantity into another (such as energy into mass); however, one cannot convert a kind of quantity into a unit of measurement.
As I noted in the previous point, atomic mass is a mass. Technically, MeV is a unit of energy, not mass; it is the magnitude (ignoring sign) of change in energy when an object with an electric charge whose magnitude is the elementary charge e (such as a proton or electron) traverses a potential difference of 1 megavolt—the energy change is the product of the charge and the potential difference, so charge e across 1 V is e(1 V), with the product e·V being turned into a measurement unit eV. The typical formal unit of mass involving MeV in nuclear and subnuclear physics is MeV/c ², where c = 299 792 458 m/s is the speed of light. This notation is based on the Einstein relationship ∆E = ∆m c ², so that an absorption [emission] of an energy ∆E into [from] a system increases [decreases] the mass of the system by ∆E/c ² (and note that a change in kinetic energy of a system does not count in this regard). Now, physicists use not only this peculiar non-SI unit MeV (and meV, eV, keV, GeV, and TeV—all of which are officially approved to be used in the context of SI without violating compliance with SI, even though they are not SI units) for energy, but because speeds close to that of light are often involved, it is easiest to scale speeds relative to c, which means effectively that a unit system is being used in which it is defined that c = 1—yes, that is simply a number without unit of measurement. As a consequence, what nominally should be MeV/c ² as a unit of mass actually has the value MeV/1² = MeV in this system. Therefore, it is very common to express amounts of mass in terms of MeV or GeV, just as you have done. Physicists also express decay times for unstable particles in terms of the mean life τ, whereas chemists use the half-life whose value is about 0.7 times the mean life. Because c = 1, distance divided by time is dimensionless, so distance and time have the same dimensions. It is rather common for physicists to refer to the mean life in terms of distance cτ instead of time.
The coherent SI unit of mass is the kilogram (symbol kg). Other (not coherent) SI units of mass are obtained by simply removing the prefix kilo to yield gram or by replacing the kilo with any of the other 19 SI scaling prefixes. The smallest SI unit of mass is the yoctogram, which conveniently is fairly close to the mass of a proton or neutron. However, for things at the atomic and subatomic level, there are two other units that are much more commonly used: (1) a unit that is alternatively known (and both names are officially approved) as the dalton (symbol Da, the more modern term) and the unified atomic mass unit (symbol u, and the word “unified” is a mandatory part of the name, as this unit replaced the atomic mass unit, which along with its symbol amu, was abrogated in 1961); (2) the electronvolt. The SI scaling prefixes may be applied to all of these units.
The dalton is used more by chemists because it corresponds closely to an atom’s mass, with the number of daltons for the mass of an atom or molecule being within about 1 % of the total number of protons in the atom or molecule. The dalton is defined so that the mass of one free ground-state ¹²C atom is exactly 12 Da. The mass of atoms is measured most accurately relative to the mass of a ¹²C atom, so atomic mass in terms of daltons can be measured very accurately, to at least 8 significant digits for atoms that are not scarce and have lifetimes of at least a few seconds. The relationship between the dalton and the kilogram is determined experimentally, with uncertainty, rather than being an exact, defined quantity:
1 Da = (1.660 539 066 60 ± 0.000 000 000 50) ×10⁻²⁷ kg.
The experimental physicists work with accelerators with design parameters including potential differences applied to electrically charged particles whose charge is an integer times the elementary charge, so the electronvolt is a very natural unit to use for energy. With the revamping of SI that took effect in 2019, the elementary charge has taken on the role of a defining quantity for SI, so its value is fixed exactly. That value is:
e = 1.602 176 634 C, so
1 MeV = 10⁶ e (1 V) = 1.602 176 634 × 10⁻¹³ J exactly in terms of coherent SI units.
The mass corresponding to this energy is this energy divided by c ²:
1 MeV/c ² = (1.602 176 634 × 10⁻¹³ J)/(299 792 458 m/s)²
= 1.782 661 921 6… × 10⁻³⁰ kg = 0.001 782 661 921 6… yg with no uncertainty.
This works out to:
1 MeV/c ² = (0.001 073 544 102 33 ± 0.000 000 000 000 32) Da and
1 Da = (931.494 102 42 ± 0.000 000 28) MeV/c ².
Therefore, to express the mass of a particle in MeV [or MeV/c ²]:
Divide the number of yoctograms by 0.001 782 661 921 6…, or
divide the number of daltons by 0.001 073 544 102 33.
For a proton this works out to:
m ₚ = (1.672 621 923 69 ± 0.000 000 000 51) yg
= (1.007 276 466 621 ± 0.000 000 000 053) Da
= (938.272 088 16 ± 0.000 000 29) MeV.
In terms of yoctograms and megaelectronvolts, the relative uncertainty of the mass of the proton is about 3.1 × 10⁻¹⁰, whereas in terms of daltons the relative uncertainty is only about 5.3 × 10⁻¹¹. Thus, the mass is known with a factor 6 improved relative uncertainty in terms of daltons as compared to yoctograms and megaelectronvolts.
Upvote ·
Momin Khalil
Lives in Delhi
·5y
Related
How can I learn atomic mass?
It's very easy
The elements with odd atomic number have to be multiplied with two and add 1 to it.
The elements with even atomic number have to be multiplied by 2.
However there are some exceptions. These exception are Hydrogen, Beryllium, Nitrogen, Chlorine and Argon.
We know that hydrogen have no neutron so its atomic mass have to be 1.
Beryllium is having an even atomic number but follows the rule of odd atomic numbers.
Nitrogen it is having an odd atomic number but follows the role of even atomic number.
Chlorine follows the normal role of on atomic number but 0.5 is added to the answer.
Argon als
Continue Reading
It's very easy
The elements with odd atomic number have to be multiplied with two and add 1 to it.
The elements with even atomic number have to be multiplied by 2.
However there are some exceptions. These exception are Hydrogen, Beryllium, Nitrogen, Chlorine and Argon.
We know that hydrogen have no neutron so its atomic mass have to be 1.
Beryllium is having an even atomic number but follows the rule of odd atomic numbers.
Nitrogen it is having an odd atomic number but follows the role of even atomic number.
Chlorine follows the normal role of on atomic number but 0.5 is added to the answer.
Argon also follows the even atomic number rule but 4 is added to the answer.
Hope you like my answer
Please vote.
Upvote ·
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Shubham Chauhan
Youtuber at Mr. Storyteller (2017–present) · Upvoted by
Swatantra Kumar Singh
, Assistant Manager in Quality Control at Gulbrandsen Technologies (2013-present) · Author has 145 answers and 1.6M answer views
·8y
Related
How do you convert MeV to Joules?
1 MeV = 10,00,000eV (eqn 1)
Also, 1J=1Volt×Coulomb{ as work=volt×charge}
Therefore
1J=1volt×6.25×10^18electron (since 1Coulomb=6.25 × 10^18e)
1J= 6.25×10^18eV
Dividing both sides by 6.25×10^18 we get,
1.6×10^-19J=1eV
Multiplying 10^6 both sides
1.6×10^-13J=10^6eV
From eqn 1
1.6×10^-13J=1MeV
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Related questions
More answers below
How can we explain relative atomic mass?
How do I calculate relative atomic mass?
Which one has units between Molar mass and Atomic mass?
How do you convert the atomic mass of an atom into grams and vice versa?
What is the difference between 35.5u and 35.5g in atomic mass?
Abob
Upvoted by
Bob Spillman
, PhD Chemistry, University of Illinois at Urbana-Champaign (1975)
·7y
Related
What is the difference between atomic mass, mass number, atomic weight, and relative atomic mass?
Here are some important points you should know that will clear all your doubts:
Atomic Number: It is the number of protons (i.e. positive charger) which is also equal to the number of electrons (i.e. negative charges) in a neutral atom. It is just a number and has no unit. Also note that a neutron has no charge.
Mass Number: Mass number is nothing but the number of protons added to the number of neutrons. We can also say that it is the number of electrons plus number of neutrons. Again, it is just a number and has no unit.
Now, if you sit in a physics class you'll see that weight is a product of
Continue Reading
Here are some important points you should know that will clear all your doubts:
Atomic Number: It is the number of protons (i.e. positive charger) which is also equal to the number of electrons (i.e. negative charges) in a neutral atom. It is just a number and has no unit. Also note that a neutron has no charge.
Mass Number: Mass number is nothing but the number of protons added to the number of neutrons. We can also say that it is the number of electrons plus number of neutrons. Again, it is just a number and has no unit.
Now, if you sit in a physics class you'll see that weight is a product of the mass and gravity and being a force, it is expressed in newtons (N). Mass however gives us an idea about the amout of substance/matter there is in a body so its S.I. Unit is Kg. Here in chemistry however these are slightly different.
Atomic Mass: It is simply the mass of a particular atom expressed in a.m.u. It does not take into consideration the various isotopes.
Now,
Mass of 1 proton = Mass of 1 neutrom = 1 a.m.u.
Also, the mass of an electron is negligible so it may/ may not be considered when calculating atomic mass.
So we have,
Atomic Mass = (No. of protons + neutrons) a.m.u. = (No. of neucleons) a.m.u. = (Mass Number) a.m.u.
Thus, atomic mass is approximately equal to the mass number (except it has a unit) and so is generally a whole number.
We also know,
1.66 × 10^-24 g = 1 a.m.u.
1 g = 1/(1.66 × 10^-24) a.m.u.
But 1 a.m.u = Actual mass of C-12 ÷ 12
Let us say that the actual mass in grams of an atom is x g.
1 g = 1/(1.66×10^-24) a.m.u = 1/(Actual mass of C-12 ÷ 12) a.m.u.
x g = x/(Actual mass of C-12 ÷12)
In other words,
Atomic mass = Actual mass of an atom/ 1 a.m.u.
That is atomic mass done.
Average Atomic Mass: It is basically a weighted average of all the atomic masses of various isotopes of an element. It is also measured in a.m.u.
Many a times atomic mass and average atomic mass are used interchangeably.
Relative isotopic mass: This is unitless/dimensionless as it is a ratio. It is exactly the same as the atomic mass expressed without a unit. This loss of unit occurs as atomic mass divided by 1/12th the mass of C-12 (or simply 1 a.m.u.). It is therefore a ratio between the two. That is why its called 'relative' as it is in relation with 1/12th the mass of C-12 which is taken as standard. Note that we are using the atomic mass here and not the average atomic mass. That is why this quantity too does not take account of the various isotopes.
Relative Atomic Mass/Atomic weight (They are the same)- It is the same as relative isotopic mass except that instead of using the atomic mass, we use the average atomic mass. So again it is a ratio, it is unitless/dimensionless and is in relation to something.
Hope this helped
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Dan Steiner
Studied at University of Wisconsin - Madison
·2y
Related
How do you convert the rest mass of a neutron into Mev?
Assuming you are starting with the neutron mass in kg, you use Einstein’s famous E=mc^2 to convert the rest mass into a rest energy. Divide that energy by the elementary charge to convert it into electron-volts, and then divide by one million to get MeV.
Upvote ·
Padmini Kalyanaraman
School Teacher · Author has 4K answers and 4.6M answer views
·7y
Related
How can we find the relative atomic masses of elements?
Atomic weight or atomic mass = Total mass of protons + Total mass of neutrons
Atom has finite Mass. Hence it can be measured in terms of kgm or gm.The radius of an atom 10^-8 cm. The size is too small & the absolute mass of an atom is very very small.
Atom Absolute weight
1.Hydrogen 1.67 x 10^-24 g
Oxygen. 26.56 x 10^-24 g
Chlorine 58.60 x 10^-24g
These weights are extremely small. Such small masses cannot be weighed by a chemical balance.Further , most of the atoms do not have free existence & as such cannot be isolated.The actual masses of atoms have been compared with mass of an atom of som
Continue Reading
Atomic weight or atomic mass = Total mass of protons + Total mass of neutrons
Atom has finite Mass. Hence it can be measured in terms of kgm or gm.The radius of an atom 10^-8 cm. The size is too small & the absolute mass of an atom is very very small.
Atom Absolute weight
1.Hydrogen 1.67 x 10^-24 g
Oxygen. 26.56 x 10^-24 g
Chlorine 58.60 x 10^-24g
These weights are extremely small. Such small masses cannot be weighed by a chemical balance.Further , most of the atoms do not have free existence & as such cannot be isolated.The actual masses of atoms have been compared with mass of an atom of some element considered as standard. It has been assigned atomic Mass or atomic weight as unity i.e 1.The atomic weights of different elements were determined on the scale of this hydrogen by comparing the mass of an atom of any element with the mass of hydrogen atom. Thereby atomic the atomic weights thus obtained are relative weights
Hydrogen as a standard , has some disadvantages, an isotope of carbon is now considered as standard
The average relative weight of atoms of the atoms compared to the weight of an atom of carbon isotope taken as standard equal to 12.
It is defined as
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Shize Liu
Physics & Biochemistry · Author has 173 answers and 275.8K answer views
·3y
Related
How do I find the mass of an isotope?
First, no matter what, isotope of an element will always have the same number of protons, and you can determine the proton of an isotope using the atomic number of the element on the periodic table.
The difference between different isotopes of an element is the number of neutrons. The mass of an isotope is determined by the sum of its protons and of its electrons plus the unit atomic mass unit (amu).
Thus, you can do a little simple algebra to find the mass of an isotope, depending on what you are given. If you are given the mass, usually indicted in the notation of the isotope (e.x. C-12 is car
Continue Reading
First, no matter what, isotope of an element will always have the same number of protons, and you can determine the proton of an isotope using the atomic number of the element on the periodic table.
The difference between different isotopes of an element is the number of neutrons. The mass of an isotope is determined by the sum of its protons and of its electrons plus the unit atomic mass unit (amu).
Thus, you can do a little simple algebra to find the mass of an isotope, depending on what you are given. If you are given the mass, usually indicted in the notation of the isotope (e.x. C-12 is carbon-12, meaning the carbon isotope has a mass of 12 amu), you can use simple algebra to find the number of neutrons, and if you are given the number of neutrons, you can also solve for the mass of the isotope.
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Peter Upton
BA in Physics&Mathematics, The Open University · Author has 14.5K answers and 10.8M answer views
·2y
Related
How do I convert gamma radiation into mass?
You could convert some gamma to mass for a short time. A gamma photon ( of sufficient energy) can create a elecron-positron pair when in close proximity to a heavy atom. It needs a heavy atom because the gamma photon has a lot of momentum, which has to be conserved. To have the same amount of momentum, the electron/ positron would need a lot of kinetic energy so the process does not happen unless there is something ( a heavy nucleus) to have the momentum but not much energy.
Have the gamma radiation pass through a container of say lead vapour. Photons will lose energy and some will undergo pair
Continue Reading
You could convert some gamma to mass for a short time. A gamma photon ( of sufficient energy) can create a elecron-positron pair when in close proximity to a heavy atom. It needs a heavy atom because the gamma photon has a lot of momentum, which has to be conserved. To have the same amount of momentum, the electron/ positron would need a lot of kinetic energy so the process does not happen unless there is something ( a heavy nucleus) to have the momentum but not much energy.
Have the gamma radiation pass through a container of say lead vapour. Photons will lose energy and some will undergo pair production.
I said a short time because the positron will rapidly annihilate producing two gamma photons. The positron would be gone as would an electron. All electrons are identical so you would not be able to distinguish between the electron that was ‘made’ and any of the others. The only way you would Know you had been successful would be the identification of two gamma photons of roughly 500 kev travelling in opposite directions.
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Tehsin Ud Din Khan Wazir
Electrical Engineer · Author has 205 answers and 3M answer views
·8y
Related
How do you convert MeV to Joules?
So however many eV's you have, you need to multiply it by 1.6x10^-19 Joules to find the amount of energy.
So in this case, we have 2 mega-electron-volts (2million eV's).
2.0 MeV = (2.0 x 10^6) x 1.6 x 10^-19
2.0 MeV = 3.2 x 10^-13 Joules.
Then sub in.
Hope this helps!
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Andrew Mason
B.A. in Mathematics and Physics, Queen's University at Kingston (Graduated 1976) · Author has 2.2K answers and 755.9K answer views
·2y
Related
How do you convert the rest mass of a neutron into Mev?
Use E=m c 2 E=m c 2
where: m=1.675×10−27 m=1.675×10−27 kg and c=2.998×10 8 c=2.998×10 8 m/sec
to get an answer in Joules. Then to convert from Joules to MeV (the energy of 1 million electrons or 10 6×1.602×10−19 10 6×1.602×10−19 Coulombs having energy of 1 Joule/Coulomb) divide by 1.602×10−13 1.602×10−13 (or multiply by: 6.242×10 12)6.242×10 12)
Upvote ·
Kimberley
Studied Physics (university level) · Author has 13.5K answers and 6.4M answer views
·2y
Related
How do you convert the rest mass of a neutron into Mev?
E = mc^
1 amu = 1.674927E-27 kg.
So plug in the values… 1amu =~1.5E-10 J (Joules are what you get when the units are combined)
To keep it simple, 1 amu = ~931 Mev.
1 neutron = 1.008665 amu
Note: officially, 1 neutron is 939.56542052 Mev/c^2.
Upvote ·
9 1
Mark Jensen
Ph.D. in Physics, North Carolina State University at Raleigh (Graduated 1981) · Author has 352 answers and 183.9K answer views
·5y
Related
How can I convert MeV/c^2 into kg?
You can work this our from first principles or got to the NIST Website Conversion Factors for Energy Equivalents and use their converter.
NIST. 2019. “Conversion Factors for Energy Equivalents.” Government. NIST Reference on Constants, Units, and Uncertainty. May 31, 2019. Conversion factors for energy equivalents.
Upvote ·
Related questions
How do you convert MeV to Joules?
How do I express Atomic mass (MA) in atomic mass unit (u)?
What is a gram atomic mass?
How do you convert atomic mass in kilograms into amu?
How do you convert mega electron volts to watts?
How can we explain relative atomic mass?
How do I calculate relative atomic mass?
Which one has units between Molar mass and Atomic mass?
How do you convert the atomic mass of an atom into grams and vice versa?
What is the difference between 35.5u and 35.5g in atomic mass?
What is the atomic mass of copper?
How do you convert joules into Mev?
What is the actual mass of oxygen atom (not gram atomic mass)?
Why is the mass of a particle sometimes measured in electron volts?
What is the difference between atomic mass, mass number, atomic weight, and relative atomic mass?
Related questions
How do you convert MeV to Joules?
How do I express Atomic mass (MA) in atomic mass unit (u)?
What is a gram atomic mass?
How do you convert atomic mass in kilograms into amu?
How do you convert mega electron volts to watts?
How can we explain relative atomic mass?
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平方根、算术平方根与立方根的区别与联系
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平方根（包括所有偶次方根）都有一正一负绝对值相等的二个根,如4的平方根可能是+2,也可能是-2.因为（-2）的平方也等于+2.
算术平方根只取正值,不取负值.
正数的立方根（包括所有奇次方根）都是正值.不可能是负值.3√8一定是+2
负数的立方根,都是负值,3√-8,一定是-2
平方根、算术平方根是实数中两个比较重要的概念.因为它们的定义相近,联系紧密,所以初学的同学很容易混淆.为帮助同学们区分这两个概念,现将它们的区别与联系总结如下.
一、区别
1.概念的区别
平方根:一般地,如果一个数的平方等于a,那么这个数叫做a的平方根,也就是说,如果x2=a,那么x叫做a的平方根.例如:32=9,3是9的平方根,(-3)2=9,-3也是9的平方根,即3和-3都是9的平方根.
算术平方根:一般地,如果一个非负数x的平方等于a,即x2=a,那么这个非负数x叫做a的算术平方根.例如:32=9,正数3是9的算术平方根.虽然(-3)2=9,但-3不是9的算术平方根.
2.书写方法的区别
平方根:一个非负数a的平方根记做±!a.例如,3的平方根记做±!3.:
算术平方根:一个非负数a的算术平方根记做!a.例如,3的算术平方根记做!3.
3.个数的区别
平方根:一个正数有两个平方根,并且它们互为相反数.例如,25的平方根有两个,一个是5,另一个是-5.:
算术平方根:一个正数的算术平方根只有一个,且这个数是正数.
二、联系
平方根立方根都是 乘方运算的逆运算,分别对应的是平方与立方.
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1022 | https://www.youtube.com/watch?v=pPoRtj1m1Yo | Gay Lussac's Law | Gas Laws | Physical Chemistry | Khan Academy
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the photo you see here is of Joseph Louis galusack and in early 1800s gay lusak was experimenting with how different volumes of gases combine and based on his calculations another chemist arrived at a relationship involving gases so this relationship which is called galusax law or the pressure law can be stated as pressure is directly proportional to the temperature but this relationship between the pressure and the temperature of the gas assumes two conditions one that the volume occupied by the gas is fixed and two that the amount of gas is also fixed so let me just write down both of these conditions here to understand this better we can consider this setup where we have this box and in this box we have a few molecules of the gas and because our box is a rigid box and the gas is within this box the volume is now fixed and also because we are not adding more gas or removing the gas from this box the amount of gas in this case is fixed so now both of these conditions are met and we know that these gas molecules will be randomly moving around and while they are randomly moving about like this they could be colliding with any of these faces of the box so now to better understand this relationship let's take another box and this box is an identical box with the same amount of gas in this box as well and in this box also the gas molecules will be moving in random directions so now that we have two identical boxes and for both of them both of these conditions are made what if we Heat this box when we are heating this lower face of the Box what happens is because of the heat the kinetic energy of these molecules increases which means that their velocities also increase and they start moving faster in the box and we know that when the kinetic energy increases the momentum will also increase and we Define force as the rate of change of momentum so effectively what's happening is when we Heat this phase as the molecules collide with this phase and their kinetic energy increases they collide with each of these faces faster and because of this increased momentum the force applied on each of these faces increases and we know that the pressure is defined as Force upon area so since the force increases the pressure also increases so what happened here was by increasing the temperature we see how the pressure also increases but this was a very qualitative description of what's happening let's see how we can express this mathematically and what new information we can get from there so I have written down our earlier relationship between pressure and temperature like this that is p is proportional to T where the p is pressure and T is temperature I can get rid of the proportional T sign and write it in the form of an equation as p is equal to some M times T where m is a constant and here you can see that m is a dimensional constant so going back to our molecules in a box if we know the temperature at which this box is kept and if we know the pressure at that point we can get the value of this m and it will be in some pressure per temperature units so now if we just rearrange this we can write this as P divided by T is constant and what this rearrangement does is that it simplifies some calculation for us so again if we go back to this box let's say we started at some temperature T1 and at that time the pressure was P1 and we then increased the temperature to T2 because of which the pressure then became P2 we know note that the ratio P by T is constant so we can write this as P1 by T1 is equal to P2 by T2 because we know that the ratios of pressures and temperature will be constant if we know any three of these quantities we can easily find the fourth one now again if we look at this equation we can see that this resembles the equation of a straight line if we think of the equation of a straight line written as Y is equal to ax plus b and let's say if this intercept is 0. so we can write the equation of the line as something like Y is equal to ax and this equation here resembles this form so if we were to graphically represent this relationship if we have pressure on the y axis and temperature on the x axis we can show this relationship as a straight line but there is one very important point to note here this temperature that we are talking about here is in the form of Kelvin and not in the form of degree solution this and not just here whenever we write this relationship in this form or in the form of a ratio everywhere the temperatures that we take are in kelvin and so if the temperatures which are known if they are given in degree celsius what we do is we add 273 to it and we can get the temperatures in kelvin which we can use in these equations so now let's go through one example problem and see how these calculations are done now let's look at this problem which is asking us to calculate the pressure change when a constant volume of gas at a pressure of 1 atm is heated from 20 degree Celsius to 30 degree celsius so if you want you can pause the video here and give this a try okay so let's write down what is known from the information given to us so first we can see that we've been told the gas has a constant volume and so if it is in a box like we saw before the amount of gas is also fixed so in that case we can apply galosac's law and we can write P1 by T1 will be equal to P2 by T2 where we know that our initial pressure is 1 atm and initial temperature is 20 degree celsius we know that the final temperature is 30 degree celsius so if we rearrange this we can write P2 as T2 divided by T1 into P1 and as we saw before the temperatures that we use here should be in kelvin and not in degree celsius so if you directly substitute 30 and 20 here instead of T2 and T1 we'll get the wrong answer so to use this relation we will first convert these temperatures into Kelvin so we will add 273 to both of these so now the initial temperature becomes 293 Kelvin and the final temperature becomes 303 Kelvin and so we can substitute these two temperatures and the initial pressure in this equation to calculate P2 which is 3 0 3 divided by 293 into 1.0 ATM and if you solve this we get P2 to be 1.03 ATM and then to calculate the pressure change what we are going to do is the pressure change will be equal to P2 minus P1 which is this 1.03 minus the initial pressure which is 1 atm and we get the pressure change to be 0.03 ATM |
1023 | https://www.biorxiv.org/content/10.1101/2022.08.09.503421v1.full-text | Selective Nigrostriatal Dopamine Excess Impairs Behaviors Linked to the Cognitive and Negative Symptoms of Psychosis | bioRxiv
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Selective Nigrostriatal Dopamine Excess Impairs Behaviors Linked to the Cognitive and Negative Symptoms of Psychosis
Nicolette A.Moya, Seongsik Yun, Stefan W.Fleps, Madison M.Martin, Jacob A.Nadel, Lisa R.Beutler, Larry S.Zweifel, Jones G.Parker
doi:
Now published in Neuropsychopharmacology doi: 10.1038/s41386-022-01492-1
Nicolette A. Moya
1 Departments of Neuroscience and Pharmacology, Northwestern University, Chicago, IL, USA
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Seongsik Yun
1 Departments of Neuroscience and Pharmacology, Northwestern University, Chicago, IL, USA
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Stefan W. Fleps
1 Departments of Neuroscience and Pharmacology, Northwestern University, Chicago, IL, USA
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Madison M. Martin
1 Departments of Neuroscience and Pharmacology, Northwestern University, Chicago, IL, USA
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Jacob A. Nadel
1 Departments of Neuroscience and Pharmacology, Northwestern University, Chicago, IL, USA
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Lisa R. Beutler
2 Department of Medicine, Division of Endocrinology, Metabolism and Molecular Medicine Northwestern University, Chicago, IL, USA
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Larry S. Zweifel
3 Departments of Psychiatry and Behavioral Sciences and Pharmacology, University of Washington, Seattle, WA, USA
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Jones G. Parker
1 Departments of Neuroscience and Pharmacology, Northwestern University, Chicago, IL, USA
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For correspondence: jones.parker@northwestern.edu
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ABSTRACT
BACKGROUND Excess dopamine release in the dorsal striatum (DS) is linked to psychosis. Antipsychotics are thought to work for positive symptoms by blocking striatal D2 dopamine receptors, but they lack efficacy for the negative and cognitive symptoms. Further, broadly increasing dopamine release improves cognitive function. These observations fueled the dogma that excess dopamine is not involved in negative and cognitive symptoms, but this has never been tested with dopamine pathway specificity.
METHODS We selectively re-expressed excitatory TRPV1 receptors in DS-projecting dopamine neurons of male and female Trpv1 knockout mice. We treated these mice with capsaicin (TRPV1 agonist) to selectively activate these neurons, validated this approach with fiber photometry, and assessed its effects on social and cognitive function. We combined this manipulation with antipsychotic treatment (haloperidol) and compared the pathway-specific manipulation to treatment with the non-selective dopamine releaser amphetamine.
RESULTS Selectively activating DS-projecting dopamine neurons increased DS (but not cortical) dopamine release and increased locomotor activity. Surprisingly, this manipulation also impaired behavioral processes linked to negative and cognitive symptoms (social drive and working memory). Haloperidol normalized locomotion, only partially rescued working memory, and had no effect on social interaction. By contrast, amphetamine increased locomotion but did not impair social interaction or working memory.
CONCLUSIONS Excess dopamine release, when restricted to the DS, causes behavioral deficits linked to negative and cognitive symptoms. Previous studies using non-selective approaches to release dopamine likely overlooked these contributions of excess dopamine to psychosis. Future therapies should address this disregarded role for excess striatal dopamine in the treatment-resistant symptoms of psychosis.
INTRODUCTION
Current antipsychotic treatments are largely ineffective for the cognitive and negative symptoms of psychosis (1). Negative symptoms such as the loss of desire for social engagement and cognitive symptoms like deficits in working memory are just as debilitating to quality of life as the hallmark positive symptoms of psychosis (e.g., hallucinations and delusions) (2). Therefore, there is an immediate need to address these treatment-resistant symptoms. A major barrier to this has been an imprecise understanding of the underlying neural substrates of psychosis and how its neuropathology maps onto different symptoms.
For decades, we have known that the neurotransmitter dopamine plays an important role in psychosis, but the mechanisms underlying this role remain unclear. Nearly all antipsychotic drugs block D2 dopamine receptors (D2Rs), and functional imaging studies show that schizophrenia patients have increased dopamine in the DS, where D2Rs are abundant (3-6). These observations fueled the prominent hypothesis that excess striatal dopamine is responsible for the D2R antagonist-responsive, positive symptoms of schizophrenia but not its D2R antagonist-resistant, negative and cognitive symptoms (7). This idea is further supported by the fact that amphetamine (a dopamine releasing drug) can acutely induce positive symptoms (8) but actually improves cognitive function in schizophrenia patients (9).
However, D2Rs are not the only dopamine receptors in the striatum, which equally expresses D1 dopamine receptors (D1Rs) (10). Because D1Rs are not a principal target of existing antipsychotics, their signaling may contribute to symptoms that are resistant to treatment with D2R antagonist-based antipsychotics. Moreover, the pro-cognitive effects of amphetamine in schizophrenia patients could stem from the fact that it increases dopamine release throughout the brain (not only in the dorsal striatum, as observed in schizophrenia), including in the prefrontal cortex (PFC) where dopamine release is actually decreased in these patients (10-12). Moreover, the dorsal striatum is anatomically interconnected with the PFC and other brain structures implicated in cognitive and social function (13-15). These circuit-level interactions and the specificity of antipsychotics for D2Rs suggest that excess dopamine signaling, when restricted to the DS, could contribute to the negative and cognitive symptoms of psychosis. This idea that has never been directly tested.
To explore this, we developed an experimental approach using the capsaicin-gated ion channel, TRPV1, to accurately recapitulate the pathway-specific increase in dopamine transmission observed in schizophrenia and examined its effects on antipsychotic treatment-responsive and resistant behavioral processes in mice. In contrast to the prevailing view, selectively driving DS dopamine transmission impaired behavioral processes related to the negative and cognitive symptoms of psychosis (social interaction and working memory). These behavioral deficits were largely unresponsive to the D2R-antagonist/antipsychotic drug haloperidol and did not occur following treatment with the non-selective dopamine releaser amphetamine. Our findings suggest that excess striatal dopamine plays a broader role in the symptomatology of psychosis than previously thought. This insight exposes a gap in our understanding of dopamine’s role in psychosis and the tools established here provide a path forward to close this gap and develop more comprehensive antipsychotic treatments.
METHODS AND MATERIALS
Animals
We housed and handled all mice according to guidelines approved by the Northwestern University Animal Care and Use Committee. We used both male and female mice housed on a reverse light cycle for all experiments. For capsaicin experiments, we crossed homozygous DAT IREScre mice (Jax #0006660) with Trpv1 knockout (KO) mice (Jax #003770) to generate double heterozygous mice that we then crossed to generate DAT cre/+; TRPV1 KO mice (16, 17). We maintained all founder mouse lines through backcrossing to C57BL/6J mice (Jax #000664), the same mouse strain we used for behavioral experiments with amphetamine. All mice were 12–24 weeks at the start of experimental testing. Detailed materials and methods are provided in the Supplementary Methods.
Drugs
We injected all drugs subcutaneously at volume of 10 mL·kg-1. We dissolved capsaicin (3.5–10 mg·kg-1; Alomone Labs) in 3.33% Tween 80 in PBS, D-Amphetamine hemisulfate (0.5–10 mg·kg-1; Sigma) in saline, and haloperidol (0.032–0.1 mg·kg-1; Sigma) in 0.3% tartaric acid.
Surgical Procedures
We stereotaxically injected either AAV2/5-hSyn-FLEX-TRPV1-mCherry or AAV2/5-hSyn-DIO-mCherry bilaterally in the substantia nigra pars compacta (SNc) of DAT cre/+; TRPV1 KO mice. For animals used for fiber photometry, we also injected AAV2/9-CAG-dLight1.3b into the dorsomedial striatum (DMS) and medial prefrontal cortex (mPFC; unilaterally), followed by fiberoptic cannula implants into the same regions (Figure 1A, 2A).
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Figure 1.Histological validation of the selective expression of functional TRPV1 in SNc dopamine neurons.
A) We virally expressed Cre-dependent mCherry (control) or TRPV1-mCherry (experimental) in DS-projecting SNc dopamine neurons of DAT cre/+; TRPV1 KO mice. B) Coronal brain sections containing SNc and VTA from a representative experimental (TRPV1-mCherry) mouse shown at 10× (left) and 16× resolution (right; blue: DAPI nuclear stain; red: α-RFP; green: α-TH). C) Mean ± s.e.m. percentage of mCherry-expressing neurons that are TH positive in control and experimental mice (N = 6 and 10 respectively; average of 3 brain slices per mouse). D) Coronal brain sections containing SNc from representative control (top) and experimental (bottom) mice immunostained for Fos (green) or RFP (red) 75 min after systemic vehicle or capsaicin treatment (10 mg·kg-1). E) Mean ± s.e.m. percentage of mCherry-expressing neurons that are Fos positive as a function of treatment and experimental group (N = 3–5 mice per group; average of 3 brain slices per mouse). In all images, white dashed lines indicate boundaries of SNc and solid lines indicate boundaries of adjacent brain structures. P< 0.01 comparing vehicle to capsaicin treatment; Unpaired t-test in C; Kruskal-Wallis unpaired one-way ANOVA in E. Details for these and all other statistical comparisons are presented in the Supplemental Table.
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Figure 2.Recordings of dLight1.3b fluorescence in mPFC and DMS following systemic capsaicin or amphetamine treatment.
A) We virally expressed mCherry (control) or TRPV1-mCherry (experimental) in DS-projecting SNc dopamine neurons of DAT cre/+; TRPV1 KO mice and the fluorescent dopamine sensor dLight1.3b in mPFC and DMS. We then implanted fiber-optic cannulas and used fiber photometry to simultaneously record dopamine transmission in mPFC and DMS. B) Representative coronal brain sections from an experimental, TRPV1 mouse expressing dLight1.3b in prefrontal cortex and dorsomedial striatum (green: α -GFP; blue: DAPI nuclear stain; scale bar: 1 mm). White dashed lines indicate the position of the implanted fiber-optic probe. C) Schematic representation of the time course of dLight1.3b recordings and drug treatments. We administered vehicle at t = 0 min, and capsaicin (10 mg·kg-1) or amphetamine (10 mg·kg-1) at t = 15 min. We recorded dLight1.3b fluorescence for 15-, 30-, or 60-min following vehicle, capsaicin, or amphetamine treatment, respectively. D) Example trace (% ΔF/F) from of dLight1.3b fluorescence in mPFC and DMS in response to capsaicin treatment in a representative TRPV1 mouse. E, F) Mean ± s.e.m. dLight1.3b fluorescence in DMS, E, and mPFC, F, calculated using area-under-the-curve (AUC) in 1-min time bins and normalized to values in the final 10 min following vehicle treatment in control (mCherry) and experimental (TRPV1) mice. Capsaicin responses are truncated to the first 10 min when changes in dopamine were most evident. G, H) Mean ± s.e.m. dLight1.3b fluorescence in DMS, G, and mPFC, H, calculated using AUC in 1-min time bins and normalized to values in the final 10-min following vehicle treatment in control and TRPV1 mice. Amphetamine responses are truncated to the first 40 min when changes in dopamine were most evident. I, J) dLight1.3b fluorescence in DMS, I, and mPFC, J, calculated using AUC in 1-min time bins, normalized to values in the final 10-min following vehicle treatment, and averaged across the 10, 1-min time bins surrounding the peak response following systemic capsaicin or amphetamine treatment in control and TRPV1 mice. Both control and experimental groups were combined in the analysis of amphetamine treatment effects (G–J). In these and all other box-and-whisker plots, the horizontal lines denote median values, boxes cover the middle two quartiles and whiskers span 1.5× the interquartile range. N = 7 mCherry, N = 11 TRPV1, and N = 18 combined amphetamine mice; P< 0.001 and P< 0.01; Holm-Sidak’s multiple comparison test.
Fiber Photometry
We used a commercial fiber photometry system and software (TDT) to record each mouse’s baseline dLight1.3b activity in the mPFC and DMS for 15 min, subcutaneously injected vehicle, recorded for an additional 15 min, and then subcutaneously injected capsaicin or amphetamine (10 mg·kg-1 each) and recorded dLight1.3b activity for 30 or 60 min, respectively (Figure 2C). We used custom MATLAB scripts to compute normalized dLight1.3b fluorescent traces (% Δ F/F) and area-under-the-curve (AUC) measurements for traces in 1-min time bins.
Open Field Locomotor Activity
We habituated mice to the open field for 20 min, injected vehicle, recorded locomotion for 30 min or 1 h, injected capsaicin or amphetamine at varying doses (one dose per day) and recorded locomotion for 30 min or 1 h, respectively (Figure 3B, 5A). For haloperidol experiments, we pretreated mice with varying doses of haloperidol (one dose per day), recorded locomotion for 30 min, injected capsaicin (3.5 mg·kg-1) and again recorded locomotion for 30 min (Figure 6A). To analyze locomotor speed, we used a video camera and custom software written in ImageJ to track each mouse’s position over time in the open field (18). The open field locomotor experiments for mice treated with vehicle followed by capsaicin (3.5 mg·kg-1) were experimentally equivalent and the data statistically indistinguishable between the first cohort of mice (Figure 3B, C) and the cohort used for the haloperidol experiments (Figures 6A, B), so we combined all mice in those treatment groups for statistical comparisons.
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Figure 3.Selective nigrostriatal dopamine transmission dose-dependently increases locomotion in experimental TRPV1, but not control mice.
A) We recorded locomotor activity in an open field arena for 30-min following vehicle then capsaicin treatment (left). Example trajectories (10-min duration) for an experimental (TRPV1) mouse following systemic treatment with vehicle (middle) or capsaicin (right; 10 mg·kg-1). B, C) Systemic capsaicin treatment dose-dependently increased locomotor speed in experimental but not control mice. Data in B are mean ± s.e.m. normalized to values following vehicle treatment and plotted in 3-min time bins; the control data are averaged across all capsaicin doses. Data in C are averaged during the first 15-min following vehicle or capsaicin treatment (N = 8 mCherry and N = 29–30 TRPV1 mice). D, E) Locomotor response to repeated capsaicin injections (3.5 mg·kg-1 every 15 min) in experimental and control mice. Data in D are mean ± s.e.m. normalized to values following vehicle treatment (t = 0–15 min) and plotted in 3-min time bins. Data in E are 15-min averages following each injection, normalized to associated values following vehicle treatment (N = 5 mCherry and N = 10 TRPV1 mice). P< 0.0001, P< 0.001, P< 0.01 and P< 0.05 comparing capsaicin and vehicle treatments; two-way ANOVA in B; Holm-Sidak’s multiple comparison test in C–E.
Juvenile Social Exploration
We habituated individual mice in their home cage for 10 min, injected vehicle, capsaicin (3.5 mg·kg-1), or amphetamine (1 mg·kg-1), waited 2 min (capsaicin) or 4 min (amphetamine), and placed a novel juvenile (postnatal day 21–35) conspecific mouse in the adult’s home cage (Figure 4A). For haloperidol experiments, we pretreated mice with either vehicle or haloperidol (0.1 mg·kg-1), waited 10 min, then injected capsaicin (3.5 mg·kg-1), waited 2 min, and placed the novel juvenile in the adult’s home cage (Figure 6C). On each day, we measured the amount of time the adult experimental mouse spent interacting with the juvenile test subject for 5 min, which included the sum duration of sniffing, grooming, approaching, or pawing initiated by the adult mouse towards the juvenile (19).
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Figure 4.Selective nigrostriatal dopamine excess disrupts social and cognitive function.
A) We recorded behavior during an assay of juvenile social exploration in which control (mCherry) and experimental (TRPV1) adult mice interacted with a juvenile, conspecific mouse following vehicle or capsaicin (3.5 mg·kg-1) treatment for 5 min. B, C) Mean ± s.e.m. interaction time, quantified as the sum duration of sniffing, grooming, approaching, or pawing initiated by the adult mouse towards the juvenile, B, or normalized to values following vehicle treatment, C, decreased following capsaicin treatment in experimental, but not control mice (N = 10 mCherry and N = 9 TRPV1 mice). D) We used an automated T-maze to evaluate spatial working memory with a delayed non-match to sample (DNMS) protocol, where mice alternated between forced and choice trials with varying delay periods (2, 10 or 60 s) in a center holding area between trials. E, F) Mean ± s.e.m. percent correct choice in control, E, and experimental, F, mice following vehicle or capsaicin treatment (3.5 mg·kg-1) in the DNMS task at different delay periods. G) Mean ± s.e.m. difference in percent correct choice following capsaicin and vehicle treatment in experimental and control mice (N = 5 mCherry and N = 8 TRPV1 mice; P< 0.01 and P< 0.05 comparing capsaicin and vehicle treatments; Wilcoxon signed-rank test in B, C; two-way ANOVA in E, F; Holm-Sidak’s multiple comparison test in G).
Spatial Working Memory
We individually housed and gradually food restricted mice to 85% of their ad libitum body weight. We then habituated mice to an automated T-maze (Maze Engineers) for 2 days. Following habituation, we pre-trained mice in sessions consisting of 10 alternating (left vs. right) forced trials from the center holding area to a baited goal arm and then back to the center holding area. The day after pre-training, we trained mice in a delayed non-match to sample (DNMS) task in which we interleaved forced and choice trials (20). During the choice trial, mice were required to choose the arm opposite to the immediately preceding forced trial to obtain a food reward. Once mice had at least three consecutive days with ≥70% correct choice trials in DNMS training, we proceeded with pharmacological testing. In each session, we injected mice with vehicle, capsaicin (3.5 mg·kg-1), or amphetamine (1 mg·kg-1), waited 2 min (capsaicin) or 4 min (amphetamine), and tested their DNMS performance with a delay period of 2, 10, or 60 s between forced and choice trials on sequential days (Figure 4E–G, 5F–G). For sessions that exceeded 15 min, we administered additional capsaicin injections every 15 min to sustain the drug’s behavioral effect (Figure 3B). For haloperidol experiments, we pretreated mice with either vehicle or haloperidol (0.1 mg·kg-1), waited 10 min, then injected capsaicin (3.5 mg·kg-1), waited 2 min, and tested their performance with the same 2, 10, and 60 s delays while administering capsaicin every 15 min until task completion (Figure 6E– G).
Histology
Following all photometry and behavioral experiments, we euthanized and intracardially perfused mice, then removed and sliced their brains for immunostaining. We used α-TH or α-Fos with α-RFP antibodies to quantify TH or Fos colocalization with mCherry in SNc. For dlight1.3b analysis, we used an α-GFP antibody in DMS and mPFC. We imaged fluorescence using a multiphoton or wide-field fluorescence microscope to verify the accuracy of injection and/or implantation (Figure 1, 2B).
Statistical Analysis
We used Prism (GraphPad) to perform all statistical tests. For comparisons between data that did not conform to a normal distribution, we used non-parametric statistical comparisons. For comparisons of more than two groups, we used one-way ANOVA. For comparisons of two or more groups across conditions or time, we used two-way repeated measures ANOVA. For all post-hoc analyses, we used Holm-Sidak’s correction for multiple comparisons. All statistical comparisons presented are in the Supplementary Table.
RESULTS
Selective Expression and Activation of TRPV1 in SNc Dopamine Neurons
To selectively drive dopamine release in the DS, we re-expressed the capsaicin-sensitive, excitatory cation channel TRPV1 in DS-projecting, substantia nigra pars compacta (SNc) dopamine neurons. Specifically, we bilaterally injected a Cre-dependent virus expressing TRPV1-mCherry (or mCherry control) into the SNc of DAT cre/+ mice that lack the endogenous Trpv1 gene (DAT cre/+; TRPV1 KO mice) (Figure 1A). This approach resulted in a high degree of overlap between neurons that express mCherry and neurons that were immunopositive for the dopamine biosynthesis enzyme tyrosine hydroxylase (TH), specifically in the SNc (Figure 1B, C). Considering these mice lack the Trpv1 gene, this allowed us to inject them systemically with the TRPV1 agonist capsaicin and selectively activate neurons at the site of virus injection without engaging endogenous TRPV1 receptors in the periphery (21, 22). To confirm that virally expressed TRPV1 was functional in SNc dopamine neurons, we systemically administered vehicle or capsaicin (10 mg·kg-1) to control and experimental (TRPV1) mice. 75-min following injection, there was a high degree of overlap between neurons expressing mCherry and the immediate early gene product Fos in experimental mice after capsaicin treatment. By contrast, there was little to no overlap of Fos with mCherry in control mice injected with capsaicin, or in TRPV1 mice injected with vehicle (Figure 1D, E). These results indicate that TRPV1 expression was restricted to SNc dopamine neurons in these mice and that systemic capsaicin treatment selectively activated these neurons.
Selective Activation of Nigrostriatal Dopamine Release In Vivo
To confirm that capsaicin treatment increases dopamine release in a pathway-specific manner in these mice, we virally expressed the fluorescent dopamine sensor dLight (AAV2/9-CAG-dLight1.3b) and implanted fiberoptic probes unilaterally into the dorsomedial striatum (DMS) and medial prefrontal cortex (mPFC) for dual-site, fiber photometry recordings (Figure 2A, B). We then recorded dLight fluorescence in the mPFC and DMS following systemic treatment with vehicle, capsaicin, or the non-selective dopamine releaser amphetamine (Figure 2C). Capsaicin treatment (10 mg·kg-1) increased dopamine transmission (Δ F/F) in the DMS of TRPV1, but not control mice (Figure 2D, E). Importantly, capsaicin treatment did not increase dopamine release in the mPFC of either TRPV1 or control mice (Figure 2D, F).
In contrast to capsaicin, amphetamine treatment (10 mg·kg-1) increased dopamine release in both the DMS and mPFC of these mice (Figure 2G, H), indicating that dLight1.3b is suitable for measuring dopamine release, even in the mPFC, where dopamine levels are lower. Taken together, these findings demonstrate that our approach selectively drives activity in DS-projecting, SNc dopamine neurons and mimics the pathway-specific excess of dopamine observed in schizophrenia patients (12) (Figure 2I, J).
Selective Nigrostriatal Dopamine Excess Increases Locomotor Activity
Locomotor hyperactivity is associated with the positive symptoms of psychosis in that both are reversed by effective antipsychotic treatments (23). We recorded the locomotor activity of control and TRPV1 mice in an open field arena following treatment with a range of capsaicin doses (Figure 3A). Capsaicin treatment dosedependently induced locomotor hyperactivity in TRPV1, but not control mice (Figure 3B, C). The effect of capsaicin on locomotion started rapidly (< ∼3 min) and lasted about 15 min after injection for the lowest dosage tested (3.5 mg·kg-1; Figure 3B). To determine whether we could repeatedly inject capsaicin without de-sensitizing its behavioral effects, we administered three successive injections separated by 15 min. These repeated capsaicin treatments consistently induced hyperlocomotion in TRPV1 mice, but not control mice, with no evidence of desensitization (Figure 3D, E). These results show that selectively driving DS dopamine release with capsaicin increases locomotor activity and establishes the time-course and dose-dependence of this effect. Given that the lowest dose tested here (3.5 mg·kg-1) only modestly increased locomotion and reliably elicited a behavioral response with no obvious desensitization, we used this dose for all subsequent behavioral assessments.
Selective Nigrostriatal Dopamine Excess Disrupts Social and Cognitive Function
The negative and cognitive symptoms of psychosis include social withdrawal and deficits in working memory. To determine if selectively driving DS dopamine transmission affects behaviors associated with these symptoms, we evaluated the effects of capsaicin treatment on juvenile social exploration (JSE) and spatial working memory (WM). Surprisingly, capsaicin treatment reduced social interaction with a juvenile conspecific mouse in TRPV1, but not control mice (Figure 4A–C). Likewise, capsaicin treatment disrupted working memory in TRPV1 but not control mice, in a T-maze, delayed non-match to sample (DNMS) task requiring mice to choose the arm opposite from the one visited on the preceding forced-choice trial (Figure 4D–G). Notably, capsaicin treatment did not affect performance in a version of the T-Maze task requiring mice to repeatedly choose the same reward arm where working memory was not required (mean ± s.e.m: 79 ± 2 and 75 ± 5 % correct choice following vehicle and capsaicin treatment, respectively; N = 4 TRPV1 mice; P = 0.75; Wilcoxon signed-rank test). These results indicate that excess dopamine, when restricted to the DS, diminishes cognitive and social function.
Non-Selective Dopamine Excess Alters Locomotor Activity but not Social or Cognitive Function
Amphetamine is a psychostimulant that is often used to approximate dopamine dysregulation in psychosis in rodents (23). In contrast to the viral-genetic approach used here, amphetamine increases dopamine release throughout the brain (not selectively in DS), including in the PFC (24). Consistent with previous studies, treating C57BL/6J mice with amphetamine dose-dependently increased locomotor activity in the open field (Figure 5A, B). A 1 mg·kg-1 amphetamine dose increased locomotor speed most comparably to the 3.5 mg·kg-1 dose of capsaicin we used for the WM and JSE experiments in TRPV1 mice (Figure 5C), so we used this dose for subsequent behavioral assays. Despite inducing similar levels of locomotion, amphetamine treatment had no effect on social interaction or working memory performance compared to vehicle treatment (Figure 5D–G). These results suggest that the behavioral deficits caused by capsaicin treatment in TRPV1 mice were not solely due to changes in their locomotor activity. Moreover, these findings highlight the differences between the effects of DS-restricted and brain wide excess in dopamine, in that only the former produces deficits in behaviors related to the negative and cognitive symptoms of psychosis (Figures 2, 4, 5).
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Figure 5.Amphetamine treatment induces hyperlocomotion but has no effect on social exploration or working memory.
A, B) Systemically treating C57BL/6J mice with amphetamine dose-dependently increased locomotor speed in the open field. Data in A are mean ± s.e.m. normalized to vehicle treatment and plotted in 5-min time bins. Data in B are 15-min averages following each injection, normalized to values following vehicle treatment. C) Amphetamine treatment (1 mg·kg-1) in C57BL/6J mice induced locomotion equivalently to capsaicin treatment (3.5 mg·kg-1) in TRPV1 mice in the 15-min following injection, normalized to values following vehicle treatment. D, E) Amphetamine treatment (1 mg·kg-1) did not significantly affect the time C57BL/6J mice spend interacting with a juvenile conspecific, D, and there was no difference, E, in social interaction time when normalized, within each mouse, to values following vehicle treatment (mean ± s.e.m.). F, G) Amphetamine treatment (1 mg·kg-1) did not significantly affect the percent of correct choices C57BL/6J mice made in the T-maze, DNMS task, F, and there was no difference, G, in the percentage of correct choices when normalized, within each mouse, to values following vehicle treatment (mean ± s.e.m.; N = 9–10 C57BL/6J mice and N = 29 TRPV1 mice; P< 0.0001, P< 0.001, and P< 0.01 comparing amphetamine, capsaicin, and vehicle treatments; Holm-Sidak’s multiple comparison test in A–C, F, G; Wilcoxon signed-rank test in D, E).
Haloperidol Normalizes Locomotor Hyperactivity, and Partially Normalizes Cognitive but not Social Deficits Caused by Selective Nigrostriatal Dopamine Excess
Classical antipsychotic drugs are largely ineffective for treating the negative and cognitive symptoms of psychosis (1). To determine whether the common antipsychotic drug haloperidol normalizes the behavioral deficits of a selective excess in DS dopamine, we tested its effects in TRPV1 mice. Pretreating these mice with a range of haloperidol doses blocked their increase in locomotor activity following systemic capsaicin treatment (Figure 6A, B). For subsequent behavioral tests in TRPV1 mice, we used the highest dose of haloperidol tested in the open field (0.1 mg·kg-1). Despite normalizing capsaicin-induced locomotion, haloperidol pretreatment failed to normalize the capsaicin-induced decrease in social interaction in TRPV1 mice (Figure 6C, D). Furthermore, haloperidol only partially rescued the disruption of working memory caused by capsaicin treatment in these mice (Figure 6E, F).
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Figure 6.Haloperidol incompletely normalizes the effects of selective nigrostriatal dopamine excess.
A, B) Pre-treating TRPV1 mice with the first-generation antipsychotic haloperidol suppressed the increase in locomotion caused by systemic capsaicin treatment (3.5 mg·kg-1). Data in A are mean ± s.e.m. normalized to values following vehicle treatment and plotted in 3-min time bins. Data in B are averaged during the first 15-min following capsaicin treatment, normalized to values following vehicle treatment (N = 26–29 TRPV1 mice). C, D) Haloperidol pre-treatment (1 mg·kg-1) failed to normalize the decrease in juvenile social interaction time in experimental mice caused by capsaicin treatment (3.5 mg·kg-1), C, or the change in social interaction time, D, between capsaicin and vehicle treatment (mean ± s.e.m; N = 12 TRPV1 mice). E, F) Mean ± s.e.m. percent correct choice, E, and change in percent choice from vehicle treatment following capsaicin treatment (3.5 mg·kg-1), F, in the DNMS T-maze task was reduced in TRPV1 mice and only partially rescued by haloperidol pre-treatment (N = 11 TRPV1 mice). G) Summary of behavioral results in the different experimental groups included in this study. Arrows indicate directionality and magnitude of effects. Holm-Sidak’s multiple comparison test compared to vehicle (P< 0.0001, P< 0.001) and vehicle + capsaicin (####P< 0.0001, ##P< 0.01 and #P< 0.05) in A, B, or to vehicle-vehicle (P< 0.001, P< 0.01 and P< 0.05) treatment in C–F).
Taken together, these data show that haloperidol normalized behavioral processes that are typically normalized by antipsychotic drugs in rodents (i.e., hyperlocomotion). However, haloperidol failed to normalize, or only partially normalized behavioral processes associated with the treatment-resistant symptoms of psychosis in patients (social interaction and working memory, respectively) (Figure 6G). Despite blocking striatally enriched D2Rs, haloperidol was only partially effective for the behavioral changes driven by a selective excess in DS dopamine.
DISCUSSION
The adverse effects of cognitive and negative symptoms in psychosis are widely recognized, yet their underlying etiology is poorly understood. The prevailing view is that excess dopamine is not involved in these symptoms, since antipsychotic drugs that block dopamine receptors do not alleviate these symptoms, and drugs that increase dopamine do not reliably induce them. In the current study, we directly tested this idea in a manner that explicitly models the pathway-specific increase in dopamine observed in schizophrenia patients.
To do this, we adapted an approach that was previously used to activate all dopamine neurons (21). Specifically, we generated a virus to selectively express excitatory TRPV1 receptors in DS-projecting, SNc dopamine neurons of DAT cre/+ mice that lack the endogenous Trpv1 gene. In contrast to the transgenic approach used in previous studies to selectively express TRPV1 (21, 22), our approach allowed us to specifically activate genetically defined neurons at the site of virus injection. This virus-mediated approach is widely applicable to manipulating circuits throughout the brain. We leveraged this regional and genetic selectivity to transiently activate DS-projecting SNc neurons (but not other dopamine neurons) using the TRPV1 agonist capsaicin and selectively induced dopamine release in the DS, but not mPFC. Capsaicin’s effects on behavior and dopamine release lasted approximately 15–20 min, consistent with earlier studies using capsaicin to activate all dopamine neurons (21). This transient duration is an advantage for experimental designs requiring a modest duration (minutes) of reversible activation, but a limitation for experiments requiring shorter or longer stimulations (subseconds or hours). Other advantages include the compatibility with traditional chemogenetic approaches (since capsaicin and DREADD ligands are distinct) and in vivo imaging approaches that are constrained by limited cranial space for implanting optogenetic fibers (18, 25, 26). A notable limitation of this approach is the requirement that the mice must be on a Trpv1 knockout background, which requires at least two generations of crosses to generate experimental animals. The lack of endogenous TRPV1 expression also has the potential to confound behavioral phenotypes. Although here we used control mice that also lacked the Trpv1 gene, this could particularly confound studies of behavioral processes such as sensation or pain that are influenced by TRPV1 (17).
Treating TRPV1 mice with capsaicin and C57BL6/J mice with amphetamine both induced hyperlocomotion. A drug’s ability to suppress amphetamine-driven locomotion in mice is commonly used to assess its antipsychotic potential in humans (23). Albeit circular, there is logic to this idea. Most antipsychotic drugs attenuate dopamine receptor signaling, and hyperlocomotion results from amphetamine-induced increases in striatal dopamine release (27). Because both amphetamine-driven locomotion in rodents and the positive symptoms of schizophrenia in patients respond to antipsychotic treatment, they are speculated to engage partly overlapping neural substrates. Although further experiments are necessary to confirm that capsaicin treatment induces behavioral changes associated with positive symptoms in TRPV1 mice, the fact that it induces haloperidol-responsive hyperlocomotion suggests it is engaging dopamine pathways related to the positive symptoms of psychosis.
Although capsaicin (3.5 mg·kg-1 in TRPV1 mice) and amphetamine (1 mg·kg-1 in C57BL6/J mice) treatments equivalently induced hyperlocomotion, only the former additionally perturbed social interaction and working memory. Social exploration and working memory are complex processes that rely on distributed brain areas (20, 28-30). The mPFC, in particular, is crucially involved in both processes (31-33), and the brains of schizophrenia patients exhibit an array of pathological changes in the prefrontal cortex that are thought to underlie cognitive and negative symptoms (34). In fact, the onset of these symptoms can precede that of the positive symptoms, further supporting distinct underlying etiologies (35, 36). However, patients with prodromal schizophrenia symptoms also have elevated dopamine signaling in the DS that correlates with neurocognitive dysfunction (37). Furthermore, the DS is both directly and indirectly connected to the brain structures most implicated in cognitive and social function. For instance, inhibiting DMS-projecting mPFC neurons during the delay period of a T-maze working memory task impairs performance, and DMS neurons are sequentially active during this delay period (13, 38). The DS is also connected through basal ganglia outputs to the thalamocortical connections thought to sustain persistent activity in the PFC during delay periods of working memory (20, 39). Striatal dysfunction has also been repeatedly implicated in social behavior deficits, particularly in the context of autismspectrum disorders (40-42). Moreover, optogenetically activating nigrostriatal dopamine neurons was recently shown to reduce social preference and decrease social interaction in mice (40). Therefore, our own findings are consistent with previous studies implicating striatal function in social interaction and working memory. Our findings add to this understanding by distinguishing nigrostriatal from brain-wide dopamine excess (i.e., following amphetamine treatment), which did not disrupt social interaction or working memory. This distinction could reflect compensation due to dopamine release in brain regions outside of the striatum (e.g., in the PFC) following amphetamine treatment, or the brain-wide competition for cognitive resources between structures when dopamine release is asymmetrically driven (15). Intriguingly, a seemingly inverse relationship between striatal and cortical dopamine signaling has long been recognized in humans and in animal studies of schizophrenia, and this circuitlevel interaction between the two structures could explain the diversity of psychosis symptoms (43-49). This interdependence and its implications for regional brain dysfunction further underscore the importance of addressing excess striatal dopamine in psychosis.
If nigrostriatal dopamine excess contributes to cognitive and negative symptoms, then why are antipsychotics that block D2 dopamine receptors largely ineffective for these symptoms? The simplest explanation is that these drugs fail to properly engage the neural substrates responsible for these symptoms (e.g., decreased dopamine, glutamate, or neuropathology in the PFC). While a lack of effects outside of the striatum undoubtedly contributes to their limited efficacy, even within the striatum, antipsychotics do not fully address the consequences of dopamine excess. Approximately half of the neurons in the striatum express D1 rather than D2Rs. Because antipsychotics do not principally target them, striatal D1Rs could contribute to the treatment-resistant symptoms. Consistent with this idea, we recently found that clozapine, a highly efficacious antipsychotic drug with some pro-cognitive effects, preferentially normalized activity in D1R-expressing spiny projection neurons (SPNs) in the DMS under hyperdopaminergic conditions (50, 51). By contrast, haloperidol affected both D1- and D2-SPN activity following amphetamine treatment (50). These data potentially explain why haloperidol partially rescued working memory in capsaicin-treated TRPV1 mice, and suggests that drugs that preferentially normalize D1-SPN activity may have greater therapeutic advantages over drugs like haloperidol that principally block D2Rs (50).
In summary, our results show that excess dopamine, when restricted to the dorsal striatum, alters behavioral processes linked to the positive, negative, and cognitive symptom subclasses of psychosis. Non-selectively driving dopamine release did not disrupt the processes associated with negative and cognitive symptoms, and haloperidol treatment only partially normalized the deficits in these behaviors in our model. Our results implicate nigrostriatal hyperdopaminegia in the etiology of psychosis symptoms that do not respond to current treatments, but other brain areas are also undoubtedly involved, particularly the PFC. Within the context of our earlier studies, treatments targeted to striatal D1-, rather than D2-SPNs may have therapeutic benefits over classical antipsychotic drugs. Further studies are necessary to unravel the intricacies of nigrostriatal dopamine excess in the behavioral processes examined here. In particular, the DS is a large, heterogeneous brain structure encompassing several subregions that contain multiple cell-types innervated by multiple dopamine cell sub-types. A better understanding of these intricacies holds the promise to develop more targeted therapies that better address the diverse symptoms of psychosis.
SUPPLEMENTARY METHODS
Viruses
We synthesized a polylinker containing a 5’ SpeI and 3’ PacI site (5’CTAGCACTATGAGCTCTTAATTAAG 3’ and 5’CTAGCTTAATTAAGAGCTCACTAGTG3’) and inserted it into the NheI site of pAAV-hSyn-FLEX-mCherry (Addgene #50459). We PCR amplified rat Trpv1 (forward primer: 5’GATATCTTAATTAAATGGAACAACGGGCTAGCTTA3’ and reverse primer: 5’GATATCACTAGTTTTCTCCCCTGGGACCATGGA3’; IDT) and inserted the amplified construct in-frame with mCherry with PacI and SpeI (NEB) using restriction digest cloning. We sent the resulting plasmid to Virovek for virus production (AAV2/5-FLEX-TRPV1-mCherry). Virovek also produced AAV2/9-CAG-dLight1.3b directly from the Addgene plasmid (#125560). We purchased a viral preparation of AAV2/5-Syn-DIO-mCherry directly from Addgene (#50459).
Surgical Procedures
For virus injections, we anesthetized mice with isoflurane (2% in O 2) and stereotaxically injected virus at a rate of 250 nL·min-1 into each region of interest using a microsyringe with a 33-gauge beveled tip needle. All anteroposterior (AP), mediolateral (ML) coordinates are reported from bregma and dorsoventral (DV) coordinates are reported from bregma or dura as specified. For all virus injections, we went 0.5-mm (DV) past the injection target and then withdrew the syringe and needle back to the target for the injection. After each injection, we left the syringe in place for 5 min, withdrew the syringe 0.1 mm, and then waited 5 additional min before slowly with-drawing the syringe to minimize viral spread. To express TRPV1-mCherry or control mCherry in SNc dopamine neurons, we injected 500 nL of AAV2/5-Syn-FLEX-TRPV1-mCherry (8.4 × 10 12 GC/mL) or AAV2/5-hSyn-DIO-mCherry (1.7 × 10 12 GC/mL) bilaterally into the SNc (AP: [bregma to lambda distance]/4.21 × -3.5 mm, ML: ±1.5 mm, and DV: -4.0 mm from bregma). To express dLight1.3b in the DMS and mPFC, we injected 500 nL of AAV2/9-CAG-dlight1.3b (1.53 × 10 12 GC/mL) unilaterally into the mPFC (AP: [bregma to lambda]/4.21 × +1.9 mm, ML= -0.25 mm and DV= -2.1 mm from dura) and DMS (AP: +0.5 mm, ML: -1.5 mm and DV: -2.7 mm from dura).
For mice only receiving virus injections, we then sutured the scalp, injected analgesic subcutaneously (Buprenorphine SR; 1 mg·kg-1) and waited 3–4 weeks for the mice to recover and for viral expression before beginning behavioral experiments. For mice undergoing fiber-optic implantation for photometry, in the same surgery we applied 4 stainless steel screws at arbitrary locations in the skull (AMS120/1; Antrin) and implanted fiberoptic cannulas (MFC_400/430-0.48_4mm; Doric) into the same AAV injection coordinates in the mPFC and DMS. After placing the fiberoptic cannulae, we applied Metabond (Parkell) to the skull and used dental acrylic (Coltene) to fix the full assembly along with a custom stainless-steel head-plate (Laser Alliance) for head-fixing mice during attachment and release of photometry patch cords. We injected analgesic and allowed the mice to recover for 3–4 weeks before recording dlight1.3b fluorescence.
Fiber Photometry
We tethered each mouse’s implanted fiberoptic cannulas to low-autofluorescence patch cords (MFP_400/430/1100-0.57_2m_LAF; Doric). We used a controller and LED sources (Thorlabs) with miniature filter cubes (FMC6_AE(400-410)_E1(460-490)_F1(500-540)_E2(550-580)_F2(600-680)_S; Doric) to deliver ∼30–70 μW of blue (470 nm) and UV (405 nm) light modulated at 211 and 511 Hz, respectively. We recorded dLight1.3b green fluorescence signals during blue and UV (isosbestic) light absorption through the same patch cords coupled to femtowatt photoreceivers (2151; Newport). We used a commercial system and software (RZ5P and Synapse; TDT) to sample (at 1.0173 kHz), de-modulate, and lock-in amplify these digital signals.
We habituated mice to recording by tethering the patch cord to their implanted fiberoptic cannulas and allowing them to explore an open field arena in two, 1-h sessions on consecutive days. For fiber photometry recordings, we recorded each mouse’s baseline dLight1.3b activity in the mPFC and DMS for 15 min, subcutaneously injected vehicle, recorded for an additional 15 min, and then subcutaneously injected capsaicin or amphetamine (10 mg·kg-1 each) and recorded dLight1.3b activity for 30 or 60 min, respectively. We then used custom MATLAB scripts to compute normalized dLight1.3b fluorescence traces (% Δ F/F) by using a polynomial fit to adjust the isosbestic (405 nm) fluorescence signal trace to that of the blue-light (470 nm) fluorescence trace and then subtracting it from the 470-nm fluorescence trace and multiplying the resulting trace by 100. We ignored the baseline recording period and removed spurious outlier points from the Δ F/F traces following vehicle and drug treatment that were greater or less than ± 3 standard deviations (SD) of the mean value for the trace during the given treatment. We then applied a 1-s median filter to temporally smooth the trace and averaged the trace to down sample its temporal resolution to 1 Hz. Next, we computed the area-under-the-curve (AUC) of the trace in 1-min time bins. We calculated the average AUC of the last 10, 1-min bins of the vehicle recording and subtracted this value from the AUC observed in every 1-min time bin following vehicle and drug treatment (Figure 2E–H). We then identified the 1-min time bin in which the normalized AUC was maximal following drug treatment and computed the average AUC for that bin and the 4, 1-min bins before and 5, 1-min bins after (average of 10-min about the maximum; Figure 2I, J). To eliminate potential false negative results due to the inability to detect dopamine signaling in the mPFC, we excluded mice where the average AUC of the 10-min bins around the maximum following amphetamine treatment did not exceed 2 SD of the mean AUC (across mice) of the 10, 1-min bins at the end of the vehicle treatment recording. This criterion resulted in 5/18 mice being excluded from the mPFC analysis due to the lack of a significant increase in dLight1.3b fluorescence following amphetamine treatment (Figure 2F, H, J). Using the same criterion, all mice had a significant increase in DMS dLight1.3b fluorescence following amphetamine treatment, so we excluded no mice from our analysis of amphetamine or capsaicin effects on DMS dopamine signaling (Figure 2E, G, I).
Open Field Locomotor Activity
We habituated mice in 1-h sessions to a clear acrylic circular open field arena (30.48-cm diameter; Tap plastics) for two days. During these habituation sessions, we also administered one subcutaneous injection of vehicle (3.33% Tween-80 in PBS or saline) and one subcutaneous injection of capsaicin (10 mg·kg-1) or amphetamine (2.5 mg·kg-1) to habituate them to drug injection. On testing days, we habituated mice to the open field for 20 min, injected vehicle, recorded locomotion for 30 min or 1 h, injected capsaicin or amphetamine at varying doses (one per day) and recorded locomotion for 30 min or 1 h (Figure 3B, 5A). For haloperidol experiments, we pretreated mice with vehicle or varying doses of haloperidol, recorded locomotion for 30 min, injected capsaicin (3.5 mg·kg-1) and again recorded locomotion for 30 min (Figure 6A). To analyze locomotor speed, we used a video camera and software (DMK 22BUC03 and IC Capture 2.4 [The Imaging Source]) with a varifocal lens (T3Z2910CS; Computar) to record 30-Hz videos of freely moving mouse behavior. We used custom software written in ImageJ to track each mouse’s position in an open field arena ( (18).
Spatial Working Memory
We individually housed mice and gradually food restricted them to 85% of their ad libitum body weight. Once at 85% body weight, we habituated mice to an automated T-maze (MazeEngineers) for 2 days. On day 1 of habituation, we placed mice in the maze and allowed them to freely explore and forage for 45 mg chocolate-flavored pellets (Bio-Serv) sporadically placed throughout the maze for 30 min. On day 2 of habituation, mice explored the maze for 30 min and received a food reward delivery each time they entered either of the maze goal arms. On the subsequent 2 days, we pre-trained mice in sessions consisting of 10 alternating (left vs. right) forced trials from the center holding area to a baited goal arm and then back to the center holding area. The day after pretraining, we trained mice in a delayed non-match to sample (DNMS) task in which we interleaved forced and choice trials (20). During the choice trial, mice were required to choose the arm opposite to the immediately preceding forced trial in order to obtain a food reward. We pseudo-randomly chose the ordering of left and right forced trials.
Once mice had at least three consecutive days with ≥70% correct choice trials, we proceeded with pharmacological testing. In each session, we injected mice with vehicle, capsaicin (3.5 mg·kg-1), or amphetamine (1 mg·kg-1), waited 2 min (capsaicin) or 4 min (amphetamine), and tested their DNMS performance with a delay period of 2, 10, or 60 s between forced and choice trials on sequential days (Figure 4E–G, 5F–G). We administered successive capsaicin injections every 15 min throughout each session to follow the time course of optimal drug effect (Figure 3B). For haloperidol experiments, we pretreated mice with either vehicle or haloperidol (0.1 mg·kg-1), waited 10 min, then injected capsaicin (3.5 mg·kg-1), waited 2 min, and tested their performance with the same 2, 10, and 60 s delays while administering capsaicin every 15-min until task completion (Figure 6E–G)
Histology
After all photometry and behavioral experiments, we euthanized and intracardially perfused mice with PBS followed by a 4% solution of paraformaldehyde in PBS. For the mice used for Fos quantification, we injected vehicle or capsaicin (10 mg·kg-1) 75 min before euthanasia. After perfusion, we extracted the brains and placed them in 4% paraformaldehyde for 1–3 days. We sliced 50–100-μM-thick coronal sections from the fixed-brain tissue using a vibratome (Leica VT1000s). For immunostaining, we used α-Tyrosine Hydroxylase (1:500; Aves TYH) or α-Fos (1:1000; Cell Signaling #2250) primary antibodies in combination with α-RFP (1:1000; Rockland #200-301-379) and Alexa 488 with 594 conjugated secondary antibodies (1:500; Jackson Immunoresearch #703-546-155 or #711-546-152 with #715-586-150, respectively) to amplify and label mCherry-expressing SNc neurons in alternating midbrain sections (Figure 1). For dlight1.3b analysis in DMS and mPFC, we used an anti-GFP antibody (1:1000, Invitrogen #A11122) and Alexa 488 conjugated secondary antibody (Figure 2B; 1:500, Jackson Immunoresearch #711-546-152). We then mounted the sections with DAPI-containing mounting media (Southern Biotech #0100-20) and imaged fluorescence using a two-photon (Bruker) or wide-field, slide-scanning fluorescence microscope (Keyence BZ-X800 or Leica THUNDER) with 10×, 16×, or 20× objectives. We used ImageJ and opensource plugins to process all data and quantify Fos or TH colocalization with mCherry in SNc neurons ( (Figure 1C, E).
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ACKNOWLEDGEMENTS AND DISCLOSURES
This work was supported by the National Institutes of Mental Health: K01 MH11313201 (JGP) and 2T32 MH067564 (NAM), Neurological Disorders and Stroke R01 NS122840 (JGP), Drug Abuse P30DA048736 (LSZ) and Diabetes and Digestive and Kidney Diseases R01 DK128477 (LRB). We thank Dr. Ben Arenkiel for providing the cDNA plasmid for Trpv1. The authors have no financial disclosures to declare.
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Selective Nigrostriatal Dopamine Excess Impairs Behaviors Linked to the Cognitive and Negative Symptoms of Psychosis
Nicolette A.Moya, Seongsik Yun, Stefan W.Fleps, Madison M.Martin, Jacob A.Nadel, Lisa R.Beutler, Larry S.Zweifel, Jones G.Parker
bioRxiv 2022.08.09.503421; doi:
Now published in Neuropsychopharmacology doi: 10.1038/s41386-022-01492-1
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Selective Nigrostriatal Dopamine Excess Impairs Behaviors Linked to the Cognitive and Negative Symptoms of Psychosis
Nicolette A.Moya, Seongsik Yun, Stefan W.Fleps, Madison M.Martin, Jacob A.Nadel, Lisa R.Beutler, Larry S.Zweifel, Jones G.Parker
bioRxiv 2022.08.09.503421; doi:
Now published in Neuropsychopharmacology doi: 10.1038/s41386-022-01492-1
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1024 | https://archive.cdc.gov/www_atsdr_cdc_gov/csem/leadtoxicity/what_lead.html | Lead (Pb) Toxicty: What is Lead? | Environmental Medicine | ATSDR
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Upon completion of this section, you will be able to
Explain what lead is.
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Lead is a soft, blue-gray metal, usually found as lead compounds, combined with other elements. Much of its presence in the environment stems from
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Elemental lead.
The chemical symbol for lead is Pb (from the Latin name plumbum). Lead has an atomic number of 82 and an atomic weight of 207.2. It is a bluish-grey metal that tarnishes easily in air to a dark grey. The density of lead is 11.34 g/cm3. It has a low melting point of 327.46 °C or 621.43 °F. Naturally occurring lead ores comprise 0.002% (15g/t) of the earth’s crust.
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This is the form of lead found in higher lead content paint, soil, dust and various consumer products. The color varies, depending on the chemical form, and the most common forms are white lead (a lead carbonate compound), yellow lead (lead chromate, lead monoxide) or red lead (lead tetraoxide). Lead acetate has a sweetish taste.
Organic lead.
Tetraethyl and tetramethyl lead is the form of lead used in leaded gasoline to increase octane rating. The combustion of organic lead when it is added to gasoline as a fuel additive results in the release of lead into the atmosphere. However, their use was phased out in the United States starting in the early 1970’s, and banned for use in gasoline for motor vehicles beginning January 1, 1996.
Organic forms of lead are extremely dangerous, as they are absorbed through the skin and are highly toxic to the brain and central nervous system, much more so than inorganic lead.
Exposure to organic lead is generally limited to an occupational context [EPA 1996]. Potential exposures to organic lead should be taken very seriously. The symptoms and treatment are different from those of inorganic lead.
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Lead is a
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Lead accumulation in the body is the result of anthropogenic (human) use, which has concentrated lead throughout the environment. Because lead is spread so widely throughout the environment, it can be found in everyone’s body today. The main exposure route is oral, especially in small children, as they have hand-to-mouth behaviors that increase their risks. The respiratory route allows exposure to lead-containing dust, especially in occupational settings and during home renovations. Some authors suggest that because environmental lead accumulation and levels found today, which can result in adverse health effects, many people are exposed to levels of lead that are “orders of magnitude” greater than that of pre-industrial levels [Jusko et al. 2008; Flegal and Smith 1995; Budd et al. 1998].
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Lead is a naturally occurring, very soft, dense, and ductile (moldable) metal.
Lead is still used in some commercial products made or imported into the United States.
Lead is very stable and accumulates in the environment.
Lead is resistant to corrosion, although acidic water may leach lead out of pipes, fittings, and solder (metal joints).
The body absorbs organic lead (as was used in leaded gasoline for “on-road” vehicles in the past in the United States, and is used in some occupational settings today) faster than inorganic lead. And, unlike inorganic lead compounds, organic lead can be readily absorbed through the skin.
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1025 | https://www.math.columbia.edu/~harris/2000/AlgebraHW5.pdf | Exercise 3.23 Show that every interval contains rational and irrational numbers.
Suppose that K is a subinterval of the form [a, b] of an arbitrary interval.
Then, let c = 1/(b-a) > 0.
Therefore, we can always find some rational number d such that 0 < c < d.
As a result, 0 < 1/d < 1/c = (b-a) () For any real number r, there exists some integers n and n+1 such that n <= r < n+1 Therefore, we know n < ad < n+1 or ad = n for some integer n.
i) n/d < a < (n+1)/d We claim that (n+1)/d is a rational number in [a, b] because (n+1)/d - a < (n+1)/d-n/d = 1/d < 1/c = (b-a) ii) if ad = n, then, a = n/d is rational in [a, b].
Therefore, every interval contains rational numbers.
To find an irrational number in K: pi = 3.141592653...... is an irrational number Consider the interval [a-pi, b-pi], we can find some rational number t in this interval. Then, t + pi is irrational and it is in [a,b] Thus, we proved 3.23. |
1026 | https://math.stackexchange.com/questions/81325/why-is-xhn-written-like-this | calculus - Why is $(x+h)^n$ written like this? - Mathematics Stack Exchange
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Why is (x+h)n(x+h)n written like this?
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I encountered this in my calculus book:
f′(x)=lim h→0 f(x+h)−f(x)h f′(x)=lim h→0 f(x+h)−f(x)h
f(x)=x n f(x)=x n
(x+h)n=(x+h)(x+h)...(x+h)=x n+n h x n−1+stuff involving h 2 as factor(x+h)n=(x+h)(x+h)...(x+h)=x n+n h x n−1+stuff involving h 2 as factor
I don't get where that n h x n−1 n h x n−1 and stuff involving h 2 h 2 as factor come from. A little help please?
calculus
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edited Nov 15, 2011 at 12:40
hmakholm left over Monica
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asked Nov 12, 2011 at 10:40
AndrewAndrew
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8 Do you know binomial theorem?Martin Sleziak –Martin Sleziak 2011-11-12 10:41:39 +00:00 Commented Nov 12, 2011 at 10:41
1 The x 2 x 2 are probably x n x n.Did –Did 2011-11-12 10:46:48 +00:00 Commented Nov 12, 2011 at 10:46
1 I corrected an x 2 x 2 to x n x n and took out a bunch of redundant material.Brian M. Scott –Brian M. Scott 2011-11-12 10:50:53 +00:00 Commented Nov 12, 2011 at 10:50
3 Just a note...it isn't necessary to pick an answer immediately. It's usually advisable to wait a day or so to see what people have to say. I'm not saying this because Paul's answer is bad, only because other answers may be better explanations for you.jprete –jprete 2011-11-12 13:57:54 +00:00 Commented Nov 12, 2011 at 13:57
@jprete thanks for your advice.Andrew –Andrew 2011-11-12 14:19:30 +00:00 Commented Nov 12, 2011 at 14:19
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Let us consider what happens when we try to expand
(x+h)n=(x+h)(x+h)⋯(x+h).(x+h)n=(x+h)(x+h)⋯(x+h).
To expand the RHS, we must pick either x x or h h from each bracket, and multiply them together to produce a term. The full expansion consists of the sum of the terms produced over all combinations of us picking x x or h h from each bracket.
What if we pick no h h's from any of the brackets? Then we pick x x every time, and the term produced is x n.x n.
Now, what if we pick h h exactly from the 1-st bracket? Then we must pick x x from the other n−1 n−1 brackets, so the term produced is h x n−1.h x n−1. Now if we pick h h exactly once, but from the 2nd bracket, same term is produced. We can pick our single h h from any of the n n brackets, so that means all the terms produced by picking exactly one h h sum to n h x n−1.n h x n−1.
Now, we've considered what happens if we pick 0 h h's (we get x n x n) and 1 h h (we get n h x n−1 n h x n−1. Everything else must involve picking h h at least twice, and if we pick h h twice the term produced has at least a factor of h 2 h 2 in it, explaining the "stuff involving h 2 h 2 as a factor" term.
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edited Nov 12, 2011 at 12:05
answered Nov 12, 2011 at 10:52
Ragib ZamanRagib Zaman
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That is the only way to explain the binomial theorem! (Well, maybe not the only way, but it's a far sight better than at least some teachers teach it :) )tom –tom 2011-11-13 11:14:47 +00:00 Commented Nov 13, 2011 at 11:14
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The binomial theorem gives another answer here :
(x+h)n=∑k=0 n(n k)x k h n−k(x+h)n=∑k=0 n(n k)x k h n−k
By expanding the first terms, we obtain
(x+h)n=x n+(n 1)h x n−1+(n 2)x n−2 h 2+⋯(x+h)n=x n+(n 1)h x n−1+(n 2)x n−2 h 2+⋯
In all the terms in the dots, the power of h h is greater than 2, so you can factorize by h 2 h 2. Moreover (n 1)=n(n 1)=n :
(x+h)n=x n+n h x n−1+h 2×(⋯)(x+h)n=x n+n h x n−1+h 2×(⋯)
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answered Nov 12, 2011 at 10:57
Paul PichaureauPaul Pichaureau
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1027 | https://www2.stat.duke.edu/courses/Fall20/sta711/lec/wk-09.pdf | STA 711: Probability & Measure Theory Robert L. Wolpert 9 Sums of Independent Random Variables We continue our study of sums of independent random variables, Sn = X1 + · · · + Xn. If each Xi is square-integrable, with mean µi = EXi and variance σ2 i = E[(Xi −µi)2], then Sn is square integrable too with mean ESn = µ≤n := P i≤n µi and variance VSn = σ2 ≤n := P i≤n σ2 i .
But what about the actual probability distribution? If the Xi have density functions fi(xi) then Sn has a density function too; for example, with n = 2, S2 = X1 + X2 has CDF F(s) and pdf f(s) = F ′(s) given by P[S2 ≤s] = F(s) = ZZ x1+x2≤s f1(x1)f2(x2) dx1dx2 = Z ∞ −∞ Z s−x2 −∞ f1(x1)f2(x2) dx1dx2 = Z ∞ −∞ F1(s −x2)f2(x2) dx2 = Z ∞ −∞ F2(s −x1)f1(x1) dx1 f(s) = F ′(s) = Z ∞ −∞ f1(s −x2)f2(x2) dx2 = Z ∞ −∞ f1(x1)f2(s −x1) dx1, the convolution f = f1 ⋆f2 of f1(x1) and f2(x2). Even if the distributions aren’t abso-lutely continuous, so no pdfs exist, S2 has a distribution measure µ given by µ(ds) = R R µ1(dx1)µ2(ds−x1). There is an analogous formula for n = 3, but it is quite messy; things get worse and worse as n increases, so this is not a promising approach for studying the distribution of sums Sn for large n.
If CDFs and pdfs of sums of independent RVs are not simple, is there some other feature of the distributions that is? The answer is Yes. What is simple about independent random variables is calculating expectations of products of the Xi, or products of any functions of the Xi; the exponential function will let us turn the partial sums Sn into products eSn = Q eXi or, more generally, ezSn = Q ezXi for any real or complex number z. Thus for independent RVs Xi and any number z ∈C we can use independence to compute the expectation EezSn = n Y i=1 EezXi, often called the “moment generating function” and denoted MX(z) = EezX for any random variable X.
For real z the function ezX becomes huge if X becomes very large (for positive z) or very negative (if z < 0), so that even for integrable or square-integrable random variables X the 1 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert expectation M(z) = EezX may be infinite. Here are a few examples of EezX for some familiar distributions (try to verify some of them): Binomial: Bi(n, p) [1 + p(ez −1)]n z ∈C Neg Bin: NB(α, p) [1 −(p/q)(ez −1)]−α z ∈C Poisson Po(λ) eλ(ez−1) z ∈C Normal: No(µ, σ2) ezµ+z2σ2/2 z ∈C Gamma: Ga(α, λ) (1 −z/λ)−α ℜ(z) < λ Cauchy: a π(a2+(x−b)2) ezb−a|z| ℜ(z) = 0 Uniform: Un(a, b) 1 z(b−a) ezb −eza z ∈C Aside from the problem that M(z) = EezX may be infinite or fail to exist for some z ∈C, the approach is promising: we can identify the probability distribution from M(z), and we can even find important features about the distribution directly from M. For example, if we can justify interchanging the limits implicit in differentiation and integration, then M′(z) = E[XezX] and M′′(z) = E[X2ezX], so (upon taking z = 0) M′(0) = EX = µ and M′′(0) = EX2 = σ2 + µ2. This lets us calculate the mean and variance (and other moments EXk = M(k)(0)) from derivatives of M(z) at zero. We have two problems to overcome: discovering how to infer the distribution of X from MX(z) = EezX, and what to do about distributions for which M(z) is infinite or doesn’t exist.
9.1 Characteristic Functions For complex numbers z = x+iy, Leonard Euler showed that the exponential ez can be given in terms of familiar real-valued transcendental functions as ex+iy = ex cos(y) + iex sin(y).
Since both sin(y) and cos(y) are bounded by one, for any real-valued random variable X and real number ω the real and imaginary parts of the complex-valued random variable eiωX are bounded and hence integrable; thus it always makes sense to define the (complex-valued) characteristic function φX(ω) = EeiωX = Z R eiωxµX(dx), ω ∈R with finite absolute value |φX(ω)| ≤1. Of course this is just φX(ω) = MX(iω) when MX exists, but φX(ω) exists even when MX does not; on the chart above you’ll notice that only the real part of z posed problems, and ℜ(z) = 0 was always OK, even for the Cauchy. For real-valued ω ∈R, some familiar distributions’ ch.f.s are: Binomial: Bi(n, p) φ(ω) = [1 + p(eiω −1)]n Neg Bin: NB(α, p) φ(ω) = [1 −(p/q)(eiω −1)]−α Poisson Po(λ) φ(ω) = eλ(eiω−1) Normal: No(µ, σ2) φ(ω) = eiωµ−ω2σ2/2 Gamma: Ga(α, λ) φ(ω) = (1 −iω/λ)−α Cauchy: a/π a2+(x−b)2 φ(ω) = eiωb−a|ω| Uniform: Un(a, b) φ(ω) = 1 iω(b−a) eiωb −eiωa Page 2 Page 2 Page 2 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert Another Perspective Fourier transforms arise frequently in engineering and signal analysis.
A recent STA711 student found a YouTube video presenting an imaginative way of visualizing them; take a look at this URL if you’re interested.
9.1.1 Uniqueness Suppose that two probability distributions µ1(A) = P[X1 ∈A] and µ2(A) = P[X2 ∈A] have the same Fourier transform ˆ µ1 ≡ˆ µ2, where: ˆ µj(ω) = E[eiωXj] = Z R eiωx µj(dx); does it follow that X1 and X2 have the same probability distributions, i.e., that µ1 = µ2?
The answer is yes; in fact, one can recover the measure µ explicitly from the function ˆ µ(ω).
Thus we regard uniqueness as a corollary of the much stronger result, the Fourier Inversion Theorem.
Resnick (1999) has lots of interesting results about characteristic functions in Chapter 9, Grimmett and Stirzaker (2001) discuss related results in their Chapter 5, and Billingsley (1995) proves several versions of this theorem in his Section 26. I’m going to take a different approach, and stress the two special cases in which µ is discrete or has a density function, trying to make some connections with other encounters you might have had with Fourier transforms.
9.1.2 Positive Definiteness Which functions φ(ω) can be characteristic functions? We know that |φ(ω)| ≤1 for every ω ∈R, and that φ(0) = 1. In a homework exercise you showed that φ(ω) must be uniformly continuous, too— is that enough?
The answer is no. Each ch.f. has the interesting property that it is “positive definite,” in the following sense: Definition 1 A function φ : R →C is positive definite if for every n ∈N, z ∈Cn, and ω ∈Rn, n X j,k=1 zjφ(ωj −ωk)¯ zk ≥0 or, equivalently, that each n × n matrix Ajk := φ(ωj −ωk) is positive-definite.
Here’s a proof that φ(ω) := R R eiωxµ(dx) is positive definite, for every distribution µ on Page 3 Page 3 Page 3 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert (R, B), with the interchange of summation & integration justified by Fubini’s Theorem: n X j,k=1 zjφ(ωj −ωk)¯ zk = n X j,k=1 Z R zjei x(ωj−ωk)µ(dx)¯ zk = Z R ( n X j=1 zjei xωj ) ( n X k=1 zkei xωk ) µ(dx) = Z R n X j=1 zjei xωj 2 µ(dx) ≥0.
Interestingly, this condition is also sufficient: Theorem 1 (Bochner) If φ : R →C is continuous at zero, satisfies φ(0) = 1, and is positive definite, then there exists a Borel probability measure µ on (R, B) such that φ(ω) = R R eiωxµ(dx) for each ω ∈R.
Here’s a proof sketch for the special (but common) case where φ ∈L1(R, dω). By positive definiteness, for any {ωj} ⊂R and {zj} ⊂C, 0 ≤ X zjφ(ωj −ωk)¯ zk and in particular, for x ∈R, ǫ > 0, and zj := exp(−ixωj −ǫω2 j/2), 0 ≤ X e−ix(ωj−ωk)−ǫ(ω2 j +ω2 k)/2φ(ωj −ωk).
Taking ωj := j/n for −n2 ≤j ≤n2, dividing by n2, and taking the limit as n →∞, 0 ≤ ZZ R2 e−ix(u−v)−ǫ(u2+v2)/2φ(u −v) du dv Now change variables from v to ω := (u −v): = ZZ R2 e−ixω−ǫ[u2+(u2−2uω+ω2)]/2φ(ω) du dω = Z R e−ixω−ǫω2/2 Z R e−ǫ(u−ω/2)2+ω2/4 du φ(ω) dω = p π/ǫ Z R e−ixω−ǫω2/4φ(ω) dω Re-scaling and then taking ǫ →0, we find that f(x) := 1 2π R R e−ixωφ(ω) dω ≥0 for every x ∈R and can verify that φ(ω) = R R eiωxµ(dx) for the absolutely-continuous distribution given by µ(dx) = f(x) dx.
Page 4 Page 4 Page 4 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert 9.1.3 Inversion: Integer-valued Discrete Case Notice that the integer-valued discrete distributions always satisfy φ(ω + 2π) = φ(ω) (and in particular are not integrable over R), while the continuous ones satisfy |φ(ω)| →0 as ω →±∞.
For integer-valued random variables X we can recover the probability mass function pk := P[X = k] by inverting the Fourier series: φ(ω) = E[eiωX] = ∞ X k=−∞ pk eikω, so (by Fubini’s thm) pk = 1 2π Z π −π e−ikω φ(ω) dω, k ∈Z.
9.1.4 Inversion: Continuous Random Variables Now let’s turn to the case of a distribution with a density function; first two preliminaries.
For any real or complex numbers a, b, c it is easy to compute (by completing the square) that Z ∞ −∞ e−a−bx−cx2 dx = rπ c e−a+b2/4c (1) if c has positive real part, and otherwise the integral is infinite. In particular, for any ǫ > 0 the function γǫ(x) := 1 √ 2πǫe−x2/2ǫ satisfies R γǫ(x) dx = 1 (it’s just the normal pdf with mean 0 and variance ǫ).
Let µ(dx) = f(x)dx be any probability distribution with density function f(x) and ch.f.
φ(ω) = ˆ µ(ω) = R eiωx f(x) dx. Then |φ(ω)| ≤1 so for any ǫ > 0 the function |e−iyω−ǫω2/2φ(ω)| is bounded above by e−ǫω2/2 and so is integrable w.r.t. ω over R. We can compute 1 2π Z R e−iyω−ǫω2/2φ(ω) dω = 1 2π Z R e−iyω−ǫω2/2 Z R eixωf(x) dx dω = 1 2π Z R2 ei(x−y)ω−ǫω2/2f(x) dx dω = 1 2π Z R Z R ei(x−y)ω−ǫω2/2 dω f(x) dx (2) = 1 2π Z R "r 2π ǫ e−(x−y)2/2ǫ # f(x) dx (3) = 1 √ 2πǫ Z R e−(x−y)2/2ǫf(x) dx = γǫ ⋆f = γǫ ⋆µ (where the interchange of orders of integration in (2) is justified by Fubini’s theorem and the calculation in (3) by equation (1)), the convolution of the normal kernel γǫ(·) with f(y). As ǫ →0 this converges Page 5 Page 5 Page 5 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert • uniformly (and in L1) to f(y) if f(·) is bounded and continuous, the most common case; • pointwise to f(y−)+f(y+) 2 if f(x) has left and right limits at x = y; and • to infinity if µ({y}) > 0, i.e., if P[X = y] > 0.
This is the Fourier Inversion Formula for f(x): we can recover the density f(x) from its Fourier transform φ(ω) = ˆ µ(ω) by f(x) = 1 2π R e−iωxφ(ω) dω, if that integral exists, or otherwise as the limit f(x) = lim ǫ→0 1 2π Z e−iωx−ǫω2/2φ(ω) dω.
There are several interesting connections between the density function f(x) and characteristic function φ(ω).
If φ(ω) “wiggles” with rate approximately ξ, i.e., if φ(ω) ≈a cos(ωξ) + b sin(ωξ) + c, then f(x) will have a spike at x = ξ and X will have a high probability of being close to ξ; if φ(ω) is very smooth (i.e., has well-behaved continuous derivatives of high order) then it does not have high-frequency wiggles and f(x) falls offquickly for large |x|, so E[|X|p] < ∞for large p. If |φ(ω)| falls offquickly as ω →±∞then φ(ω) doesn’t have large low-frequency components and f(x) must be rather tame, without any spikes. Thus φ and f both capture information about the distribution, but from different perspectives. This is often useful, for the vague descriptions of this paragraph can be made precise: Theorem 2 If R R |ˆ µ(ω)| dω < ∞then µǫ := µ ⋆γǫ converges a.s as ǫ →0 to an L1 function f(x), ˆ µǫ(ω) := R eiωxµǫ(dx) converges uniformly to ˆ f(ω) := R eiωxf(x) dx, and µ(A) = R A f(x) dx for each Borel A ⊂R. Also f(x) = 1 2π R R e−iωxˆ µ(ω) dω for almost-every x.
Theorem 3 For any distribution µ and real numbers a < b, µ(a, b) + 1 2µ {a, b} = lim T→∞ Z T −T e−iωa −e−iωb 2πiω ˆ µ(ω) dω.
Theorem 4 If R R |x|k µ(dx) < ∞for an integer k ≥0 then ˆ µ(ω) has continuous derivatives of order k given by ˆ µ(k)(ω) = Z R (ix)keiωx µ(dx).
(1) Conversely, if ˆ µ(ω) has a derivative of finite even order k at ω = 0, then R R |x|k µ(dx) < ∞ and EXk = Z R xk µ(dx) = (−1)k/2 ˆ µ(k)(0).
(2) Page 6 Page 6 Page 6 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert To prove (1) first note it’s true by definition for k = 0, then apply induction: ˆ µ(k+1)(ω) = lim ǫ→0 Z R (ix)k eiǫx −1 ǫ eiωx µ(dx) = Z R (ix)k+1eiωx µ(dx) by LDCT since |eiǫx −1| ≤ǫ|x|.
By Theorem 4 the first few moments of the distribution, if they exist, can be determined from derivatives of the characteristic function or its logarithm log φ(z) at z = 0: φ(0) = 1, φ′(0) = iE[X], φ′′(0) = −E[X2], so [log φ] (0) = log φ(0) = log 1 = 0 Mean: [log φ]′ (0) = φ′(0)/φ(0) = iE[X] = iµ Variance: [log φ]′′ (0) = φ′′(0)φ(0)−(φ′(0))2 φ(0)2 = E[X]2 −E[X2] = −σ2 Etc.: [log φ]′′′ (0) = −iE[X3] −3σ2µ −µ3 ≤c E|X|3 for some c < ∞, so by Taylor’s theorem we have:1 log φ(ω) = 0 + iµω −σ2ω2/2 + O(ω3) (3) φ(ω) ≈eiµω−σ2ω2/2+O(ω3) 9.1.5 Convergence in Distribution In the Week 6 Notes we defined convergence in distribution of a sequence of distributions {µn} to a limiting distribution µ on a measurable space (X , E) (written µn ⇒µ): ∀φ ∈Cb(X ) lim n→∞ Z X φ(x) µn(dx) = Z X φ(x) µ(dx) (4) In fact requiring this convergence for all bounded continuous functions φ is more than what is necessary. For X = Rd, for example, it is enough to verify (4) for infinitely-differentiable C∞ b , or even just for complex exponentials φω(x) = exp(iω′x) for ω ∈Rd, i.e., Theorem 5 Let {µn(dx)} and µ(dx) be distributions on Euclidean space (Rd, B).
Then µn ⇒µ if and only if the characteristic functions converge pointwise, i.e., if φn(ω) := Z Rd eiω′x µn(dx) →φ(ω) := Z Rd eiω′x µ(dx) (5) for all ω ∈Rd.
How would you prove this?
1The “big oh” notation “f = O(g) at a” means that for some M < ∞and ǫ > 0, |f(x)| ≤Mg(x) whenever |x −a| < ǫ— roughly, that lim supx→a |f(x)/g(x)| < ∞. Here (implicitly) a = 0.
Page 7 Page 7 Page 7 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert Examples Un: Let Xn have the discrete uniform distribution on the points j/n, for 1 ≤j ≤n. Then its ch.f. is φn(ω) = 1 n n X j=1 eiωj/n = eiω/n −ei(n+1)ω/n n(1 −eiω/n) = 1 −eiω n(e−iω/n −1) →1 −eiω −iω = eiω −1 iω = Z 1 0 eiωx dx, the ch.f. of the Un(0, 1) distribution.
Po: Let Xn have Binomial Bi(n, pn) distributions with success probabilities pn ≍λ/n, so that n pn →λ for some λ > 0 as n →∞. Then the ch.f.s satisfy φn(ω) = n X k=0 n k eiωk pk n (1 −pn)n−k = 1 + pn(eiω −1) n →e(eiω−1)λ, the ch.f. of the Po(λ) distribution. This is an example of a “law of small numbers”.
No: Let Xn have Binomial Bi(n, pn) distributions with success probabilities pn such that σ2 n := n pn(1 −pn) →∞as n →∞, and set µn := npn. Then the ch.f.s of Zn := (Xn −µn)/σn satisfy φn(ω) = 1 + pn(eiω/σn −1) ne−iωµn/σn ≈exp µn(eiω/σn −1) −pnµn(eiω/σn −1)2/2 −iωµn/σn ≈exp iµnω/σn −µnω2/2σ2 n −pnµn(−ω2/σ2 n)/2 −iωµn/σn →e−ω2/2, the ch.f. of the No(0, 1) distribution. This result is called the “DeMoivre-Laplace” theorem, a pre-cursor (and special case) of the Central Limit Theorem.
9.2 Limits of Partial Sums and the Central Limit Theorem Let {Xi} be iid and L2, with common mean µ and variance σ2, and set Sn := Pn i=1 Xi for n ∈N. We’ll need to center and scale the distribution of Sn before we can hope to make sense of Sn’s distribution for large n, so we’ll need some facts about characteristic functions Page 8 Page 8 Page 8 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert of linear combinations of independent RVs. For independent X and Y , and real numbers α, β, γ, φα+βX+γY (ω) = Eeiω(α+βX+γY ) = EeiωαEeiωβXEeiωγY = eiωα φX(ωβ) φY (ωγ).
In particular, for iid L3 random variables {Xi} with characteristic function φ(t), the nor-malized sum [Sn −nµ]/ √ nσ2 has characteristic function φn(ω) = n Y j=1 h φ ω/ √ nσ2 e−iωµ/ √ nσ2 i Setting s := ω/ √ nσ2, this is = φ(s)e−isµn = en[log φ(s)−isµ] with logarithm log φn(ω) = n log φ(s) −isµ = n 0 + iµs −σ2s2/2 + O(s3) −nisµ (by (3)) = −n σ2(ω2/nσ2)/2 + O(n−3/2) (since s2 = ω2/nσ2) = −ω2/2 + O(n−1/2), so φn(ω) →e−ω2/2 for all ω ∈R and hence Zn := [Sn −nµ]/ √ nσ2 ⇒No(0, 1), the Central Limit Theorem.
Note: We assumed Xi were iid with finite third moment γ := E|Xi|3 < ∞. Under those con-ditions one can prove the uniform “Berry-Ess´ een” bound supx |Fn(x) −Φ(x)| ≤γ/ 2σ3√n for the CDF Fn of Zn. Another version of the CLT for iid {Xi} asserts weak convergence of Zn to No(0, 1) assuming only E[X2 i ] < ∞(i.e., no L3 requirement), but this version gives no bound on the difference of the CDFs. Another famous version, due to Lindeberg and Feller, asserts that Sn sn = ⇒No(0, 1) for partial sums Sn = X1 + · · · + Xn of independent mean-zero L2 random variables Xj that need not be identically distributed, but whose variances σ2 j = V[Xj] aren’t too extreme. The specific condition, for s2 n := σ2 1 + · · · + σ2 n, is 1 s2 n n X j=1 E X2 j 1{|Xj|>tsn} →0 Page 9 Page 9 Page 9 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert as n →∞for each t > 0. This follows immediately for iid {Xj} ⊂L2 (where it becomes 1 σ2E |X1|2 1{|X1|2>nt2σ2} which tends to zero as n →∞by the DCT), but for applications it’s important to know that independent but non-iid summands still lead to a CLT.
This “Lindeberg Condition” implies both of max j≤n σ2 j s2 n →0 max j≤n P {|Xj|/sn > ǫ} →0 as n →∞, for any ǫ > 0; roughly, no single Xj is allowed to dominate the sum Sn. This condition follows from the easier-to-verify Liapunov Condition, (∃δ > 0) s.t.: s−2−δ n n X j=1 E|Xj|2+δ →0 as n →∞.
Other versions of the CLT apply to non-identically distributed or nonindependent {Xj}, but Sn cannot converge to a normally-distributed limit if E[X2] = ∞; ask for details (or read Gnedenko and Kolmogorov (1968)) if you’re interested.
More recently an interesting new approach to proving the Central Limit Theorem and related estimates with error bounds was developed by Charles Stein (Stein, 1972, 1986; Barbour and Chen, 2005), described later in these notes.
9.3 Failure of Central Limit Theorem The CLT only applies to square-integrable random variables {Xj} ⊂L2. Some contemporary statistical work, both theoretical and applied, entails heavier-tailed distributions that do not have a finite variance (or, often, even a finite mean). In these cases, sums and averages of independent random variables do not have have approximate normal distributions, and may not even be concentrated around a central value.
For example, if {Xi} iid ∼Ca(m, s) are IID Cauchy random variables with pdf and ch.f.
f(x) = s/π s2 + (x −m)2 χ(ω) = exp imω −s|ω| then the sample mean ¯ Xn ∼Ca(m, s) also has the same Cauchy distribution— and, in particular, no weak or strong LLN applies and no CLT applies.
Worse— if {ζi} iid ∼No(0, 1) are IID standard Normals, then the random variables Xi := 1/ζi|ζi| have the symmetric α-stable St( 1 2, 0, 1, 0) distribution with ch.f. χ1(ω) := exp(−|ω|1/2), so the sample average ¯ Xn has ch.f.
χn(ω) = h χ1(ω/n) in = exp −|nω|1/2 , the same distribution as nX1. The average of several independent replicates has a much wider distribution than the individual terms.
Heavy-tailed distributions like the Fr´ echet, α-Stable, and Pareto arise when modeling income distribution, weather extremes, volcanic flows, and many other phenomena.
Page 10 Page 10 Page 10 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert 10 Distributional Limits of Heavy-Tailed Sums Although one cannot expect ¯ Xn to have an approximate normal distribution for {Xi} ̸⊂L2, other distributional limit theorems may still apply. Before we introduce those, we introduce some tools useful for studying heavy-tailed distributions.
10.1 Compound Poisson Distributions Let Xj have independent Poisson distributions with means νj and let uj ∈R; then the ch.f.
for Y := P uj Xj is φY (ω) = Y exp νj(eiωuj −1) = exp h X (eiωuj −1)νj i = exp h Z R (eiωu −1)ν(du) i for the discrete measure ν(du) = P νjδuj(du) that assigns mass νj to each point uj. Evidently we could take a limit using a sequence of discrete measures that converges to a continuous measure ν(du) so long as the integral makes sense, i.e., R R |eiωu−1|ν(du) < ∞; this will follow from the requirement that R R(1 ∧|u|)ν(du) < ∞. Such a distribution is called Compound Poisson, at least when ν+ := ν(R) < ∞; in that case we can also write represent it in the form Y = N X i=1 Xi, N ∼Po(ν+), Xi iid ∼ν(dx)/ν+.
We’ll now see that it includes an astonishingly large set of distributions, each with ch.f. of the form exp R (eiωu −1)ν(du) with “L´ evy measure” ν(du) as given: Distribution Log Ch Function φ(ω) L´ evy Measure ν(du) Poisson Po(λ) λ(eiω −1) λδ1(du) Gamma: Ga(α, λ) −α log(1−iω/λ) αe−λuu−1 1{u>0}du Normal: No(0, σ2) −ω2σ2/2 −1 2σ2δ′′ 0(du) Neg Bin: NB(α, p) −α log[1 −p q(eiω −1)] P∞ k=1 αpk k δk(du) Cauchy: Ca(γ, 0) −γ|ω| γ π u−2 du Stable: StA(α, β, γ) −γ|ω|α[1 −iβ tan πα 2 sgn(ω)] γcα [1 + β sgn u] α|u|−1−α du, where cα := 1 πΓ(α) sin πα 2 . Try to verify the measures ν(du) for the Negative Binomial and Cauchy distributions. All these distributions share the property called infinite divisibility (“ID” for short), that for every integer n ∈N each can be written as a sum of n independent Page 11 Page 11 Page 11 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert identically distributed terms. In 1936 the French probabilist Paul L´ evy and Russian proba-bilist Alexander Ya. Khinchine discovered that every distribution with this property must have a ch.f. of a very slightly more general form than that given above, log φ(ω) = iaω −σ2 2 ω2 + Z R eiωu −1 −iω h(u) ν(du), where a ∈R and σ ∈R+ are constants and where h(u) is any bounded Borel function that acts like u for u close to zero (for example, h(u) = arctan(u) or h(u) = sin(u) or h(u) = u/(1 + u2)). The measure ν(du) need not quite be finite, but we must have u2 integrable near zero and 1 integrable away from zero... one way to write this is to require that R (1 ∧u2) ν(du) < ∞, another is to require R u2 1+u2 ν(du) < ∞. Some authors consider the finite measure κ(du) = u2 1+u2ν(du) and write log φ(ω) = iaω + Z R eiωu −1 −iω h(u) 1 + u2 u2 κ(du), where now the Gaussian component −σ2ω2 2 arises from a point mass for κ(du) of size σ2 at u = 0.
If u is locally integrable, i.e., if R ǫ −ǫ |u| ν(du) < ∞for some (and hence every) ǫ > 0, then the term “−iω h(u)” is unnecessary (it can be absorbed into iaω). This always happens if ν(R−) = 0, i.e., if ν is concentrated on the positive half-line. Every increasing stationary independent-increment stochastic process Xt (or subordinator) has increments which are infinitely divisible with ν concentrated on the positive half-line and no Gaussian component (σ2 = 0), so has the representation log φ(ω) = iaω + Z ∞ 0 eiωu −1 ν(du) for some a ≥0 and some measure ν on R+ with R ∞ 0 (1 ∧u) ν(du) < ∞. In the compound Poisson example, ν(du) = P νjδuj(du) was the sum of point masses of size νj at the possible jump magnitudes uj. This interpretation extends to help us understand all ID distributions: every ID random variable X may be viewed as the sum of a constant, a Gaussian random variable, and a compound Poisson random variable, the sum of independent Poisson jumps of sizes u ∈E ⊂R with rates ν(E).
10.2 Stable Limit Laws Let Sn = X1+· · ·+Xn be the partial sum of iid random variables. IF the random variables are all square integrable, THEN the Central Limit Theorem applies and necessarily Sn−nµ √ nσ2 = ⇒ No(0, 1). But what if each Xn is not square integrable? We have already seen that the CLT fails for Cauchy variables Xj. Denote by F(x) = P[Xn ≤x] the common CDF of the {Xn}.
Page 12 Page 12 Page 12 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert Theorem 6 (Stable Limit Law) Let Sn = P j≤n Xj be the sum of iid random variables.
There exist constants An > 0 and Bn ∈R and a non-trivial distribution G for which the scaled and centered partial sums converge in distribution Sn −Bn An = ⇒G to a non-trivial limit G if and only if {Xj} ⊂L2 (in which case An ≍√n, Bn = nµ+O(√n), and G = No µ, σ2) is the Normal distribution) or there are constants 0 < α < 2, M−≥0, and M+ ≥0, with M−+ M+ > 0, such that as x →∞the following limits hold for every ξ > 0: M+ = lim x→∞xα[1 −F(x)] and, if M+ > 0, 1 −F(xξ) 1 −F(x) →ξ−α (6) M−= lim x→∞xα[F(−x)] and, if M−> 0, F(−xξ) F(−x) →ξ−α In this case the limit is the α-Stable Distribution, with index α, with characteristic function E eiωY = exp iδω −γ|ω|α 1 −iβ tan πα 2 sgn(ω) (7a) in the StA parametrization, where β = M+−M− M−+M+, γ = (M−+ M+), and δ ∈R is arbitrary.
The sequences An, Bn must be (see Section (10.5)) An = LA n n1/α Bn = ( LB n βγ tan πα 2 −δ n1/α −n α ̸= 1 LB n 2βγ π n log n α = 1 (8) for constants or sequences LA n, LB n that, like (log n)p for any power p ∈R but not like np for any p ̸= 0, are “slowly varying” in the sense that lim n→∞ Lc n Ln = 1 for every c > 0. Don’t get mesmerized by this— think of the Lns as proportionality constants.
For α ∈(1, 2) the sample means converge at rate n(1−α)/α to EY , more slowly (much more slowly, if α is close to one) than in the L2 case where the central limit theorem applies and ¯ Xn →µ at rate n−1/2. No means exist for α ≤1. The limits in (6) above are equivalent to the requirement that F(x) ∝|x|−αL−(x) as x →−∞and (1 −F(x)) ∝x−αL+(x) as x →+∞for slowly varying functions L±— roughly, that F(−x) and 1 −F(x) are both ∝x−α (or zero) as x →∞, i.e., that the pdf falls offlike |x|−α−1 in both directions.
Page 13 Page 13 Page 13 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert The simplest case is the symmetric α-stable (SαS). For 0 < α ≤2 and 0 < γ < ∞, the St(α, 0, γ, 0) has ch.f.
φ(ω) = e−γ|ω|α This includes the centered Cauchy Ca(0, s) (with α = 1, γ = s) and the centered Normal No(0, σ2) (with α = 2, γ = σ2/2). The SαS family interpolates between these (for 1 < α < 2) and extends them (for 0 < α < 1) to distributions with even heavier tails.
Although each Stable distribution has an absolutely continuous distribution with continuous unimodal probability density function f(y), these two cases and the “inverse Gaussian” or “L´ evy” distribution with α = 1/2 and β = ±1 are the only ones whose pdf is available in closed form. Perhaps that’s the reason these are less studied than normal distributions; still, they are very useful for problems with “heavy tails”, i.e., where P[X > u] does not die off quickly with increasing u. The symmetric (SαS) ones all have bell-shaped pdfs.
Moments are easy enough to compute but, for α < 2, moments E|X|p are only finite for p < α. In particular, means only exist for α > 1 and none of them has a finite variance.
The Cauchy has finite moments of order p < 1, but (despite its symmetry) does not have a well-defined mean.
Equation (6) says that each tail must be fall offlike a power (sometimes called Pareto tails), and the powers must be identical; it also gives the tail ratio. A common special case is M−= 0 (or equivalently β = 1) , the “one-sided” or “fully skewed” Stable. For 0 < α < 1 these take only values in [δ, ∞) (R+ if δ = 0). For example, random variables Xn with the Pareto distribution (often used to model income) given by P[Xn > t] = (1 + t/k)−α for t ∈R+ and some α > 0, k > 0 will have a stable limit for their partial sums if α < 2, and (by CLT) a normal limit if α ≥2. There are close connections between the theory of Stable random variables and the more general theory of statistical extremes. Ask me for references if you’d like to learn more about this exciting area.
Expression (7a) for the α-stable ch.f.
(called the “A-parametrization” StA(α, β, γ, δ) by Zolotarev (1986)) behaves badly as α →1 if β ̸= 0, because the tangent function has a pole at π/2— so, for β > 0, the imaginary part of log E eiωY converges to +∞as α ր 1 and to −∞as α ց 1 (the signs reverse for β < 0). For α ≈1 the complex part of the log ch.f. is: ℑ log E[eiωY ] = iδω + iβγ tan πα 2 |ω|α sgn(ω) = iδω + iβγ tan πα 2 |ω|α−1ω = iω δ + βγ tan πα 2 −iβγ tan πα 2 ω 1 −|ω|α−1 where the last term is bounded as α →1, so (following V. M. Zolotarev, 1986) the α-stable is often parametrized differently in the “M-parametrization” StM(α, β, γ, δ) as E[eiωY ] = exp −γ|ω|α + iδ∗ω −iβγ tan πα 2 ω 1 −|ω|α−1 (9) for α ̸= 1 with shifted “drift” term δ∗= δ + βγ tan(πα/2). You can find out more details by asking me or reading Breiman (1968, Chapter 9).
Page 14 Page 14 Page 14 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert 10.3 Examples Let {Un} iid ∼Un(0, 1), and set Xn := 1/Un. How would you describe the distribution of Sn := P j≤n Xj, or of ¯ Xn := 1 nSn? Evidently EXn = R 1 0 1 u du = ∞, so 1 nSn →∞a.s.; to get any interesting limit we will have to re-scale or recenter or both.
By Theorem 6, there exist sequences {An}, {Bn} such that (Sn −Bn)/An ⇒G for a non-trivial distribution G if and only if Xn ∈L2 (which fails in this case) or if the CDF F(z) = P[Xn ≤z] = P[Un ≥1/z] = ( 1 −1/z z > 1 0 z ≤1 satisfies for some 0 < α < 2 the conditions M+ = limx→∞xα[1 −F(x)] ≥0, M−= limx→∞xα[F(−x)] ≥0, (M−+ M+) > 0 and, if M+ > 0, 1 −F(xξ) 1 −F(x) = 1/xξ 1/x →ξ−α.
Evidently the conditions hold for α = 1, M−= 0, and M+ = 1, so β = M+/(M−+ M+) = 1 and γ = (M−+ M+) = 1. By (8), with α = β = γ = 1, suitable sequences are given by An = n and Bn = 2 πn log n, so Sn −2 πn log n n = Sn n −2 π log n ≈StM(α = 1, β = 1, γ = 1, δ = 0), or Sn n ≈StM(α = 1, β = 1, γ = 1, δ = 2 π log n).
Sample averages ¯ Xn of Xj := (1/Uj) will grow to infinity at approximate rate (2/π) log n, and ¯ Xn −(2/π) log n will have an asymptotic StM(1, 1, 1, 0) distribution. What do you think happens to ¯ Xn for Xn := U−1/α n for other values of α > 0?
10.4 Key Idea of the Stable Limit Laws The stable limit law of Theorem 6 says that if there exist nonrandom sequences An > 0 and Bn ∈R and a nondegenerate distribution G such that the partial sums Sn := P j≤n Xj of iid random variables {Xj} satisfy Sn −Bn An = ⇒G (10) then G must be either the normal distribution or an α-stable distribution for some 0 < α < 2.
The key idea behind the theorem is that if a distribution µ with cdf G satisfies (10) then also for any n the distribution of the sum Sn of n independent random variables with cdf G Page 15 Page 15 Page 15 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert must also (after suitable shift and scale changes) have cdf G— i.e., that cn Sn + dn ∼G for some constants cn > 0 and dn ∈R, so the characteristic function χ(ω) := R eiωx G(dx) and log ch.f. ψ(ω) := log χ(ω) must satisfy χ(ω) = E exp {iω(cn Sn + dn)} = exp(iωdn) χ(ωcn)n ψ(ω) = iωdn + nψ(cnω) (11) whose only solutions are the normal and α-stable distributions. Here’s a sketch of the proof for the symmetric (SαS) case, where ψ(−ω) = ψ(ω) and so dn = 0. Set γ := −ψ(1) and note that (11) with ω = ck n for k = 0, 1, . . . implies successively: ψ(cn) = −γ n ψ(c2 n) = ψ(cn) 1 n = −γ n2 . . .
ψ(ck n) = −γ nk .
Results from complex analysis imply this must hold for all k ≥0, not just integers. Thus, with |w| = ck n and k = log |w|/ log cn, ψ(w) = −γn−k = −γ exp {−(log |w|)(log n)/(log cn)} = −γ|w|−(log n)/(log cn) = −γ|w|α, where α is the constant value of −log n log cn . It follows that cn = n−1/α (i.e., Sn/n1/α ∼G) and that χ(ω) = e−γ|ω|α, the ch.f. for SαS for 0 < α < 2 and for No(0, 2γ) for α = 2.
10.5 The constants An and Bn Here we show where the constants given in (8) come from. Let φ(ω) be the ch.f. of iid random variables Yi ∼StM(α, β, γ, δ) (given in (9)), and let ψ(ω) := log φ(ω) be its logarithm. For the partial sums Sn := Pn j=1 Yj to have the property that (Sn −Bn)/An ∼StM(α, β, γ, δ) has the same distribution as Yj for each n, we need ψ(ω) = −γ|ω|α + iδ∗ω −iβγ tan πα 2 ω 1 −|ω|α−1 = nψ(ω/An) −iωBn/An = − h n Anα i γ|ω|α + iω n An δ∗−βγ tan πα 2 −Bn/n (12) + h n Anα i iβγ tan πα 2 ω|ω|α−1.
From the real part of (12) we find An = n1/α, and from the imaginary part we find Bn = βγ tan πα 2 −δ n1/α −n for α ̸= 1. Find the value of Bn for α = 1 as a limit.
Page 16 Page 16 Page 16 STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert STA 711 Week 9 R L Wolpert References Barbour, A. D. and Chen, L. H. Y., eds. (2005), An Introduction to Stein’s Method, Lecture Note Series, Institute for Mathematical Sciences, volume 4, Singapore: National University of Singapore.
Billingsley, P. J. (1995), Probability and Measure, Wiley Series in Probability and Mathe-matical Statistics, New York, NY: John Wiley & Sons, third edition.
Breiman, L. (1968), Probability, Reading, MA: Addison-Wesley.
Gnedenko, B. V. and Kolmogorov, A. N. (1968), Limit Distributions for Sums of Independent Random Variables (revised), Reading, MA: Addison-Wesley.
Grimmett, G. R. and Stirzaker, D. R. (2001), Probability and Random Processes, Oxford, UK: Oxford University Press, third edition.
Khinchine, A. Y. and L´ evy, P. (1936), “Sur les lois stables,” Comptes rendus hebdomadaires des seances de l’Acad´ emie des sciences. Acad´ emie des science (France). Serie A. Paris, 202, 374–376.
Resnick, S. I. (1999), A Probability Path, New York, NY: Birkh¨ auser.
Stein, C. (1972), “A bound for the error in the normal approximation to the distribution of a sum of dependent random variables,” in Proc. Sixth Berkeley Symp. Math. Statist. Prob., eds. L. M. Le Cam, J. Neyman, and E. L. Scott, Berkeley, CA: University of California Press, volume 2, pp. 583–602.
Stein, C. (1986), Approximate computation of expectations, IMS Lecture Notes-Monograph Series, volume 7, Hayward, CA: Institute of Mathematical Statistics.
Zolotarev, V. M. (1986), One-dimensional Stable Distributions, Translations of Mathematical Monographs, volume 65, Providence, RI: American Mathematical Society.
Last edited: October 8, 2021 Page 17 Page 17 Page 17 |
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Monodactylus argenteus
Monodactylus argenteus(Linnaeus, 1758)
Silver moony
Silvery moony,Akoli,Chandlak,Purrandee,Silver-batfish,Vavalmeen,चाँदळाक,புரண்டி,புரண்டீ,അകൊലി,ആകോലി,വവാല് മീന്,വാവല് മീന്
photo by Ramani Shirantha
Family:Monodactylidae (Moonyfishes or fingerfishes)
Max. size:27 cm SL (male/unsexed)
Environment:pelagic-neritic; freshwater; brackish; marine; depth range 0 - 12 m
Distribution:Indo-West Pacific: Red Sea and east coast of Africa south to the Breë River (Ref. 7293, 52193); also the Persian Gulf (Ref. 96822) to Samoa, north to the Yaeyamas, south to New Caledonia and Australia (Ref. 4959). Known from the freshwater tidal zone of the Mekong delta (Ref. 12693).
Diagnosis:Dorsal spines (total): 7-8; Dorsal soft rays (total): 27-31; Anal spines: 3-3; Anal soft rays: 27-32. Description: Body diamond-shaped, depth 1.2-1.6 times in standard length (Ref. 52193). Eyes large; mouth small; maxilla width 3.4-5.0 in orbit diameter (Ref. 52193, 90102). Dorsal fin with 8 spines and 27-30 soft rays; anal fin with 3 spines and 27-30 soft rays; pelvic fins small or absent (Ref. 52193). Scales small, 52-58 in lateral line; scales over base of dorsal and anal fins (Ref. 52193).
Colouration: Adults bright silver with yellow and dusky dorsal fin tip; small juveniles more colourful with yellow over most of the dorsal fin and two vertical black bands over the head, one through the eye and the second behind the head (Ref. 48636, 52193). Fins yellow except pectoral translucent. (Ref. 90102).
Biology:Found in bays, mangrove estuaries, tidal creeks, and lower reaches of freshwater streams (Ref. 2847, 44894, 48636); occasionally in silty coastal reefs (Ref. 9710). It feeds on plankton and detritus (Ref. 5213, 44894). Commonly seen in large schools (Ref. 44894, 52193); small juveniles either solitary or in small aggregations (Ref. 48635). Usually only juveniles enter freshwater (Ref. 52193). Highly territorial (Ref. 9710). Caught with throw nets (Ref. 30573).
IUCN Red List Status:Least Concern (LC); Date assessed: 28 June 2016 Ref. (130435)
Threat to humans:harmless
Country info:Type locality (Ref. 48636). Known from Kerala coast (Ref. 45255). Also found in Chilka Lake (Ref. 29108). Also Ref. 4833.
Source and more info: www.fishbase.org. For personal, classroom, and other internal use only. Not for publication. |
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Combinatorial identity
Contents
1 Pascal's Identity
1.1 Proof
1.2 Alternate Proofs
2 Vandermonde's Identity
2.1 Video Proof
2.2 Combinatorial Proof
2.3 Algebraic proof
3 Hockey-Stick Identity
3.1 Proof
4 Another Identity
4.1 Hat Proof
4.2 Proof 2
5 Even Odd Identity
6 Examples
7 See also
Pascal's Identity
Pascal's Identity is a very important formula for olympiad math and it states that
for any positive integers and . Here, is the binomial coefficient .
This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things.
Proof
If then and so the result is trivial. So assume . Then
Alternate Proofs
Here, we prove this using committee forming.
Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways.
Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways.
But we already know they can be picked in ways, so
Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, .
Vandermonde's Identity
Vandermonde's Identity states that , which can be proven combinatorially by noting that any combination of objects from a group of objects must have some objects from group and the remaining from group .
Video Proof
Combinatorial Proof
Think of the right hand side as picking people from men and women. Think of the left hand side as picking men from the total men and picking women from the total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true.
~avn
Algebraic proof
For all The coefficients of on both sides must be the same, so using the Binomial Theorem,
Hockey-Stick Identity
For .
This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical.
Proof
Inductive Proof
This identity can be proven by induction on .
Base Case Let .
.
Inductive Step Suppose, for some , . Then .
Algebraic Proof
It can also be proven algebraically with Pascal's Identity, . Note that
, which is equivalent to the desired result.
Combinatorial Proof 1
Imagine that we are distributing indistinguishable candies to distinguishable children. By a direct application of Balls and Urns, there are ways to do this. Alternatively, we can first give candies to the oldest child so that we are essentially giving candies to kids and again, with Balls and Urns, , which simplifies to the desired result.
Combinatorial Proof 2
We can form a committee of size from a group of people in ways. Now we hand out the numbers to of the people. We can divide this into disjoint cases. In general, in case , , person is on the committee and persons are not on the committee. This can be done in ways. Now we can sum the values of these disjoint cases, getting
Algebraic Proof 2
Apply the finite geometric series formula to : Then expand with the Binomial Theorem and simplify: Finally, equate coefficients of on both sides: Since for , , this simplifies to the hockey stick identity.
Algebraic Proof 3
Consider the number of solutions to the equation ++++++.......+ ≤ N where each is a non-negative integer for 1≤i≤r.
METHOD 1 We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer.
Therfore, ++++++.......+ = 0,1,2......N. Consider each case seperately.
++++++.......+ =0 by Stars-and-bars the equation has solutions.
++++++.......+ =1 by Stars-and-bars the equation has solutions.
++++++.......+ =2 by Stars-and-bars the equation has solutions.
........... ++++++.......+ =N by Stars-and-bars the equation has solutions.
Hence, the equation has + + +.... = (where k=N+r-1) SOLUTIONS.
METHOD 2 Since, ++++++.......+ ≤ N Therefore we may say ++++++.......+ = N -m where m is another non-negative integer. 0 ≤++++++.......+ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction.
Now, ++++++.......+ +m = N. Again by Stars-and-bars this has solutions.
Therefore, the equation has = solutions(As N+r-1 =k).
Since, both methods yeild the same answer ⇒ = . Taking r-1= p redirects to the honeystick identity.
~SANSGANKRSNGUPTA
Another Identity
Hat Proof
We have different hats. We split them into two groups, each with k hats: then we choose hats from the first group and hats from the second group. This may be done in ways. Evidently, to generate all possible choices of hats from the hats, we must choose hats from the first and the remaining hats from the second ; the sum over all such is the number of ways of choosing hats from . Therefore , as desired.
Proof 2
This is a special case of Vandermonde's identity, in which we set .
Even Odd Identity
Examples
1986 AIME Problem 11
2000 AIME II Problem 7
2013 AIME I Problem 6 (Solution 2)
2015 AIME I Problem 12
2018 AIME I Problem 10
2020 AIME I Problem 7
2016 AMC 10A Problem 20
2021 AMC 12A Problem 15
1981 IMO Problem 2
2022 AIME I Problem 12
See also
Pascal's Triangle
Combinatorics
Pascal's Identity
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1030 | https://cplusplus.com/doc/tutorial/operators/ | TUTORIALS
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Operators
Operators
Once introduced to variables and constants, we can begin to operate with them by using operators. What follows is a complete list of operators. At this point, it is likely not necessary to know all of them, but they are all listed here to also serve as reference.
Assignment operator (=)
The assignment operator assigns a value to a variable.
| | |
--- |
| 1 | x = 5; |
This statement assigns the integer value 5 to the variable x. The assignment operation always takes place from right to left, and never the other way around:
| | |
--- |
| 1 | x = y; |
This statement assigns to variable x the value contained in variable y. The value of x at the moment this statement is executed is lost and replaced by the value of y. Consider also that we are only assigning the value of y to x at the moment of the assignment operation. Therefore, if y changes at a later moment, it will not affect the new value taken by x. For example, let's have a look at the following code - I have included the evolution of the content stored in the variables as comments:
| | |
--- |
| // assignment operator #include using namespace std; int main () { int a, b; // a:?, b:? a = 10; // a:10, b:? b = 4; // a:10, b:4 a = b; // a:4, b:4 b = 7; // a:4, b:7 cout << "a:"; cout << a; cout << " b:"; cout << b; } | a:4 b:7 |
This program prints on screen the final values of a and b (4 and 7, respectively). Notice how a was not affected by the final modification of b, even though we declared a = b earlier. Assignment operations are expressions that can be evaluated. That means that the assignment itself has a value, and -for fundamental types- this value is the one assigned in the operation. For example:
| | |
--- |
| 1 | y = 2 + (x = 5); |
In this expression, y is assigned the result of adding 2 and the value of another assignment expression (which has itself a value of 5). It is roughly equivalent to:
| | |
--- |
| 12 | x = 5; y = 2 + x; |
With the final result of assigning 7 to y. The following expression is also valid in C++:
| | |
--- |
| 1 | x = y = z = 5; |
It assigns 5 to the all three variables: x, y and z; always from right-to-left.
Arithmetic operators ( +, -, , /, % )
The five arithmetical operations supported by C++ are:
| operator | description |
--- |
| + | addition |
subtraction |
| | multiplication |
| / | division |
| % | modulo |
Operations of addition, subtraction, multiplication and division correspond literally to their respective mathematical operators. The last one, modulo operator, represented by a percentage sign (%), gives the remainder of a division of two values. For example:
| | |
--- |
| 1 | x = 11 % 3; |
results in variable x containing the value 2, since dividing 11 by 3 results in 3, with a remainder of 2.
Compound assignment (+=, -=, =, /=, %=, >>=, <<=, &=, ^=, |=)
Compound assignment operators modify the current value of a variable by performing an operation on it. They are equivalent to assigning the result of an operation to the first operand:
| expression | equivalent to... |
--- |
| y += x; | y = y + x; |
| x -= 5; | x = x - 5; |
| x /= y; | x = x / y; |
| price = units + 1; | price = price (units+1); |
and the same for all other compound assignment operators. For example:
| | | |
---
| 1234567891011 | // compound assignment operators #include using namespace std; int main () { int a, b=3; a = b; a+=2; // equivalent to a=a+2 cout << a; } | 5 |
Increment and decrement (++, --)
Some expression can be shortened even more: the increase operator (++) and the decrease operator (--) increase or reduce by one the value stored in a variable. They are equivalent to +=1 and to -=1, respectively. Thus:
| | |
--- |
| 123 | ++x; x+=1; x=x+1; |
are all equivalent in its functionality; the three of them increase by one the value of x. In the early C compilers, the three previous expressions may have produced different executable code depending on which one was used. Nowadays, this type of code optimization is generally performed automatically by the compiler, thus the three expressions should produce exactly the same executable code. A peculiarity of this operator is that it can be used both as a prefix and as a suffix. That means that it can be written either before the variable name (++x) or after it (x++). Although in simple expressions like x++ or ++x, both have exactly the same meaning; in other expressions in which the result of the increment or decrement operation is evaluated, they may have an important difference in their meaning: In the case that the increase operator is used as a prefix (++x) of the value, the expression evaluates to the final value of x, once it is already increased. On the other hand, in case that it is used as a suffix (x++), the value is also increased, but the expression evaluates to the value that x had before being increased. Notice the difference:
| Example 1 | Example 2 |
--- |
| x = 3; y = ++x; // x contains 4, y contains 4 | x = 3; y = x++; // x contains 4, y contains 3 |
In Example 1, the value assigned to y is the value of x after being increased. While in Example 2, it is the value x had before being increased.
Relational and comparison operators ( ==, !=, >, <, >=, <= )
Two expressions can be compared using relational and equality operators. For example, to know if two values are equal or if one is greater than the other. The result of such an operation is either true or false (i.e., a Boolean value). The relational operators in C++ are:
| operator | description |
--- |
| == | Equal to |
| != | Not equal to |
| < | Less than |
| > | Greater than |
| <= | Less than or equal to |
| >= | Greater than or equal to |
Here there are some examples:
| | |
--- |
| 12345 | (7 == 5) // evaluates to false (5 > 4) // evaluates to true (3 != 2) // evaluates to true (6 >= 6) // evaluates to true (5 < 5) // evaluates to false |
Of course, it's not just numeric constants that can be compared, but just any value, including, of course, variables. Suppose that a=2, b=3 and c=6, then:
| | |
--- |
| 1234 | (a == 5) // evaluates to false, since a is not equal to 5 (ab >= c) // evaluates to true, since (23 >= 6) is true (b+4 > ac) // evaluates to false, since (3+4 > 26) is false ((b=2) == a) // evaluates to true |
Be careful! The assignment operator (operator =, with one equal sign) is not the same as the equality comparison operator (operator ==, with two equal signs); the first one (=) assigns the value on the right-hand to the variable on its left, while the other (==) compares whether the values on both sides of the operator are equal. Therefore, in the last expression ((b=2) == a), we first assigned the value 2 to b and then we compared it to a (that also stores the value 2), yielding true.
Logical operators ( !, &&, || )
The operator ! is the C++ operator for the Boolean operation NOT. It has only one operand, to its right, and inverts it, producing false if its operand is true, and true if its operand is false. Basically, it returns the opposite Boolean value of evaluating its operand. For example:
| | |
--- |
| 1234 | !(5 == 5) // evaluates to false because the expression at its right (5 == 5) is true !(6 <= 4) // evaluates to true because (6 <= 4) would be false !true // evaluates to false !false // evaluates to true |
The logical operators && and || are used when evaluating two expressions to obtain a single relational result. The operator && corresponds to the Boolean logical operation AND, which yields true if both its operands are true, and false otherwise. The following panel shows the result of operator && evaluating the expression a&&b:
| && OPERATOR (and) |
| a | b | a && b |
| true | true | true |
| true | false | false |
| false | true | false |
| false | false | false |
The operator || corresponds to the Boolean logical operation OR, which yields true if either of its operands is true, thus being false only when both operands are false. Here are the possible results of a||b:
| || OPERATOR (or) |
| a | b | a || b |
| true | true | true |
| true | false | true |
| false | true | true |
| false | false | false |
For example:
| | |
--- |
| 12 | ( (5 == 5) && (3 > 6) ) // evaluates to false ( true && false ) ( (5 == 5) || (3 > 6) ) // evaluates to true ( true || false ) |
When using the logical operators, C++ only evaluates what is necessary from left to right to come up with the combined relational result, ignoring the rest. Therefore, in the last example ((5==5)||(3>6)), C++ evaluates first whether 5==5 is true, and if so, it never checks whether 3>6 is true or not. This is known as short-circuit evaluation, and works like this for these operators:
| operator | short-circuit |
--- |
| && | if the left-hand side expression is false, the combined result is false (the right-hand side expression is never evaluated). |
| || | if the left-hand side expression is true, the combined result is true (the right-hand side expression is never evaluated). |
This is mostly important when the right-hand expression has side effects, such as altering values:
| | |
--- |
| 1 | if ( (i<10) && (++i/.../ } // note that the condition increments i |
Here, the combined conditional expression would increase i by one, but only if the condition on the left of && is true, because otherwise, the condition on the right-hand side (++i) is never evaluated.
Conditional ternary operator ( ? )
The conditional operator evaluates an expression, returning one value if that expression evaluates to true, and a different one if the expression evaluates as false. Its syntax is: condition ? result1 : result2 If condition is true, the entire expression evaluates to result1, and otherwise to result2.
| | |
--- |
| 1234 | 7==5 ? 4 : 3 // evaluates to 3, since 7 is not equal to 5. 7==5+2 ? 4 : 3 // evaluates to 4, since 7 is equal to 5+2. 5>3 ? a : b // evaluates to the value of a, since 5 is greater than 3. a>b ? a : b // evaluates to whichever is greater, a or b. |
For example:
| | | |
---
| 1234567891011121314 | // conditional operator #include using namespace std; int main () { int a,b,c; a=2; b=7; c = (a>b) ? a : b; cout << c << '\n'; } | 7 |
In this example, a was 2, and b was 7, so the expression being evaluated (a>b) was not true, thus the first value specified after the question mark was discarded in favor of the second value (the one after the colon) which was b (with a value of 7).
Comma operator ( , )
The comma operator (,) is used to separate two or more expressions that are included where only one expression is expected. When the set of expressions has to be evaluated for a value, only the right-most expression is considered. For example, the following code:
| | |
--- |
| 1 | a = (b=3, b+2); |
would first assign the value 3 to b, and then assign b+2 to variable a. So, at the end, variable a would contain the value 5 while variable b would contain value 3.
Bitwise operators ( &, |, ^, ~, <<, >> )
Bitwise operators modify variables considering the bit patterns that represent the values they store.
| operator | asm equivalent | description |
---
| & | AND | Bitwise AND |
| | | OR | Bitwise inclusive OR |
| ^ | XOR | Bitwise exclusive OR |
| ~ | NOT | Unary complement (bit inversion) |
| << | SHL | Shift bits left |
| >> | SHR | Shift bits right |
Explicit type casting operator
Type casting operators allow to convert a value of a given type to another type. There are several ways to do this in C++. The simplest one, which has been inherited from the C language, is to precede the expression to be converted by the new type enclosed between parentheses (()):
| | |
--- |
| 123 | int i; float f = 3.14; i = (int) f; |
The previous code converts the floating-point number 3.14 to an integer value (3); the remainder is lost. Here, the typecasting operator was (int). Another way to do the same thing in C++ is to use the functional notation preceding the expression to be converted by the type and enclosing the expression between parentheses:
| | |
--- |
| 1 | i = int (f); |
Both ways of casting types are valid in C++.
sizeof
This operator accepts one parameter, which can be either a type or a variable, and returns the size in bytes of that type or object:
| | |
--- |
| 1 | x = sizeof (char); |
Here, x is assigned the value 1, because char is a type with a size of one byte. The value returned by sizeof is a compile-time constant, so it is always determined before program execution.
Other operators
Later in these tutorials, we will see a few more operators, like the ones referring to pointers or the specifics for object-oriented programming.
Precedence of operators
A single expression may have multiple operators. For example:
| | |
--- |
| 1 | x = 5 + 7 % 2; |
In C++, the above expression always assigns 6 to variable x, because the % operator has a higher precedence than the + operator, and is always evaluated before. Parts of the expressions can be enclosed in parenthesis to override this precedence order, or to make explicitly clear the intended effect. Notice the difference:
| | |
--- |
| 12 | x = 5 + (7 % 2); // x = 6 (same as without parenthesis) x = (5 + 7) % 2; // x = 0 |
From greatest to smallest priority, C++ operators are evaluated in the following order:
| Level | Precedence group | Operator | Description | Grouping |
--- ---
| 1 | Scope scope qualifier | Left-to-right |
| 2 | Postfix (unary) | ++ -- | postfix increment / decrement | Left-to-right |
| () | functional forms |
| [] | subscript |
| . -> | member access |
| 3 | Prefix (unary) | ++ -- | prefix increment / decrement | Right-to-left |
| ~ ! | bitwise NOT / logical NOT |
| + - | unary prefix |
| & | reference / dereference |
| new delete | allocation / deallocation |
| sizeof | parameter pack |
| (type) | C-style type-casting |
| 4 | Pointer-to-member | . -> | access pointer | Left-to-right |
| 5 | Arithmetic: scaling | / % | multiply, divide, modulo | Left-to-right |
| 6 | Arithmetic: addition | + - | addition, subtraction | Left-to-right |
| 7 | Bitwise shift | << >> | shift left, shift right | Left-to-right |
| 8 | Relational | < > <= >= | comparison operators | Left-to-right |
| 9 | Equality | == != | equality / inequality | Left-to-right |
| 10 | And | & | bitwise AND | Left-to-right |
| 11 | Exclusive or | ^ | bitwise XOR | Left-to-right |
| 12 | Inclusive or | | | bitwise OR | Left-to-right |
| 13 | Conjunction | && | logical AND | Left-to-right |
| 14 | Disjunction | || | logical OR | Left-to-right |
| 15 | Assignment-level expressions | = = /= %= += -= >>= <<= &= ^= |= | assignment / compound assignment | Right-to-left |
| ?: | conditional operator |
| 16 | Sequencing | , | comma separator | Left-to-right |
When an expression has two operators with the same precedence level, grouping determines which one is evaluated first: either left-to-right or right-to-left. Enclosing all sub-statements in parentheses (even those unnecessary because of their precedence) improves code readability.
| | | |
---
| Previous:Constants | Index | Next:Basic Input/Output |
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From Wikipedia, the free encyclopedia
Distance between the nearest and the furthest objects that are in focus in an image
For other uses, see Depth of field (disambiguation).
The depth of field (DOF) is the distance between the nearest and the farthest objects that are in acceptably sharp focus in an image captured with a camera. See also the closely related depth of focus.
Factors affecting depth of field
[edit]
For cameras that can only focus on one object distance at a time, depth of field is the distance between the nearest and the farthest objects that are in acceptably sharp focus in the image. "Acceptably sharp focus" is defined using a property called the "circle of confusion".
The depth of field can be determined by focal length, distance to subject (object to be imaged), the acceptable circle of confusion size, and aperture. The approximate depth of field can be given by:
for a given maximum acceptable circle of confusion diameter c, focal length f, f-number N, and distance to subject u.
As distance or the size of the acceptable circle of confusion increases, the depth of field increases; however, increasing the size of the aperture (i.e., reducing f-number) or increasing the focal length reduces the depth of field. Depth of field changes linearly with f-number and circle of confusion, but changes in proportion to the square of the distance to the subject and inversely in proportion to the square of the focal length. As a result, photos taken at extremely close range (i.e., so small u) have a proportionally much smaller depth of field.
Rearranging the DOF equation shows that it is the ratio between distance and focal length that affects DOF;
Note that is the transverse magnification which is the ratio of the lateral image size to the lateral subject size.
Image sensor size affects DOF in counterintuitive ways. Because the circle of confusion is directly tied to the sensor size, decreasing the size of the sensor while holding focal length and aperture constant will decrease the depth of field (by the crop factor). The resulting image however will have a different field of view. If the focal length is altered to maintain the field of view, while holding the f-number constant, the change in focal length will counter the decrease of DOF from the smaller sensor and increase the depth of field (also by the crop factor). However, if the focal length is altered to maintain the field of view, while holding the aperture diameter constant, the DOF will remain constant.
Effect of lens aperture
[edit]
For a given subject framing and camera position, the DOF is controlled by the lens aperture diameter, which is usually specified as the f-number (the ratio of lens focal length to aperture diameter). Reducing the aperture diameter (increasing the f-number) increases the DOF because only the light travelling at shallower angles passes through the aperture so only cones of rays with shallower angles reach the image plane. In other words, the circles of confusion are reduced or increasing the DOF.
For a given size of the subject's image in the focal plane, the same f-number on any focal length lens will give the same depth of field. This is evident from the above DOF equation by noting that the ratio u/f is constant for constant image size. For example, if the focal length is doubled, the subject distance is also doubled to keep the subject image size the same. This observation contrasts with the common notion that "focal length is twice as important to defocus as f/stop", which applies to a constant subject distance, as opposed to constant image size.
Motion pictures make limited use of aperture control; to produce a consistent image quality from shot to shot, cinematographers usually choose a single aperture setting for interiors (e.g., scenes inside a building) and another for exteriors (e.g., scenes in an area outside a building), and adjust exposure through the use of camera filters or light levels. Aperture settings are adjusted more frequently in still photography, where variations in depth of field are used to produce a variety of special effects.
Aperture = f/1.4. DOF = 0.8 cm
Aperture = f/4.0. DOF = 2.2 cm
Aperture = f/22. DOF = 12.4 cm
Depth of field for different values of aperture using 50 mm objective lens and full-frame DSLR camera. Focus point is on the first blocks column.[better source needed]
Effect of circle of confusion
[edit]
Precise focus is only possible at an exact distance from a lens;[a] at that distance, a point object will produce a small spot image. Otherwise, a point object will produce a larger or blur spot image that is typically and approximately a circle. When this circular spot is sufficiently small, it is visually indistinguishable from a point, and appears to be in focus. The diameter of the largest circle that is indistinguishable from a point is known as the acceptable circle of confusion, or informally, simply as the circle of confusion.
The acceptable circle of confusion depends on how the final image will be used. The circle of confusion as 0.25 mm for an image viewed from 25 cm away is generally accepted.
For 35 mm motion pictures, the image area on the film is roughly 22 mm by 16 mm. The limit of tolerable error was traditionally set at 0.05 mm (0.0020 in) diameter, while for 16 mm film, where the size is about half as large, the tolerance is stricter, 0.025 mm (0.00098 in). More modern practice for 35 mm productions set the circle of confusion limit at 0.025 mm (0.00098 in).
Camera movements
[edit]
See also: View camera and Tilt–shift photography
The term "camera movements" refers to swivel (swing and tilt, in modern terminology) and shift adjustments of the lens holder and the film holder. These features have been in use since the 1800s and are still in use today on view cameras, technical cameras, cameras with tilt/shift or perspective control lenses, etc. Swiveling the lens or sensor causes the plane of focus (POF) to swivel, and also causes the field of acceptable focus to swivel with the POF; and depending on the DOF criteria, to also change the shape of the field of acceptable focus. While calculations for DOF of cameras with swivel set to zero have been discussed, formulated, and documented since before the 1940s, documenting calculations for cameras with non-zero swivel seem to have begun in 1990.
More so than in the case of the zero swivel camera, there are various methods to form criteria and set up calculations for DOF when swivel is non-zero. There is a gradual reduction of clarity in objects as they move away from the POF, and at some virtual flat or curved surface the reduced clarity becomes unacceptable. Some photographers do calculations or use tables, some use markings on their equipment, some judge by previewing the image.
When the POF is rotated, the near and far limits of DOF may be thought of as wedge-shaped, with the apex of the wedge nearest the camera; or they may be thought of as parallel to the POF.
Object-field calculation methods
[edit]
Traditional depth-of-field formulas can be hard to use in practice. As an alternative, the same effective calculation can be done without regard to the focal length and f-number.[b] Moritz von Rohr and later Merklinger observe that the effective absolute aperture diameter can be used for similar formula in certain circumstances.
Moreover, traditional depth-of-field formulas assume equal acceptable circles of confusion for near and far objects. Merklinger[c] suggested that distant objects often need to be much sharper to be clearly recognizable, whereas closer objects, being larger on the film, do not need to be so sharp. The loss of detail in distant objects may be particularly noticeable with extreme enlargements. Achieving this additional sharpness in distant objects usually requires focusing beyond the hyperfocal distance, sometimes almost at infinity. For example, if photographing a cityscape with a traffic bollard in the foreground, this approach, termed the object field method by Merklinger, would recommend focusing very close to infinity, and stopping down to make the bollard sharp enough. With this approach, foreground objects cannot always be made perfectly sharp, but the loss of sharpness in near objects may be acceptable if recognizability of distant objects is paramount.
Other authors such as Ansel Adams have taken the opposite position, maintaining that slight unsharpness in foreground objects is usually more disturbing than slight unsharpness in distant parts of a scene.
Overcoming DOF limitations
[edit]
Some methods and equipment allow altering the apparent DOF, and some even allow the DOF to be determined after the image is made. These are based or supported by computational imaging processes. For example, focus stacking combines multiple images focused on different planes, resulting in an image with a greater (or less, if so desired) apparent depth of field than any of the individual source images. Similarly, in order to reconstruct the 3-dimensional shape of an object, a depth map can be generated from multiple photographs with different depths of field. Xiong and Shafer concluded, in part, "... the improvements on precisions of focus ranging and defocus ranging can lead to efficient shape recovery methods."
Another approach is focus sweep. The focal plane is swept across the entire relevant range during a single exposure. This creates a blurred image, but with a convolution kernel that is nearly independent of object depth, so that the blur is almost entirely removed after computational deconvolution. This has the added benefit of dramatically reducing motion blur.
Light Scanning Photomacrography (LSP) is another technique used to overcome depth of field limitations in macro and micro photography. This method allows for high-magnification imaging with exceptional depth of field. LSP involves scanning a thin light plane across the subject that is mounted on a moving stage perpendicular to the light plane. This ensures the entire subject remains in sharp focus from the nearest to the farthest details, providing comprehensive depth of field in a single image. Initially developed in the 1960s and further refined in the 1980s and 1990s, LSP was particularly valuable in scientific and biomedical photography before digital focus stacking became prevalent.
Other technologies use a combination of lens design and post-processing: Wavefront coding is a method by which controlled aberrations are added to the optical system so that the focus and depth of field can be improved later in the process.
The lens design can be changed even more: in colour apodization the lens is modified such that each colour channel has a different lens aperture. For example, the red channel may be f/2.4, green may be f/2.4, whilst the blue channel may be f/5.6. Therefore, the blue channel will have a greater depth of field than the other colours. The image processing identifies blurred regions in the red and green channels and in these regions copies the sharper edge data from the blue channel. The result is an image that combines the best features from the different f-numbers.
At the extreme, a plenoptic camera captures 4D light field information about a scene, so the focus and depth of field can be altered after the photo is taken.
Diffraction and DOF
[edit]
Diffraction causes images to lose sharpness at high f-numbers (i.e., narrow aperture stop opening sizes), and hence limits the potential depth of field. (This effect is not considered in the above formula giving approximate DOF values.) In general photography this is rarely an issue; because large f-numbers typically require long exposure times to acquire acceptable image brightness, motion blur may cause greater loss of sharpness than the loss from diffraction. However, diffraction is a greater issue in close-up photography, and the overall image sharpness can be degraded as photographers are trying to maximize depth of field with very small apertures.
Hansma and Peterson have discussed determining the combined effects of defocus and diffraction using a root-square combination of the individual blur spots. Hansma's approach determines the f-number that will give the maximum possible sharpness; Peterson's approach determines the minimum f-number that will give the desired sharpness in the final image and yields a maximum depth of field for which the desired sharpness can be achieved.[d] In combination, the two methods can be regarded as giving a maximum and minimum f-number for a given situation, with the photographer free to choose any value within the range, as conditions (e.g., potential motion blur) permit. Gibson gives a similar discussion, additionally considering blurring effects of camera lens aberrations, enlarging lens diffraction and aberrations, the negative emulsion, and the printing paper.[e] Couzin gave a formula essentially the same as Hansma's for optimal f-number, but did not discuss its derivation.
Hopkins, Stokseth, and Williams and Becklund have discussed the combined effects using the modulation transfer function.
DOF scales
[edit]
Many lenses include scales that indicate the DOF for a given focus distance and f-number; the 35 mm lens in the image is typical. That lens includes distance scales in feet and meters; when a marked distance is set opposite the large white index mark, the focus is set to that distance. The DOF scale below the distance scales includes markings on either side of the index that correspond to f-numbers. When the lens is set to a given f-number, the DOF extends between the distances that align with the f-number markings.
Photographers can use the lens scales to work backwards from the desired depth of field to find the necessary focus distance and aperture. For the 35 mm lens shown, if it were desired for the DOF to extend from 1 m to 2 m, focus would be set so that index mark was centered between the marks for those distances, and the aperture would be set to f/11.[f]
On a view camera, the focus and f-number can be obtained by measuring the depth of field and performing simple calculations. Some view cameras include DOF calculators that indicate focus and f-number without the need for any calculations by the photographer.
Detail from a lens set to f/11. The point half-way between the 1 m and 2 m marks, the DOF limits at f/11, represents the focus distance of approximately 1.33 m (the reciprocal of the mean of the reciprocals of 1 and 2 being 4/3).
DOF scale on Tessina focusing dial
Hyperfocal distance
[edit]
This section is an excerpt from Hyperfocal distance.[edit]
In optics and photography, hyperfocal distance is a distance from a lens beyond which all objects can be brought into an "acceptable" focus. As the hyperfocal distance is the focus distance giving the maximum depth of field, it is the most desirable distance to set the focus of a fixed-focus camera. The hyperfocal distance is entirely dependent upon what level of sharpness is considered to be acceptable.
The hyperfocal distance has a property called "consecutive depths of field", where a lens focused at an object whose distance from the lens is at the hyperfocal distance H will hold a depth of field from H/2 to infinity, if the lens is focused to H/2, the depth of field will be from H/3 to H; if the lens is then focused to H/3, the depth of field will be from H/4 to H/2, etc.
Thomas Sutton and George Dawson first wrote about hyperfocal distance (or "focal range") in 1867. Louis Derr in 1906 may have been the first to derive a formula for hyperfocal distance. Rudolf Kingslake wrote in 1951 about the two methods of measuring hyperfocal distance.
Some cameras have their hyperfocal distance marked on the focus dial. For example, on the Minox LX focusing dial there is a red dot between 2 m and infinity; when the lens is set at the red dot, that is, focused at the hyperfocal distance, the depth of field stretches from 2 m to infinity. Some lenses have markings indicating the hyperfocal range for specific f-stops, also called a depth-of-field scale.
Near:far distribution
[edit]
The DOF beyond the subject is always greater than the DOF in front of the subject. When the subject is at the hyperfocal distance or beyond, the far DOF is infinite, so the ratio is 1:∞; as the subject distance decreases, near:far DOF ratio increases, approaching unity at high magnification. For large apertures at typical portrait distances, the ratio is still close to 1:1.
DOF formulae
[edit]
This section covers some additional formula for evaluating depth of field; however they are all subject to significant simplifying assumptions: for example, they assume the paraxial approximation of Gaussian optics. They are suitable for practical photography, lens designers would use significantly more complex ones.
Focus and f-number from DOF limits
[edit]
For given near and far DOF limits DN and DF, the required f-number is smallest when focus is set to
the harmonic mean of the near and far distances. In practice, this is equivalent to the arithmetic mean for shallow depths of field. Sometimes, view camera users refer to the difference vN − vF as the focus spread.
Foreground and background blur
[edit]
If a subject is at distance s and the foreground or background is at distance D, let the distance between the subject and the foreground or background be indicated by
The blur disk diameter b of a detail at distance xd from the subject can be expressed as a function of the subject magnification ms, focal length f, f-number N, or alternatively the aperture d, according to
The minus sign applies to a foreground object, and the plus sign applies to a background object.
The blur increases with the distance from the subject; when b is less than the circle of confusion, the detail is within the depth of field.
See also
[edit]
Angle of view
Bokeh
Camera angle
Depth of focus
Depth-of-field adapter
Frazier lens (very deep depth of field)
Light-field camera
Miniature faking
Numerical aperture
Perspective distortion
Explanatory notes
[edit]
^ Strictly, at an exact distance from a plane.
^ Notwithstanding that the f-number is derived from the focal length.
^ Englander describes a similar approach in his paper Apparent Depth of Field: Practical Use in Landscape Photography; Conrad discusses this approach, under Different Circles of Confusion for Near and Far Limits of Depth of Field, and The Object Field Method, in Depth of Field in Depth.
^ Peterson does not give a closed-form expression for the minimum f-number, though such an expression obtains from simple algebraic manipulation of his Equation 3.
^ The analytical section at the end of Gibson (1975) was originally published as "Magnification and Depth of Detail in Photomacrography" in the Journal of the Photographic Society of America, Vol. 26, No. 6, June 1960.
^ The focus distance to have the DOF extend between given near and far object distances is the harmonic mean of the object conjugates. Most helicoid-focused lenses are marked with image plane-to-subject distances,[citation needed] so the focus determined from the lens distance scale is not exactly the harmonic mean of the marked near and far distances.
References
[edit]
Citations
[edit]
^ Salvaggio & Stroebel 2009, pp. 110-.
^ Barbara London; Jim Stone; John Upton (2005). Photography (8th ed.). Pearson. p. 58. ISBN 978-0-13-448202-6.
^ Elizabeth Allen; Sophie Triantaphillidou (2011). The Manual of Photography. Taylor & Francis. pp. 111–. ISBN 978-0-240-52037-7.
^ "Depth of field". Stanford Computer Graphics Laboratory.
^ Hecht, Eugene (2017). "5.2.3 Thin Lenses". Optics (5th ed.). Pearson. p. 172. ISBN 978-1-292-09693-3.
^ Nasse, H. H. (March 2010). "Depth of Field and Bokeh" (PDF). Zeiss Lenspire (Whitepaper).
^ "Digital Camera Sensor Sizes: How it Influences Your Photography". Cambridge In Colour.
^ Malan, Francois (6 April 2018). "Sensor Size, Perspective and Depth of Field". Photography Life.
^ Vinson, Jason (22 January 2016). "The Smaller the Sensor Size, the Shallower Your Depth of Field". Fstoppers.
^ "Why Does a Small Aperture Increase Depth of Field?". 18 May 2012.[self-published source]
^ Reichmann, Michael (13 January 2009). "DOF2". Luminous Landscape.
^ "Ken Rockwell".[self-published source]
^ "photoskop: Interactive Photography Lessons". April 25, 2015.
^ Savazzi 2011, p. 109.
^ Film and Its Techniques. University of California Press. 1966. p. 56. Retrieved 24 February 2016.
^ Thomas Ohanian and Natalie Phillips (2013). Digital Filmmaking: The Changing Art and Craft of Making Motion Pictures. CRC Press. p. 96. ISBN 9781136053542. Retrieved 24 February 2016.
^ Merklinger 1993, pp. 49–56.
^ Tillmanns 1997, p. 71.
^ Jump up to: a b Merklinger 1992.
^ Adams 1980, p. 51.
^ Xiong, Yalin, and Steven A. Shafer. "Depth from focusing and defocusing." Computer Vision and Pattern Recognition, 1993. Proceedings CVPR'93., 1993 IEEE Computer Society Conference on. IEEE, 1993.
^ Bando et al. "Near-Invariant Blur for Depth and 2D Motion via Time-Varying Light Field Analysis." ACM Transactions on Graphics, Vol. 32, No. 2, Article 13, 2013.
^ Root, N. (January 1991) "A simplified unit for making deep-field (scanning) Macrographs". Journal of Biological Photography, Vol. 59, No. 1, pp. 3-8.
^ Clarke, T. "Constructing a Scanning Light Photomacrography System." The McCrone Group (accessed July 7, 2024).
^ Mary, D.; Roche, M.; Theys, C.; Aime, C. (2013). "Introduction to Wavefront Coding for Incoherent Imaging" (PDF). EAS Publications Series. 59: 77–92. Bibcode:2013EAS....59...77R. doi:10.1051/eas/1359005. ISSN 1633-4760. S2CID 120502243. Archived (PDF) from the original on 2022-06-14.
^ Kay, Mather & Walton 2011.
^ Jump up to: a b Gibson 1975, p. 64.
^ Gibson 1975, p. 53.
^ Lefkowitz 1979, p. 84.
^ Hansma 1996.
^ Peterson 1996.
^ Couzin 1982.
^ Hopkins 1955.
^ Stokseth 1969.
^ Williams & Becklund 1989.
^ "Depth of Field in Depth", Jeff Conrad
^ "Photographic Lenses Tutorial", David M. Jacobson, 26 October 1996
^ Ray 1994, p. 315.
^ Tillmanns 1997, p. 67-68.
^ Ray 1994, p. 230-231.
^ Kingslake, Rudolf (1951). Lenses in Photography: The Practical Guide to Optics for Photographers. Garden City, NY: Garden City Press.
^ Sutton, Thomas; Dawson, George (1867). A Dictionary of Photography. London: Sampson Low, Son & Marston.
^ Minolta (1985). Minolta MD Zoom Lenses owner's manual. p. 9.
^ Conrad, Jeff. "Depth of Field in Depth" (PDF). Large Format Photography Info. Retrieved 10 February 2023.
^ Hansma 1996, p. 55.
General and cited references
[edit]
Adams, Ansel (1980). The Camera. New York Graphic Society. ISBN 9780821210925. Adams, Ansel. 1980. The Camera.
Couzin, Dennis (November 1982). "Depths of Field". SMPTE Journal: 1096–1098. doi:10.5594/J00204. Available in PDF online.
Gibson, H. Lou (1975). Close-Up Photography and Photomacrography. Kodak Publication No. N-16. Vol. II: Photomacrography (2nd combined ed.). Rochester, NY: Eastman Kodak Company, Professional and Finishing Markets Division. ISBN 0-87985-160-0. OCLC 1991448.
Hansma, Paul K. (March–April 1996). "View Camera Focusing in Practice". Photo Techniques. pp. 54–57. "Available as GIF images on the Large Format page".
Hopkins, H. H. (1955). "The frequency response of a defocused optical system". Proceedings of the Royal Society A. 231 (1184): 91–103. Bibcode:1955RSPSA.231...91H. doi:10.1098/rspa.1955.0158. S2CID 121289809.
Kay, Andrew; Mather, Jonathan; Walton, Harry (1 December 2011). "Extended Depth of Field by Colored Apodization". Optics Letters. 36 (23): 4614–4616. Bibcode:2011OptL...36.4614K. doi:10.1364/OL.36.004614. ISSN 0146-9592. PMID 22139260.
Lefkowitz, Lester (1979). The Manual of Close-Up Photography. Garden City, NY: Amphoto. ISBN 0-8174-2456-3. OCLC 4883084.
Merklinger, Harold M. (1992). The INs and OUTs of FOCUS: An Alternative Way to Estimate Depth-of-Field and Sharpness in the Photographic Image (v. 1.0.3 ed.). Bedford, Nova Scotia: Seaboard Printing Limited. ISBN 0-9695025-0-8. OCLC 23651332. Version 1.03e available in PDF at .
Merklinger, Harold M. (1993). Focusing the View Camera: A Scientific Way to Focus the View Camera and Estimate Depth of Field (v. 1.0 ed.). Bedford, Nova Scotia: Seaboard Printing Limited. ISBN 0-9695025-2-4. OCLC 1072495227. Version 1.6.1 available in PDF at
Peterson, Stephen (March–April 1996). "Image Sharpness and Focusing the View Camera". Photo Techniques: 51–53. Available as GIF images on the Large Format page.
Ray, Sidney F. (1994). Photographic Lenses and Optics. Oxford: Focal Press. ISBN 0-240-51387-8. OCLC 31648843.
Ray, Sidney F. (2000). "The Geometry of Image Formation". In Jacobson, Ralph E.; Ray, Sidney F.; Atteridge, Geoffrey G.; Axford, Norman R. (eds.). The Manual of Photography: Photographic and Digital Imaging (9th ed.). Oxford: Focal Press. ISBN 0-240-51574-9. OCLC 44267873.
Ray, Sidney F. (2002). Applied Photographic Optics (3rd ed.). Oxford: Focal Press. ISBN 0-240-51540-4. OCLC 47868399.
Salvaggio, Nanette; Stroebel, Leslie (2009). Basic Photographic Materials and Processes. Taylor & Francis. pp. 110–. ISBN 978-0-240-80984-7.
Savazzi, Enrico (2011). Digital Photography for Science (Hardcover). Lulu.com. ISBN 978-0-557-91133-2.[self-published source?]
Shipman, Carl (1977). SLR Photographers Handbook. Tucson, AZ: H.P. Books. ISBN 0-912656-59-X. OCLC 3403925.
Stokseth, Per A. (October 1969). "Properties of a Defocused Optical System". Journal of the Optical Society of America. 59 (10): 1314–1321. Bibcode:1969JOSA...59.1314S. doi:10.1364/JOSA.59.001314.
Stroebel, Leslie (1976). View Camera Technique (3rd ed.). London: Focal Press. ISBN 0-240-50901-3. OCLC 3856606.
Tillmanns, Urs (1997). Creative Large Format: Basics and Applications (2nd ed.). Feuerthalen, Switzerland: Sinar AG. ISBN 3-7231-0030-9. OCLC 1193525959.
von Rohr, Moritz (1906). Die optischen Instrumente. Leipzig, Germany: B. G. Teubner. OCLC 750993595.
Williams, Charles S.; Becklund, Orville (1989). Introduction to the Optical Transfer Function. New York: Wiley. pp. 293–300. Reprinted 2002, Bellingham, WA: SPIE Press, ISBN 0-8194-4336-0.
Williams, John B. (1990). Image Clarity: High-Resolution Photography. Boston: Focal Press. ISBN 0-240-80033-8. OCLC 19514912.
Further reading
[edit]
Hummel, Rob, ed. (2001). American Cinematographer Manual (8th ed.). Hollywood, Cal.: The ASC Press. ISBN 0-935578-15-3. OCLC 49686312.
External links
[edit]
Wikimedia Commons has media related to Depth of field.
The Wikibook Maple has a page on the topic of: Depth of field for optical lens
Depth of Field in Photography—Beginner's Guide
Online Depth of Field Calculator Simple depth of field and hyperfocal distance calculator
photoskop: Interactive Photography Lessons—Interactive depth of field
Bokeh simulator and depth of field calculator—Interactive depth of field calculator with background blur simulation feature
Lens Comparison: Nikon 50mm f/1.4D vs. 50mm f/1.4G—Demonstration of varying apertures on depth of field
Depth of Field for Beginners—A quick explainer video for DOF
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Help Writing Nim/Take away Proof
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I'm new to proof by induction and found this: A Take-Away Game.
"Consider the following game: there are n n chips on the table. You are challenged by your arch-enemy to play the following game: you both have to alternate moves, where in each move, you can take away either 1, 2, or 3 chips from the table. The person who has no more chips to remove from the table chips loses."
It's very similar to the proof found here: Variation of Nim: Player who takes last match loses...except player who takes last match wins.
Although I understand what is being said/explained in the proof on the first page, I'm not sure it's shown mathematically (like it is in the second link)?
For example, it says by inductive hypothesis - but I'm not sure what that would be? Is it k=4 a+b k=4 a+b? and why does this verify, mathematically, that the first player wins if n is not a multiple of 4? Would it be better explained using n=0 n=0(mod 4), and so on?
I'm trying to combine the two proofs from the links above to make a nice, easy to follow proof...but I'm coming up short.
Here is what I have so far:
n=4a+b ∀n∈N∀n∈N where a,b∈Z a,b∈Z and b={0,1,2,3}.b={0,1,2,3}.
The first player will win if and only if n≠4 a n≠4 a.
Base:
The first player wins if and only if n≢0 n≢0 (mod 4).
Let n=0 n=0
n=4⋅0+0=4.n=4⋅0+0=4.
The first player can pick up 1- 4 chips, and second player can pick up 1, 2 or 3 chips respectively (whatever is leftover), and the second player would win.
Let n=1 n=1
n=4⋅0+1=1.n=4⋅0+1=1.
The first player wins by picking up all of the chips.
Let n=2 n=2
n=4⋅0+2=2.n=4⋅0+2=2.
The first player wins by picking up all of the chips.
Let n=3 n=3
n=4⋅0+3=3.n=4⋅0+3=3.
The first player wins by picking up all of the chips.
Hypothesis-
Assume that in a game with k k or fewer matches, the first player will win for all b=1,2,3 b=1,2,3 such that i∤4 i∤4, and lose for all b=0 b=0 such that i∣4 i∣4, ∀i∈N∀i∈N, 0≤i≤k 0≤i≤k for some k∈N k∈N such that k≥3 k≥3.
Conclusion-
Suppose we have a game with k+1 k+1 matches with k≥3 k≥3, therefore k+1≥4 k+1≥4.
If k+1=4 a+1 k+1=4 a+1, the first player can pick up 1 chip, leaving k=4 a k=4 a chips for the second player. By the inductive hypothesis, the first player will win.
This is where I don't know where to go, because from here I'm left with:
If k+1=4 a+2 k+1=4 a+2, the first player picks up 2 chips, leaving k=4 a−1 k=4 a−1 sticks for the second player...
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You should give the rules, not a link, at the start. In your base case you say one player can take 1 1 to 4 4 while the other takes 1 1 to 3 3. Usually the game is symmetric. Which is correct?Ross Millikan –Ross Millikan 2016-12-03 06:21:30 +00:00 Commented Dec 3, 2016 at 6:21
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You are claiming that Alice, the first player, wins if the pile is 1,2,1,2, or 3(mod 4)3(mod 4) and Bob, the second player, wins if the pile is 0(mod 4)0(mod 4). your base case is that 0 0 matches are a second player win (because Bob just took the last match) and 1 1 to 3 3 matches are a first player win (because Alice takes them all). Then your induction says suppose we have verified it up through k k where k≥3 k≥3. Then we want to verify it for k+1 k+1. If k+1≡0(mod 4)k+1≡0(mod 4), Bob wins by taking 4 4 less whatever Alice takes and getting to k−3≡0(mod 4)k−3≡0(mod 4) which we know is a win for him. Otherwise, Alice takes k+1(mod 4)k+1(mod 4) and gets to a second player win.
It is easier to use the language of P P positions and N N positions from the Sprague-Grundy theorem for impartial games.
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answered Dec 3, 2016 at 6:29
Ross MillikanRoss Millikan
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I think I'm still confused about how to connect the conclusion to the hypothesis. Is there a way to break down each case (1, 2 or 3 mod 4) in the conclusion?Math1 –Math1 2016-12-03 19:43:46 +00:00 Commented Dec 3, 2016 at 19:43
Yes, when k+1≡0(mod 4)k+1≡0(mod 4) you could list all three choices for Alice and the corresponding move for Bob, showing that in all three you get to k−3 k−3 which we have already proved is a win for Bob. Then when it is not you could list the cases and show that Alice can leave a multiple of 4 4 and win.Ross Millikan –Ross Millikan 2016-12-03 20:26:15 +00:00 Commented Dec 3, 2016 at 20:26
Okay, so let's say k+1≡2 k+1≡2 (mod 4). Alice can remove 2, but that leaves k−1≡0 k−1≡0 (mod 4) for Bob...but that would be a win for him, so that's counter to what I want to show.Math1 –Math1 2016-12-03 20:31:09 +00:00 Commented Dec 3, 2016 at 20:31
...I think. I'm also new to mod so I'm learning this as well. Would it not rearrange to k≡1 k≡1 (mod 4)?Math1 –Math1 2016-12-03 20:32:26 +00:00 Commented Dec 3, 2016 at 20:32
mod is just the remainder, so k≡2(mod 4)k≡2(mod 4) means k=2,6,10,14…k=2,6,10,14…. Yes, if k+1≡2(mod 4)k+1≡2(mod 4) Alice takes 2 2 and leaves 0(mod 4)0(mod 4) for Bob. If you receive 0(mod 4)0(mod 4) you are losing. If you leave 0(mod 4)0(mod 4) you are winning.Ross Millikan –Ross Millikan 2016-12-03 21:08:09 +00:00 Commented Dec 3, 2016 at 21:08
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Uniform and Pointwise convergence Almost Everywhere
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In class our professor wants us to give an example of a function f f on an interval [a,b][a,b] and a sequence {f n f n} converging to f f almost uniformly so that there is no set E of measure zero so that the sequence {f n f n} converges uniformly to f f on [a,b]∖E[a,b]∖E.
I am a little confused about what he's asking; does this mean we want a function and sequence that does NOT converge uniformly almost everywhere to f f? Or do we want a function in which uniform convergence (not just uniform convergence almost everywhere) holds?
convergence-divergence
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edited Mar 16, 2016 at 2:07
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Your first paragraph seems completely clear to me.zhw. –zhw. 2016-03-16 01:03:10 +00:00 Commented Mar 16, 2016 at 1:03
If you want the symbol \ inside the dollar signs, you need \backslash. Or you can put \ outside the dollars.DanielWainfleet –DanielWainfleet 2016-03-16 02:10:54 +00:00 Commented Mar 16, 2016 at 2:10
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"almost uniformly" means that for any ϵ>0 ϵ>0, there exists a set E E, μ(E)<ϵ μ(E)<ϵ such that f n→f f n→f uniformly on E c E c. The name may be a little misleading. By Egorov's theorem, if μ(X)<∞μ(X)<∞, then a.s. convergence implies almost uniform convergence.
What you're asked to show that almost uniform convergence does not imply the stronger uniform convergence almost everywhere.
A simple example for this is the sequence f n=x n f n=x n on [0,1][0,1]. This converges pointwise to the function 0 0 on (0,1)(0,1) and to 1 1 at 0 0, that is, it converges a.e. to the constant function f=0 f=0. The sequence clearly converges almost uniformly to f f, because it converges uniforly to 0 0 on every interval of the form [0,1−ϵ][0,1−ϵ].
What about uniform convergence except on a set of measure zero ?
Let E E have zero measure. Observe that for any m m, there exists a point in the interval [1−1/m,1−1/(m+1)][1−1/m,1−1/(m+1)] not in E E (othwise E E would have positive measure). Therefore (using monotonicity of f n x n f n x n):
sup x∈[0,1]∖E|f n(x)−f(x)|≥(1−1/m)n−0,sup x∈[0,1]∖E|f n(x)−f(x)|≥(1−1/m)n−0,
for any m m. Letting m→∞m→∞, we see that
sup x∈[0,1]∖E|f n(x)−f(x)|≥1−0=1.sup x∈[0,1]∖E|f n(x)−f(x)|≥1−0=1.
(It is of course equal to 1 1).
Thus, no matter which measure zero set E E we choose, we will never converge uniformly to the (a.e.) limit f=0 f=0 outside E E.
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edited Oct 16, 2021 at 23:06
answered Mar 16, 2016 at 1:27
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I think you got a typo, it should be f n→f f n→f uniformly on E c E c.RJ Acuña –RJ Acuña 2021-10-13 07:11:58 +00:00 Commented Oct 13, 2021 at 7:11
1 Corrected. Thanks.Fnacool –Fnacool 2021-10-16 23:07:08 +00:00 Commented Oct 16, 2021 at 23:07
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You might want to look up almost uniform convergence. It sounds like you want a function f f and a sequence of functions {f n}{f n} for which:
Given any δ>0 δ>0, there exists a set A A, m(A)<δ m(A)<δ, such that f n→f f n→f uniformly on [a,b]∖A[a,b]∖A
Given any measure 0 0 set E E, the functions f n f n do not converge uniformly on [a,b]∖E[a,b]∖E.
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answered Mar 16, 2016 at 1:16
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1 Yes, I think that is what the professor asks Friedrich Philipp –Friedrich Philipp 2016-03-16 01:18:54 +00:00 Commented Mar 16, 2016 at 1:18
Ah thanks! That makes a lot more sense.Min –Min 2016-03-16 01:24:05 +00:00 Commented Mar 16, 2016 at 1:24
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Zarankiewicz’s problem for semilinear hypergraphs
8 September, 2020 in math.CO, math.LO, paper | Tags: Abdul Basit, Artem Chernikov, Chieu-Minh Tran, dyadic models, hypergraphs, incidence geometry, Sergei Starchenko, Szemeredi-Trotter theorem | by Terence Tao | 12 comments
Abdul Basit, Artem Chernikov, Sergei Starchenko, Chiu-Minh Tran and I have uploaded to the arXiv our paper Zarankiewicz’s problem for semilinear hypergraphs. This paper is in the spirit of a number of results in extremal graph theory in which the bounds for various graph-theoretic problems or results can be greatly improved if one makes some additional hypotheses regarding the structure of the graph, for instance by requiring that the graph be “definable” with respect to some theory with good model-theoretic properties.
A basic motivating example is the question of counting the number of incidences between points and lines (or between points and other geometric objects). Suppose one has points and lines in a space. How many incidences can there be between these points and lines? The utterly trivial bound is , but by using the basic fact that two points determine a line (or two lines intersect in at most one point), a simple application of Cauchy-Schwarz improves this bound to . In graph theoretic terms, the point is that the bipartite incidence graph between points and lines does not contain a copy of (there does not exist two points and two lines that are all incident to each other). Without any other further hypotheses, this bound is basically sharp: consider for instance the collection of points and lines in a finite plane , that has incidences (one can make the situation more symmetric by working with a projective plane rather than an affine plane). If however one considers lines in the real plane , the famous Szemerédi-Trotter theorem improves the incidence bound further from to . Thus the incidence graph between real points and lines contains more structure than merely the absence of .
More generally, bounding on the size of bipartite graphs (or multipartite hypergraphs) not containing a copy of some complete bipartite subgraph (or in the hypergraph case) is known as Zarankiewicz’s problem. We have results for all and all orders of hypergraph, but for sake of this post I will focus on the bipartite case.
In our paper we improve the bound to a near-linear bound in the case that the incidence graph is “semilinear”. A model case occurs when one considers incidences between points and axis-parallel rectangles in the plane. Now the condition is not automatic (it is of course possible for two distinct points to both lie in two distinct rectangles), so we impose this condition by fiat:
Theorem 1Suppose one has points and axis-parallel rectangles in the plane, whose incidence graph contains no ‘s, for some large .
(i) The total number of incidences is .
(ii) If all the rectangles are dyadic, the bound can be improved to .
(iii) The bound in (ii) is best possible (up to the choice of implied constant).
We don’t know whether the bound in (i) is similarly tight for non-dyadic boxes; the usual tricks for reducing the non-dyadic case to the dyadic case strangely fail to apply here. One can generalise to higher dimensions, replacing rectangles by polytopes with faces in some fixed finite set of orientations, at the cost of adding several more logarithmic factors; also, one can replace the reals by other ordered division rings, and replace polytopes by other sets of bounded “semilinear descriptive complexity”, e.g., unions of boundedly many polytopes, or which are cut out by boundedly many functions that enjoy coordinatewise monotonicity properties. For certain specific graphs we can remove the logarithmic factors entirely. We refer to the preprint for precise details.
The proof techniques are combinatorial. The proof of (i) relies primarily on the order structure of to implement a “divide and conquer” strategy in which one can efficiently control incidences between points and rectangles by incidences between approximately points and boxes. For (ii) there is additional order-theoretic structure one can work with: first there is an easy pruning device to reduce to the case when no rectangle is completely contained inside another, and then one can impose the “tile partial order” in which one dyadic rectangle is less than another if and . The point is that this order is “locally linear” in the sense that for any two dyadic rectangles , the set is linearly ordered, and this can be exploited by elementary double counting arguments to obtain a bound which eventually becomes after optimising certain parameters in the argument. The proof also suggests how to construct the counterexample in (iii), which is achieved by an elementary iterative construction.
An incidence theorem in higher dimensions
17 March, 2011 in math.AG, math.CO, paper | Tags: Ham sandwich theorem, incidence geometry, Jozsef Solymosi, Szemeredi-Trotter theorem | by Terence Tao | 11 comments
Jozsef Solymosi and I have just uploaded to the arXiv our paper “An incidence theorem in higher dimensions“, submitted to Discrete and Computational Geometry. In this paper we use the polynomial Ham Sandwich method of Guth and Katz (as discussedpreviouslyonthis blog) to establish higher-dimensional versions of the Szemerédi-Trotter theorem, albeit at the cost of an epsilon loss in exponents.
Recall that the Szemerédi-Trotter theorem asserts that given any finite set of points and lines in the plane , the number of incidences has the upper bound
Apart from the constant factor, this bound is sharp. As discussed in this previous blog post, this theorem can be proven by the polynomial method, and the strategy can be rapidly summarised as follows. Select a parameter . By the polynomial Ham Sandwich theorem, one can divide into cell interiors, each with points, and incident to lines on the average, plus a boundary set which is an algebraic curve of degree . To handle the contribution of each cell interior, one uses a more elementary incidence bound (such as the bound coming from the fact that two points determine at most one line); to handle the contribution on the cell boundary, one uses algebraic geometry tools such as Bezout’s theorem. One then combines all the bounds and optimises in to obtain the result.
As a general rule, the contribution of the cell interiors becomes easier to handle as increases, while the contribution of the cell boundaries become easier as decreases. As such, the optimal value of is often an intermediate one (in the case of Szemerédi-Trotter, the choice is typical). Among other things, this requires some control of moderately high degree algebraic sets, though in this planar case , one only needs to control algebraic curves in the plane, which are very well understood.
In higher dimensions, though, the complexity of the algebraic geometry required to control medium degree algebraic sets increases sharply; compare for instance the algebraic geometry of ruled surfaces appearing in the three-dimensional work of Guth and Katzas discussed here, compared with the algebraic geometry of curves in the two-dimensional Szemerédi-Trotter theorem discussed here.
However, Jozsef and I discovered that it is also possible to use the polynomial method with a non-optimised value of , and in particular with a bounded value of , which makes the algebraic geometry treatment of the boundary significantly easier. The drawback to this is that the cell interiors can no longer be adequately controlled by trivial incidence estimates. However, if instead one controls the cell interiors by an induction hypothesis, then it turns out that in many cases one can recover a good estimate. For instance, let us consider the two-dimensional Szemerédi-Trotter theorem in the most interesting regime, namely when the term on the RHS is dominant. If we perform the cell decomposition with some small parameter , then we obtain cells with points and lines on the average; applying the Szemerédi-Trotter theorem inductively to each of these cells, we end up with a total contribution of
for the cell interiors, and so provided that one can also control incidences on the (low degree) cell boundary, we see that we have closed the induction (up to changes in the implied constants in the notation).
Unfortunately, as is well known, the fact that the implied constants in the notation degrade when we do this prevents this induction argument from being rigorous. However, it turns out this method does work nicely to give the weaker incidence bound
for any fixed ; the point being that this extra epsilon of room in the exponents means that the induction hypothesis gains a factor of or so over the desired conclusion, allowing one to close the induction with no loss of implied constants if all the parameters are chosen properly. While this bound is weaker than the full Szemerédi-Trotter theorem, the argument is simpler in the sense that one only needs to understand bounded degree algebraic sets, rather than medium degree algebraic sets. As such, this argument is easier to extend to higher dimensions.
Indeed, as a consequence of this strategy (and also a generic projection argument to cut down the ambient dimension, as used in this previous paper with Ellenberg and Oberlin, and an induction on dimension), we obtain in our paper a higher-dimensional incidence theorem:
Theorem 1 (Higher-dimensional Szemerédi-Trotter theorem) Let , and let and be finite sets of points and -planes in , such that any two -planes in intersect in at most one point. Then we have for any .
Something like the transversality hypothesis that any two -planes in intersect in at most one point is necessary, as can be seen by considering the example in which the points are all collinear, and the -planes in all contain the common line. As one particular consequence of this result, we recover (except for an epsilon loss in exponents) the complex Szemerédi-Trotter theorem of Toth, whose proof was significantly more complicated; it also gives some matrix and quaternionic analogues, in particular giving new sum-product theorems in these rings. Note that the condition is natural, because when the ambient dimension is less than , it is not possible for two -planes to intersect in just one point. Using the generic projection trick, one can easily reduce to the critical case .
Actually, for inductive reasons, we prove a more general result than Theorem 1, in which the -planes are replaced instead by -dimensional real algebraic varieties. The transversality condition is now that whenever a point is incident to two such varieties , that the tangent cones of and at only intersect at . (Also for technical reasons, it is convenient to consider a partial subset of incidences in , rather than the full set of incidences. Indeed, it seems most natural to consider the triplet
as a single diagram, rather than to consider just the sets and .)
The reason for working in this greater level of generality is that it becomes much easier to use an induction hypothesis to deal with the cell boundary; one simply intersects each variety in with the cell boundary, which usually lowers the dimension of the variety by one. (There are some exceptional cases in which is completely trapped inside the boundary, but due to the transversality hypothesis, this cannot contribute many incidences if the ambient dimension is (so the cell boundary is only dimensional).)
As one application of this more general incidence result, we almost extend a classic result of Spencer, Szemerédi, and Trotter asserting that points in the plane determine unit distances from to , at the cost of degrading to .
Recent progress on the Kakeya conjecture
11 May, 2009 in math.AG, math.AP, math.AT, math.CO, talk, travel | Tags: additive combinatorics, heat flow, incidence geometry, Kakeya conjecture, polynomial method | by Terence Tao | 34 comments
Below the fold is a version of my talk “Recent progress on the Kakeya conjecture” that I gave at the Fefferman conference.
Read the rest of this entry »
The crossing number inequality
18 September, 2007 in expository, math.CO | Tags: additive combinatorics, amplification, crossing numbers, graph theory, incidence geometry, sum-product estimates | by Terence Tao | 58 comments
Today I’d like to discuss a beautiful inequality in graph theory, namely the crossing number inequality. This inequality gives a useful bound on how far a given graph is from being planar, and has a number of applications, for instance to sum-product estimates. Its proof is also an excellent example of the amplification trick in action; here the main source of amplification is the freedom to pass to subobjects, which is a freedom which I didn’t touch upon in the previous post on amplification. The crossing number inequality (and its proof) is well known among graph theorists but perhaps not among the wider mathematical community, so I thought I would publicise it here.
In this post, when I talk about a graph, I mean an abstract collection of vertices V, together with some abstract edges E joining pairs of vertices together. We will assume that the graph is undirected (the edges do not have a preferred orientation), loop-free (an edge cannot begin and start at the same vertex), and multiplicity-free (any pair of vertices is joined by at most one edge). More formally, we can model all this by viewing E as a subset of , the set of 2-element subsets of V, and we view the graph G as an ordered pair G = (V,E). (The notation is set up so that .)
Now one of the great features of graphs, as opposed to some other abstract maths concepts, is that they are easy to draw: the abstract vertices become dots on a plane, while the edges become line segments or curves connecting these dots. [To avoid some technicalities we do not allow these curves to pass through the dots, except if the curve is terminating at that dot.] Let us informally refer to such a concrete representation D of a graph G as a drawing of that graph. Clearly, any non-trivial graph is going to have an infinite number of possible drawings. In some of these drawings, a pair of edges might cross each other; in other drawings, all edges might be disjoint (except of course at the vertices, where edges with a common endpoint are obliged to meet). If G has a drawing D of the latter type, we say that the graph G is planar.
Given an abstract graph G, or a drawing thereof, it is not always obvious as to whether that graph is planar; just because the drawing that you currently possess of G contains crossings, does not necessarily mean that all drawings of G do. The wonderful little web game “Planarity” illustrates this point excellently. Nevertheless, there are definitely graphs which are not planar; in particular the complete graph on five vertices, and the complete bipartite graph on two sets of three vertices, are non-planar.
There is in fact a famous theorem of Kuratowski that says that these two graphs are the only “source” of non-planarity, in the sense that any non-planar graph contains (a subdivision of) one of these graphs as a subgraph. (There is of course the even more famous four-colour theorem that asserts that every planar graphs is four-colourable, but this is not the topic of my post today.)
Intuitively, if we fix the number of vertices |V|, and increase the number of edges |E|, then the graph should become “increasingly non-planar”; conversely, if we keep the same number of edges |E| but spread them amongst a greater number of vertices |V|, then the graph should become “increasingly planar”. Is there a quantitative way to measure the “non-planarity” of a graph, and to formalise the above intuition as some sort of inequality?
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1035 | https://opencurriculum.org/6492/8g-g-co-origami-silver-rectangle/ | 8.G, G-CO Origami Silver Rectangle ‹ OpenCurriculum
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8.G, G-CO Origami Silver Rectangle
Illustrative Mathematics's files
This task examines the mathematics behind an origami construction of a rectangle whose sides have the ratio (2–√:1)(2:1). Such a rectangle is called a silver rectangle.
Beginning with a square piece of paper, first fold and unfold it leaving the diagonal crease as shown here:
Next fold the bottom right corner up to the diagonal:
After unfolding then fold the left hand side of the rectangle over to the crease from the previous fold:
Here is a picture, after the last step has been unfolded, with all folds shown and some important points marked. In the picture T T is the reflection of S S about ℓ ℓ.
Suppose s s is the side length of our square. Show that |P T|=s|P T|=s.
Show that △P Q T△P Q T is a 45-45-90 isosceles triangle.
Calculate |P Q||P Q| and conclude that P Q R S P Q R S is a silver rectangle.
Commentary
The purpose of this task is to apply geometry in order analyze the shape of a rectangle obtained by folding paper. The central geometric ideas involved are reflections (used to model the paper folds), analysis of angles in triangles, and the Pythagorean Theorem. The task is appropriate either at the 8th grade level or in high school: the only difference would be the level of rigor expected in the explanation. The solution given is appropriate for either level.
The silver rectangle is one of three rectangles identified in ancient times as having important properties: the bronze rectangle has a side ratio of (3–√,1)(3,1) and the golden rectangle has a side ratio of (1+5–√:2)(1+5:2). Each of these three rectangles can be constructed by folding paper. Further properties of silver rectangles are examined in www.illustrativemathematics.org/illustrations/1489. Much more interesting information about the silver rectangle can be found here:
This task could be given in a much more open ended form, particularly if done in high school. Teachers could go through the different steps of the construction and then prompt students to examine the ratio of side lengths in the final rectangle. There are many possible solution paths revolving around the different triangles in the picture that are similar: half of the original square, △T Q P△T Q P, and the small triangle in the lower left of the last picture. These are all 45-45-90 triangles and the dimensions of the rectangle can be found by calculating the scale factor between any pair of these triangles.
At least two of the standards of mathematical practice are relevant for this task. First, students will ''Use Appropriate Tools Strategically'' (MP5) if they are given paper squares to fold and experiment with before writing down a formal argument. Students will also ''Reason Abstractly and Quantitatively'' (MP2) as they need to represent the paper folding geometrically in order to calculate angles and side lengths.
Solutions
Solution: 1
Since reflection about ℓ ℓ maps P P to itself and maps S S to T T this means that segments P S¯¯¯¯¯¯¯P S¯ and P T¯¯¯¯¯¯¯P T¯ are interchanged by reflection about ℓ ℓ. Hence
|P S|=|P T||P S|=|P T| because reflections preserve lengths of segments. We are given that |P S|=s|P S|=s since it is one side of the square so |P T|=s|P T|=s.
2. We know that m(∠Q P T)=45 m(∠Q P T)=45 because the diagonal of a square bisects the 90 degree angles. Alternatively, reflection over the diagonal P T←→P T↔ is a symmetry of the square: this reflection interchanges angles Q P T Q P T and S P T S P T and so these must each measure 45 degrees. Angle T Q P T Q P is a right angle because Q R¯¯¯¯¯¯¯¯Q R¯ is a crease resulting from a horizontal fold of the paper. Since
m(∠Q T P)+m(∠Q P T)+m(∠T Q P)=180 m(∠Q T P)+m(∠Q P T)+m(∠T Q P)=180 this means that m(∠Q T P)=45 m(∠Q T P)=45. Thus △P Q T△P Q T is a 45-45-90 triangle: it is isosceles because m(∠Q T P)=m(∠Q P T)m(∠Q T P)=m(∠Q P T).
3. To find |P Q||P Q| we use the fact that △P Q T△P Q T is a right isosceles triangle. Thus we know that |T Q|=|P Q||T Q|=|P Q| and, from the Pythagorean Theorem,
|P Q|2+|T Q|2=|P T|2.|P Q|2+|T Q|2=|P T|2. Substituting |P Q||P Q| for |T Q||T Q| this is equivalent to 2|P Q|2=|P T|2.2|P Q|2=|P T|2. From part (a) this means that |P Q|2=s 2 2.|P Q|2=s 2 2. So |P Q|=s 2–√.|P Q|=s 2.
We can now check that P Q R S P Q R S is a silver rectangle. Our calculations show that |P SP Q||P SP Q| is s:s 2√s:s 2. But the ratio s:s 2√s:s 2 is equivalent to the ratio 2–√:1 2:1 as we can see using a scaling factor of 2√s 2 s.
G-CO Origami Silver Rectangle by illustrativemathematics, used under CC-BY-NC-SA 3.0
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1036 | https://www.engineeringtoolbox.com/specific-speed-pump-fan-d_637.html | Pumps - Specific Speed
Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
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Pumps - Specific Speed
Characterizing of impeller types in pumps in a unique and coherent manner.
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Specific Speed is a number characterizing the type of impeller in pump in a unique and coherent manner. Specific speed is determined independent of the pump size and can be useful when comparing different pump designs.
Specific Speed identifies the geometrically similarity of pumps
Specific speed can be expressed as
_N s = n q 1/2 / h 3/4 (1)_
_where_
_N s = specific speed_
_n = pump shaft rotational speed (rpm)_
_q = flow rate (m 3/h, l/s, l/min, m 3/min, U.S. gpm, British gpm) at Best Efficiency Point (BEP)_
_h = head rise (m, ft)_
Online Specific Speed Pump Calculator
The calculator below can used to calculate the specific speed:
_n_ _- pump shaft rotational speed (rpm)_
_q - flow rate (m 3/h, l/s, l/min, m 3/min, U.S. gpm, British gpm)_
_h = head rise (m, ft)_
Note! When comparing pumps and their documentation - be aware of the units used.
Typical values for specific speed - _N s_ - for different designs in U.S. units _(US gpm, ft)_
radial flow pump design - _500 < N s< 4000_ - typical for centrifugal impeller pumps with radial vanes - double and single suction. Francis vane impellers in the upper range
mixed flow pump design -_2000 < N s< 8000_ - more typical for mixed impeller single suction pumps
axial flow pump design - _7000 < N s< 20000_- typical for propellers and axial fans
Convert Specific Speed between U.S. units _(gpm)_ and Metric units _(m 3/h)_
_N s(US gpm, ft) = 0.861 N s(m 3/h, m)_
_N s(British gpm, ft) = 0.787 N s(m 3/h, m)_
when
_1 U.S. gpm = 0.2271 m 3/h_
_1 British gpm = 0.2728 m 3/h_
_1 ft = 0.3048 m_
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Example - Specific Speed for a Pump with different units
A pump has the capacity _1500 U.S. gal/min (1249 British gpm, 340 m 3/h, 5667 liters/min)_ at _100 ft(30.5 m)_ of head and is rotating at _1760 rev/min_.
US gpm, ft
_N s(US gpm, ft) = (1760 rev/min) (1500 gal/min)1/2 / (100 ft)3/4_
_= 2156_
British gpm, ft
_N s(British gpm, ft) = (1760 rev/min) (1249 gal/min)1/2 / (100 ft)3/4_
_= 1967_
m 3/h, m
_N s(m3/h, m) = (1760 rev/min) (340 m 3/h)1/2 / (30.5 m)3/4_
_= 2500_
liter/min, m
_N s(liters/min, m) = (1760 rev/min) (5667 liters/min)1/2 / (30.5 m)3/4_
_= 10209_
Specific Speed Hydro-power Turbines
Specific speed is also used when evaluating hydro-power turbines. Typical range of turbine specific speeds:
Francis turbine : _70 - 500_
Kaplan or propeller turbine : _100 - 350_
Pelton, Turgo or Cross-flow turbines : _10 - 80_
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Given two hourglasses of 4 minutes and 7 minutes, the task is to measure 9 minutes.
Check if you were right - full answer with solution below.
Solution:
1. At 0 minutes: Start both hourglasses at the same time.
2. At 4 minutes: 4 minutes hourglass runs out and flip it. 7 minutes hourglass is left with 3 minutes.
3. At 7 minutes: 4 minutes hourglass is left with 1 minute. 7 minutes hourglass runs out and flip it.
4. At 8 minutes: 4 minutes hourglass runs out and 7 is filled with 6 minutes and 1 minute on the other side. Flip it as the sand is left with 1 minute.
5. At 9 minutes: 7 minutes hourglass becomes empty from above side.
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1038 | https://mathoverflow.net/questions/259152/efficient-method-to-write-number-as-a-sum-of-four-squares | Skip to main content
Efficient method to write number as a sum of four squares?
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Wikipedia states that there randomized polynomial-time algorithms for writing n as a sum of four squares
n=x21+x22+x23+x24
in expected running time O(log2n).
My question is can someone give the efficient algorithm( O(log2n) ) to represent n as sum of four squares.
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asked Jan 9, 2017 at 10:48
nequitnequit
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Not only does Wikipedia state this, but it gives a precise reference for where you can read about it.
– Kevin Buzzard
Commented
Jan 9, 2017 at 10:50
1
...although my reading of the paper is that the O(log2n) method they give assumes ERH, and the other method (which doesn't) runs in time O(log2n.loglog(n)) (so it seems to me that the statement in Wikipedia is slightly inaccurate, unless one argues that the truth of ERH is "expected" :-) )
– Kevin Buzzard
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Jan 9, 2017 at 10:59
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If you have access to a library, you can see whether they can get it for you by interlibrary loan.
– Gerry Myerson
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Negative attention to this question and its answers may be explained by the fact that potentially this is related to the current CodeChef challenge. Your browser settings permitting you can follow the links here and judge yourself.
– Jyrki Lahtonen
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Jan 10, 2017 at 22:10
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Nothing, nequit. I came here because some helpful users at another related Math.SE question (trackable to that on-going competition) referred the asker here. Not accusing you of anything. It is just that the timing was a bit unfortunate. I apologize for implying anything else.
– Jyrki Lahtonen
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One of their methods for n=4k+2 is as follows:
Randomly select an even number a and an odd number b such that a2+b2<n. Then, we hope p=n−a2−b2 is a prime (you can show there's about a 1/(Alognloglogn) chance of p being prime ); p is of the form 4r+1, so if p is prime there's a solution to c2+d2=p.
To find that, we try to solve m2+1≡0(modp); I'm actually going to describe a slightly different method. Select x at random from 1 to p−1; then, x2r=±1 depending on whether x is a quadratic residue so calculate xr by repeated squaring, and with a 1/2 chance if p is prime (a smaller one if p is composite) you'll find a valid m.
Given such an m, m+i is a Gaussian integer with norm divisible by p but smaller than p2; use the Euclidean algorithm on the Gaussian integers with p and get c+di with norm p, and a2+b2+c2+d2=n.
For an odd number n, solve for a2+b2+c2+d2=2n; note that by mod 4 considerations exactly two of a,b,c,d must be odd, and without loss of generality assume a,b are odd and c,d are even. Then,(12(a+b))2+(12(a−b))2+(12(c+d))2+(12(c−d))2=n.
For n a multiple of 4, solve for n/4 recursively, and multiply all values by 4.
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answered May 26, 2017 at 23:15
W. Cadegan-SchlieperW. Cadegan-Schlieper
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This seems to be addressed in this paper by Bumby.
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answered Jan 9, 2017 at 12:44
Igor RivinIgor Rivin
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Is he related to Grothendieck's ex-wife?
– Sylvain JULIEN
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May 26, 2017 at 17:20
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That seems to rely on factorization, which I would not call efficient.
– W. Cadegan-Schlieper
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May 26, 2017 at 22:47
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@SylvainJULIEN By marriage, yes.
– Igor Rivin
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Why the downvote?
– Igor Rivin
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I am going to provide you with 4 links and I hope it will help you.
this one is an online calculator based on an algorithm from math overflow.
the math overflow algorithm can be found here.
and then there is also this algorithm from cs stackexchange
and if you want to do a bit of work, then there is this algorithm that can find the sum of 2 squares for a given integer. So you would have to divide your number in two ( not necessarily equal part ) and then add up the results to get a 4 square representation.
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edited May 23, 2017 at 12:37
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Why all the negative votes I wonder, for what look like more than adequate posts? Also, in view of Euler's quaternion norm product identity, doesn't finding multiple essentially distinct four square representations of an integer allow non-trivial factorizations of it? If so then finding four square forms can't be that easy or else factoring would be also! (Or maybe this readily works only for two square forms?)
– John R Ramsden
Commented
Jan 10, 2017 at 10:42
This answer also appears at cs.stackexchange.com/a/68536/755.
– D.W.
Commented
Jan 20, 2017 at 5:41
@JohnRRamsden I don't see how this answer addresses the issues of runtime which were explicitly mentioned in the original questoin
– Yemon Choi
Commented
Jan 23, 2017 at 3:17
2
@YemonChoi, good point. you could maybe tackle the question yourself and derive an algorithm that is efficient to answer the original question or review the different algorithms known to produce a 4-square representation of an integer and show us which one is the most efficient. we will all be grateful to you.
– user25406
Commented
Jan 23, 2017 at 15:04
John, it's only the counting problem that would provide a way to factor RSA numbers, and the efficient algorithms don't provide any way to get a count.
– W. Cadegan-Schlieper
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Dec 5, 2024 at 17:54
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Two conics have exactly one intersection point
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$\begingroup$
We have two conics $Q_1,Q_2$ on $\mathbb{P}_2$ over some algebraically closed field. Also $Q_1$ and $Q_2$ are supposed to be smooth.I've just discovered Bezout's theorem, which states that two algebraic curves with degrees $m$ and $k$ have exactly $km$ intersection points (of course some of them can be the same). So, I'd like to build some examples when two conics have exactly $1,2,3$ and $4$ distinct intersection points. All cases but not the $1$st are quite easy. For example, $q_1(x,y,z)=x^2+y^2-z^2$ and $q_2(x,y,x)=x^2+2y^2-z^2$ are two polynomials which define smooth conics on $\mathbb{P}_2$ and they have exactly $2$ intersection points $(1:0:1)$ and $(1:0:-1)$. When we need three or four points, there are similar examples. But what about $1$ point? I cannot build an example. Is it possible? I'd like to note that conics are smooth.
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asked Jan 8, 2014 at 21:08
user74574user74574
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The conics $x^2-yz=0$ and $x^2+z^2-yz=0$ are smooth (in all characteristics) and intersect only in one point, namely $[0:1:0]$.
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edited Jan 8, 2014 at 22:00
answered Jan 8, 2014 at 21:25
Georges ElencwajgGeorges Elencwajg
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$\endgroup$
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$\begingroup$ What is the method to finding the intersection? you follow the classic way? $\endgroup$
Bill
– Bill
2021-12-20 16:32:41 +00:00
Commented Dec 20, 2021 at 16:32
$\begingroup$ @Bill: For $z=1$ you must solve $x^2-y=0, x^2+1-y=0$ (no solution) and for $z=0$ you must solve $z=0, x^2=0$ whose only solution is $[0:1:0]$ . $\endgroup$
Georges Elencwajg
– Georges Elencwajg
2021-12-20 17:17:04 +00:00
Commented Dec 20, 2021 at 17:17
$\begingroup$ Thank you so much, I have two conics in $\mathbb{P]^2$, is there anyway to find the intersection of the conics by their symmetric matrices? $\endgroup$
Bill
– Bill
2021-12-20 17:24:05 +00:00
Commented Dec 20, 2021 at 17:24
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1
$\begingroup$
Take any two pairs of intersecting lines with different slope: $x^2+a_1y^2=0$ and $x^2+a_2y^2=0$ for $a_1 \neq a_2$. They only cross at the origin $(0:0:1)$
Also, take the double y axis $x^2=0$ and the double x axis $y^2=0$ (or any pair of double-lines with different slopes), which again only cross at the origin.
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answered Jan 8, 2014 at 21:37
rewrittenrewritten
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$\begingroup$ Dear rewritten, your conics are not smooth. $\endgroup$
Georges Elencwajg
– Georges Elencwajg
2014-01-08 21:40:21 +00:00
Commented Jan 8, 2014 at 21:40
$\begingroup$ You are completely right. $\endgroup$
rewritten
– rewritten
2014-01-08 21:41:05 +00:00
Commented Jan 8, 2014 at 21:41
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Home : Support : Online Help : System : Error Message Guide : wrong kind of parameters in piecewise
Error, (in PiecewiseTools:-Is) wrong kind of parameters in piecewise Description Examples Description You can express piecewise-continuous functions using the piecewise function. The piecewise function takes as its parameters a condition (or conditions) and an associated expression (or expressions), should the condition evaluate to true. For more information on the piecewise function, see the piecewisehelp page. The conditions must be an equality or inequality, or a Boolean combination of inequalities. The wrong kind of parameters in piecewise error occurs when any of the conditional arguments are not Boolean-valued (they do not evaluate to true or false). You can also encounter this error when using the 2-D math notation for a piecewise expression. Again, it indicates that the conditions are not Boolean-valued. See the second example. Examples Example 1: In this case, the unassigned x is not an equality, inequality, or some Boolean combination of inequalities. >piecewise x,y; Error, (in PiecewiseTools:-Is) wrong kind of parameters in piecewise Solution: Replace the first argument with a valid condition. One possible solution is to express the function as >piecewise x>0,y; {y 0b≔−x x<2 x 2 x+1 x>2 Error, (in PiecewiseTools:-Is) wrong kind of parameters in piecewise Solution: x=2. >b≔−x x<2 x x=2 x+1 x>2 b:={−x x<2 x x=2 x+1 2<x(2.2)See Also piecewise
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August 23, 2011|By:D. Jack Li
Aa)Aa)Aa)
Artificial Photosynthesis: Adapting Nature’s Energy Production for Our Own
During our generation and after, the issue of energy sustainability will be of paramount importance. To many scientists, the key to meeting surging energy demands amidst declining non-renewable resources and the deteriorating environment lies in learning from plants. Although these ubiquitous forms of life predate us by hundreds of millions of years, they have only recently been recognized as models of energy sustainability. Relying on photosynthesis, they require only water, carbon dioxide (CO2), and sunlight - all abundant resources - to produce their own carbohydrate fuel and oxygen. Adapting photosynthesis for our own use would represent the pinnacle of energy sustainability: We could produce carbon-based fuels using a waste product of fuel consumption, CO2, also a major contributor to the green house effect. The problem is that it is not feasible to harness natural photosynthesis, as plants convert only 3% of the solar energy they receive into sugars, which in turn have to be further processed for industrial use.1 We need to make photosynthesis better.
However, mimicking photosynthesis, let alone optimizing its energy conversion, is a daunting task. Such an ingenious natural process centers around an intricate network of protein complexes (called Photosystems II and I) that absorb photons to excite their electrons, which when transported by other proteins to lower energy levels (like boulders rolling down a hill), help power the enzymatic reactions involved in reducing CO2 into carbohydrates. At the same time, water molecules replenish the electrons that Photosystem II loses in the (electron-transfer) process, and as a result are "split" into oxygen gas and hydrogen ions. All this occurs in an exquisitely coordinated manner in spatially defined centers of the plant's chloroplast.
While the complexity of this system is apparent, the wholesale artificial recapitulation of photosynthesis is not necessary to harvest its benefits. Synthetic approaches that mimic specific steps can be turned to human use. Photosynthesis consists of three major phases: light absorption, water splitting, and the light independent conversion of CO2 into organic molecules.
CO2 Conversion Phase:
The last phase in photosynthesis is of particular interest. Manipulating how CO2 is fixed from the atmosphere can yield fuels like methane (CH4), methanol (CH3OH), and carbon monoxide (CO) for industrial use and energy production.2 Many current artificial systems for CO2 reduction employ photocatalysts like titanium dioxide (TiO2).2 Photocatalysts are materials that absorb light to speed up a chemical reaction. When TiO2 is enriched with other metal oxides, it can - like Photosystems I and II - absorb light in the visible range to excite its electrons. These high energy electrons in an aqueous environment create hydroxide (OH) radicals (molecules with unpaired electrons) and free electrons, which can go on to reduce CO2 into
organic fuels. The ratio of CH4, CH3OH, and CO yield depends on the proportion of H2O and CO2, the type of photocatalyst, and the reaction temperature.2
Light Absorption Phase:
In order to better capture solar energy, scientists have also tried to mimic plants' pigment-mediated sunlight absorption. Currently, solar cells manufactured from thin semiconducting material can convert sunlight into electricity, but at high material costs.3 Dye-sensitized solar cells (DSCs) may help reduce those costs while maintaining similar energy conversion efficiencies.3 DSCs consist of two thin sheets of conductors that sandwich a space filled with an organic electrolyte (e.g. I3-/I- pair). One of these sheets (transparent electrode) is also coated on the inside with a layer of titanium dioxide (TiO2) nanoparticles covered with dye molecules that mimic chlorophyll in plants. When exposed to light of the appropriate wavelength, electrons from the dye migrate through TiO2 to the conducting sheet, thereby creating an electric current. The organic electrolyte then donates its electrons to the excited dye, thus allowing for the dye to be continuously excited. The electrolyte is in turn replenished (bottom green arrow) by the electrons that travel from the dye-sensitized sheet to the opposite conducting sheet (counter electrode) via an external connection, thus creating a closed circuit. Through such a device, scientists are using biomimicry based on natural processes in plant leaves to create electricity from sunlight.
Water Splitting Phase:
Due to high thermodynamic requirements, the most difficult part of photosynthesis to mimic is using light to split water into its constituent elements and subsequently making them into gaseous O2 and H2. O2 and H2 gases are useful due to the massive energy release that occurs when they react to form H2O, a process that can be made to transpire in a controlled fashion inside oxygen-hydrogen fuel cells. These fuel cells can convert the heat evolved from the chemical reaction into electricity. Scientists have coupled metal catalysts with DSCs in tandem cells to artificially induce water splitting.3 To obtain O2, sunlight is allowed to hit a thin surface of semiconductor like tungsten trioxide (WO3) or hematite (rust, Fe2O3) riddled with small pores that help excited electrons escape recombination with the electron-deficient semiconductor atoms. The electron-deficient semiconductor atoms then become strong enough oxidizers to evolve O2 gas from H2O. Wired to this cell is a DSC providing an electric current to a platinum (Pt) catalyst in the gas production cell. The electric current from DSC provides the extra energetic push required to decompose H2O into H2 gas. Since O2 and H2 are produced at different sites, they can be collected separately and later reacted to yield energy.
Although these various plant photosynthesis biomimicry methods have been proven successful, one major challenge is scaling up these artificial processes to sustain our daily energy demands. Recently, a team at MIT has made progress in doing so by making an artificial "leaf" composed of silicon, electronics, and catalysts (image to the right).4 The size of a poker card, this leaf splits H2O into O2 and H2 for fuel cell storage. When exposed to sunlight and immersed in a gallon of water, this leaf could lead to the production of enough electricity to power a house in a developing country for an entire day.4 The next step is finding an efficient way to collect O2 and H2 for electricity production. Given all the current advances in synthetic photosynthesis, it is not hard to imagine a future where we can put to use one of the most important lessons nature can teach us - how to live sustainably for generations to come.
References:
Sanderson, K. Chemistry: the photon trap. Nature 452, 400-402 (2008).
Dey, G.R. Chemical Reduction of Co2 to Different Products during Photo Catalytic Reaction on TiO2 under Diverse Conditions: an Overview. Journal of Natural Gas Chemistry 16, 217-226 (2007).
Kalyanasundaram, K., & Graetzel, M. Artificial photosynthesis: biomimetic approaches to solar energy conversion and storage. Current Opinion in Biotechnology 21, 298-310 (2010).
American Chemical Society. Debut of the first practical "artificial leaf". (press release, 2011).
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1042 | https://www.reddit.com/r/HomeworkHelp/comments/l3mwh0/college_statsprob_urn_problem/ | [College Stats/Prob: Urn Problem] : r/HomeworkHelp
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jhould007
[College Stats/Prob: Urn Problem]
Further Mathematics
"An urn contains 10 chips, of which 5 are blue and 5 are red. Randomly select 5 chips, one at a time without replacement. Let X be the absolute difference between the numbers of blue and red chips that have been selected."
a) Find the pmf of X. Show your work.
b) What value of X is most likely?
So here's what I've been thinking so far:
Since 5 chips are being selected without replacement (and order doesn't appear to matter for this problem since we're only concerned with the relative numbers of red and blue chips), the total number of outcomes appears to be 10C5 = 252.
If 5 chips are selected, then these are the possibilities: 3 of one color and 2 of the other, 4 of one color and 1 of the other, or 5 of one color and 0 of the other.
The probability of X=0 would be 0, since if an odd number of chips is being selected, it's impossible to have an even difference between the number of red and blue chips selected.
Probability of X=1 is 2(5C3/252), since we're trying to find how many ways there could be three of the same color in the selection of 5, dividing that by 252 (the total number of possible outcomes) to find the probability, and multiplying the whole thing by 2 because you could either have 3 blue and 2 red, or 3 red and 2 blue, and X would still be equal to 1. I ended up getting 0.08.
P(X=2) = 0.
P(X=3): P(4 blue and 1 red) + P(4 red and 1 blue) = 2(5C4/252) = 10/252 = 0.04.
P(X=4) = 0.
P(X=5) = P(5 blue and 0 red) + P(5 red and 0 blue) = 2(5C5/252) = 2/252 = 0.008.
Because X has to be between 0 and 5 (inclusive), the sum of the probabilities must equal 1, but obviously (0.08 + 0.04 + 0.008) < 1. I'm not sure what I'm doing wrong here, but if someone could point me in the right direction I'd greatly appreciate it.
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14 1/21/08
14.1 Euclidean algorithm
15 1/8/08-1/20/08
16 1/7/08
16.1 Math books
17 1/6/08
17.1 Limit
18 1/5/08
18.1 American Mathematics Competitions
19 1/4/08
19.1 Physics
20 1/3/08
20.1 United States of America Mathematical Olympiad
21 12/13/07 - 1/2/08
22 12/22/07
22.1 Leonhard Euler
23 12/21/07
23.1 Zermelo-Fraenkel Axioms
24 12/20/07
24.1 Joining an ARML team
25 12/19/07
25.1 Zorn's Lemma
26 12/18/07
27 12/17/07
27.1 Diophantine equation
28 12/16/07
28.1 Fibonacci sequence
29 12/15/07
29.1 Cauchy-Schwarz inequality
30 12/14/07
30.1 Rearrangement inequality
31 12/13/07
32 12/12/07
32.1 Trigonometric identities
33 12/11/07
33.1 MATHCOUNTS
34 12/10/07
34.1 Polynomial
35 12/9/07
35.1 Pascal's identity
36 12/8/07
36.1 Canada/USA Mathcamp
37 12/7/07
37.1 Prime number
38 12/6/07
38.1 Calculus
39 12/5/07
39.1 Pi
40 12/4/07
40.1 Isaac Newton
41 12/3/07
41.1 Logarithm
42 12/2/07
42.1 American Invitational Mathematics Examination
43 12/1/07
43.1 Inequality
44 11/30/07
44.1 Asymptote (geometry)
45 11/27/07
45.1 Complex number
46 11/25/07
46.1 Set
2/3/08 - 2/29/08
AotD will be back in March as Article of the Week!
2/2/08
Law of Tangents
The Law of Tangents is a useful trigonometric identity that, along with the law of sines and law of cosines, can be used to determine angles in a triangle. Note that the law of tangents usually cannot determine sides, since only angles are involved in its statement.... [more]
2/1/08
Law of Sines
The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function... [more]
1/31/08
Law of Cosines
This article was also a AoPSWiki word of the week
The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. It can be derived in several different ways, the most common of which are listed in the "proofs" section ... [more]
1/30/08
Euler's totient function
Euler's totient function applied to a positive integer is defined to be the number of positive integers less than or equal to that are relatively prime to . is read "phi of n."... [more]
1/29/08
There was no AotD for January 1st, 2008.
1/28/08
Expected value
This article was also a AoPSWiki word of the week
Given an event with a variety of different possible outcomes, the expected value is what one should expect to be the average outcome if the event were to be repeated many times. Note that this is not the same as the "most likely outcome."
More formally, we can define expected value as follows: if we have an event whose outcomes have a discreteprobability distribution, the... [more]
1/27/08
Introductory modular arithmetic
Modular arithmetic is a special type of arithmetic that involves only integers. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using... [more]
1/26/08
Rational approximation of famous numbers
Rational approximation is the application of Dirichlet's theorem which shows that, for each irrational number , the inequality has infinitely many solutions. On the other hand, sometimes it is useful to know that cannot be approximated by rationals too well, or, more precisely, that is not a Liouvillian number, i.e., that for some power , the inequality [more]
1/25/08
Power set
The power set of a given set is the set of all subsets of that set This is denoted, other than by the common $\math{P}(S)$ (Error compiling LaTeX. Unknown error_msg), by (which has to do with the number of elements in the power set of a... [more]
1/24/08
Function
A function is a rule that maps one set of values to another set of values. For instance, one function may map 1 to 1, 2 to 4, 3 to 9, 4 to 16, and so on. This function has the rule that it takes its input value, and squares it to get an output value. One can... [more]
1/23/08
There was no AotD for January 23rd, 2008.
1/22/08
Permutation
A permutation of a set of objects is any rearrangement (linear ordering) of the objects. There are (the factorial of ) permutations of a set with distinct objects.
One can also consider permutations of infinite sets. In this case, a permutation of a set is simply a bijection between and... [more]
1/21/08
Euclidean algorithm
This article was also a AoPSWiki word of the week
The Euclidean algorithm (also known as the Euclidean division algorithm or Euclid's algorithm) is an algorithm that finds the greatest common divisor (GCD) of two elements of a Euclidean domain, the most common of which is the nonnegativeintegers, without factoring... [more]
1/8/08-1/20/08
On vacation.
1/7/08
Math books
These math books are recommended by Art of Problem Solving administrators and members of the index.php AoPS-MathLinks Community.
Levels of reading and math ability are loosely defined as follows:
Elementary is for elementary school students up through possibly early middle school.
Getting Started is recommended for students grades 6 to 9.
Intermediate is recommended for students grades 9 to 12.
Olympiad is recommended for high school students... [more]
1/6/08
Limit
For a realfunction and some value , (said, "the limit of at as goes to ) equals iff for every there exists a such that if , then ... [more]
1/5/08
American Mathematics Competitions
The American Mathematics Competitions (AMC) consist of a series of increasingly difficult tests for students in middle school and high school. The AMC sets the standard in the United States for talented high school students of mathematics. The AMC curriculum is both comprehensive and modern. AMC exams are so well designed that some top universities such as MIT now ask students for their AMC scores. "AMC" is also used as an abbreviation for American Math Contest, used to refer to the AMC 8, AMC 10, and AMC 12... [more]
1/4/08
Physics
The study of energy is known as physics. Everything concerning energy in some form or the other is covered by physics.
Physics as was known till the end of the nineteenth century is known now as classical physics. It is broadly classified into the following branches:
Mechanics
Acoustics
Optics
Thermodynamics
Electromagnetism
[more]
1/3/08
United States of America Mathematical Olympiad
The United States of America Mathematical Olympiad (USAMO) is the third test in a series of exams used to challenge bright students on the path toward choosing the team that represents the United States at the International Mathematics Olympiad (IMO).
The USAMO is administered by the American Mathematics Competitions (AMC). Art of Problem Solving (AoPS) is a proud sponsor of the AMC and of the recent expansion of USAMO participants from around 250 to around 400. [more]
12/13/07 - 1/2/08
AotD is on vacation.
12/22/07
Leonhard Euler
Leonhard Euler (1707-1735, pronounced Oiler) was a famous Swiss mathematician and physicist. He made numerous contributions to many fields of mathematics and science. Euler is often considered to be one of the greatest mathematicians of all time, along with Isaac Newton, Archimedes, and Carl Friedrich Gauss.
Euler was born on April 15, 1707 in Basel, Switzerland. Euler's parents were Paul Euler, a pastor of the Reformed Church, and Marguerite Brucker, a pastor's
12/21/07
Zermelo-Fraenkel Axioms
The Zermelo-Fraenkel Axioms are a set of axioms that compiled by Ernst Zermelo and Abraham Fraenkel that make it very convenient for set theorists to determine whether a given collection of objects with a given property describable by the language of set theory could be called a set. As shown by paradoxes such as Russell's Paradox, some restrictions must be put on which collections to call sets.
This axiom establishes the... [more]
12/20/07
Joining an ARML team
Team selection for the American Regions Mathematics League varies from team to team.
Florida ARML sends two teams to ARML each year. The selection criteria for the Florida ARML team takes into consideration several factors:
AMC and AIME performance
Past AMC, AIME, and USAMO scores
Past ARML performance
FAMAT-designated competitions
An annual statewide tryout test
The organizers are Jason Wiggins of... [more]
12/19/07
Zorn's Lemma
Zorn's Lemma is a set theoretic result which is equivalent to the Axiom of Choice.
Let be a partially ordered set.
We say that is inductively ordered if every totally orderedsubset of has an upper bound, i.e., an element such that for all , . We say that is strictly inductively ordered if every totally ordered subset of has a least upper bound, i.e., an upper bound so that if is an upper bound of , then .
An element is maximal if the relation implies . (Note that a set may have several maximal... [more]
12/18/07
There was no AotD for December eighteenth.
12/17/07
Diophantine equation
This article was also a AoPSWiki word of the week
A Diophantine equation is an multi-variable equation for which integer solutions (or sometimes natural number or whole number solutions) are to be found.
Finding the solution or solutions to a Diophantine equation is closely tied to modular arithmetic and number theory. Often, when a Diophantine equation has infinitely many solutions, parametric form is used to express the relation between the variables of the equation.
Diophantine equations are named for the ancient Greek/Alexandrian mathematician Diophantus.
A Diophantine equation in the form is known as a linear combination. If two relatively prime integers and are written in this form with , the equation will have an infinite number of solutions. More generally, there will always be an... [more]
12/16/07
Fibonacci sequence
This article was also a AoPSWiki word of the week
The Fibonacci sequence is a sequence of integers in which the first and second terms are both equal to 1 and each subsequent term is the sum of the two preceding it. The first few terms are .
The Fibonacci sequence can be written recursively as and for . This is the simplest nontrivial... [more]
12/15/07
Cauchy-Schwarz inequality
The Cauchy-Schwarz Inequality (which is known by other names, including Cauchy's Inequality, Schwarz's Inequality, and the Cauchy-Bunyakovsky-Schwarz Inequality) is a well-known inequality with many elegant applications... [more]
12/14/07
Rearrangement inequality
The Rearrangement Inequality states that, if is a permutation of a finiteset (in fact, multiset) of real numbers and is a permutation of another finite set of real numbers, the quantity is maximized when and are similarly sorted (that is, if is greater than or equal to exactly of the other members of , then is also greater than or equal to exactly of the other members of ). Conversely, is minimized when and are oppositely sorted (that is, if is less than or equal
12/13/07
There was no AotD for December thirteenth.
12/12/07
Trigonometric identities
Trigonometric identities are used to manipulate trigonometryequations in certain ways. Here is a list of them:
The six basic trigonometric functions can be defined using [more]
12/11/07
MATHCOUNTS
MATHCOUNTS is a large national mathematics competition and mathematics coaching program that has served millions of middle school students since 1984. Sponsored by the CNA Foundation, National Society of Professional Engineers, the National Council of Teachers of Mathematics, and others, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.
MATHCOUNTS curriculum... [more]
12/10/07
Polynomial
A polynomial is a function in one or more variables that consists of a sum of variables raised to nonnegative, integral powers and multiplied by coefficients (usually integral, rational, real or complex, but in abstract algebra often coming from an arbitrary field).
For example, these are... [more]
12/9/07
Pascal's identity
Pascal's identity is a common and useful theorem in the realm of combinatorics dealing with combinations (also known as binomial coefficients), and is often used to reduce large, complicated combinations.
Pascal's identity is also known as Pascal's rule, Pascal's formula, and occasionally... [more]
12/8/07
Canada/USA Mathcamp
This article was also a AoPSWiki word of the week
The Canada/USA Mathcamp is an intensive 5-week-long summer program for mathematically talented high school students, designed to inspire and motivate mathematically talented high school students by exposing them to the beauty and variety of mathematics, to impart valuable knowledge and skills for the pursuit of mathematics in high school, university, and beyond, and to provide a supportive and fun environment for interaction among students who love mathematics.
The environment of Mathcamp tends to be relaxed in terms of rules; in fact it officially only has four rules (generally involving common sense and respect). Originally it had been stricter, though by 1997 most of the mentors found the structure too... [more]
12/7/07
Prime number
A prime number (or simply prime) is a positive integer whose only positive divisors are 1 and itself. Note that is usually defined as being neither prime nor composite because it is its only factor among the natural numbers. The Sieve of Eratosthenes is a relatively quick method for... [more]
12/6/07
Calculus
The discovery of the branch of mathematics known as calculus was motivated by two classical problems: how to find the slope of the tangent line to a curve at a point and how to find the area bounded by a curve. What is surprising is that these two problems are fundamentally connected and, together with the notion of limits, can be used to analyse instantaneous rates of change, accumulations of change, volumes of irregular solids, and... [more]
12/5/07
Pi
Pi is an irrational number (in fact, transcendental number, as proved by Lindeman in 1882) denoted by the greek letter .
Pi is the ratio of the circumference (perimeter) of a given circle to its diameter. It is approximately equal to 3.141592653. The number pi is one of the most important constants in all of mathematics and appears in some of the most surprising places, such as in the sum . Some common... [more]
12/4/07
Isaac Newton
Isaac Newton (1643 – 1727) was a famous British physicist and mathematician. His most famous work in mathematics was the compilation of calculus.
Isaac Newton was born on January 4, 1643 in Lincolnshire, England. Newton was born very shortly after the death of his father. He did very well... [more]
12/3/07
Logarithm
This article was also a AoPSWiki word of the week
Logarithms and exponents are very closely related. In fact, they are inversefunctions. This means that logarithms can be used to reverse the result of exponentiation and vice versa, just as addition can be used to reverse the result of subtraction. Thus, if we have , then taking the logarithm with base on both sides will give us .
We would read this as "the logarithm of b, base a, is x". For example, we know that . To express the same fact... [more]
12/2/07
American Invitational Mathematics Examination
The American Invitational Mathematics Examination (AIME) is the second exam in the series of exams used to challenge bright students on the path toward choosing the team that represents the United States at the International Mathematics Olympiad (IMO). While most AIME participants are high school students, some bright middle school students also qualify each year.
High scoring AIME students are invited to take the prestigious United States of America Mathematics Olympiad (USAMO).
The AIME is administered by... [more]
12/1/07
Inequality
The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads.
Inequalities are arguably a branch of... [more]
11/30/07
Asymptote (geometry)
This article was also a AoPSWiki word of the week
An asymptote is a line or curve that a certain function approaches.
Linear asymptotes can be of three different kinds: horizontal, vertical or slanted (oblique).
The vertical asymptote can be found by finding values of that make the function undefined, generally because it results in a division by zero, which is undefined... [more]
11/27/07
Complex number
The complex numbers arise when we try to solve equations such as .
We know (from the trivial inequality) that the square of a real number cannot be negative, so this equation has no solutions in the real numbers. However, it is possible to define a number, , such that . If we add this new number to the reals, we will have solutions to . It turns out that in the system that results... [more]
11/25/07
Set
The notion of a set is one of the fundamental notions in mathematics that is difficult to precisely define. Of course, we have plenty of synonyms for the word "set," like collection, ensemble, group, etc., but those names really do not define the meaning of the word set; all they can do is replace it in various sentences. So, instead of defining what sets are, one has to define what can be done with them or, in other words, what axioms the sets satisfy. These axioms are chosen to agree with our intuitive concept of a set, on one hand, and to allow various, sometimes quite sophisticated, mathematical constructions on the other hand. For the full collection...[more]
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1044 | https://www.youtube.com/watch?v=oJmz799P42Y | TVM Calculations using BAII Plus Financial Calculator
Introduction to Finance
919 subscribers
1290 likes
Description
67915 views
Posted: 20 Sep 2021
Time value of money calculations using the BA II Plus from Texas Instruments
00:00 Set Decimal Places
00:33 Clear Previous Stored TVM Values
01:13 Single Cash Flow
05:18 Uneven Cash Flows
10:14 Annuities
11:00 Annuity Due
13:48 Non-Annual Compounding
36 comments
Transcript:
Set Decimal Places so the first thing you want to do when working with the financial calculator is set your number of decimal periods so typically when you first get it and turn it on it's going to be defaulting to two decimal places and you want to do more than that so to reset it you're going to go to second which is the yellow button the period right now mine's set to eight so i'm just going to switch this to six so i'm going to type six then press enter now i can do a second compute so the yellow button then the button right above it and that does the quit option up there so now i'm just back at the home page Clear Previous Stored TVM Values now in this type of calculator whenever you solve a new time value of money function the first thing you want to do is you want to clear out whatever you solved before so there's two places you want to clear so first you want to click second the yellow button and then future value over here on the right hand side and what that does is it clears all your time value of money calculations so it clears anything that was previously set into these buttons the second thing you want to do is you want to clear your work so you want to do second and then the clear button down here in the bottom corner that clears out anything you're currently calculating so those are the two buttons you want to clear whenever you start one of these problems Single Cash Flow starting with a single cash flow if we look at problem one so in this problem we have 41 is our n five percent five thousand dollars is our present value and then we're calculating what our future value is so the buttons we're going to be using are these white ones right across this row and the way you input them is you type in the number first and then you press on the button so for instance for 41 i type in 41 then i would press my n value next step i'm going to do 5 and in these calculators you type it in as a percentage not as a decimal so i type in 5 then press on i over y next up i'm doing my present value so i'm going to type in 5000 and here again either your present value or your future value needs to be negative so i like to think of it as i'm investing five thousand dollars at the beginning so i'm losing five thousand dollars then at the end i'm getting back whatever that future value is so that would be a positive number so type in 5000 then i'm going to press this plus minus sign down here at the bottom of the number keypad that makes it negative then set it as my present value now i'm going to clear out of this menu i just leave the payment option exactly as it is i don't set anything for that so to get out of this menu i'm going to do second compute that quits me out of this menu then i'm going to press on the compute button and now when i click on future value it's going to give me the future value of that cash flow so it's 36 959 let's look at another example so problem two was the college education problem so we know we need a future value of two hundred thousand dollars in ten years with a rate of seven percent so what is our present value so again before i solve this problem i'm going to quit out of my previous one so we do second quit and then i'm gonna clear those two menus so i'm gonna do second click on future value so clear my time value of money then i'm gonna do a second clear work so i've cleared everything out for the last problem now i'm going to do 10 as my number of periods 7 is my percentage present value is what we're solving for payment i don't have a payment in this option so i'm just going to leave that button alone don't type anything in there and then i'm going to do 200 000 for my future value now again i'm going to quit out of this menu so i'm going to do a second compute then click on the compute button one more time and now when i click on present value it gives me the value of negative 101 669 so that's the amount i would have to invest today to end up with the future value of 200. okay perfect so now oh yeah so if i just hit future value which i just did i accidentally just set the future value to be that number so don't do that so now we're gonna quit out of this menu i'm gonna do a second clear time value of money second clear work so for this last one here you've recently sold a home that you purchased seven years ago purchase price was 250 000 selling price was 315 000 so what is the annual interest rate so again clear time value of money clear work seven is my number of time periods the interest rate is what i'm solving for i have 250 000 is my present value and i'm gonna make that one negative then set it as my present value i leave my payment alone and then my future value is going to be 315 000. now i'm going to quit out of that menu hit compute and then click on my percentage so i get 3.36 Uneven Cash Flows i'm going to quit out of this menu again and i always make sure to clear all those things so next up for uneven cash flows this is probably the trickiest thing to solve in these financial or graphing calculators so the first option for how to solve these is that you could discount each one individually so i could say i have a future value of 200 discounted back one year at a rate of five percent then add it to 100 discounted back two years and use this menu to solve each basically cash flow of the problem then add them all together the other option is you can use the npv formula so the way you're going to get here for these problems is you're going to click on this cash flow button so right next to the second button and it's gonna say cash flow zero so that's our initial cash flow button so we don't have anything there so we're just leaving it as zero right we have no cash flow at time zero so i'm just going to press enter then down so i'm pressing this down arrow over here now it's asking me what cash flow one is so my cash flow one is two hundred dollars i'm going to press enter then down f01 stands for the frequency of that first cash flow so how many times in a row does that first cash flow happen in our case that first cash flow of 200 only happens one time in a row so the frequency is one so i'm just press enter down now it's asking for the second cash flow so in this case i type in 100 enter down that cash flow only happens one time in the series in a row so here i'm going to do enter down cash flow 3 150 enter it down only happens one time cash flow 4 400 enter down only happens one time enter down and now you don't have to go beyond there in this menu so what you can do is now you're just gonna quit out of this menu so i'm gonna do second compute now i'm gonna click on npv the button right next to cash flow my i value is five percent so i'm going to type in five enter down and now it says npv equals so when i see this green i want to press compute and i get the value of 739.86 so there are a couple tricky things here so first let me go back into my cashflow menu and clear this one out so i'm going to clear everything in here so when i do that make sure i clear both those things specifically when i do the second clear work that's going to clear out all my cash flows that were previously in there so you want to make sure you do that before you solve any of these so a couple things to keep in mind when you're typing these in first is that so in this example right we had a cash flow every single period so we had a cash flow one a cash flow two cash flow three and a clash so four but what happens if let's say we had two hundred dollars the first year 100 the second year 150 the third year then we skipped a year so we got zero in year four then 400 in year five in that case we would have to put in the zero as cash flow four so i would have 200 is my cash flow one happens one time cash flow two would be 100 enter down happens one time 150 is my cash flow three happens one time then i would have zero as this cash flow enter down happens one time then cash flow five would be 400. then i can quit out of that menu again go over to npv 5 is my interest rate enter down compute that value so that's something to keep in mind if you don't have a cash flow in a period you still have to type it in in the series of cash flows right you have to put in a zero as that value the next thing to keep in mind that's really tricky about these problems are these frequencies so you can only use the frequency option if you have two cash flows that are exactly the same in periods right next to each other so let's say on my timeline 200 was my first payment and 200 was my second payment in that case i could do 200 frequency two that would then say what it thinks of that would be two hundred dollars the first year two hundred dollars the second year then when i go down to the next cash flow and it says cash flow two that's essentially cash flow 3. so it can be a bit tricky when you have multiple cash flows that are the same and you're using that frequency option so make sure to keep that in mind Annuities now the next thing we can do first i'm just going to clear this out is we can solve annuities on these calculators and how we're going to do that is we're going to use the payment option right here so if we look at chapter 5 part 1 problem 6 this was again the mortgage problem so we could afford to pay 25 000 per year we had a mortgage rate of 4 30 years so first all we're just going to do is basically incorporate in the payment button so i have 30 years that's my n value 4 is my interest rate my payment each period is going to be 25 000. then i just quit out of the menu compute present value and i get that negative 432 000. Annuity Due so we can also solve annuity dues on these calculators but one thing i would caution you of is that if you reset the calculator where the payments begin at the beginning of each period so it is an annuity do make sure when you're done solving that problem you switch it back so the payment goes to the end of each period because you're basically switching up like the way the calculator is actually solving all problems from then on when you set this feature so the way you switch the feature is first you're gonna go to let me find it real quick okay so the begin is right above payment so i'm gonna do a second payment and here we can say it's see it's currently set to end so if i want to change that i'm going to do a second yellow button enter and now it's going to say beginning and you'll see now in the top corner of my screen it also says bgn so that tells me it's set to the beginning of each period so even if i quit out of this menu and i go back to my home screen it still says begin so if your computer or your calculator says begin in that top right corner that means you're solving everything as a do or as the payment occurs in the beginning of each period if it doesn't say anything up there in the top corner then it's solving things at the end of each period as an ordinary annuity so in this case in this example we have 500 payments we have a series of four of them and we have a discount rate of 10 so i just type in 4 as my n value 10 as my percentage 500 is going to be my payment and then i want to quit out of this menu hit compute and then i can calculate what my present value is so in this case it's negative and forty three 1743. now now that i'm done solving that i'm gonna quit out of this menu i'm gonna clear all my work and then i wanna switch this back so it's the end of each period i don't wanna leave it as the beginning so to switch it back i'm going to do the do again second payment so now it says begin second enter now it says end and i can just quit out so that beginning is gone in that top corner of the screen so now we can solve everything again as an ordinary annuity the last two things i wanted to mention is first that there is no built-in function for a growing annuity so those ones can really only be solved using the formula same with a growing perpetuity and then for perpetuities in general again there is no built-in function for perpetuities you can do the same thing as in excel where you assume a large value for your number of periods or you can just simply do the payment divided by the rate Non-Annual Compounding now lastly i just wanted to go through non-annual compounding so the way we're going to do this is exactly the same to the way we did this in excel where we basically want to take into account the compounding before we type in what the different variables are so first let me show this with a single cash flow so this was our problem for my class where we invest 50 today and we keep it invested for three years and the rate we're earning is going to be 12 compounded semi-annually and it's asking for the future value so my number of time periods is going to be three years times two compounds per year so first i'm going to do 3 times 2 equals 6. so that's going to be what i type in for my number of periods then next for my rate i have 12 percent compounded semi-annually so my six month rate is going to be 12 divided by 2 equals then i set that as my i over y value next step i'm just going to do 50. as a cash outflow i'm investing that today it's my present value now i can quit out of this menu hit compute and here my future value is going to be 70.93 so next up here we have an annuity with non-annual payments so this is our example where we could afford to pay two thousand dollars per month for the next thirty years towards a mortgage and the interest rate was four percent compounded monthly so again just like the single cash flow as we go across this row and we type in our inputs we want to take into account the number of payments and also the compounds on the rate before we type these in right so we want to take into account and then set it so first let's start off with the number of time periods or in this case the number of payments so we have 30 years 12 payments per year so i'm going to do 30 times 12 equals 360. i'm gonna set that as my end value now next up for my rate i have four percent compounded monthly so i'm going to take 4 divided by 12 equals set that as my i over y two thousand dollars is my payment now i can quit out hit compute and here i'm going to get a present value of negative hundred and eighteen thousand nine hundred and twenty two so that's how to take into account compounding when we're using the same time value of money functions first you wanna take into account that n times m or the i over m and then use those values as your inputs |
1045 | https://www.aafp.org/pubs/afp/issues/2004/0115/p436.html | ACOG Releases Bulletin on Managing Cervical Insufficiency | AAFP
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Practice Guidelines
ACOG Releases Bulletin on Managing Cervical Insufficiency
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GENEVIEVE W. RESSEL
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Am Fam Physician. 2004;69(2):436-439
The American College of Obstetricians and Gynecologists (ACOG) recently published a clinical management guideline on cervical insufficiency. The complete guideline, ACOG Practice Bulletin No. 48, appeared in the November 2003 issue of Obstetrics and Gynecology. This report includes evidence for screening asymptomatic at-risk women and offers management guidelines.
According to the National Center for Health Statistics, 23,000 discharge records from short-stay hospitals included the diagnosis of cervical incompetence in 2000. Diagnostic criteria remain elusive, and several surgical and nonsurgical treatments have been proposed.
Clinical Considerations and Recommendations
Is there a role for routine ultrasound screening of the cervix? Serial ultrasound assessments of the cervix in low-risk women have demonstrated low sensitivity and low positive-predictive values, meaning ultrasonography lacks enough discriminatory power to recommend routine use.
What is the role of ultrasonography in evaluating women who have had a previous pregnancy loss? Study results suggest that serial transvaginal ultrasound may be considered in women with a history of second- or early third-trimester deliveries. Because the upper portion of the cervix is not easily distinguished from the lower uterine segment in early pregnancy, these assessments should not begin before 16 to 20 weeks of gestation. According to ACOG, there is no reason to perform ultrasound screening for cervical insufficiency in women with a history of first-trimester pregnancy losses.
In whom is a cerclage indicated? In the past, patient selection for elective cerclage has been based on congenital or acquired visible defects in the ectocervix or classic features of cervical incompetence, which include history of two or more second-trimester pregnancy losses (excluding those resulting from preterm labor or abruption); history of losing each pregnancy at an earlier gestational age; history of painless cervical dilation of up to 4 to 6 cm; absence of clinical findings consistent with placental abruption; and history of cervical trauma caused by cone biopsy, intrapartum cervical lacerations, and excessive, forced cervical dilation during pregnancy termination.
Based on limited clinical information, elective cerclage for historical factors generally should be confined to patients with three or more otherwise unexplained second-trimester pregnancy losses or preterm deliveries. Cerclage should be performed at 13 to 16 weeks of gestation after ultrasound evaluation has demonstrated the presence of a live fetus with no apparent anomalies.
Urgent, or therapeutic, cerclage often is recommended for women who have ultrasonographic changes consistent with a short cervix or evidence of funneling. Management of this group remains speculative because of the limited number of well-designed randomized trials. The decision to proceed with cerclage should be made with caution.
In the past, women who present with advanced cervical dilation in the absence of labor and abruption have been candidates for emergency cerclage. No randomized trials have been done in this area, and retrospective studies are limited by selection bias, inadequate patient numbers, and inconsistent selection criteria.
In the second trimester, how should a short cervix be treated? According to ACOG, if transvaginal ultrasonography before 16 to 20 weeks of gestation identifies a short cervix, the examination should be repeated because of the inability to adequately distinguish the cervix from the lower uterine segment in early pregnancy. Identification of a short cervix at or after 20 weeks of gestation should prompt assessment of the fetus for anomalies, uterine activity to rule out preterm labor, and maternal factors to rule out chorioamnionitis. Regular evaluations may be performed (particularly in patients with pelvic pressure, backache, or increased mucoid discharge) every few days to avoid missing rapid changes in cervical dilation or until the trend in cervical length can be characterized.
In patients with a history of fewer than three second-trimester pregnancy losses, urgent cerclage is not supported by evidence-based studies, and further transvaginal ultrasound surveillance may be the more judicious approach. Management for cervical shortening or funneling is unclear, and the decision to proceed with urgent cerclage should be made with caution. Cervical change noted before fetal viability is a better indication for cerclage than if it is identified after fetal viability has been achieved. Emergency cerclage may be considered in women if clinical chorioamnionitis or signs of labor are not present.
In the third trimester, how should a short cervix be treated? If the patient's cervical length is below the 10th percentile (25 mm) for gestational age at or after fetal viability, evaluation should include ultrasound assessment of fetal anatomy to exclude anomalies, tocodynamometry to detect the presence of uterine contractions, and assessment of maternal factors to exclude chorioamnionitis. If the patient is in labor, tocolytic therapy may delay delivery long enough to promote fetal lung maturation with maternal glucocorticoid therapy. The presence of chorioamnionitis is grounds for immediate delivery and the use of broad-spectrum antibiotics. If labor or chorioamnionitis is not present, modification of activity, pelvic rest, tobacco cessation, and expectant management may be considered. Cerclage in the treatment of women with cervical insufficiency after determining fetal viability has not been adequately assessed.
Is there a role for scheduled early or first-trimester cerclage in patients with a suspicious clinical history? The evidence-based risk-benefit ratio does not support first-trimester cerclage, even with transabdominal procedures.
Is cerclage placement associated with an increase in morbidity? Suture displacement, rupture of membranes, and chorioamnionitis are the most common complications associated with cerclage placement, and incidence varies widely in relation to the timing and indications for the cerclage. Urgent and emergency cerclages are associated with a higher incidence of morbidity as a result of cervical shortening and exposure of the fetal membranes to the vaginal ecosystem.
Transabdominal cerclage can be complicated by rupture of membranes and chorioamnionitis. It carries the added risk of intraoperative hemorrhage from the uterine veins when the cerclage band is tunneled between the bifurcation of the uterine artery, as well as the known risks associated with laparotomy. Life-threatening complications of uterine rupture and maternal septicemia are extremely rare but have been reported with all types of cerclage.
Should perioperative antibiotics and tocolytics be used in association with cerclage placement? Studies using perioperative antibiotics have been small, nonrandomized, and inconclusive. The use of unnecessary antibiotics may lead to the development of resistant strains of bacteria and other morbidity for the patient and her fetus. No randomized studies have shown that use of tocolytic therapy after cerclage is effective. The lack of clear benefit for these adjunctive treatments suggests that these drugs should be used with caution.
Does a patient who was exposed to diethylstilbestrol require cerclage? To date, no definitive epidemiologic studies have proved that cervical insufficiency is more frequent in women exposed to diethylstilbestrol than in comparable control subjects. There are no randomized trials of cerclage in these patients. A woman who has been exposed may be evaluated using the same criteria as a nonexposed patient.
When is removal of cerclage indicated in a patient with preterm labor or preterm rupture of membranes? Because the available studies are small and nonrandomized, the optimal timing of cerclage removal is unclear, according to ACOG.
Coverage of guidelines from other organizations does not imply endorsement by AFP or the AAFP.
This series is coordinated by Michael J. Arnold, MD, MHPE, Assistant Medical Editor.
A collection of Practice Guidelines published in AFP is available at
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The Mating Embrace
Part of the Frogs: A Chorus of Colors exhibition.
© Scott Egan
Wood frogs in amplexus
When reptiles, birds, and mammals mate, the male fertilizes the eggs inside the female's body. Like fish, almost all frogs fertilize the eggs on the outside. The male holds the female around the waist in a mating hug called amplexus. He fertilizes the eggs as the female lays them. Amplexus can last hours or days—one pair of Andean toads stayed in amplexus for four months! |
1047 | https://artofproblemsolving.com/articles/files/SatoNT.pdf?srsltid=AfmBOooMhtnz8HAlts2dLYMG96vdNusVXnndXNyyKIPqaM_3vftyb6iM | Number Theory
Naoki Sato sato@artofproblemsolving.com
0 Preface
This set of notes on number theory was originally written in 1995 for students at the IMO level. It covers the basic background material that an IMO student should be familiar with. This text is meant to be a reference, and not a replacement but rather a supplement to a number theory textbook; several are given at the back. Proofs are given when appropriate, or when they illustrate some insight or important idea. The problems are culled from various sources, many from actual contests and olympiads, and in general are very difficult. The author welcomes any corrections or suggestions.
1 Divisibility
For integers a and b, we say that a divides b, or that a is a divisor (or
factor ) of b, or that b is a multiple of a, if there exists an integer c such that b = ca , and we denote this by a | b. Otherwise, a does not divide b, and we denote this by a - b. A positive integer p is a prime if the only divisors of
p are 1 and p. If pk | a and pk+1 - a where p is a prime, i.e. pk is the highest power of p dividing a, then we denote this by pk‖a.Useful Facts
• If a, b > 0, and a | b, then a ≤ b.
• If a | b1, a | b2, . . . , a | bn, then for any integers c1, c2, . . . , cn,
a |
n
∑
i=1
bici.
Theorem 1.1 . The Division Algorithm . For any positive integer a and integer b, there exist unique integers q and r such that b = qa + r and 0 ≤ r < a , with r = 0 iff a | b.1Theorem 1.2 . The Fundamental Theorem of Arithmetic . Every integer greater than 1 can be written uniquely in the form
pe1
1
pe2
2
· · · pek
k
,
where the pi are distinct primes and the ei are positive integers.
Theorem 1.3 . (Euclid) There exist an infinite number of primes.
Proof . Suppose that there are a finite number of primes, say p1, p2, . . . ,
pn. Let N = p1p2 · · · pn + 1. By the fundamental theorem of arithmetic, N
is divisible by some prime p. This prime p must be among the pi, since by assumption these are all the primes, but N is seen not to be divisible by any of the pi, contradiction.
Example 1.1 . Let x and y be integers. Prove that 2 x + 3 y is divisible by 17 iff 9 x + 5 y is divisible by 17.
Solution . 17 | (2 x + 3 y) ⇒ 17 | [13(2 x + 3 y)], or 17 | (26 x + 39 y) ⇒
17 | (9 x + 5 y), and conversely, 17 | (9 x + 5 y) ⇒ 17 | [4(9 x + 5 y)], or 17 | (36 x + 20 y) ⇒ 17 | (2 x + 3 y).
Example 1.2 . Find all positive integers d such that d divides both n2 +1 and ( n + 1) 2 + 1 for some integer n.
Solution . Let d | (n2 + 1) and d | [( n + 1) 2 + 1], or d | (n2 + 2 n + 2). Then d | [( n2 + 2 n + 2) − (n2 + 1)], or d | (2 n + 1) ⇒ d | (4 n2 + 4 n + 1), so
d | [4( n2+2 n+2) −(4 n2+4 n+1)], or d | (4 n+7). Then d | [(4 n+7) −2(2 n+1)], or d | 5, so d can only be 1 or 5. Taking n = 2 shows that both of these values are achieved.
Example 1.3 . Suppose that a1, a2, . . . , a2n are distinct integers such that the equation (x − a1)( x − a2) · · · (x − a2n) − (−1) n(n!) 2 = 0 has an integer solution r. Show that
r = a1 + a2 + · · · + a2n
2n .
(1984 IMO Short List)
Solution . Clearly, r 6 = ai for all i, and the r − ai are 2 n distinct integers, so
|(r − a1)( r − a2) · · · (r − a2n)| ≥ | (1)(2) · · · (n)( −1)( −2) · · · (−n)| = ( n!) 2,
2with equality iff
{r − a1, r − a2, . . . , r − a2n} = {1, 2, . . . , n, −1, −2, . . . , −n}.
Therefore, this must be the case, so (r − a1) + ( r − a2) + · · · + ( r − a2n)= 2 nr − (a1 + a2 + · · · + a2n)= 1 + 2 + · · · + n + ( −1) + ( −2) + · · · + ( −n) = 0
⇒ r = a1 + a2 + · · · + a2n
2n .
Example 1.4 . Let 0 < a 1 < a 2 < · · · < a mn +1 be mn + 1 integers. Prove that you can select either m + 1 of them no one of which divides any other, or n + 1 of them each dividing the following one. (1966 Putnam Mathematical Competition)
Solution . For each i, 1 ≤ i ≤ mn + 1, let ni be the length of the longest sequence starting with ai and each dividing the following one, among the integers ai, ai+1 , . . . , amn +1 . If some ni is greater than n then the problem is solved. Otherwise, by the pigeonhole principle, there are at least m + 1 values of ni that are equal. Then, the integers ai corresponding to these ni
cannot divide each other. Useful Facts
• Bertrand’s Postulate . For every positive integer n, there exists a prime
p such that n ≤ p ≤ 2n.
• Gauss’s Lemma . If a polynomial with integer coefficients factors into two polynomials with rational coefficients, then it factors into two poly-nomials with integer coefficients. Problems 1. Let a and b be positive integers such that a | b2, b2 | a3, a3 | b4, b4 | a5,. . . . Prove that a = b.2. Let a, b, and c denote three distinct integers, and let P denote a poly-nomial having all integral coefficients. Show that it is impossible that
P (a) = b, P (b) = c, and P (c) = a.(1974 USAMO) 33. Show that if a and b are positive integers, then
(
a + 12
)n
+
(
b + 12
)n
is an integer for only finitely many positive integers n.(A Problem Seminar , D.J. Newman) 4. For a positive integer n, let r(n) denote the sum of the remainders when
n is divided by 1, 2, . . . , n respectively. Prove that r(k) = r(k − 1) for infinitely many positive integers k.(1981 K¨ ursch´ ak Competition) 5. Prove that for all positive integers n,0 <
n
∑
k=1
g(k)
k − 2n
3 < 23,
where g(k) denotes the greatest odd divisor of k.(1973 Austrian Mathematics Olympiad) 6. Let d be a positive integer, and let S be the set of all positive integers of the form x2 + dy 2, where x and y are non-negative integers. (a) Prove that if a ∈ S and b ∈ S, then ab ∈ S.(b) Prove that if a ∈ S and p ∈ S, such that p is a prime and p | a,then a/p ∈ S.(c) Assume that the equation x2 + dy 2 = p has a solution in non-negative integers x and y, where p is a given prime. Show that if
d ≥ 2, then the solution is unique, and if d = 1, then there are exactly two solutions.
2 GCD and LCM
The greatest common divisor of two positive integers a and b is the great-est positive integer that divides both a and b, which we denote by gcd( a, b ), and similarly, the lowest common multiple of a and b is the least positive 4integer that is a multiple of both a and b, which we denote by lcm( a, b ). We say that a and b are relatively prime if gcd( a, b ) = 1. For integers a1, a2,. . . , an, gcd( a1, a 2, . . . , a n) is the greatest positive integer that divides all of
a1, a2, . . . , an, and lcm( a1, a 2, . . . , a n) is defined similarly. Useful Facts
• For all a, b, gcd( a, b ) · lcm( a, b ) = ab .
• For all a, b, and m, gcd( ma, mb ) = m gcd( a, b ) and lcm( ma, mb ) =
mlcm( a, b ).
• If d | gcd( a, b ), then gcd
(ad , bd
)
= gcd( a, b )
d .
In particular, if d = gcd( a, b ), then gcd( a/d, b/d ) = 1; that is, a/d and
b/d are relatively prime.
• If a | bc and gcd( a, c ) = 1, then a | b.
• For positive integers a and b, if d is a positive integer such that d | a,
d | b, and for any d′, d′ | a and d′ | b implies that d′ | d, then d =gcd( a, b ). This is merely the assertion that any common divisor of a
and b divides gcd( a, b ).
• If a1a2 · · · an is a perfect kth power and the ai are pairwise relatively prime, then each ai is a perfect kth power.
• Any two consecutive integers are relatively prime.
Example 2.1 . Show that for any positive integer N , there exists a multiple of N that consists only of 1s and 0s. Furthermore, show that if N
is relatively prime to 10, then there exists a multiple that consists only of 1s.
Solution . Consider the N + 1 integers 1, 11, 111, . . . , 111...1 ( N + 1 1s). When divided by N , they leave N + 1 remainders. By the pigeonhole princi-ple, two of these remainders are equal, so the difference in the corresponding integers, an integer of the form 111...000, is divisible by N . If N is relatively prime to 10, then we may divide out all powers of 10, to obtain an integer of the form 111...1 that remains divisible by N .5Theorem 2.1 . For any positive integers a and b, there exist integers x
and y such that ax + by = gcd( a, b ). Furthermore, as x and y vary over all integers, ax + by attains all multiples and only multiples of gcd( a, b ).
Proof . Let S be the set of all integers of the form ax +by , and let d be the least positive element of S. By the division algorithm, there exist integers q
and r such that a = qd + r, 0 ≤ r < d . Then r = a − qd = a − q(ax + by ) = (1 − qx )a − (qy )b, so r is also in S. But r < d , so r = 0 ⇒ d | a, and similarly, d | b, so d | gcd( a, b ). However, gcd( a, b ) divides all elements of S,so in particular gcd( a, b ) | d ⇒ d = gcd( a, b ). The second part of the theorem follows.
Corollary 2.2 . The positive integers a and b are relatively prime iff there exist integers x and y such that ax + by = 1.
Corollary 2.3 . For any positive integers a1, a2, . . . , an, there exist integers x1, x2, . . . , xn, such that a1x1+a2x2+· · · +anxn = gcd( a1, a 2, . . . , a n).
Corollary 2.4 . Let a and b be positive integers, and let n be an integer. Then the equation
ax + by = n
has a solution in integers x and y iff gcd( a, b ) | n. If this is the case, then all solutions are of the form (x, y ) =
(
x0 + t · bd , y 0 − t · ad
)
,
where d = gcd( a, b ), ( x0, y 0) is a specific solution of ax + by = n, and t is an integer.
Proof . The first part follows from Theorem 2.1. For the second part, as stated, let d = gcd( a, b ), and let ( x0, y 0) be a specific solution of ax + by = n,so that ax 0 + by 0 = n. If ax + by = n, then ax + by − ax 0 − by 0 = a(x − x0) +
b(y − y0) = 0, or a(x − x0) = b(y0 − y), and hence (x − x0) · ad = ( y0 − y) · bd.
Since a/d and b/d are relatively prime, b/d must divide x − x0, and a/d must divide y0 − y. Let x − x0 = tb/d and y0 − y = ta/d . This gives the solutions described above. 6Example 2.2 . Prove that the fraction 21 n + 4 14 n + 3 is irreducible for every positive integer n. (1959 IMO)
Solution . For all n, 3(14 n + 3) − 2(21 n + 4) = 1, so the numerator and denominator are relatively prime.
Example 2.3 . For all positive integers n, let Tn = 2 2n
Show that if
m 6 = n, then Tm and Tn are relatively prime.
Solution . We have that
Tn − 2 = 2 2n
− 1 = 2 2n−1·2 − 1= ( Tn−1 − 1) 2 − 1 = T 2
n−1
− 2Tn−1
= Tn−1(Tn−1 − 2) = Tn−1Tn−2(Tn−2 − 2) = · · ·
= Tn−1Tn−2 · · · T1T0(T0 − 2) = Tn−1Tn−2 · · · T1T0,
for all n. Therefore, any common divisor of Tm and Tn must divide 2. But each Tn is odd, so Tm and Tn are relatively prime.
Remark . It immediately follows from this result that there are an infinite number of primes.
The Euclidean Algorithm . By recursive use of the division algorithm, we may find the gcd of two positive integers a and b without factoring either, and the x and y in Theorem 2.1 (and so, a specific solution in Corollary 2.4). For example, for a = 329 and b = 182, we compute 329 = 1 · 182 + 147 ,
182 = 1 · 147 + 35 ,
147 = 4 · 35 + 7 ,
35 = 5 · 7,
and stop when there is no remainder. The last dividend is the gcd, so in our example, gcd(329,182) = 7. Now, working through the above equations 7backwards, 7 = 147 − 4 · 35 = 147 − 4 · (182 − 1 · 147) = 5 · 147 − 4 · 182 = 5 · (329 − 182) − 4 · 182 = 5 · 329 − 9 · 182 .
Remark . The Euclidean algorithm also works for polynomials.
Example 2.4 . Let n be a positive integer, and let S be a subset of n + 1 elements of the set {1, 2, . . . , 2n}. Show that (a) There exist two elements of S that are relatively prime, and (b) There exist two elements of S, one of which divides the other.
Solution . (a) There must be two elements of S that are consecutive, and thus, relatively prime. (b) Consider the greatest odd factor of each of the n + 1 elements in
S. Each is among the n odd integers 1, 3, . . . , 2 n − 1. By the pigeon-hole principle, two must have the same greatest odd factor, so they differ (multiplication-wise) by a power of 2, and so one divides the other.
Example 2.5 . The positive integers a1, a2, . . . , an are such that each is less than 1000, and lcm( ai, a j ) > 1000 for all i, j, i 6 = j. Show that
n
∑
i=1
1
ai
< 2.
(1951 Russian Mathematics Olympiad)
Solution . If 1000
m+1
< a ≤ 1000
m
, then the m multiples a, 2 a, . . . , ma do not exceed 1000. Let k1 the number of ai in the interval ( 1000 2 , 1000], k2 in (1000 3 , 1000 2 ], etc. Then there are k1 + 2 k2 + 3 k3 + · · · integers, no greater than 1000, that are multiples of at least one of the ai. But the multiples are distinct, so
k1 + 2 k2 + 3 k3 + · · · < 1000
⇒ 2k1 + 3 k2 + 4 k3 + · · · = ( k1 + 2 k2 + 3 k3 + · · · ) + ( k1 + k2 + k3 + · · · )
< 1000 + n< 2000 .
8Therefore,
n
∑
i=1
1
ai
≤ k1
21000 + k2
31000 + k3
41000 + · · ·
= 2k1 + 3 k2 + 4 k3 + · · ·
1000
< 2.
Note: It can be shown that n ≤ 500 as follows: Consider the greatest odd divisor of a1, a2, . . . , a1000 . Each must be distinct; otherwise, two differ, multiplication-wise, by a power of 2, which means one divides the other, contradiction. Also, there are only 500 odd numbers between 1 and 1000, from which the result follows. It also then follows that
n
∑
i=1
1
ai
< 32.
Useful Facts
• Dirichlet’s Theorem . If a and b are relatively prime positive integers, then the arithmetic sequence a, a + b, a + 2 b, . . . , contains infinitely many primes. Problems 1. The symbols ( a, b, . . . , g ) and [ a, b, . . . , g ] denote the greatest common divisor and lowest common multiple, respectively of the positive inte-gers a, b, . . . , g. Prove that [a, b, c ]2
[a, b ][ a, c ][ b, c ] = (a, b, c )2
(a, b )( a, c )( b, c ).
(1972 USAMO) 2. Show that gcd( am − 1, a n − 1) = agcd( m,n ) − 1 for all positive integers
a > 1, m, n.93. Let a, b, and c be positive integers. Show that lcm( a, b, c ) = abc · gcd( a, b, c )gcd( a, b ) · gcd( a, c ) · gcd( b, c ).
Express gcd( a, b, c ) in terms of abc , lcm( a, b, c ), lcm( a, b ), lcm( a, c ), and lcm( b, c ). Generalize. 4. Let a, b be odd positive integers. Define the sequence ( fn) by putting
f1 = a, f2 = b, and by letting fn for n ≥ 3 be the greatest odd divisor of fn−1 + fn−2. Show that fn is constant for n sufficiently large and determine the eventual value as a function of a and b.(1993 USAMO) 5. Let n ≥ a1 > a 2 > · · · > a k be positive integers such that lcm( ai, a j ) ≤
n for all i, j. Prove that ia i ≤ n for i = 1, 2, . . . , k.
3 Arithmetic Functions
There are several important arithmetic functions, of which three are pre-sented here. If the prime factorization of n > 1 is pe1
1
pe2
2
· · · pek
k
, then the number of positive integers less than n, relatively prime to n, is
φ(n) =
(
1 − 1
p1
) (
1 − 1
p2
)
· · ·
(
1 − 1
pk
)
n
= pe1−11 pe2−12 · · · pek −1
k
(p1 − 1)( p2 − 1) · · · (pk − 1) ,
the number of divisors of n is
τ (n) = ( e1 + 1)( e2 + 1) · · · (ek + 1) ,
and the sum of the divisors of n is
σ(n) = ( pe1
1
pe1−11 + · · · + 1)( pe2
2
pe2−12 + · · · + 1)
· · · (pek
k
pek −1
k
· · · + 1) =
(pe1+1 1 − 1
p1 − 1
) ( pe2+1 2 − 1
p2 − 1
)
· · ·
(pek +1
k
− 1
pk − 1
)
.
Also, φ(1), τ (1), and σ(1) are defined to be 1. We say that a function
f is multiplicative if f (mn ) = f (m)f (n) for all relatively prime positive 10 integers m and n, and f (1) = 1 (otherwise, f (1) = 0, which implies that
f (n) = 0 for all n).
Theorem 3.1 . The functions φ, τ , and σ are multiplicative. Hence, by taking the prime factorization and evaluating at each prime power, the formula above are found easily.
Example 3.1 . Find the number of solutions in ordered pairs of positive integers ( x, y ) of the equation 1
x + 1
y = 1
n,
where n is a positive integer.
Solution . From the given, 1
x + 1
y = 1
n ⇔ xy = nx + ny ⇔ (x − n)( y − n) = n2.
If n = 1, then we immediately deduce the unique solution (2,2). For
n ≥ 2, let n = pe1
1
pe2
2
· · · pek
k
be the prime factorization of n. Since x, y > n ,there is a 1-1 correspondence between the solutions in ( x, y ) and the factors of n2, so the number of solutions is
τ (n2) = (2 e1 + 1)(2 e2 + 1) · · · (2 ek + 1) .
Example 3.2 . Let n be a positive integer. Prove that
∑
d|n
φ(d) = n.
Solution . For a divisor d of n, let Sd be the set of all a, 1 ≤ a ≤ n, such that gcd( a, n ) = n/d . Then Sd consists of all elements of the form b · n/d ,where 0 ≤ b ≤ d, and gcd( b, d ) = 1, so Sd contains φ(d) elements. Also, it is clear that each integer between 1 and n belongs to a unique Sd. The result then follows from summing over all divisors d of n.Problems 1. Let n be a positive integer. Prove that
n
∑
k=1
τ (k) =
n
∑
k=1
⌊nk
⌋
.
11 2. Let n be a positive integer. Prove that
∑
d|n
τ 3(d) =
∑
d|n
τ (d)
2
.
Prove that if σ(N ) = 2 N + 1, then N is the square of an odd integer. (1976 Putnam Mathematical Competition)
4 Modular Arithmetic
For a positive integer m and integers a and b, we say that a is congruent to
b modulo m if m | (a − b), and we denote this by a ≡ b modulo m, or more commonly a ≡ b (mod m). Otherwise, a is not congruent to b modulo m,and we denote this by a 6 ≡ b (mod m) (although this notation is not used often). In the above notation, m is called the modulus , and we consider the integers modulo m.
Theorem 4.1 . If a ≡ b and c ≡ d (mod m), then a + c ≡ b + d (mod m)and ac ≡ bd (mod m).
Proof . If a ≡ b and c ≡ d (mod m), then there exist integers k and l
such that a = b + km and c = d + lm . Hence, a + c = b + d + ( k + l)m, so
a + c ≡ b + d (mod m). Also,
ac = bd + dkm + blm + klm 2
= bd + ( dk + bl + klm )m,
so ac ≡ bd (mod m). Useful Facts
• For all integers n,
n2 ≡
{ 01
}
(mod 4)
{ if n is even ,
if n is odd .
• For all integers n,
n2 ≡
041
(mod 8)
if n ≡ 0 (mod 4) ,
if n ≡ 2 (mod 4) ,
if n ≡ 1 (mod 2) .
12 • If f is a polynomial with integer coefficients and a ≡ b (mod m), then
f (a) ≡ f (b) (mod m).
• If f is a polynomial with integer coefficients of degree n, not identically zero, and p is a prime, then the congruence
f (x) ≡ 0 (mod p)has at most n solutions modulo p, counting multiplicity.
Example 4.1 . Prove that the only solution in rational numbers of the equation
x3 + 3 y3 + 9 z3 − 9xyz = 0 is x = y = z = 0. (1983 K¨ ursch´ ak Competition)
Solution . Suppose that the equation has a solution in rationals, with at least one non-zero variable. Since the equation is homogeneous, we may obtain a solution in integers ( x0, y 0, z 0) by multiplying the equation by the cube of the lowest common multiple of the denominators. Taking the equa-tion modulo 3, we obtain x30 ≡ 0 (mod 3). Therefore, x0 must be divisible by 3, say x0 = 3 x1. Substituting, 27 x31 + 3 y30 + 9 z30 − 27 x1y0z0 = 0
⇒ y30 + 3 z30 + 9 x31 − 9x1y0z0 = 0 .
Therefore, another solution is ( y0, z 0, x 1). We may then apply this reduction recursively, to obtain y0 = 3 y1, z0 = 3 z1, and another solution ( x1, y 1, z 1). Hence, we may divide powers of 3 out of our integer solution an arbitrary number of times, contradiction.
Example 4.2 . Does one of the first 10 8 + 1 Fibonacci numbers terminate with 4 zeroes?
Solution . The answer is yes. Consider the sequence of pairs ( Fk, F k+1 )modulo 10 4. Since there are only a finite number of different possible pairs (10 8 to be exact), and each pair is dependent only on the previous one, this sequence is eventually periodic. Also, by the Fibonacci relation, one can find the previous pair to a given pair, so this sequence is immediately periodic. But F0 ≡ 0 (mod 10 4), so within 10 8 terms, another Fibonacci number divisible by 10 4 must appear. 13 In fact, a computer check shows that 10 4 | F7500 , and ( Fn) modulo 10 4
has period 15000, which is much smaller than the upper bound of 10 8.If ax ≡ 1 (mod m), then we say that x is the inverse of a modulo m,denoted by a−1, and it is unique modulo m.
Theorem 4.2 . The inverse of a modulo m exists and is unique iff a is relatively prime to m.
Proof . If ax ≡ 1 (mod m), then ax = 1+ km for some k ⇒ ax −km = 1. By Corollary 2.2, a and m are relatively prime. Now, if gcd( a, m ) = 1, then by Corollary 2.2, there exist integers x and y such that ax + my = 1 ⇒ ax =1 − my ⇒ ax ≡ 1 (mod m). The inverse x is unique modulo m, since if x′ is also an inverse, then ax ≡ ax ′ ≡ 1 ⇒ xax ≡ xax ′ ≡ x ≡ x′.
Corollary 4.3 . If p is a prime, then the inverse of a modulo p exists and is unique iff p does not divide a.
Corollary 4.4 . If ak ≡ bk (mod m) and k is relatively prime to m, then
a ≡ b (mod m).
Proof . Multiplying both sides by k−1, which exists by Theorem 4.2, yields the result. We say that a set {a1, a 2, . . . , a m} is a complete residue system modulo
m if for all i, 0 ≤ i ≤ m−1, there exists a unique j such that aj ≡ i (mod m).
Example 4.3 . Find all positive integers n such that there exist complete residue systems {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} modulo n for which {a1 +
b1, a 2 + b2, . . . , a n + bn} is also a complete residue system.
Solution . The answer is all odd n. First we prove necessity. For any complete residue system {a1, a 2, . . . , a n} modulo n, we have that
a1 + a2 + · · · + an ≡ n(n + 1) /2 (mod n). So, if all three sets are complete residue systems, then a1 +a2 +· · · +an +b1 +b2 +· · · +bn ≡ n2 +n ≡ 0 (mod n)and a1 + b1 + a2 + b2 + · · · + an + bn ≡ n(n + 1) /2 (mod n), so n(n + 1) /2 ≡ 0(mod n). The quantity n(n + 1) /2 is divisible by n iff ( n + 1) /2 is an integer, which implies that n is odd. Now assume that n is odd. Let ai = bi = i for all i. Then ai + bi = 2 i
for all i, and n is relatively prime to 2, so by Corollary 4.4, {2, 4, . . . , 2n} is a complete residue system modulo n.
Theorem 4.5 . Euler’s Theorem . If a is relatively prime to m, then
aφ(m) ≡ 1 (mod m). 14 Proof . Let a1, a2, . . . , aφ(m) be the positive integers less than m that are relatively prime to m. Consider the integers aa 1, aa 2, . . . , aa φ(m). We claim that they are a permutation of the original φ(m) integers ai, modulo
m. For each i, aa i is also relatively prime to m, so aa i ≡ ak for some k. Since
aa i ≡ aa j ⇔ ai ≡ aj (mod m), each ai gets taken to a different ak under multiplication by a, so indeed they are permuted. Hence,
a1a2 · · · aφ(m) ≡ (aa 1)( aa 2) · · · (aa φ(m))
≡ aφ(m)a1a2 · · · aφ(m)
⇒ 1 ≡ aφ(m) (mod m).
Remark . This gives an explicit formula for the inverse of a modulo m:
a−1 ≡ aφ(m)−2 (mod m). Alternatively, one can use the Euclidean algorithm to find a−1 ≡ x as in the proof of Theorem 4.2.
Corollary 4.6 . Fermat’s Little Theorem (FLT) . If p is a prime, and p
does not divide a, then ap−1 ≡ 1 (mod p).
Example 4.4 . Show that if a and b are relatively prime positive integers, then there exist integers m and n such that am + bn ≡ 1 (mod ab ).
Solution . Let S = am + bn, where m = φ(b) and n = φ(a). Then by Euler’s Theorem, S ≡ bφ(a) ≡ 1 (mod a), or S − 1 ≡ 0 (mod a), and
S ≡ aφ(b) ≡ 1 (mod b), or S − 1 ≡ 0 (mod b). Therefore, S − 1 ≡ 0, or S ≡ 1(mod ab ).
Example 4.5 . For all positive integers i, let Si be the sum of the products of 1, 2, . . . , p − 1 taken i at a time, where p is an odd prime. Show that
S1 ≡ S2 ≡ · · · ≡ Sp−2 ≡ 0 (mod p).
Solution . First, observe that (x − 1)( x − 2) · · · (x − (p − 1)) = xp−1 − S1xp−2 + S2xp−3 − · · · − Sp−2x + Sp−1.
This polynomial vanishes for x = 1, 2, . . . , p − 1. But by Fermat’s Little Theorem, so does xp−1 − 1 modulo p. Taking the difference of these two polynomials, we obtain another polynomial of degree p − 2 with p − 1 roots modulo p, so it must be the zero polynomial, and the result follows from comparing coefficients. 15 Remark . We immediately have that ( p − 1)! ≡ Sp−1 ≡ − 1 (mod p), which is Wilson’s Theorem. Also, xp − x ≡ 0 (mod p) for all x, yet we cannot compare coefficients here. Why not?
Theorem 4.7 . If p is a prime and n is an integer such that p | (4 n2 + 1), then p ≡ 1 (mod 4).
Proof . Clearly, p cannot be 2, so we need only show that p 6 ≡ 3 (mod 4). Suppose p = 4 k + 3 for some k. Let y = 2 n, so by Fermat’s Little Theorem,
yp−1 ≡ 1 (mod p), since p does not divide n. But, y2 + 1 ≡ 0, so
yp−1 ≡ y4k+2 ≡ (y2)2k+1 ≡ (−1) 2k+1 ≡ − 1 (mod p),
contradiction. Therefore, p ≡ 1 (mod 4).
Remark . The same proof can be used to show that if p is a prime and
p | (n2 + 1), then p = 2 or p ≡ 1 (mod 4).
Example 4.6 . Show that there are an infinite number of primes of the form 4 k + 1 and of the form 4 k + 3.
Solution . Suppose that there are a finite number of primes of the form 4k + 1, say p1, p2, . . . , pn. Let N = 4( p1p2 · · · pn)2 + 1. By Theorem 4.7, N
is only divisible by primes of the form 4 k + 1, but clearly N is not divisible by any of these primes, contradiction. Similarly, suppose that there are a finite number of primes of the form 4k + 3, say q1, q2, . . . , qm. Let M = 4 q1q2 · · · qm − 1. Then M ≡ 3 (mod 4), so M must be divisible by a prime of the form 4 k + 3, but M is not divisible by any of these primes, contradiction.
Example 4.7 . Show that if n is an integer greater than 1, then n does not divide 2 n − 1.
Solution . Let p be the least prime divisor of n. Then gcd( n, p − 1) = 1, and by Corollary 2.2, there exist integers x and y such that nx +( p−1) y = 1. If p | (2 n − 1), then 2 ≡ 2nx +( p−1) y ≡ (2 n)x(2 p−1)y ≡ 1 (mod p) by Fermat’s Little Theorem, contradiction. Therefore, p - (2 n − 1) ⇒ n - (2 n − 1).
Theorem 4.8 . Wilson’s Theorem . If p is a prime, then ( p − 1)! ≡ − 1(mod p). (See also Example 4.5.)
Proof . Consider the congruence x2 ≡ 1 (mod p). Then x2 − 1 ≡ (x −
1)( x + 1) ≡ 0, so the only solutions are x ≡ 1 and −1. Therefore, for each i,2 ≤ i ≤ p − 2, there exists a unique inverse j 6 = i of i, 2 ≤ j ≤ p − 2, modulo 16 p. Hence, when we group in pairs of inverses, (p − 1)! ≡ 1 · 2 · · · (p − 2) · (p − 1)
≡ 1 · 1 · · · 1 · (p − 1)
≡ − 1 (mod p).
Example 4.8 . Let {a1, a 2, . . . , a 101 } and {b1, b 2, . . . , b 101 } be complete residue systems modulo 101. Can {a1b1, a 2b2, . . . , a 101 b101 } be a complete residue system modulo 101?
Solution . The answer is no. Suppose that {a1b1, a 2b2, . . . , a 101 b101 } is a complete residue system modulo 101. Without loss of generality, assume that a101 ≡ 0 (mod 101). Then b101 ≡ 0 (mod 101), because if any other
bj was congruent to 0 modulo 101, then aj bj ≡ a101 b101 ≡ 0 (mod 101), contradiction. By Wilson’s Theorem, a1a2 · · · a100 ≡ b1b2 · · · b100 ≡ 100! ≡−1 (mod 101), so a1b1a2b2 · · · a100 b100 ≡ 1 (mod 101). But a101 b101 ≡ 0(mod 101), so a1b1a2b2 · · · a100 b100 ≡ 100! ≡ − 1 (mod 101), contradiction.
Theorem 4.9 . If p is a prime, then the congruence x2 + 1 ≡ 0 (mod p)has a solution iff p = 2 or p ≡ 1 (mod 4). (Compare to Theorem 7.1)
Proof . If p = 2, then x = 1 is a solution. If p ≡ 3 (mod 4), then by the remark to Theorem 4.7, no solutions exist. Finally, if p = 4 k + 1, then let
x = 1 · 2 · · · (2 k). Then
x2 ≡ 1 · 2 · · · (2 k) · (2 k) · · · 2 · 1
≡ 1 · 2 · · · (2 k) · (−2k) · · · (−2) · (−1) (multiplying by 2 k −1s)
≡ 1 · 2 · · · (2 k) · (p − 2k) · · · (p − 2) · (p − 1)
≡ (p − 1)! ≡ − 1 (mod p).
Theorem 4.10 . Let p be a prime such that p ≡ 1 (mod 4). Then there exist positive integers x and y such that p = x2 + y2.
Proof . By Theorem 4.9, there exists an integer a such that a2 ≡ − 1(mod p). Consider the set of integers of the form ax − y, where x and y
are integers, 0 ≤ x, y < √p. The number of possible pairs ( x, y ) is then (b√pc + 1) 2 > (√p)2 = p, so by pigeonhole principle, there exist integers 0 ≤ x1, x 2, y 1, y 2 < √p, such that ax 1−y1 ≡ ax 2−y2 (mod p). Let x = x1−x2
and y = y1 − y2. At least one of x and y is non-zero, and ax ≡ y ⇒ a2x2 ≡
17 −x2 ≡ y2 ⇒ x2 + y2 ≡ 0 (mod p). Thus, x2 + y2 is a multiple of p, and 0 < x 2 + y2 < (√p)2 + ( √p)2 = 2 p, so x2 + y2 = p.
Theorem 4.11 . Let n be a positive integer. Then there exist integers
x and y such that n = x2 + y2 iff each prime factor of n of the form 4 k + 3 appears an even number of times.
Theorem 4.12 . The Chinese Remainder Theorem (CRT) . If a1, a2, . . . ,
ak are integers, and m1, m2, . . . , mk are pairwise relatively prime integers, then the system of congruences
x ≡ a1 (mod m1),x ≡ a2 (mod m2),
...
x ≡ ak (mod mk)has a unique solution modulo m1m2 · · · mk.
Proof . Let m = m1m2 · · · mk, and consider m/m 1. This is relatively prime to m1, so there exists an integer t1 such that t1 · m/m 1 ≡ 1 (mod m1). Accordingly, let s1 = t1 · m/m 1. Then s1 ≡ 1 (mod m1) and s1 ≡ 0(mod mj ), j 6 = 1. Similarly, for all i, there exists an si such that si ≡ 1(mod mi) and si ≡ 0 (mod mj ), j 6 = i. Then, x = a1s1 + a2s2 + · · · + aksk is a solution to the above system. To see uniqueness, let x′ be another solution. Then x − x′ ≡ 0 (mod mi) for all i ⇒ x − x′ ≡ 0 (mod m1m2 · · · mk).
Remark . The proof shows explicitly how to find the solution x.
Example 4.9 . For a positive integer n, find the number of solutions of the congruence x2 ≡ 1 (mod n).
Solution . Let the prime factorization of n be 2 epe1
1
pe2
2
· · · pek
k
. By CRT,
x2 ≡ 1 (mod n) ⇔ x2 ≡ 1 (mod pei
i
) for all i, and x2 ≡ 1 (mod 2 e). We consider these cases separately. We have that x2 ≡ 1 (mod pei
i
) ⇔ x2 − 1 = ( x − 1)( x + 1) ≡ 0 (mod pei
i
). But pi cannot divide both x − 1 and x + 1, so it divides one of them; that is,
x ≡ ± 1 (mod pei
i
). Hence, there are two solutions. Now, if ( x − 1)( x + 1) ≡ 0 (mod 2 e), 2 can divide both x − 1 and x + 1, but 4 cannot divide both. For e = 1 and e = 2, it is easily checked that there are 1 and 2 solutions respectively. For e ≥ 3, since there is at most one factor 18 of 2 in one of x − 1 and x + 1, there must be at least e − 1 in the other, for their product to be divisible by 2 e. Hence, the only possibilities are x − 1 or
x + 1 ≡ 0, 2 e−1 (mod 2 e), which lead to the four solutions x ≡ 1, 2 e−1 − 1, 2e−1 + 1, and 2 e − 1. Now that we know how many solutions each prime power factor con-tributes, the number of solutions modulo n is simply the product of these, by CRT. The following table gives the answer:
e Number of solutions 0, 1 2k
2 2k+1
≥ 3 2k+2
Theorem 4.11 . Let m be a positive integer, let a and b be integers, and let k = gcd( a, m ). Then the congruence ax ≡ b (mod m) has k solutions or no solutions according as k | b or k - b.Problems 1. Prove that for each positive integer n there exist n consecutive positive integers, none of which is an integral power of a prime. (1989 IMO) 2. For an odd positive integer n > 1, let S be the set of integers x,1 ≤ x ≤ n, such that both x and x + 1 are relatively prime to n. Show that ∏
x∈S
x ≡ 1 (mod n).
Find all positive integer solutions to 3 x + 4 y = 5 z .(1991 IMO Short List) 4. Let n be a positive integer such that n + 1 is divisible by 24. Prove that the sum of all the divisors of n is divisible by 24. (1969 Putnam Mathematical Competition) 5. (Wolstenholme’s Theorem) Prove that if 1 + 12 + 13 + · · · + 1
p − 119 is expressed as a fraction, where p ≥ 5 is a prime, then p2 divides the numerator. 6. Let a be the greatest positive root of the equation x3 − 3x2 + 1 = 0. Show that ba1788 c and ba1988 c are both divisible by 17. (1988 IMO Short List) 7. Let {a1, a 2, . . . , a n} and {b1, b 2, . . . , b n} be complete residue systems modulo n, such that {a1b1, a 2b2, . . . , a nbn} is also a complete residue system modulo n. Show that n = 1 or 2. 8. Let m, n be positive integers. Show that 4 mn − m − n can never be a square. (1984 IMO Proposal)
5 Binomial Coefficients
For non-negative integers n and k, k ≤ n, the binomial coefficient (nk
) is defined as n!
k!( n − k)! ,
and has several important properties. By convention, (nk
) = 0 if k > n .In the following results, for polynomials f and g with integer coefficients, we say that f ≡ g (mod m) if m divides every coefficient in f − g.
Theorem 5.1 . If p is a prime, then the number of factors of p in n! is
⌊np
⌋
+
⌊ np2
⌋
+
⌊ np3
⌋
· · · .
It is also n − sn
p − 1 ,
where sn is the sum of the digits of n when expressed in base p.
Theorem 5.2 . If p is a prime, then
(pi
)
≡ 0 (mod p)20 for 1 ≤ i ≤ p − 1.
Corollary 5.3 . (1 + x)p ≡ 1 + xp (mod p).
Lemma 5.4 . For all real numbers x and y, bx + yc ≥ b xc + byc.
Proof . x ≥ b xc ⇒ x + y ≥ b xc + byc ∈ Z, so bx + yc ≥ b xc + byc.
Theorem 5.5 . If p is a prime, then
(pk
i
)
≡ 0 (mod p)for 1 ≤ i ≤ pk − 1.
Proof . By Lemma 5.4,
k
∑
j=1
(⌊ ipj
⌋
+
⌊pk − ipj
⌋)
≤
k
∑
j=1
⌊pk
pj
⌋
,
where the LHS and RHS are the number of factors of p in i!( pk − i)! and
pk! respectively. But,
⌊ ipk
⌋
=
⌊pk −ipk
⌋
= 0 and
⌊pk
pk
⌋
= 1, so the inequality is strict, and at least one factor of p divides (pk
i
).
Corollary 5.6 . (1 + x)pk
≡ 1 + xpk
(mod p).
Example 5.1 . Let n be a positive integer. Show that the product of n
consecutive positive integers is divisible by n!.
Solution . If the consecutive integers are m, m + 1, . . . , m + n − 1, then
m(m + 1) · · · (m + n − 1)
n! =
(m + n − 1
n
)
.
Example 5.2 . Let n be a positive integer. Show that (n + 1) lcm
(( n
0
)
,
(n
1
)
, . . . ,
(nn
))
= lcm(1 , 2, . . . , n + 1) .
(AMM E2686)
Solution . Let p be a prime ≤ n + 1 and let α (respectively β) be the highest power of p in the LHS (respectively RHS) of the above equality. Choose r so that pr ≤ n + 1 < p r+1 . Then clearly β = r. We claim that if pr ≤ m < p r+1 , then pr+1 -
(mk
)
for 0 ≤ k ≤ m. (∗)21 Indeed, the number of factors of p in (mk
) is
γ =
r
∑
s=1
(⌊ mps
⌋
−
⌊ kps
⌋
−
⌊m − kps
⌋)
.
Since each summand in this sum is 0 or 1, we have γ ≤ r; that is, () holds. For 0 ≤ k ≤ n, let
ak = ( n + 1)
(nk
)
= ( n − k + 1)
(n + 1
k
)
= ( k + 1)
(n + 1
k + 1
)
.
By ( ∗), pr+1 does not divide any of the integers (nk
), (n+1
k
), or (n+1
k+1
). Thus,
pr+1 can divide ak only if p divides each of the integers n + 1, n − k + 1, and
k + 1. This implies that p divides ( n + 1) − (n − k + 1) − (k + 1) = −1, contradiction. Therefore, pr+1 - ak. On the other hand, for k = pr − 1, we have that k ≤ n and ak = ( k + 1) (n+1
k+1
) is divisible by pr. Therefore,
β = r = α.
Theorem 5.7 . Lucas’s Theorem . Let m and n be non-negative integers, and p a prime. Let
m = mkpk + mk−1pk−1 + · · · + m1p + m0, and
n = nkpk + nk−1pk−1 + · · · + n1p + n0
be the base p expansions of m and n respectively. Then
(mn
)
≡
(mk
nk
)( mk−1
nk−1
)
· · ·
(m1
n1
)( m0
n0
)
(mod p).
Proof . By Corollary 5.6, (1 + x)m ≡ (1 + x)mk pk +mk−1pk−1+··· +m1p+m0
≡ (1 + x)pk mk (1 + x)pk−1mk−1 · · · (1 + x)pm 1 (1 + x)m0
≡ (1 + xpk
)mk (1 + xpk−1
)mk−1 · · · (1 + xp)m1 (1 + x)m0 (mod p).
By base p expansion, the coefficient of xn on both sides is
(mn
)
≡
(mk
nk
)( mk−1
nk−1
)
· · ·
(m1
n1
)( m0
n0
)
(mod p).
22 Corollary 5.8 . Let n be a positive integer. Let A(n) denote the number of factors of 2 in n!, and let B(n) denote the number of 1s in the binary expansion of n. Then the number of odd entries in the nth row of Pascal’s Triangle, or equivalently the number of odd coefficients in the expansion of (1 + x)n, is 2 B(n). Furthermore, A(n) + B(n) = n for all n.Useful Facts
• For a polynomial f with integer coefficients and prime p,[f (x)] pn
≡ f (xpn
) (mod p).
Problems 1. Let a and b be non-negative integers, and p a prime. Show that
(pa pb
)
≡
(ab
)
(mod p).
Let an be the last non-zero digit in the decimal representation of the number n!. Is the sequence a1, a2, a3, . . . eventually periodic? (1991 IMO Short List) 3. Find all positive integers n such that 2 n | (3 n − 1). 4. Find the greatest integer k for which 1991 k divides 1990 1991 1992
1992 1991 1990
.
(1991 IMO Short List) 5. For a positive integer n, let a(n) and b(n) denote the number of binomial coefficients in the nth row of Pascal’s Triangle that are congruent to 1 and 2 modulo 3 respectively. Prove that a(n) − b(n) is always a power of 2. 6. Let n be a positive integer. Prove that if the number of factors of 2 in
n! is n − 1, then n is a power of 2. 23 7. For a positive integer n, let
Cn = 1
n + 1
(2nn
)
,
and Sn = C1 + C2 + · · · + Cn.Prove that Sn ≡ 1 (mod 3) if and only if there exists a 2 in the base 3 expansion of n + 1.
6 Order of an Element
We know that if a is relatively prime to m, then there exists a positive integer
n such that an ≡ 1 (mod m). Let d be the smallest such n. Then we say that d is the order of a modulo m, denoted by ord m(a), or simply ord( a) if the modulus m is understood.
Theorem 6.1 . If a is relatively prime to m, then an ≡ 1 (mod m) iff ord( a) | n. Furthermore, an0 ≡ an1 iff ord( a) | (n0 − n1).
Proof . Let d = ord( a). It is clear that d | n ⇒ an ≡ 1 (mod m). On the other hand, if an ≡ 1 (mod m), then by the division algorithm, there exist integers q and r such that n = qd + r, 0 ≤ r < d . Then an ≡ aqd +r ≡
(ad)qar ≡ ar ≡ 1 (mod m). But r < d , so r = 0 ⇒ d | n. The second part of the theorem follows.
Remark . In particular, by Euler’s Theorem, ord( a) | φ(m).
Example 6.1 . Show that the order of 2 modulo 101 is 100.
Solution . Let d = ord(2). Then d | φ(101), or d | 100. If d < 100, then d divides 100/2 or 100/5; that is, d is missing at least one prime factor. However, 250 ≡ 1024 5 ≡ 14 5 ≡ 196 · 196 · 14 ≡ (−6) · (−6) · 14 ≡ − 1 (mod 101) ,
and 220 ≡ 1024 2 ≡ 14 2 ≡ − 6 (mod 101) ,
so d = 100.
Example 6.2 . Prove that if p is a prime, then every prime divisor of 2p − 1 is greater than p.24 Solution . Let q | (2 p − 1), where q is a prime. Then 2 p ≡ 1 (mod q), so ord(2) | p. But ord(2) 6 = 1, so ord(2) = p. And by Fermat’s Little Theorem, ord(2) | (q − 1) ⇒ p ≤ q − 1 ⇒ q > p .In fact, for p > 2, q must be of the form 2 kp + 1. From the above, ord(2) | (q − 1), or p | (q − 1) ⇒ q = mp + 1. Since q must be odd, m must be even.
Example 6.3 . Let p be a prime that is relatively prime to 10, and let n
be an integer, 0 < n < p . Let d be the order of 10 modulo p.(a) Show that the length of the period of the decimal expansion of n/p is
d.(b) Prove that if d is even, then the period of the decimal expansion of n/p
can be divided into two halves, whose sum is 10 d/ 2 − 1. For example, 1/7 = 0 .142857, so d = 6, and 142 + 857 = 999 = 10 3 − 1.
Solution . (a) Let m be the length of the period, and let n/p =0.a 1a2 . . . a m. Then 10 mnp = a1a2 . . . a m.a 1a2 . . . a m
⇒ (10 m − 1) np = a1a2 . . . a m,
an integer. Since n and p are relatively prime, p must divide 10 m − 1, so d
divides m. Conversely, p divides 10 d −1, so (10 d −1) n/p is an integer, with at most d digits. If we divide this integer by 10 d − 1, then we obtain a rational number, whose decimal expansion has period at most d. Therefore, m = d.(b) Let d = 2 k, so n/p = 0 .a 1a2 . . . a kak+1 . . . a 2k. Now p divides 10 d −1 = 10 2k − 1 = (10 k − 1)(10 k + 1). However, p cannot divide 10 k − 1 (since the order of 10 is 2 k), so p divides 10 k + 1. Hence, 10 knp = a1a2 . . . a k.a k+1 . . . a 2k
⇒ (10 k + 1) np = a1a2 . . . a k + 0 .a 1a2 . . . a k + 0 .a k+1 . . . a 2k
is an integer. This can occur iff a1a2 . . . a k +ak+1 . . . a 2k is a number consisting only of 9s, and hence, equal to 10 k − 1. Problems 25 1. Prove that for all positive integers a > 1 and n, n | φ(an − 1). 2. Prove that if p is a prime, then pp−1 has a prime factor that is congruent to 1 modulo p.3. For any integer a, set na = 101 a−100 ·2a. Show that for 0 ≤ a, b, c, d ≤
99 , n a + nb ≡ nc + nd (mod 10100) implies {a, b } = {c, d }.(1994 Putnam Mathematical Competition) 4. Show that if 3 ≤ d ≤ 2n+1 , then d - (a2n
1) for all positive integers a.
7 Quadratic Residues
Let m be an integer greater than 1, and a an integer relatively prime to m. If
x2 ≡ a (mod m) has a solution, then we say that a is a quadratic residue
of m. Otherwise, we say that a is a quadratic non-residue . Now, let p be an odd prime. Then the Legendre symbol
(ap
)
is assigned the value of 1 if a is a quadratic residue of p. Otherwise, it is assigned the value of −1.
Theorem 7.1 . Let p be an odd prime, and a and b be integers relatively prime to p. Then (a)
(ap
)
≡ a(p−1) /2 (mod p), and (b)
(ap
)( bp
)
=
(ab p
)
.
Proof . If the congruence x2 ≡ a (mod p) has a solution, then a(p−1) /2 ≡
xp−1 ≡ 1 (mod p), by Fermat’s Little Theorem. If the congruence x2 ≡ a
(mod p) has no solutions, then for each i, 1 ≤ i ≤ p − 1, there is a unique
j 6 = i, 1 ≤ j ≤ p − 1, such that ij ≡ a. Therefore, all the integers from 1 to
p − 1 can be arranged into ( p − 1) /2 such pairs. Taking their product,
a(p−1) /2 ≡ 1 · 2 · · · (p − 1) ≡ (p − 1)! ≡ − 1 (mod p),
26 by Wilson’s Theorem. Part (b) now follows from part (a).
Remark . Part (a) is known as Euler’s criterion.
Example 7.1 . Show that if p is an odd prime, then
(1
p
)
+
(2
p
)
· · · +
(p − 1
p
)
= 0 .
Solution . Note that 1 2, 2 2, . . . , (( p − 1) /2) 2 are distinct modulo p,and that (( p + 1) /2) 2, . . . , ( p − 1) 2 represent the same residues, simply in reverse. Hence, there are exactly ( p−1) /2 quadratic residues, leaving ( p−1) /2quadratic non-residues. Therefore, the given sum contains ( p − 1) /2 1s and (p − 1) /2 −1s.
Theorem 7.2 . Gauss’s Lemma . Let p be an odd prime and let a be relatively prime to p. Consider the least non-negative residues of a, 2 a, . . . , (( p − 1) /2) a modulo p. If n is the number of these residues that are greater than p/ 2, then (ap
)
= ( −1) n.
Theorem 7.3 . If p is an odd prime, then
(−1
p
)
= ( −1) (p−1) /2; that is,
(−1
p
)
=
{ 1 if p ≡ 1 (mod 4) ,
−1 if p ≡ 3 (mod 4) .
Proof . This follows from Theorem 4.9 (and Theorem 7.1).
Theorem 7.4 . If p is an odd prime, then
(2
p
)
= ( −1) (p2−1) /8; that is,
(2
p
)
=
{ 1 if p ≡ 1 or 7 (mod 8) ,
−1 if p ≡ 3 or 5 (mod 8) .
27 Proof . If p ≡ 1 or 5 (mod 8), then 2(p−1) /2
(p − 12
)
! ≡ 2 · 4 · 6 · · · (p − 1)
≡ 2 · 4 · 6 · · ·
(p − 12
)
·
(
−p − 32
)
· · · (−5) · (−3) · (−1)
≡ (−1) (p−1) /4
(p − 12
)
!
⇒ 2(p−1) /2 ≡ (−1) (p−1) /4 (mod p).
By Theorem 7.1,
(2
p
)
= ( −1) (p−1) /4. Hence,
(2
p
)
= 1 or −1 according as
p ≡ 1 or 5 (mod 8). Similarly, if p ≡ 3 or 7 (mod 8), then 2(p−1) /2
(p − 12
)
! ≡ 2 · 4 · 6 · · ·
(p − 32
)
·
(
−p − 12
)
· · · (−5) · (−3) · (−1)
≡ (−1) (p+1) /4
(p − 12
)
!
⇒ 2(p−1) /2 ≡ (−1) (p+1) /4 (mod p).
Hence,
(2
p
)
= 1 or −1 according as p ≡ 7 or 3 (mod 8).
Example 7.2 . Prove that if n is an odd positive integer, then every prime divisor of 2 n − 1 is of the form 8 k ± 1. (Compare to Example 6.2)
Solution . Let p | (2 n − 1), where p is prime. Let n = 2 m + 1. Then 2n ≡ 22m+1 ≡ 2(2 m)2 ≡ 1 (mod p) ⇒
(2
p
)
= 1 ⇒ p is of the form 8 k ± 1.
Theorem 7.5 . The Law of Quadratic Reciprocity . For distinct odd primes p and q, (pq
) ( qp
)
= ( −1) p−12 · q−12 .
Example 7.3 . For which primes p > 3 does the congruence x2 ≡ − 3(mod p) have a solution?
Solution . We seek p for which
(−3
p
)
=
(−1
p
) ( 3
p
)
= 1. By quadratic reciprocity, (3
p
) ( p
3
)
= ( −1) (p−1) /2 =
(−1
p
)
,
28 by Theorem 7.3. Thus, in general,
(−3
p
)
=
(−1
p
) ( 3
p
)
=
(p
3
) ( −1
p
)2
=
(p
3
)
.
And, ( p
3
) = 1 iff p ≡ 1 (mod 3). Since p 6 ≡ 4 (mod 6), we have that x2 ≡ − 3(mod p) has a solution iff p ≡ 1 (mod 6).
Example 7.4 . Show that if p = 2 n + 1, n ≥ 2, is prime, then 3 (p−1) /2 + 1 is divisible by p.
Solution . We must have that n is even, say 2 k, for otherwise p ≡ 0(mod 3). By Theorem 7.1,
(3
p
)
≡ 3(p−1) /2 (mod p).
However, p ≡ 1 (mod 4), and p ≡ 4k + 1 ≡ 2 (mod 3) ⇒ (p
3
) = −1, and by quadratic reciprocity,
(3
p
) ( p
3
)
= ( −1) (p−1) /2 = 1 ,
so (3
p
)
= −1 ⇒ 3(p−1) /2 + 1 ≡ 0 (mod p).
Useful Facts
• (a) If p is a prime and p ≡ 1 or 3 (mod 8), then there exist positive integers x and y such that p = x2 + 2 y2.(b) If p is a prime and p ≡ 1 (mod 6), then there exist positive integers
x and y such that p = x2 + 3 y2.Problems 1. Show that if p > 3 is a prime, then the sum of the quadratic residues among the integers 1, 2, . . . , p − 1 is divisible by p.2. Let Fn denote the nth Fibonacci number. Prove that if p > 5 is a prime, then
Fp ≡
(p
5
)
(mod p).
29 3. Show that 16 is a perfect 8 th power modulo p for any prime p.4. Let a, b, and c be positive integers that are pairwise relatively prime, and that satisfy a2 − ab + b2 = c2. Show that every prime factor of c is of the form 6 k + 1. 5. Let p be an odd prime and let ζ be a primitive pth root of unity; that is,
ζ is a complex number such that ζp = 1 and ζk 6 = 1 for 1 ≤ k ≤ p − 1. Let Ap and Bp denote the set of quadratic residues and non-residues modulo p, respectively. Finally, let α = ∑
k∈Ap
ζk and β = ∑
k∈Bp
ζk.For example, for p = 7, α = ζ + ζ2 + ζ4 and β = ζ3 + ζ5 + ζ6. Show that α and β are the roots of
x2 + x +1 −
(−1
p
)
p
4 = 0 .
8 Primitive Roots
If the order of g modulo m is φ(m), then we say that g is a primitive root
modulo m, or simply of m.
Example 8.1 . Show that 2 is a primitive root modulo 3 n for all n ≥ 1.
Solution . The statement is easily verified for n = 1, so assume the result is true for some n = k; that is, 2 φ(3 k ) ≡ 22·3k−1
≡ 1 (mod 3 k). Now, let d be the order of 2 modulo 3 k+1 . Then 2 d ≡ 1 (mod 3 k+1 ) ⇒ 2d ≡ 1 (mod 3 k), so 2 · 3k−1 | d. However, d | φ(3 k+1 ), or d | 2 · 3k. We deduce that d is either 2 · 3k−1 or 2 · 3k. Now we require the following lemma: Lemma. 2 2·3n−1
≡ 1 + 3 n (mod 3 n+1 ), for all n ≥ 1. This is true for n = 1, so assume it is true for some n = k. Then by assumption, 22·3k−1
= 1 + 3 k + 3 k+1 m for some integer m
⇒ 22·3k
= 1 + 3 k+1 + 3 k+2 M for some integer M (obtained by cubing)
⇒ 22·3k
≡ 1 + 3 k+1 (mod 3 k+2 ).
By induction, the lemma is proved. Therefore, 2 2·3k−1
≡ 1 + 3 k 6 ≡ 1 (mod 3 k+1 ), so the order of 2 modulo 3 k+1
is 2 · 3k, and again by induction, the result follows. 30 Corollary 8.2 . If 2 n ≡ − 1 (mod 3 k), then 3 k−1 | n.
Proof . The given implies 2 2n ≡ 1 (mod 3 k) ⇒ φ(3 k) | 2n, or 3 k−1 | n.
Theorem 8.3 . If m has a primitive root, then it has φ(φ(m)) (distinct) primitive roots modulo m.
Theorem 8.4 . The positive integer m has a primitive root iff m is one of 2, 4, pk, or 2 pk, where p is an odd prime.
Theorem 8.5 . If g is a primitive root of m, then gn ≡ 1 (mod m) iff
φ(m) | n. Furthermore, gn0 ≡ gn1 iff φ(m) | (n0 − n1).
Proof . This follows directly from Theorem 6.1.
Theorem 8.6 . If g is a primitive root of m, then the powers 1, g, g2,. . . , gφ(m)−1 represent each integer relatively prime to m uniquely modulo m.In particular, if m > 2, then gφ(m)/2 ≡ − 1 modulo m.
Proof . Clearly, each power gi is relatively prime to m, and there are
φ(m) integers relatively prime to m. Also, if gi ≡ gj (mod m), then gi−j ≡
1 ⇒ φ(m) | (i − j) by Theorem 8.6, so each of the powers are distinct modulo m. Hence, each integer relatively prime to m is some power gi
modulo m. Furthermore, there is a unique i, 0 ≤ i ≤ φ(m) − 1, such that
gi ≡ − 1 ⇒ g2i ≡ 1 ⇒ 2i = φ(m), or i = φ(m)/2.
Proposition 8.7 . Let m be a positive integer. Then the only solutions to the congruence x2 ≡ 1 (mod m) are x ≡ ± 1 (mod m) iff m has a primitive root.
Proof . This follows from Example 4.9.
Example 8.2 . For a positive integer m, let S be the set of positive integers less than m that are relatively prime to m, and let P be the product of the elements in S. Show that P ≡ ± 1 (mod m), with P ≡ − 1 (mod m)iff m has a primitive root.
Solution . We use a similar strategy as in the proof of Wilson’s Theorem. The result is clear for m = 2, so assume that m ≥ 3. We partition S as follows: Let A be the elements of S that are solutions to the congruence
x2 ≡ 1 (mod m), and let B be the remaining elements. The elements in B
can be arranged into pairs, by pairing each with its distinct multiplicative inverse. Hence, the product of the elements in B is 1 modulo m.The elements in A may also be arranged into pairs, by pairing each with 31 its distinct additive inverse, i.e. x and m − x. These must be distinct, because otherwise, x = m/ 2, which is not relatively prime to m. Note that their product is x(m − x) ≡ mx − x2 ≡ − 1 (mod m). Now if m has a primitive root, then by Proposition 8.7, A consists of only the two elements 1 and −1, so P ≡ − 1 (mod m). Otherwise, by Example 4.9, the number of elements of A is a power of two that is at least 4, so the number of such pairs in A is even, and P ≡ 1 (mod m).
Remark . For m prime, this simply becomes Wilson’s Theorem.
Theorem 8.8 .(1) If g is a primitive root of p, p a prime, then g or g + p is a primitive root of p2, according as gp−1 6 ≡ 1 (mod p2) or gp−1 ≡ 1 (mod p2). (2) If g is a primitive root of pk, where k ≥ 2 and p is prime, then g is a primitive root of pk+1 .By Theorem 8.6, given a primitive root g of m, for each a relatively prime to m, there exists a unique integer i modulo φ(m) such that gi ≡ a (mod m). This i is called the index of a with respect to the base g, denoted by ind g(a)(i is dependent on g, so it must be specified). Indices have striking similarity to logarithms, as seen in the following properties: (1) ind g(1) ≡ 0 (mod φ(m)), ind g(g) ≡ 1 (mod φ(m)), (2) a ≡ b (mod m) ⇒ ind g(a) ≡ ind g(b) (mod φ(m)), (3) ind g(ab ) ≡ ind g(a) + ind g(b) (mod φ(m)), (4) ind g(ak) ≡ k ind g(a) (mod φ(m)).
Theorem 8.9 . If p is a prime and a is not divisible by p, then the con-gruence xn ≡ a (mod p) has gcd( n, p − 1) solutions or no solutions according as
a(p−1) / gcd( n,p −1) ≡ 1 (mod p) or a(p−1) / gcd( n,p −1) 6 ≡ 1 (mod p).
Proof . Let g be a primitive root of p, and let i be the index of a with respect to g. Also, any solution x must be relatively prime to p, so let u be the index of x. Then the congruence xn ≡ a becomes gnu ≡ gi (mod p) ⇔
nu ≡ i (mod p − 1). Let k = gcd( n, p − 1). Since g is a primitive root of p,
k | i ⇔ gi(p−1) /k ≡ a(p−1) /k ≡ 1. The result now follows from Theorem 4.11. 32 Remark . Taking p to be an odd prime and n = 2, we deduce Euler’s criterion.
Example 8.3 Let n ≥ 2 be an integer and p = 2 n + 1. Show that if 3(p−1) /2 + 1 ≡ 0 (mod p), then p is a prime. (The converse to Example 7.4.)
Solution . From 3 (p−1) /2 ≡ 32n−1
≡ − 1 (mod p), we obtain 3 2n
≡ 1(mod p), so the order of 3 is 2 n = p−1, but the order also divides φ(p) ≥ p−1. Therefore, φ(p) = p − 1, and p is a prime.
Example 8.4 . Prove that if n = 3 k−1, then 2 n ≡ − 1 (mod 3 k). (A partial converse to Corollary 8.2.)
Solution . By Example 8.1, 2 is a primitive root of 3 k. Therefore, 2 has order φ(3 k) = 2 · 3k−1 = 2 n ⇒ 22n ≡ 1 ⇒ (2 n − 1)(2 n + 1) ≡ 0 (mod 3 k). However, 2 n − 1 ≡ (−1) 3k−1
− 1 ≡ 1 6 ≡ 0 (mod 3), so 2 n + 1 ≡ 0 (mod 3 k).
Example 8.5 . Find all positive integers n > 1 such that 2n + 1
n2
is an integer. (1990 IMO)
Solution . Clearly, n must be odd. Now assume that 3 k‖n; that is, 3 k
is the highest power of 3 dividing n. Then 3 2k | n2 | (2 n + 1) ⇒ 2n ≡ − 1(mod 3 2k) ⇒ 32k−1 | n, by Corollary 8.2 ⇒ 2k − 1 ≤ k ⇒ k ≤ 1, showing that n has at most one factor of 3. We observe that n = 3 is a solution. Suppose that n has a prime factor greater than 3; let p be the least such prime. Then p | (2 n +1) ⇒ 2n ≡ − 1 (mod p). Let d be the order of 2 modulo
p. Since 2 2n ≡ 1, d | 2n. If d is odd, then d | n ⇒ 2n ≡ 1, contradiction, so d is even, say d = 2 d1. Then 2 d1 | 2n ⇒ d1 | n. Also, d | (p − 1), or 2d1 | (p − 1) ⇒ d1 ≤ (p − 1) /2 < p . But d1 | n, so d1 = 1 or d1 = 3. If d1 = 1, then d = 2, and 2 2 ≡ 1 (mod p), contradiction. If d1 = 3, then d = 6, and 26 ≡ 1 (mod p), or p | 63 ⇒ p = 7. However, the order of 2 modulo 7 is 3, which is odd, again contradiction. Therefore, no such p can exist, and the only solution is n = 3. Useful Facts
• All prime divisors of the Fermat number 2 2n
1, n > 1, are of the form 2n+2 k + 1. 33 Problems 1. Let p be an odd prime. Prove that 1i + 2 i + · · · + ( p − 1) i ≡ 0 (mod p)for all i, 0 ≤ i ≤ p − 2. 2. Show that if p is an odd prime, then the congruence x4 ≡ − 1 (mod p)has a solution iff p ≡ 1 (mod 8). 3. Show that if a and n are positive integers with a odd, then a2n
≡ 1(mod 2 n+2 ). 4. The number 142857 has the remarkable property that multiplying it by 1, 2, 3, 4, 5, and 6 cyclically permutes the digits. What are other numbers that have this property? Hint: Compute 142857 × 7.
9 Dirichlet Series
Despite the intimidating name, Dirichlet series are easy to work with, and can provide quick proofs to certain number-theoretic identities, such as Example 3.2. Let α be a function taking the positive integers to the integers. Then we say that
f (s) =
∞
∑
n=1
α(n)
ns = α(1) + α(2) 2s + α(3) 3s + · · ·
is the Dirichlet series generating function (Dsgf) of the function α,which we denote by f (s) ↔ α(n). Like general generating functions, these generating functions are used to provide information about their correspond-ing number-theoretic functions, primarily through manipulation of the gen-erating functions. Let 1 denote the function which is 1 for all positive integers; that is, 1( n) = 1 for all n. Let δ1(n) be the function defined by
δ1(n) =
{ 1 if n = 1 ,
0 if n > 1.
It is easy to check that 1 and δ1 are multiplicative. 34 Now, let α and β be functions taking the positive integers to the integers. The convolution of α and β, denoted α ∗ β, is defined by (α ∗ β)( n) = ∑
d|n
α(d)β(n/d ).
Note that convolution is symmetric; that is, α ∗ β = β ∗ α.
Theorem 9.1 . Let f (s) ↔ α(n) and g(s) ↔ β(n). Then ( f · g)( s) ↔
(α ∗ β)( n). We now do three examples. The Dsgf of 1( n) is the well-known Riemann Zeta function ζ(s):
ζ(s) =
∞
∑
n=1
1
ns = 1 + 12s + 13s + · · · ,
so ζ(s) ↔ 1( n). This function will play a prominent role in this theory. What makes this theory nice to work with is that we may work with these functions at a purely formal level; no knowledge of the analytic properties of
ζ(s) or indeed of any other Dsgf is required. By Theorem 9.1, the number-theoretic function corresponding to ζ2(s) is
∑
d|n
1( d)1( n/d ) = ∑
d|n
1 = τ (n).
Hence, ζ2(s) ↔ τ (n). Finally, it is clear that 1 ↔ δ1(n). If α is a multiplicative function, then we can compute the Dsgf corre-sponding to α using the following theorem.
Theorem 9.2 . Let α be a multiplicative function. Then
∞
∑
n=1
α(n)
ns = ∏
p
∞
∑
k=0
α(pk)
pks = ∏
p
[1 + α(p)p−s + α(p2)p−2s + α(p3)p−3s + · · · ],
where the product on the right is taken over all prime numbers. 35 As before, if we take α = 1, then we obtain
ζ(s) = ∏
p
(1 + p−s + p−2s + p−3s + · · · )= ∏
p
( 11 − p−s
)
= 1
∏
p
(1 − p−s),
an identity that will be useful. We say that a positive integer n > 1 is square-free if n contains no repeated prime factors; that is, p2 - n for all primes p. With this in mind, we define the M¨ obius function μ as follows:
μ(n) =
1 if n = 1 ,
0 if n is not square-free, and (−1) k if n is square-free and has k prime factors .
It is easy to check that μ is multiplicative. By Theorem 9.2, the corresponding Dsgf is given by ∏
p
(1 − p−s) = 1
ζ(s).
Hence, 1 /ζ (s) ↔ μ(n), and this property makes the the seemingly mysterious function μ very important, as seen in the following theorem.
Theorem 9.3 . (M¨ obius Inversion Formula) Let α and β be functions such that
β(n) = ∑
d|n
α(d).
Then
α(n) = ∑
d|n
μ(n/d )β(d).
Proof . Let f (s) ↔ α(n) and g(s) ↔ β(n). The condition is equivalent to
β = α ∗ 1, or g(s) = f (s)ζ(s), and the conclusion is equivalent to α = β ∗ μ,or f (s) = g(s)/ζ (s).
Theorem 9.4 . Let f (s) ↔ α(n). Then for any integer k, f (s − k) ↔
nkα(n). 36 For more on Dirichlet series, and generating functions in general, see H. Wilf, Generatingfunctionology .Problems 1. Let α, β, and γ be functions taking the positive integers to the integers. (a) Prove that α ∗ δ1 = α.(b) Prove that ( α ∗ β) ∗ γ = α ∗ (β ∗ γ). (c) Prove that if α and β are multiplicative, then so is α ∗ β.2. Prove that the following relations hold:
ζ(s − 1)
ζ(s) ↔ φ(n),ζ(s)ζ(s − 1) ↔ σ(n),ζ(s)
ζ(2 s) ↔ | μ(n)|.
Let the prime factorization of a positive integer n > 1 be pe1
1
pe2
2
· · · pek
k
.Define the functions λ and θ by λ(n) = ( −1) e1+e2+··· +ek and θ(n) = 2 k.Set λ(1) = θ(1) = 1. Show that λ and θ are multiplicative, and that
ζ(2 s)
ζ(s) ↔ λ(n) and ζ2(s)
ζ(2 s) ↔ θ(n).
For all positive integers n, let
f (n) =
n
∑
m=1
n
gcd( m, n ).
(a) Show that f (n) = ∑
d|n
dφ (d). (b) Let n = pe1
1
pe2
2
· · · pek
k
1 be the prime factorization of n. Show that
f (n) =
(p2e1+1 1 + 1
p1 + 1
) ( p2e2+1 2 + 1
p2 + 1
)
· · ·
(p2ek +1 1 + 1
pk + 1
)
.
Verify Example 3.2 in one calculation. 37 6. Let id denote the identity function; that is, id( n) = n for all n. Verify each of the following identities in one calculation: (a) φ ∗ τ = σ.(b) μ ∗ 1 = δ1.(c) μ ∗ id = φ.(d) φ ∗ σ = id · τ .(e) σ ∗ id = 1 ∗ (id · τ ). 7. Let a1, a2, . . . , be the sequence of positive integers satisfying
∑
d|n
ad = 2 n
for all n. Hence, a1 = 2, a2 = 2 2 − 2 = 2, a3 = 2 3 − 2 = 6, a4 =24 − 2 − 2 = 12, and so on. Show that for all n, n | an.Hint: Don’t use the Dsgf of ( an)∞
1
; use the M¨ obius Inversion Formula. Bigger Hint: Consider the function f : [0 , 1] → [0 , 1] defined by f (x) =
{2x}, where {x} = x − b xc is the fractional part of x. Find how the formula in the problem relates to the function f (n) = f ◦ f ◦ · · · ◦ f
︸ ︷︷ ︸
n
.8. For all non-negative integers k, let σk be the function defined by
σk(n) = ∑
d|n
dk.
Thus, σ0 = τ and σ1 = σ. Prove that
ζ(s)ζ(s − k) ↔ σk(n).
10 Miscellaneous Topics
10.1 Pell’s Equations
Pell’s equations (or Fermat’s equations, as they are rightly called) are diophantine equations of the form x2 − dy 2 = N , where d is a positive non-square integer. There always exist an infinite number of solutions when
N = 1, which we characterize. 38 Theorem 10.1.1 . If ( a, b ) is the lowest positive integer solution of x2 −
dy 2 = 1, then all positive integer solutions are of the form (xn, y n) =
(
(a + b√d)n + ( a − b√d)n
2 , (a + b√d)n − (a − b√d)n
2√d
)
.
We will not give a proof here, but we will verify that every pair indicated by the formula is a solution. The pair ( xn, y n) satisfy the equations
xn + yn
√d = ( a + b√d)n, and
xn − yn
√d = ( a − b√d)n.
Therefore,
x2
n
− dy 2
n
= ( xn + yn
√d)( xn − yn
√d)= ( a + b√d)n(a − b√d)n
= ( a2 − db 2)n
= 1 ,
since ( a, b ) is a solution.
Remark . The sequences ( xn), ( yn) satisfy the recurrence relations xn =2ax n−1 − xn−2, yn = 2 ay n−1 − yn−2.For x2 − dy 2 = −1, the situation is similar. If ( a, b ) is the least positive solution, then the ( xn, y n) as above for n odd are the solutions of x2 − dy 2 =
−1, and the ( xn, y n) for n even are the solutions of x2 − dy 2 = 1.
Example 10.1.1 Find all solutions in pairs of positive integers ( x, y ) to the equation x2 − 2y2 = 1.
Solution . We find that the lowest positive integer solution is (3,2), so all positive integer solutions are given by (xn, y n) =
(
(3 + 2 √2) n + (3 − 2√2) n
2 , (3 + 2 √2) n − (3 − 2√2) n
2√2
)
.
The first few solutions are (3,2), (17,12), and (99,70). 39 Example 10.1.2 . Prove that the equation x2 −dy 2 = −1 has no solution in integers if d ≡ 3 (mod 4).
Solution . It is apparent that d must have a prime factor of the form 4 k +3, say q. Then x2 ≡ − 1 (mod q), which by Theorem 4.9 is a contradiction. Problems 1. In the sequence 12, 53, 11 8 , 27 19 , . . . ,
the denominator of the nth term ( n > 1) is the sum of the numerator and the denominator of the ( n − 1) th term. The numerator of the nth
term is the sum of the denominators of the nth and ( n−1) th term. Find the limit of this sequence. (1979 Atlantic Region Mathematics League) 2. Let x0 = 0, x1 = 1, xn+1 = 4 xn − xn−1, and y0 = 1, y1 = 2, yn+1 =4yn − yn−1. Show for all n ≥ 0 that y2
n
= 3 x2
n
(1988 Canadian Mathematical Olympiad) 3. The polynomials P , Q are such that deg P = n, deg Q = m, have the same leading coefficient, and P 2(x) = ( x2 − 1) Q2(x) + 1. Show that
P ′(x) = nQ (x). (1978 Swedish Mathematical Olympiad, Final Round)
10.2 Farey Sequences
The nth Farey sequence is the sequence of all reduced rationals in [0,1], with both numerator and denominator no greater than n, in increasing order. Thus, the first 5 Farey sequences are: 0/1, 1/1, 0/1, 1/2, 1/1, 0/1, 1/3, 1/2, 2/3, 1/1, 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1, 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1. Properties of Farey sequences include the following: 40 (1) If a/b and c/d are consecutive fractions in the same sequence, in that order, then ad − bc = 1. (2) If a/b , c/d , and e/f are consecutive fractions in the same sequence, in that order, then a + eb + f = cd.
(3) If a/b and c/d are consecutive fractions in the same sequence, then among all fractions between the two, ( a + c)/(b + d) (reduced) is the unique fraction with the smallest denominator. For proofs of these and other interesting properties, see Ross Honsberger, “Farey Sequences”, Ingenuity in Mathematics .Problems 1. Let a1, a2, . . . , am be the denominators of the fractions in the nth Farey sequence, in that order. Prove that 1
a1a2
1
a2a3
· · · + 1
am−1am
= 1 .
10.3 Continued Fractions
Let a0, a1, . . . , an be real numbers, all positive, except possibly a0. Then let
〈a0, a 1, . . . , a n〉 denote the continued fraction
a0 + 1
a1 + · · · + 1
an−1 + 1
an
.
If each ai is an integer, then we say that the continued fraction is simple .Define sequences ( pk) and ( qk) as follows:
p−1 = 0 , p0 = a0, and pk = akpk−1 + pk−2,q−1 = 0 , q0 = 1 , and qk = akpk−1 + qk−2, for k ≥ 1.
Theorem 10.3.1 . For all x > 0 and k ≥ 1,
〈a0, a 1, . . . , a k−1, x 〉 = xp k−1 + pk−2
xq k−1 + qk−2
.
41 In particular,
〈a0, a 1, . . . , a k〉 = pk
qk
.
Theorem 10.3.2 . For all k ≥ 0, (1) pkqk−1 − pk−1qk = ( −1) k−1,(2) pkqk−2 − pk−2qk = ( −1) kak.Define ck to be the kth convergence 〈a0, a 1, . . . , a k〉 = pk/q k.
Theorem 10.3.3 . c0 < c 2 < c 4 < · · · < c 5 < c 3 < c 1.For a nice connection between continued fractions, linear diophantine equations, and Pell’s equations, see Andy Liu, “Continued Fractions and Diophantine Equations”, Volume 3, Issue 2, Mathematical Mayhem .Problems 1. Let a = 〈1, 2, . . . , 99 〉 and b = 〈1, 2, . . . , 99 , 100 〉. Prove that
|a − b| < 199!100! .
(1990 Tournament of Towns) 2. Evaluate
8
√√√√√2207 − 12207 − 12207 − · · ·
.
Express your answer in the form a+b√cd , where a, b, c, d are integers. (1995 Putnam)
10.4 The Postage Stamp Problem
Let a and b be relatively prime positive integers greater than 1. Consider the set of integers of the form ax + by , where x and y are non-negative integers. The following are true: (1) The greatest integer that cannot be written in the given form is ( a −
1)( b − 1) − 1 = ab − a − b.42 (2) There are 12 (a − 1)( b − 1) positive integers that cannot be written in the given form. (3) For all integers t, 0 ≤ t ≤ ab − a − b, t can be written in the given form iff ab − a − b − t cannot be. (If you have not seen or attempted this enticing problem, it is strongly suggested you have a try before reading the full solution.) Before presenting the solution, it will be instructive to look at an example. Take a = 12 and b = 5. The first few non-negative integers, in rows of 12, with integers that cannot be written in the given form in bold, are shown: 0 1 2 3 4 5 6 7 8 9 10 11
12 13 14 15 16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 With this arrangement, one observation should become immediately ap-parent, namely that bold numbers in each column end when they reach a multiple of 5. It should be clear that when reading down a column, once one hits an integer that can be written in the given form, then all successive integers can be as well, since we are adding 12 for each row we go down. It will turn out that this one observation is the key to the solution.
Proof . Define a grapefruit to be an integer that may be written in the given form. For each i, 0 ≤ i ≤ a − 1, let mi be the least non-negative integer such that b | (i + am i). It is obvious that for k ≥ mi, i + ak is a grapefruit. We claim that for 0 ≤ k ≤ mi − 1, i + ak is not a grapefruit. It is sufficient to show that i + a(mi − 1) is not a grapefruit, if mi ≥ 1. Let i + am i = bn i, ni ≥ 0. Since i + a(mi − b) = b(ni − a), mi must be strictly less than b; otherwise, we can find a smaller mi. Then i + a(mi − b) ≤
a−1−a = −1, so ni < a , or ni ≤ a−1. Suppose that ax +by = i+a(mi −1) =
bn i − a, for some non-negative integers x and y. Then a(x + 1) = b(ni − y), so ni − y is positive. Since a and b are relatively prime, a divides ni − y.However, ni ≤ a − 1 ⇒ ni − y ≤ a − 1, contradiction. Therefore, the greatest non-grapefruit is of the form bn i − a, ni ≤ a − 1. The above argument also shows that all positive integers of this form are also non-grapefruits. Hence, the greatest non-grapefruit is b(a−1) −a = ab −a−b,proving (1). 43 Now, note that there are mi non-grapefruits in column i. The above tells us the first grapefruit appearing in column i is nib. Since 0, b, 2 b, . . . , ( a−1) b
appear in different columns (because a and b are relatively prime), and there are a columns, we conclude that as i varies from 0 to a − 1, ni takes on 0, 1, . . . , a − 1, each exactly once. Therefore, summing over i, 0 ≤ i ≤ a − 1,
∑
i
(i + am i) = ∑
i
i + ∑
i
am i = a(a − 1) 2 + a ∑
i
mi
= ∑
i
bn i = a(a − 1) b
2
⇒ a ∑
i
mi = a(a − 1)( b − 1) 2
⇒ ∑
i
mi = (a − 1)( b − 1) 2 ,
proving (2). Finally, suppose that ax 1 + by 1 = t, and ax 2 + by 2 = ab − a − b − t, for some non-negative integers x1, x2, y1, and y2. Then a(x1 + x2) + b(y1 + y2) =
ab − a − b, contradicting (1). So, if we consider the pairs ( t, ab − a − b − t), 0 ≤ t ≤ (a − 1)( b − 1) /2 − 1, at most one element in each pair can be written in the given form. However, we have shown that exactly ( a − 1)( b − 1) /2 integers cannot be written in the given form, which is the number of pairs. Therefore, exactly one element of each pair can be written in the given form, proving (3).
Remark . There is a much shorter proof using Corollary 2.4. Can you find it? For me, this type of problem epitomizes problem solving in number theory, and generally mathematics, in many ways. If I merely presented the proof by itself, it would look artificial and unmotivated. However, by looking at a specific example, and finding a pattern, we were able to use that pattern as a springboard and extend it into a full proof. The algebra in the proof is really nothing more than a translation of observed patterns into formal notation. (Mathematics could be described as simply the study of pattern.) Note also that we used nothing more than very elementary results, showing how powerful basic concepts can be. It may have been messy, but one should never be afraid to get one’s hands dirty; indeed, the deeper you go, the 44 more you will understand the importance of these concepts and the subtle relationships between them. By trying to see an idea through to the end, one can sometimes feel the proof almost working out by itself. The moral of the story is: A simple idea can go a long way. For more insights on the postage stamp problem, see Ross Honsberger, “A Putnam Paper Problem”, Mathematical Gems II .Problems 1. Let a, b, and c be positive integers, no two of which have a common divisor greater than 1. Show that 2 abc − ab − bc − ca is the largest integer that cannot be expressed in the form xab + yca + zab , where x,
y, and z are non-negative integers. (1983 IMO)
References
A. Adler & J. Coury, The Theory of Numbers , Jones and Bartlett I. Niven & H. Zuckerman, An Introduction to the Theory of Numbers ,John Wiley & Sons c© First Version October 1995 c© Second Version January 1996 c© Third Version April 1999 c© Fourth Version May 2000 Thanks to Ather Gattami for an improvement to the proof of the Postage Stamp Problem. This document was typeset under L ATEX, and may be freely distributed provided the contents are unaltered and this copyright notice is not removed. Any comments or corrections are always welcomed. It may not be sold for profit or incorporated in commercial documents without the express permis-sion of the copyright holder. So there. 45 |
1048 | https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-probability-statistics/cc-7th-basic-prob/v/basic-probability | Theoretical probability | Statistics and Probability (video) | Khan Academy
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7th grade math
Course: 7th grade math>Unit 7
Lesson 1: Basic probability
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Basic probability
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Intro to theoretical probability
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We give you an introduction to probability through the example of flipping a quarter and rolling a die.Created by Sal Khan.
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Lyndsey Schimmelpenningh 13 years ago Posted 13 years ago. Direct link to Lyndsey Schimmelpenningh's post “what does p(h) mean?” more what does p(h) mean?
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ace 8 years ago Posted 8 years ago. Direct link to ace's post “P(H) means the probabilit...” more P(H) means the probability of heads, or probability x heads.
P = Probability H = Heads
2 comments Comment on ace's post “P(H) means the probabilit...”
(14 votes)
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Malay Kedia 7 years ago Posted 7 years ago. Direct link to Malay Kedia's post “At 2:38, if we assume tha...” more At 2:38 , if we assume that the coin can land on a corner and stand straight, will the probability become 1/3?
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Gabrielle Stagg 5 years ago Posted 5 years ago. Direct link to Gabrielle Stagg's post “I would assume that if y...” more I would assume that if you can stand the coin straight up, without flipping it, you would have a chance of landing it on it's corner, it would just be very unlikely. Therefore, it would not be 1/3, because it does not have the same probability as the heads/tails. I would assume that the probability of it landing straight up could be about as low as 11/100, just because the coin is very thin, and (assuming it is not a sphere) your faces of the coin are wider.
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Anoushka B. 9 years ago Posted 9 years ago. Direct link to Anoushka B.'s post “1. Why does a larger numb...” more 1. Why does a larger number of experiments bring the percentage of say getting heads closer to fifty percent?
How would you describe the probability of getting all tails in the flipping coin experiment?
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Luis 5 years ago Posted 5 years ago. Direct link to Luis's post “Is the Law of Large Numbe...” more Is the Law of Large Numbers:
If you flip a coin #1 time you can have:
{[H] or [T]}
If you flip repet it 2 times you can have:
{[H,H],[H,T], or [T,H], [T,T]}
Now for the important part. If you don't care about the order you could say that the event [H,T] is equal to the event [T,H] so it'd be the same as:
{[H,H], 2[H,T], [T,T]}
the probability of each event would be:
P([H,H]) = 1/4
P([H,T]) = P([T,H]) = 2/4 =1/2
P([T,T]) = 1/4
So flip the coin 100 times and you would see that there are more combinations of HEADS & TAILS that add up to 50% each than any other.
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💙Rohita💙 9 years ago Posted 9 years ago. Direct link to 💙Rohita💙's post “Alan, Beth Carlos and Dia...” more Alan, Beth Carlos and Diana were discussing their possible grades in mathematics class this grading period. Alan said, "If I get an A, then Beth will get an A." Beth said, "If I get an A, then Carlos will get an A." Carlos said, "If I get an A, then Diana will get an A." All of these statements were true, but only two of the students received an A. Which two received A's?
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Judith Gibson 9 years ago Posted 9 years ago. Direct link to Judith Gibson's post “Take this one person at a...” more Take this one person at a time.
If Alan gets an A, how many other people get an A?
If Beth gets an A, how many other people get an A?
If Carlos gets an A, how many other people get an A?
That should lead you to the correct answer.
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Interstellar 2 years ago Posted 2 years ago. Direct link to Interstellar's post “It basically means the pr...” more It basically means the probability (p) of the coin landing on heads (h).
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KSS 2 years ago Posted 2 years ago. Direct link to KSS's post “Theoretically, if I flipp...” more Theoretically, if I flipped a coin 1,000 times, it would land on heads 500 times and on tails 500 times. But that's just theoretical. What if I flipped a coin 1,000 times, and every single time it landed on heads? Is there a way to calculate the chances of that happening? (even if it is very small?)
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joshua 2 years ago Posted 2 years ago. Direct link to joshua's post “First, what's the probabi...” more First, what's the probability of getting a head per flip? It is 1 / 2 = 0.5.
Now, what's the probability of getting 2 heads in 2 flip? You have to flip head for the first time, and for the second time as well, so it will be 0.5 0.5 = 0.25.
You can expand this to 1000 times, which is 0.5^1000.
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☯Foxy&AtCore☯ 2 years ago Posted 2 years ago. Direct link to ☯Foxy&AtCore☯'s post “With the dice, can't you ...” more With the dice, can't you still get a 2 and a 3 by rolling a 5, or maybe a 6? You would still have a 2 and a 3.
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joshua 2 years ago Posted 2 years ago. Direct link to joshua's post “I don't think you are sur...” more I don't think you are sure what you are even asking. Please explain how by rolling a 5 or 6 you can get both 2 and 3 on one dice? It's impossible.
P(2 and 3) means getting both the number 2 and 3 on the dice. As you can only get 1 integer from 1 to 6 inclusively from the dice per throw, you cannot get 2 integer at the same time.
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manogna.dyagala 2 years ago Posted 2 years ago. Direct link to manogna.dyagala's post “If I had a condition such...” more If I had a condition such that; I were to flip a coin, and if it landed on heads, I would get to roll the die, and if it landed on tails, I toss the coin again. What is the possibility that I'd get a head and a 6, and three tails in a row??
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MysteriousEuler 2 years ago Posted 2 years ago. Direct link to MysteriousEuler's post “You can basically just mu...” more You can basically just multiply the chances out: 1/2 1/6 1/2 1/2 1/2 = 1/96
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kellan 2 years ago Posted 2 years ago. Direct link to kellan's post “when he says that if you ...” more when he says that if you flip a quarter it would be 50/50 that is not true actually since the heads side is slightly bulges more then te tails side it is a 51/49 chance
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30kmahapatro 3 years ago Posted 3 years ago. Direct link to 30kmahapatro's post “Try to think about a simp...” more Try to think about a simple question, like Joe needs to win the car lottery. If there are 19 people in the lottery, the chance that Joe will win is 1/19. The less the total, the more the probability.
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What I want to do in this video is give you at least a basic overview of probability. Probability, a word that you've probably heard a lot of, and you are probably a little bit familiar with it. But hopefully, this will give you a little deeper understanding. Let's say that I have a fair coin over here. And so when I talk about a fair coin, I mean that it has an equal chance of landing on one side or another. So you can maybe view it as the sides are equal, their weight is the same on either side. If I flip it in the air, it's not more likely to land on one side or the other. It's equally likely. And so you have one side of this coin. So this would be the heads I guess. Try to draw George Washington. I'll assume it's a quarter of some kind. And the other side, of course, is the tails. So that is heads. The other side right over there is tails. And so if I were to ask you, what is the probability-- I'm going to flip a coin. And I want to know what is the probability of getting heads. And I could write that like this-- the probability of getting heads. And you probably, just based on that question, have a sense of what probability is asking. It's asking for some type of way of getting your hands around an event that's fundamentally random. We don't know whether it's heads or tails, but we can start to describe the chances of it being heads or tails. And we'll talk about different ways of describing that. So one way to think about it, and this is the way that probability tends to be introduced in textbooks, is you say, well, look, how many different, equally likely possibilities are there? So how many equally likely possibilities. So number of equally-- let me write equally-- of equally likely possibilities. And of the number of equally possibilities, I care about the number that contain my event right here. So the number of possibilities that meet my constraint, that meet my conditions. So in the case of the probability of figuring out heads, what is the number of equally likely possibilities? Well, there's only two possibilities. We're assuming that the coin can't land on its corner and just stand straight up. We're assuming that it lands flat. So there's two possibilities here, two equally likely possibilities. You could either get heads, or you could get tails. And what's the number of possibilities that meet my conditions? Well, there's only one, the condition of heads. So it'll be 1/2. So one way to think about it is the probability of getting heads is equal to 1/2. If I wanted to write that as a percentage, we know that 1/2 is the same thing as 50%. Now, another way to think about or conceptualize probability that will give you this exact same answer is to say, well, if I were to run the experiment of flipping a coin-- so this flip, you view this as an experiment. I know this isn't the kind of experiment that you're used to. You know, you normally think an experiment is doing something in chemistry or physics or all the rest. But an experiment is every time you do, you run this random event. So one way to think about probability is if I were to do this experiment, an experiment many, many, many times-- if I were to do it 1,000 times or a million times or a billion times or a trillion times-- and the more the better-- what percentage of those would give me what I care about? What percentage of those would give me heads? And so another way to think about this 50% probability of getting heads is if I were to run this experiment tons of times, if I were to run this forever, an infinite number of times, what percentage of those would be heads? You would get this 50%. And you can run that simulation. You can flip a coin. And it's actually a fun thing to do. I encourage you to do it. If you take 100 or 200 quarters or pennies, stick them in a big box, shake the box so you're kind of simultaneously flipping all of the coins, and then count how many of those are going to be heads. And you're going to see that the larger the number that you are doing, the more likely you're going to get something really close to 50%. And there's always some chance-- even if you flipped a coin a million times, there's some super-duper small chance that you would get all tails. But the more you do, the more likely that things are going to trend towards 50% of them are going to be heads. Now, let's just apply these same ideas. And while we're starting with probability, at least kind of the basic, this is probably an easier thing to conceptualize. But a lot of times, this is actually a helpful one, too, this idea that if you run the experiment many, many, many, many times, what percentage of those trials are going to give you what you're asking for. In this case, it was heads. Now, let's do another very typical example when you first learn probability. And this is the idea of rolling a die. So here's my die right over here. And of course, you have, you know, the different sides of the die. So that's the 1. That's the 2. And that's the 3. And what I want to do-- and we know, of course, that there are-- and I'm assuming this is a fair die. And so there are six equally likely possibilities. When you roll this, you could get a 1, a 2, a 3, a 4, a 5, or a 6. And they're all equally likely. So if I were to ask you, what is the probability given that I'm rolling a fair die-- so the experiment is rolling this fair die, what is the probability of getting a 1? Well, what are the number of equally likely possibilities? Well, I have six equally likely possibilities. And how many of those meet my conditions? Well, only one of them meets my condition, that right there. So there is a 1/6 probability of rolling a 1. What is the probability of rolling a 1 or a 6? Well, once again, there are six equally likely possibilities for what I can get. There are now two possibilities that meet my conditions. I could roll a 1 or I could roll a 6. So now there are two possibilities that meet my constraints, my conditions. There is a 1/3 probability of rolling a 1 or a 6. Now, what is the probability-- and this might seem a little silly to even ask this question, but I'll ask it just to make it clear. What is the probability of rolling a 2 and a 3? And I'm just talking about one roll of the die. Well, in any roll of the die, I can only get a 2 or a 3. I'm not talking about taking two rolls of this die. So in this situation, there's six possibilities, but none of these possibilities are 2 and a 3. None of these are 2 and a 3. 2 and a 3 cannot exist. On one trial, you cannot get a 2 and a 3 in the same experiment. Getting a 2 and a 3 are mutually exclusive events. They cannot happen at the same time. So the probability of this is actually 0. There's no way to roll this normal die and all of a sudden, you get a 2 and a 3, in fact. And I don't want to confuse you with that, because it's kind of abstract and impossible. So let's cross this out right over here. Now, what is the probability of getting an even number? So once again, you have six equally likely possibilities when I roll that die. And which of these possibilities meet my conditions, the condition of being even? Well, 2 is even, 4 is even, and 6 is even. So 3 of the possibilities meet my conditions, meet my constraints. So this is 1/2. If I roll a die, I have a 1/2 chance of getting an even number.
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1049 | https://math.libretexts.org/Bookshelves/Algebra/Algebra_and_Trigonometry_1e_(OpenStax)/12%3A_Analytic_Geometry/12.03%3A_The_Parabola/12.3E%3A_The_Parabola_(Exercises) | 12.3E: The Parabola (Exercises) - Mathematics LibreTexts
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12.3: The Parabola
12: Analytic Geometry
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Thu, 26 Dec 2024 22:31:01 GMT
12.3E: The Parabola (Exercises)
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Contents
1. Home
2. Bookshelves
3. Algebra
4. Algebra and Trigonometry 1e (OpenStax)
5. 12: Analytic Geometry
6. 12.3: The Parabola
7. 12.3E: The Parabola (Exercises)
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Algebra and Trigonometry 1e (OpenStax)
Front Matter
1: Prerequisites
2: Equations and Inequalities
3: Functions
4: Linear Functions
5: Polynomial and Rational Functions
6: Exponential and Logarithmic Functions
7: The Unit Circle - Sine and Cosine Functions
8: Periodic Functions
9: Trigonometric Identities and Equations
10: Further Applications of Trigonometry
11: Systems of Equations and Inequalities
12: Analytic Geometry
13: Sequences, Probability, and Counting Theory
Back Matter
12.3E: The Parabola (Exercises)
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12.4: Rotation of Axes
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For the following exercises, write the equation of the parabola in standard form. Then give the vertex, focus, and directrix.
y 2=12x
(x+2)2=1 2(y−1)
y 2−6y−6x−3=0
x 2+10x−y+23=0
For the following exercises, graph the parabola, labeling vertex, focus, and directrix.
x 2+4y=0
(y−1)2=1 2(x+3)
x 2−8x−10y+46=0
2y 2+12y+6x+15=0
For the following exercises, write the equation of the parabola using the given information.
Focus at (-4,0) ; directrix is x=4
Focus at (2,9 8); directrix is y=7 8
A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 5 feet across at its opening and 1.5 feet deep.
This page titled 12.3E: The Parabola (Exercises) 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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12.1: The EllipseThe key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other...
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Question: Ten students namely viz Jaidev, Jaya, Jeeva, Janaki, Jyothi, Jaspal, Jamuna, Jagadish, Jayanthi and Jasmine of ten different colleges but not necessarily in the same order have an exam on five different days starting from Monday to Friday of the same week. Each student has an exam at two different time slots, i.e 09.00 AM or 11.00 A.M. The number of people who have an exam between Jamuna and Janaki is the same as the number of people who have an exam between Jeeva and Jagadish. Only two people have an exam between Jaspal and Jasmine. Neither Jyothi nor Jamuna does not have an exam on Friday. Jayanthi has an exam on Tuesday at 09.00 A.M. Jagadish does not have an exam at 11.00 AM. Janaki does not have an exam on any one of the days after Jyothi. Jayanthi does not have an exam on any of the days before Jamuna. Jaspal does not have an exam on any of the days after Jagadish. Jaya has exam immediately before Jayanthi. The one who has an exam at 09.00 A.M. immediately before Jasmine. Janaki has exam immediately after the day of one who has an exam on Monday. Jaspal does not have an exam at 11.00 A.M. Only three people have an exam between Jamuna and Jyothi.
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Transcript:
ten students namely Jaidev Jaya Jeeva Johnny you t just Paul Jamuna jugdish gentian Jasmine of 10 different colleges but not necessarily in the same order have examined five different days starting from Monday to Friday of the same week each student have exam at two different time slots 9:00 a.m. or 11 a.m. okay so if you try and understand Monday - there are like how many students ten students names are being given but there only five this Monday to Friday and then there are two time stores so it looks like five days two slots per day total total ten slots and you have to figure out which person has taken the exam in which slot like Monday 9:00 a.m. Monday lavinia Tuesday 9:00 a.m. to 11:00 a.m. and so on right total 10 slots there and I think the clue starts from this point onwards the number of people who have exam between Jamuna and John Key's same as the number of people who have example in GUI and jugdish only two people have exam between Jess just Pauline Jasmine neither Duty nor Jim not does not have examined Oh neither duty nor Jamuna does not have exam on Friday this doesn't look a proper English statement neither duty nor Jamna have exam on Friday right what does not have exam on Friday neither Jyoti not Jamuna I have exam on Friday means what these two girls will not have exams on Friday but neither Jyoti no Tom not does not have exam on Friday it's like negative negative makes positive right neither nor is negative does not is negative sentence becomes positive which means Jo teen Jamuna have exam on faculty so they mean that you have to clear this part I'm highlighting it this sentence is not very clear neither to t-norm nor does not have excellent right here I mean assuming that you mean that Jew theorem does not have exam on Friday we can try it out let's see the next statement says J and T has examined used at 9:00 a.m. so this is something which is straight given to us please does not have exam at leve named John he does not have exam on any of the days after duty so which means John Key has exam before duty J and he does not have examine any of the days before Jamuna just Paul does not have examining in the days after Jagdish jr. has exam immediately before JN t the one who has exam at 9:00 a.m. immediately before what the one who has exam at 9:00 a.m. immediately before just mean what is this the one who has exam at 9:00 a.m. immediately before this looks incomplete to me the one who has exam at 9:00 a.m. immediately before just means what does it mean John he has exam immediately after the day of the one who has exam on Monday is fine which means Johnny key is on Tuesday right John Key has exam immediately after the day of one who has exam on Monday so John gives on Tuesday just Paul does not have exam at 11:00 a.m. only three people have examined Jamuna and Jyoti so I think couple of confusions here but let's give it a shot so I think it looks like this Monday Tuesday we have Wednesday Thursday and Friday these are the slots I mean is that the days and then there are two slots right 9:00 a.m. and 11:00 a.m. so what do we do anything that is given directly I think one point which was given directly here is Shanthi has examined Tuesday at 9:00 a.m. so Jane thing is on Tuesday nine a m-- hey Tuesday 9:00 aim that Scott is done so Jane this monk Chuck DS does not have exam at 11:00 a.m. now what does it mean I mean of course I've started from middle but yeah because some direct stick museum junkies does not have exam at 11:00 it means what if it does not have exam at 11:00 a.m. it has to be 9:00 a.m. so jugdish I can say it is at nine a m-- I mean I don't know which day so Judy said my name just put it around that okay the number of people who have exam between Jamuna and John Key's same as the number of people who have between g1 jugdish to this point anyway cannot be used immediately right number of people between German and John Key's same as those between Gy and generation only two people have examined we just pearl and just mean only two people have exam between just polish ask me this also cannot be use immediately then John Key does not have exam on any of the days after duty John Key does not have examine any of the days after duty which means John Key has exam before duty but not useful immediately JT does not have exam on any of the days before Jamuna get this this is interesting JMP does not have exam on any of the days before Jamuna so Jane this exam is not before Jamuna Jane D does not have exam on any day before Jamuna which means gently has exam after Jamuna are able to follow this statement indirectly means that Jen chieh in these after Jamuna Chanti does not have her exam before Jamuna so if it is not before samurai it has to be after gemenon gentillesse exam is after Jamuna now we know that Jane T is on Tuesday so clearly Jamuna has to be on Monday so which slog on Monday 9 a male named you don't know so I'll just write Jamuna here Jamuna will come on Monday 9 or 11 that we don't know right just Paul does not have an exam on any of their is after check these so just Paul will be before check these but we don't know where Jagdish is Jaya has exam immediately before Jane D now look at this point also Jaya has exam immediately before J and E so Jaya has exam immunity before chain theme is what Jane these on Tuesday so ji should be on Monday right so J will come here on Monday so we know who are the two persons for Monday Jaya and Jamuna but there's lots are not clear it can be 9 a.m. or 11 a.m. I mean between those two but maybe if you say Jaya has exam immediately before GNTC jmt is 9 a.m. on Tuesday immediately before J and T is lemon a.m. on Monday so maybe ji is at 11 a.m. on Monday here able to follow see when you say immediately before is it immunity before only with respect to days or also with respect to time it is also with respect to time then Jane these on 9 a.m. Tuesday Jay has to be immediately before J and E so immediately before Jane these what slot 11 a.m. Monday so I think let's fix that let us see what happens ji is at 11 a.m. on monday yamuna is at 9 a.m. on monday right in the worst case these two have to be reversed the one who has examined my name immediately before just mean this is incomplete John Key has exam immediately after the day of the one who has examined Monday this point you see John Key has exam immediately after the day of the one who has exam on Monday so John Key has exam after the one who has taken his exam on so clearly junkie if it is immediately after Monday it should be Tuesday on Tuesday we have already got jnt' so that's what should be for junkie so this is junkie you're getting it johnny has the exam immediately after Monday then just pile does not have exam at 11:00 a.m. and just Paul does not have exam at Levin iam so you just Paul does not have exam at 11:00 a.m. it should be my name so just Paul we come for my name just Paul will have is exam at 9:00 a.m. right and only three people have exam between Jamuna and Jyoti only three people have exam between so can we use this only three people have exam between Jamuna and Jyoti my confusion here is when you say three people yeah I think with respect to slot only so between Java and Jyoti they'd have to be three people so who are the people between John and Joe TC Gemini is here 9:00 a.m. Monday so one two three so I think Jyoti will come here here able to follow only three people have the exam between Jamuna and Jyoti means what Germany has taken the 9:00 a.m. slot on Monday what are the three spots after that 11:00 a.m. on Monday 9 a.m. on Tuesday at 11:00 a.m. on Monday or Tuesday so after these three slots there should be duty so she comes on Wednesday checking five persons have been mugged already let's try and fix the remaining five now read the statements once again the first one here the number of people who have example in Jamuna and Joong Ki how many people have example in German and Janki see the number of people who have example in German Janki between German and Janki we have two persons right so the number of people who are example in German and John Key is same as the number of people who have exam between G Y and jugdish so between German and John Key there are two persons so between G Y and Jake this also there will be two persons right between gia and junk this also there have to be two persons keep that in mind only two people have examples even just pile and just mean only two people have examine just pollen Jasmine so between just pollen Jasmine also there have to be two persons so between GYN Jack these two persons and between just paren Jasmine there have to be two persons right okay now I think one point that you can see here is we know that Jagdish will take the 9:00 a.m. slot right check this will take the 9:00 a.m. slot so if jugdish comes on Thursday 9:00 a.m. when will gia come after two slots or before two slots before two sorts is not possible it should be only after two starts so then gia will come here and just be here in gy or there's one more possibility if you jug these takes up the fight a lot then gia will come to spots before so gia will come here I think two possibilities let's leave it Jamuna and junkie also yeah and other one is just pollen jasmine they also have got two possibilities PK let's move to the next point I think whatever is underlined has been used look at this point Joong Ki does not have examine any of the days after duty oh this is already done they Joong Ki does not have exam or any of the days after duty here John Key's before Julie so see this point which is given here I'll highlight this point that has been given here is redundant we don't need it John he does not have exam on any of the days after Jyoti you know we already understood that junkies exam is before Jyoti right something that is not needed just file does not have okay just Paul does not have exam on any of the days after Judy just Paul is not after Japanese just Paul should be before check T this is the key just Paul does not have exam on any of that is after Japanese so did you you know jug beach there are two possibilities jug these will come either yeah I think they are done see we know that just pile inch of these modes will take on the nine M slots so the only possibilities are for the in Friday and now we also had just Paul will not have exam after jettisons just pulse should be before jug please so obviously just Paul will take 9 a.m. on Thursday and check this will take 9 a.m. on Friday yes or no just pollen check these both have to take 9 m slots we know that from the given point that just Paul will not be afters of these just Paul has to be be fortunately so fixed now we can do the remaining part of it what is that we know that there are two people between gy and jugdish remember this point the number of people G between German and John Key's same as the number of people between g1 jugdish so if there are two people between GYN jugdish we will G become C jug this is 9 a.m. on Friday then Jeeva cannot come after this because there is only one slot available after that so leave two slots before jug these two shorts before check this is 11 a.m. Thursday or 9 a.m. Thursday and then this one so I think G bar will come here you're getting it two people wouldn't even check this one - these are two people using this so this point is also done I'll underline it and the other point that we had seen is only two people between just pollen just mean if there are only two between just bow and just mean the only possibilities just Paul is here 1 & 2 and Jasmine would come here who's left out who is that person was left out Jaya is done right Jaidev J they would come here done I think interesting question but not very complex so these are the this is the arrangement of these guys Monday to Friday 9 a.m. in 11 M slots right and that's order so Jamuna Jaya Jane thejungie Jyoti Jeeva just Paul Jay Dave Jagdish and jasmine |
1051 | https://math.stackexchange.com/questions/3272127/calculate-the-sum-of-series-with-square-roots | calculus - Calculate the sum of series with square roots - Mathematics Stack Exchange
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Calculate the sum of series with square roots
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Calculate the sum of the following series using partial sums:
∑n=1∞n+1−−−−−√−n−−√n−−√n+1−−−−−√∑n=1∞n+1−n n n+1
I rationalized the upper part of the fraction but I got lost. Could you please help me showing the steps of the how to transform the fraction into a partial sum? Thanks in advance.
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edited Jun 23, 2019 at 23:11
ArsenBerk
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asked Jun 23, 2019 at 22:58
Rolando GonzálezRolando González
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As mentioned in the answers, this is a telescoping series. If you struggle with the details you might find this helpful: math.stackexchange.com/questions/3250988/…Cornman –Cornman 2019-06-23 23:03:44 +00:00 Commented Jun 23, 2019 at 23:03
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HINT:
n+1−−−−−√−n−−√n−−√n+1−−−−−√=1 n−−√−1 n+1−−−−−√n+1−n n n+1=1 n−1 n+1
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answered Jun 23, 2019 at 23:00
ArsenBerkArsenBerk
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Thank you very much but how can I get there. Did you multiplied the fraction by n+1√+n√n+1√+n√n+1+n n+1+n ?Rolando González –Rolando González 2019-06-23 23:03:48 +00:00 Commented Jun 23, 2019 at 23:03
2 No, you don't have to make this too complicated. Just write it as n+1√n√n+1√−n√n√n+1√n+1 n n+1−n n n+1.ArsenBerk –ArsenBerk 2019-06-23 23:06:43 +00:00 Commented Jun 23, 2019 at 23:06
Oh, I completely forgot about that! Thanks a lot !!!!Rolando González –Rolando González 2019-06-23 23:17:18 +00:00 Commented Jun 23, 2019 at 23:17
1 You're welcome, good luck!ArsenBerk –ArsenBerk 2019-06-23 23:19:54 +00:00 Commented Jun 23, 2019 at 23:19
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1052 | https://math.stackexchange.com/questions/2265700/from-a-square-of-unit-length-pieces-from-the-corners-are-removed-to-form-a-regul | From a square of unit length,pieces from the corners are removed to form a regular octagon.Then, the value of area removed is? - Mathematics Stack Exchange
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From a square of unit length,pieces from the corners are removed to form a regular octagon.Then, the value of area removed is? [duplicate]
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Find the area of the following (1 answer)
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I'm having so much problem calculating the area, can anyone help?
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asked May 4, 2017 at 15:22
Iti ShreeIti Shree
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Let x x be the length of the octagon. Then
x+x cos 45∘+x cos 45∘=1 x+x cos45∘+x cos45∘=1
So x=1 1+2–√=2–√−1 x=1 1+2=2−1.
The area removed is equal to the area of a square with side x x.
The removed area is x 2=3−2 2–√x 2=3−2 2.
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answered May 4, 2017 at 15:26
CY AriesCY Aries
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Thank you so much, I forgot that I could take the angle 45\degree 45\degree.Iti Shree –Iti Shree 2017-05-04 15:35:11 +00:00 Commented May 4, 2017 at 15:35
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The drag polar or drag curve is the relationship between the lift on an aircraft and its drag, expressed in terms of the dependence of the drag coefficient on the lift coefficient. It may be described by an equation or displayed in a diagram called a polar plot.
drag coefficient
drag
polar
1. The Drag Polar
Drag and lift coefficients for NACA 633618 airfoil. Full curves are lift, dashed drag; red curves have Re = 3·106, blue 9·106.
Drag polar for the NACA 633618 airfoil, colour-coded as opposite plot.
The significant aerodynamic properties of aircraft wings are summarised by two dimensionless quantities, the lift and drag coefficients CL and CD. Like other such aerodynamic quantities, they are functions only of the angle of attack α, the Reynolds number Re and the Mach number M. CL and CD are often presented individually, plotted against α, but an alternative graph plots CL as a function of CD, using α parametrically. Similar plots can be made for other components or for whole aircraft; in all cases they are referred to as drag polars. The drag polar of an aircraft contains almost all the information required to analyse its performance and hence to begin a design.
Since the lift and the drag forces, L and D are scaled by the same factor to get CL and CD, L/D = CL/CD. As L and D are at right angles, the latter parallel to the free stream velocity or relative velocity of the surrounding, distant, air, the resultant force R lies at the same angle to that direction as the line from the origin of the polar plot to the corresponding CL, CD point does to the CD axis. If, in a wind tunnel or whirling arm system an aerodynamic surface is held at a fixed angle of attack and both the magnitude and direction of the resulting force measured, they can be plotted using polar coordinates. When this measurement is repeated at different angles of attack the drag polar is obtained. Lift and drag data was gathered in this way in the 1880s by Otto Lilienthal and around 1910 by Gustav Eiffel, though not presented in terms of the more recent coefficients. Eiffel was the first to use the name drag polar.
Because of the Reynolds and Mach number dependence of the coefficients, families of drag polars may be plotted together. The design of a fighter will involve a set at different Mach numbers, whereas gliders, which spend their time either flying slowly in thermals or rapidly between them may require polars at different Reynolds numbers but are unaffected by compressibility effects. During the evolution of the design the drag polar will be refined. A particular aircraft may have different polar plots even at the same Re and M values, depending for example on whether undercarriage and flaps are deployed.
Drag polar for light aircraft. CD0= 0.017, K = 0.075 and CL0 = 0.1. The tangent gives the maximum L/D point.
The accompanying diagram shows a drag polar for a typical light aircraft. Such diagrams identify a minimum CD point at the left-most point on the plot, where the drag is locally independent of lift; to the right, the drag is lift related. One component here is the induced drag of the wing, an unavoidable companion of the wing's lift, though one that can be reduced by increasing the aspect ratio. Prandtl's lifting line theoretical work shows that this increases as CL2. The other drag mechanisms, parasitic and wave drag, have both constant components, totalling CD0 say, and CL dependent contributions that are often assumed to increase as CL2. If so, then
: : : : : : : : : : CD = CD0 + K.(CL - CL0)2.
The effect of CL0 is to lift the polar curve upwards; physically this is caused by some vertical asymmetry, such as a cambered wing or a finite angle of incidence, which ensures the minimum drag attitude produces lift and increases the maximum lift to drag ratio.
2. Power Required Plots
One example of the way the polar is used in the design process is the calculation of the power required (PR) curve, which plots the power needed for steady, level flight over the operating speed range. The forces involved are obtained from the coefficients by multiplication with (ρ/2).S V2, where ρ is the density of the atmosphere at the flight altitude, S is the wing area and V is the speed. In level flight, lift equals weight W and thrust equals drag, so
PR curve for the light aircraft with the drag polar above and weighing 2000 kg, with a wing area of 15 m² and a propeller efficiency of 0.8.
: : : : : : : : : : W = (ρ/2).S.V2.CL and
: : : : : : : : : : PR = (ρ/2η).S.V3.CD.
The extra factor of V/η, with η the propeller efficiency, in the second equation enters because PR= (required thrust)×V/η. Power rather than thrust is appropriate for a propeller driven aircraft, since it is roughly independent of speed; jet engines produce constant thrust. Since the weight is constant, the first of these equations determines how CL falls with increasing speed. Putting these CL values into the second equation with CD from the drag polar produces the power curve. The low speed region shows a fall in lift induced drag, through a minimum followed by an increase in profile drag at higher speeds. The minimum power required, at a speed of 195 km/h (121 mph) is about 86 kW (115 hp); 135 kW (181 hp) is required for a maximum speed of 300 km/h (186 mph). Flight at the power minimum will provide maximum endurance; the speed for greatest range is where the tangent to the power curve passes through the origin, about 240 km/h (150 mph).)
If an analytical expression for the polar is available, useful relationships can be developed by differentiation. For example the form above, simplified slightly by putting CL0 = 0, has a maximum CL/CD at CL2 = CD0/K. For a propeller aircraft this is the maximum endurance condition and gives a speed of 185 km/h (115 mph). The corresponding maximum range condition is the maximum of CL3/2/CD, at CL2 = 3.CD0/K, and so the optimum speed is 244 km/h (152 mph). The effects of the approximation CL0 = 0 are less than 5%; of course, with a finite CL0 = 0.1, the analytic and graphical methods give the same results.
3. Rate of Climb
For an aircraft to climb at an angle θ and at speed V its engine must be developing more power, P say, than that PR required to balance the drag experienced at that speed in level flight and shown on the power required plot. In level flight PR/V = D but in the climb there is the additional weight component to include, that is
: : : : : : : : : : P/V = D + W.sin θ = PR/V + W.sin θ.
Hence the climb rate RC = V.sin θ = (P - PR)/W. Supposing the 135 kW engine required for a maximum speed at 300 km/h is fitted, the maximum excess power is 135 - 87 = 48 Kw at the minimum of PR and the rate of climb 2.4 m/s. This suggests a more powerful engine is called for.
4. Glider Polars
The same aircraft, without power. The tangent defines the minimum glide angle.
Without power, a gliding aircraft has only gravity to propel it. At a glide angle of θ, the weight has two components, W.cos θ at right angles to the flight line and W.sin θ parallel to it. These are balanced by the force and lift components respectively, so
: : : : : : : : : : W.cos θ = (ρ/2).S.V2.CL and
: : : : : : : : : : W. sin θ = (ρ/2).S.V2.CD.
Dividing one equation by the other shows that the glide angle is given by tan θ = CD/CL. The performance characteristics of most interest in unpowered flight are the speed across the ground, Vg say, and the sink speed Vs; these are displayed by plotting V.sin θ = Vs against V.cos θ = Vg. Such plots are generally termed polars, and to produce them the glide angle as a function of V is required.
One way of finding solutions to the two force equations is to square them both then add together; this shows the possible CL, CD values lie on a circle of radius 2.W / S.ρ.V2. When this is plotted on the drag polar, the intersection of the two curves locates the solution and its θ value read off. Alternatively, bearing in mind that glides are usually shallow, the approximation cos θ ≃ 1, good for θ less than 10°, can be used in the lift equation and the value of CL for a chosen V calculated, finding CL from the drag polar and then calculating θ.
The example polar here shows the gliding performance of the aircraft analysed above, assuming its drag polar is not much altered by the stationary propeller. A straight line from the origin to some point on the curve has a gradient equal to the glide angle at that speed, so the corresponding tangent shows the best glide angle tan−1(CD/CL)min ≃ 3.3°. This is not the lowest rate of sink but provides the greatest range, requiring a speed of 240 km/h (149 mph); the minimum sink rate of about 3.5 m/s is at 180 km/h (112 mph), speeds seen in the previous, powered plots.
References
Anderson, John D. Jnr. (1999). Aircraft Performance and Design. Cambridge: WCB/McGraw-Hill. ISBN 0-07-116010-8.
Abbott, Ira H.; von Doenhoff, Albert E. (1958). Theory of wing sections. New York: Dover Publications. pp. 57–70;129–142. ISBN 0-486-60586-8.
Aircraft Performance and Design. pp. 414–5.
Aircraft Performance and Design. pp. 199–252, 293–309.
Aircraft Performance and Design. pp. 265–270.
Aircraft Performance and Design. pp. 282–7.
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Table of Contents
1. The Drag Polar
2. Power Required Plots
3. Rate of Climb
4. Glider Polars
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1054 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_5?srsltid=AfmBOor87j9G4LX2m7KOA45HV6ckmYC6TYfdYGf6Mq244AvQg-Vv0Y1S | Art of Problem Solving
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1971 IMO Problems/Problem 5
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Remarks (added by pf02, January 2025)
5 Definitions and terminology
6 Solution 1, rewritten
7 Solution 2, rewritten
8 Solution 3
9 See Also
Problem
Prove that for every natural number , there exists a finite set of points in a plane with the following property: For every point in , there are exactly points in which are at unit distance from .
Solution 1
I shall prove a more general statement about the unit distance graph(, adjacency iff the Euclidean distance between the points is ) and this will follow as a consequence. if occur as unit distance graphs, then so does ( here is described as or ). this is seen by describing the vertices by complex numbers. suppose there is an embedding of by the complex numbers and for by the numbers . we claim that for some choice of , will do the job(a suitable rotation). what we need is that iff either or . clearly if the condition holds then the adjacency is satisfied. suppose and that the corresponding complex numbers are at a distance from one another. Then this gives a quadratic in and hence can take only finitely many values here.ruling this out for each such set of hence rules out finitely many values of and therefore a suitable choice exists. for the given problem we need a unit distance graph which is regular of degree for any given . since we can form the graph , we can form (the unit cube) and that solves the problem.
This solution was posted and copyrighted by seshadri. The original thread for this problem can be found here:
Solution 2
Suppose has the formwhere is unknown set of distinct unit vectors in . We can build inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected: 1. All vectors in are distinct; 2. Two vector sums are at unit distance from one another if and only if they differ in presence of exactly one summand (i.e. one and only one coefficient in the sum changes from to ). If these two conditions are kept, then each of points at will be at unit distance from exactly points corresponding to sums at which one and only one of coefficients differs from coefficients of this point. However, respecting these conditions is not hard because and for each new vector being added to there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.
This solution was posted and copyrighted by Bandera. The original thread for this problem can be found here:
Remarks (added by pf02, January 2025)
On the original thread at there is a third solution, by Pgm03B. It is simpler than the two solutions above. It has a flaw in the argument, and it is not presented in a nice way, but it is a nice idea, well worth presenting here. As a public service, I will add an edited version of it below.
Solutions 1 and 2 above are based on good ideas, but they are presented very poorly. If this page was a reviewed publication, and I were a reviewer, I would reject both of them saying "rewrite and resubmit".
2.1. Solution 1 suffers from undefined notation and terminology, from minor errors, and from unacceptable amount of hand waving replacing explanations of details.
2.2. Solution 2 suffers from poor explanation of details and from what seems to me to be an error (starting an inductive proof for a property of vectors from an empty set).
As a public service I will rewrite these proofs below, in what I hope is a much more presentable style.
Definitions and terminology
It will be helpful (though not necessary) to imagine the set of points as a graph in the plane. Specifically, the vertices of the graph are the points in , and the edges of the graph are the line segments of length joining the points. All points at distance are joined, and only the points at distance are joined.
We will call such a graph -regular iff every has exactly lines of the graph with one end at , in other words, there are exactly points in at unit distance from . Using this definition, the problem can be reformulated as "prove that for any natural number , there exists an -regular graph."
For the purposes of solution 1, it will be useful to think of the points as points in the complex plane. For the purposes of solution 2, it will be useful to think of the plane as having an origin . We will work with vectors in the plane. The points will be the end points of vectors. We will try to differentiate between vectors and their end-points, although at times, when there is no danger of confusion, we may be sloppy about it.
Solution 1, rewritten
We will give a proof by induction on . First some notation:
Given two finite sets of points in the complex plane, and given ![Image 80: $\theta \in 0, 2\pi)$, define .
: If is -regular and is -regular, and , then we can choose so that is -regular.
: Let be the neighboring points in at distance from , and be the neighboring points in at distance from Clearly is at distance from and is at distance from . So has neighbors at distance if we choose such that for all .
But the equations are just a finite number of linear equations in , so we just have to avoid choosing giving a solution to any of these equations. Thus, with such a choice of there are definitely at least points in at distance from .
We have to show that we can choose so that there are no more than points at distance . If we had more points from at distance from , we would have for some and .
This would imply ,
or ,
or ,
or .
This becomes an equation of degree in . So as long as we choose not to equal to any of the solutions of these equations, we can be sure that none of the points in are at distance from .
This proves the proposition because we have to choose ![Image 139: $\theta \in 0, 2\pi)$ different from finite number of values.
Now the problem is very simple to prove by induction.
For , take , which is a -regular set. For the induction step, assume we have an -regular graph. Using the proposition, it follows that we can choose such that is an -regular graph, where consists of two points in the plane at distance , none of which is in .
(Just for the sake of completeness, let us mention that when we choose , it is a value in ![Image 152: $0, 2\pi)$ which differs from a finite set of values modulo , dependent on the points in .)
Solution 2, rewritten
Given a set of vectors in the plane, define
Let the set of end points of all .
Note that in the case for all , the sum , which yields the point (the origin) in the plane. In the notation this corresponds to . (Just as a curiosity, note that if contains vectors, then contains points. We are not going to use this fact.)
We will prove by induction on that there is a set of unit vectors with the starting point at , such that the vector sums in are all distinct, and is -regular. If , take , where is a unit vector starting at . Then consists of , and consists of the origin and the end point of , so it is clearly a -regular set.
Now assume that we have a set of unit vectors starting at , , such that the sums in the corresponding are all distinct, and is -regular. We want to show that we can find a unit vector starting at , such that if we take , then the vector sums in are all distinct, and is -regular. Note that and .
Prove that we can choose such that the sums in are all distinct. If then . So we just need to make sure that we choose such that for all . This means that has to avoid satisfying a finite number of equations. Since there are infinitely many possibilities for the unit vector , we can definitely choose it such that it is not satisfying a finite number of equations.
We will now prove that we can choose such that is -regular. First, let correspond to . Then, the same viewed as a point in has at least neighbors at unit distance, namely the neighbors it has in and the point corresponding to . It would have more neighbors at unit distance if we there would exist an (where and not at distance from ) such that the end points of and would be at distance . So, the end point of the unit vector should not be at distance from the end point of the vector , or in other words, should not be on the circle of radius centered at the end of . This circle intersects the unit circle centered at (on which the end point of is) in at most points. This gives yet another finite set of conditions should avoid satisfying.
Second, take corresponding to for a . Let be the vectors which give the points in at unit distance from the point corresponding to . Then and are vectors whose end points are at unit distance from . Again, we have to show that we can choose so that there are no more points in at distance from the end point of . There would be more points in at distance from if we had vectors (where and not at distance from ) such that the end points of and , or the end points of and were at distance . The former is already avoided by the choice of made so far, and the latter is avoided because of the induction hypothesis.
Since there are only finite number of equations the unit vector with starting point at has to avoid satisfying, it follows that we can find such that is -regular.
Solution 3
We will give a proof by induction on . For consider the graph consisting of two points at distance joined by a line segment.
Assume that we have a graph which is -regular. Let be a direction along which we make a translation of of length (that is, the distance of translation ). Denote by the translated graph, and let be the graph whose vertices are , and whose edges are those of , together with those of , and the edges obtained when we join each with the corresponding (where is the result of translating ). We will prove that we can choose the direction of translation such that is -regular.
First, let us make sure we choose such that no point in is a point of . This means that has to be different from a finite number of values, which is possible since to begin with, we have infinitely many choices for .
Let , and the translated image of . Let be the points in at distance from , and let the coresponding points in . The point has neighbors in at distance , namely , and the point has neighbors in at distance , namely . So every point in has at least neighbors at distance .
We have to show that we can choose the direction so that no point has more than -neighbors at distance . For this, we have to show that for all and the corresponding we can choose such that the distance between and is . This means that should be such that is not at the intersection of the two circles of radius centered at and . Since these two circles intersect in at most points, and we have finitely many pairs , this imposes that the direction does not equal finitely many values. This proves the induction step, and the problem.
(Note that as far as the construction goes, this solution is essentially the same as solution 1, but the formalism and the point of view are so different that I think it should be viewed as a different solution.)
This solution is based on the idea by Pgm03B from the discussion thread .
See Also
1971 IMO (Problems) • Resources
Preceded by
Problem 41•2•3•4•5•6Followed by
Problem 6
All IMO Problems and Solutions
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1055 | https://www.quora.com/The-sum-of-the-first-n-terms-of-an-arithmetic-sequence-is-1-2-5n%C2%B2-n-Determine-the-nth-term-of-the-sequence-Hence-what-is-the-150th-term | Something went wrong. Wait a moment and try again.
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5
The sum of the first n terms of an arithmetic sequence is 1/2 (5n² - n). Determine the nth term of the sequence. Hence, what is the 150th term?
Eki Victory Ogheneobukohwo
Founder at Victorious Farms & Enterprises (2019–present)
·
3y
Arithmetic Progression
Sum=1/2 (5n^2-n)
when n = 1
nth term = 1
first term “a” = 1/2(5n^2-n)
a = 1/2(5(1)^2–1)
a = 1/2(5–1)
a = 1/2(4)
a1 = 2
when n = 2
nth term = 2
first term “a” = 1/2(5n^2-n)
a = 1/2(5(2)^2–2)
a = 1/2(54–2)
a = 1/2(20–2)
a = 1/2(18)
a = 9
a2 = 9 - 2
a2 = 7
when n = 3
nth term = 3
first term “a” = 1/2(5n^2-n)
a = 1/2(5(3)^2–3)
a = 1/2(59–3)
a = 1/2(45–3)
a = 1/2(42)
a = 21
a3 = 21 - 9
a3 = 12
Series 2, 7, 12
common difference = 2nd term - 1st term
d = 7 -2
d = 5
Tn = a + (n-1)d
= 2 + (150–1)5
= 2 + (149)5
= 2 + 745
= 747
T150 = 747
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Related questions
The sum of the first nth terms of a linear sequence (A.P) is 2n²-n-1. What is its nth term?
How do I find the sum of the first n terms of an arithmetic progression n (5n+1) /2 and also find the formula for the general term of the series?
The sum of n terms of an arithmetic sequence is n² + 2n. What is the sum of the first 10 terms of the sequence?
In a sequence, t₆ = 50 and
t
n
=
2
t
n
−
1
+
5
n
²
+
n
. What are the first five terms of the sequence?
What are the first 5 terms of the sequence given the nth term an=n(2-n)?
Manjunath Subramanya Iyer
I am a retired bank officer teaching maths
·
Author has 7.2K answers and 10.4M answer views
·
3y
Given Sn = 1/2 (5n² - n).
S(n+1) = 1/2{5(n+1)² - (n+1)
= 1/2{ 5(n² + 2n + 1) - (n+1)
= 1/2{5 n² + 10n + 5 - n - 1}
= 1/2{5n² + 9n + 4}
an = S(n+1) - Sn
= 1/2{5n² + 9n + 4} - 1/2 (5n² - n).
=1/2{10n + 4) = 5n + 2
nth term of the AP = 5n + 2.
So 150th term = 5(150) + 2 = 752.
Assistant
Bot
·
1y
To find the nth term of the arithmetic sequence whose sum of the first n terms is given by
Sn=12(5n2−n),
we start by using the formula for the sum of the first n terms of an arithmetic sequence, which can also be expressed as:
Sn=n2(2a+(n−1)d),
where a is the first term and d is the common difference.
Step 1: Find the nth Term
The nth term an can be expressed in terms of the sum:
an=Sn−Sn−1.
First, we compute Sn−1:
Sn−1=12(5(n−1)2−(n−1)).
Calculating Sn−1:
S_{n-1} = \frac{1}{2} \left( 5(n^2 - 2n + 1) - n + 1 \right) = \frac{1
To find the nth term of the arithmetic sequence whose sum of the first n terms is given by
Sn=12(5n2−n),
we start by using the formula for the sum of the first n terms of an arithmetic sequence, which can also be expressed as:
Sn=n2(2a+(n−1)d),
where a is the first term and d is the common difference.
Step 1: Find the nth Term
The nth term an can be expressed in terms of the sum:
an=Sn−Sn−1.
First, we compute Sn−1:
Sn−1=12(5(n−1)2−(n−1)).
Calculating Sn−1:
Sn−1=12(5(n2−2n+1)−n+1)=12(5n2−10n+5−n+1)=12(5n2−11n+6).
Now we can find an:
an=Sn−Sn−1=12(5n2−n)−12(5n2−11n+6).
Combining the terms:
an=12((5n2−n)−(5n2−11n+6))=12(−n+11n−6)=12(10n−6)=5n−3.
Step 2: Find the 150th Term
To find the 150th term, we substitute n=150 into the formula for the nth term:
a150=5(150)−3=750−3=747.
Final Answer
Thus, the nth term of the arithmetic sequence is
an=5n−3,
and the 150th term is
Chris Manning
B. Math. (Hons.) in Mathematics, The University of Newcastle (Australia) (Graduated 1981)
·
Author has 1.8K answers and 4.2M answer views
·
3y
The sum of the first n terms of an arithmetic sequence is
na + n(n - 1)d/2 = (d/2)n^2 + (a - d/2)n
where a = the first term and d = the common difference.
(1/2)(5n^2 - n) = (5/2)n^2 + (-1/2)n
So we require
d/2 = 5/2 and a - d/2 = -1/2
d = 5 and a = 2
The nth term = t(n) = a + (n - 1)d = 2 + 5(n - 1) = 5n - 3
The 150th term = t(150) = (5 x 150) - 3 = 747
Related questions
What is the specified term of the arithmetic sequence 1, 1/2, 0…5th term?
What are the five terms of the sequence of the given nth term an= n + n/2?
Consider the arithmetic sequence whose rule is a = 6+(n-1) 18. What is the sum of the first 15 terms of the sequence?
What is the sum of the first n terms of the sequence 10, 6, 2…?
In an arithmetic sequence, the sum of the first 20 terms is 950 and the sums of the second 20 terms are 0. What is the general term of the sequence?
Murali Krishna
Former Retired Senior Lecturer in DIET at Government (General) (1990–2004)
·
Author has 6.5K answers and 6.7M answer views
·
3y
AP
Sn=1/2 (5n^2-n)
Put n=1 a1 = 1/2 (4) =2
Put n=2 s2 =1/2 (20-2)=9
a2 =9--2=7
Put n=3 s3 =1/2 (42) =21
a3=21–(9)=12
Series
2 7 12 ---
a1=2 d common difference is 7-2=5
Tn=2+(n-1)5=5n--3
T150=2+149×5 =2+745=747
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Ajay Gupta
B.Tech. from IIT D (Graduated 1986)
·
Author has 465 answers and 85.1K answer views
·
3y
nth term is the difference of sum of n terms & sum of (n-1) terms. Hence nth term equals
1/2(5[n^2-{n-1}^2]-n+n-1)
1/2(5[2n-1]—1)=1/2(10n-6)=5n-3.
150th term would be 747.
Ansh Keer
Former Owner
·
Author has 6K answers and 1.7M answer views
·
2y
Tn = Sn — Sn-1 = 2.5 n^2 — 0.5 n — 1/2 (5n^2 -10n +5 -n+1)
= 5.5n —0.5
T150 = 11/2150 —0.5 = 824.5
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.8M answer views
·
3y
Related
The sum of the first 2n terms of a series is 18n - 12n². What is the sum ofthe first n terms and the nth term of this series? Show that this series is an arithmetic series.
This looks quite confusing!
This is clearly an arithmetic series with first term a = 6 and the common difference is d = – 6
The nth term is a + (n – 1)d = 6 – 6(n – 1) = 12 – 6n
This looks quite confusing!
This is clearly an arithmetic series with first term a = 6 and the common difference is d = – 6
The nth term is a + (n – 1)d = 6 – 6(n – 1) = 12 – 6n
Luqman Khan
Author has 771 answers and 784.1K answer views
·
4y
Related
The sum of the first n terms of an arithmetic sequence is given by Sn=n4 (3n−1). Which expression is the general term of the sequence?
Please format the question properly
You’ve said Sn=n4 (3n-1)
I don’t know what that ‘4’ was supposed to be
Arithmetic series are linear and their sum sequences are quadratic. Therefore if you meant n^4, then this isn’t possible.
If you mean 4n(3n-1), that would make more sense)
First we find the formula for the sum of an arithmetic sequence
Now we make them identical to each other and see what we get
Now we compare coefficients to form some simultaneous equations
divide by common factor of ‘n’
Now or the first equation 1/2 d = 12 d=24
now substitute into the (a-1/2d=-4)
we get a=8
Now we know ‘a’ and ‘d’
N
Please format the question properly
You’ve said Sn=n4 (3n-1)
I don’t know what that ‘4’ was supposed to be
Arithmetic series are linear and their sum sequences are quadratic. Therefore if you meant n^4, then this isn’t possible.
If you mean 4n(3n-1), that would make more sense)
First we find the formula for the sum of an arithmetic sequence
Now we make them identical to each other and see what we get
Now we compare coefficients to form some simultaneous equations
divide by common factor of ‘n’
Now or the first equation 1/2 d = 12 d=24
now substitute into the (a-1/2d=-4)
we get a=8
Now we know ‘a’ and ‘d’
Now we just need the formula for a arithmetic sequence.
now we just substitute the values we have found
simplify this.
And this would the arithmetic sequence formula.
Hope this helped, please upvote if it did
Thank you
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Jan van Delden
MSc Math and still interested
·
Author has 4.8K answers and 6.5M answer views
·
1y
Related
Let an = 1/ ((n+1) (n+2)) be the nth sequence. What is the nTH partial sum of this sequence?
I suppose you wish to find a shorter description of
where I assume that you wish to count from onwards (but that’s easily corrected if you feel that the natural numbers start at ).
If you have some experience in simplifying sequences then you should immediately think partial fractions. If not, you should hope that partial fractions might come to the rescue, or if all else fails, compute a few of these and try to discover a simplification of the resulting numbers and proof this by using induction.
I spot
I suppose you wish to find a shorter description of
where I assume that you wish to count from onwards (but that’s easily corrected if you feel that the natural numbers start at ).
If you have some experience in simplifying sequences then you should immediately think partial fractions. If not, you should hope that partial fractions might come to the rescue, or if all else fails, compute a few of these and try to discover a simplification of the resulting numbers and proof this by using induction.
I spot
and this turns the given series into a telescoping series
This means that we have
My advice would be to just write down a few terms of and you’ll discover this ‘rule’ yourself.
If you didn’t come up with partial fractions and a telescoping series, then my advice is still to compute a few terms
and I see a clear pattern emerging. Realize that this is not a proof, this time you need to show that this pattern fits the bill.
Enrico Gregorio
Associate professor in Algebra
·
Author has 18.4K answers and 16M answer views
·
Aug 4
Related
The sum of the first 20 terms of an arithmetic sequence is 760. What is the first term if the last term is 95? What is its common difference?
The sum of the first 20 terms of an arithmetic sequence is 760. What is the first term if the last term is 95? What is its common difference?
As few formulas as possible. Call your first term. Then you know that the terms are
where is the common difference. Then we see that
You surely know how to compute the sum in parentheses, don’t you? It’s
Now the second equation simplifies to
and subtracting yields
and therefore
Bruce Stevens
Author has 2.8K answers and 1.5M answer views
·
Updated 1y
Related
An arithmetic series has a first term a and its common difference of 10. The sum of the first n terms is 10,000. If the 'n-th' term is less than 500, then what is the largest possible value for n?
Set the question into AS equations:
Sum(n) = 10000 = ( an + (((n(n—1))/2)10) ),
500 > t(n) = ( a + ((n — 1) 10) ).
Isolate the 1st term, a, of the nth term. Substitute the 1st term into the summation equation, and solve for the number of terms, n:
500 > ( a + 10n — 10 ), → 510 > ( a + 10n ),
a < ( 510 — 10n ).
Sum(n) = 10000 = ( an + (((n(n—1))/2)10) ),
20000 = ( 2an + 10n^2 — 10n ),
20000 = ((2 (510 — 10n) n) + 10n^2 — 10n),
20000 = ( 1020n — 20n^2 + 10n^2 — 10n ),
Simplifying, and rearranging into quadratic equation form:
(n^2) — 101n + 2000 = 0,
Using the quadratic equation formula, of:
n = (
Set the question into AS equations:
Sum(n) = 10000 = ( an + (((n(n—1))/2)10) ),
500 > t(n) = ( a + ((n — 1) 10) ).
Isolate the 1st term, a, of the nth term. Substitute the 1st term into the summation equation, and solve for the number of terms, n:
500 > ( a + 10n — 10 ), → 510 > ( a + 10n ),
a < ( 510 — 10n ).
Sum(n) = 10000 = ( an + (((n(n—1))/2)10) ),
20000 = ( 2an + 10n^2 — 10n ),
20000 = ((2 (510 — 10n) n) + 10n^2 — 10n),
20000 = ( 1020n — 20n^2 + 10n^2 — 10n ),
Simplifying, and rearranging into quadratic equation form:
(n^2) — 101n + 2000 = 0,
Using the quadratic equation formula, of:
n = ( ( — b ± √( (b^2) — (4 a c) ) ) / 2 ),
where a = 1, b = — 101, c = 2000,
simplifies to:
n = ((101 — √(2201)) / 2) = 27.0425…
n = ((101 + √(2201)) / 2) = 73.9574…
NOTE: The two qualities of the number of terms, n, are: 1. n must be positive, 2. n must be a whole number.
So, the largest possible value for n is: 73.
Solve for the value, of the 1st term, a:
Substituting n = 73 into:
a < ( 510 — 10n ), → a < ( 510 — 10(73) ),
a < ( 510 — 730 ), a < — 220.
Sum(73) = 10000 = ( 73a + 2626d ),
10000 = ( 73a + (2628 10) ),
10000 = ( 73a + 26280 ),
73a = ( 10000 — 26280 ), → 73a = — 16280,
a = — 223.0136… , a = — (16280 / 73).
The arithmetic sequence is:
—(16280/73), —(15550/73), —(14820/73),
—(14090/73), —(13360/73), —(12630/73),
—(11900/73), —(11170/73), —(10440/73), …
The general term rule, of this arithmetic sequence, when n = 1, 2, 3, …, is:
t(n) = ( 10n — (17010 / 73) )
OR
t(n) = ( (10 (73n — 1701)) / 73 ).
Used: Microsoft QuickC Version 2.00
```
include int main(void){register int n=0x0000;signed long int sum_N=0x0L;double a=0.0; / The number of terms, n: 1. MUST be positive, 2. MUST be a whole number. 1 to 100 inclusive. / for(n=0x0001; n < 0x0065; n++){ / 2000 = ( 101n — (n^2) ). / sum_N = ( (101L n) — (n n) ); / sum_N > 1990 AND sum_N < 2100 / if( (sum_N > 0x076CL) && (sum_N < 0x0834L) ){ / a < ( 510 — 10n ) but, using: a = ( 510 — 10n ) / a = (double)(510.0 — (10.0 n)); printf("Sum(n) = %ld, t(1) = %.3lf, n = %d\n", sum_N, a, n ); / reset / a = 0.0; } / end if / else{} / reset / sum_N = 0x0L; } / end for / printf("\n"); printf("Program complete."); flush(stdout); return(0x0000);} / end main /
```
Sum(n) = 1950, t(1) = 250.000, n = 26
Sum(n) = 1998, t(1) = 240.000, n = 27
Sum(n) = 2044, t(1) = 230.000, n = 28
Sum(n) = 2088, t(1) = 220.000, n = 29
Sum(n) = 2088, t(1) = — 210.000, n = 72
Sum(n) = 2044, t(1) = — 220.000, n = 73
Sum(n) = 1998, t(1) = — 230.000, n = 74
Sum(n) = 1950, t(1) = — 240.000, n = 75
Program complete.
NOTE: From the output, Sum(n) = 2000, lies between:
a < — 220.000 AND a > — 230.000,
when, n = 73 (rounded down).
Neupane Damodar
Studied Mathematics (Graduated 2008)
·
Author has 696 answers and 403.4K answer views
·
2y
Related
The sum of the first n terms of the sequence is given by the formula n² - 1. What is the sum of all the sums?
Let, tn = (n^2- 1) ….(1)
Put n = 1, 2, 3, 4, ..... in (1)
tn = 0, 3, 8, 15, 24, 35, 48, .. ..
d = 3, 5, 7, 9, 11, 13, .. ..
Sn=(1×1-1)+(2×2-1)+(3×3-1)+(4×4-1)+(5×5-1) +(6×6-1)+(7×7-1)+ .. + ..
Or, Sn = (1^2+2^2+3^2+4^2+5^2+6^2+7^2+ .. ..) -(1+1+1+1+1+1+1+ .. ..)
Sn = [{n(n+1)(2n+1)}/6 -n]
S1 = [{1×(1+1)(2×1+1)}/6 -1] = 0
S2 = [{2×(2+1)(2×2+1)}/6 -2] = 3 (t1+t2)
S3 = [{3×(3+1)(2×3+1)}/6 -3] = 11 (t1+t2+t3)
S4 = [{4×(4+1)(2×4+1)}/6 -4] = 26 (t1+t2+t3+t4)
Related questions
The sum of the first nth terms of a linear sequence (A.P) is 2n²-n-1. What is its nth term?
How do I find the sum of the first n terms of an arithmetic progression n (5n+1) /2 and also find the formula for the general term of the series?
The sum of n terms of an arithmetic sequence is n² + 2n. What is the sum of the first 10 terms of the sequence?
In a sequence, t₆ = 50 and
. What are the first five terms of the sequence?
What are the first 5 terms of the sequence given the nth term an=n(2-n)?
What is the specified term of the arithmetic sequence 1, 1/2, 0…5th term?
What are the five terms of the sequence of the given nth term an= n + n/2?
Consider the arithmetic sequence whose rule is a = 6+(n-1) 18. What is the sum of the first 15 terms of the sequence?
What is the sum of the first n terms of the sequence 10, 6, 2…?
In an arithmetic sequence, the sum of the first 20 terms is 950 and the sums of the second 20 terms are 0. What is the general term of the sequence?
If the nth partial sum geometric sequence is (1/2) ^n-1, what is the general term or anth?
In the sequence 2, 4, 6, 8, . . ., what is the nth term rule, 2n, n + 1, 2 + n, or none?
In an arithmetic sequence the 7th term is 17. the sum of the 1st and 2nd terms is 12. Determine the first three terms of the sequence?
What is the approximate sum of the first 100 terms of the sequence (1/n) ^2?
The sum of the first n terms of an arithmetic series is: Sn= 3(n^2) - 8n. Determine the first 4 terms of the Series?
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Rapid and fast have the same meaning. Rapidly is usually past tense.
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rapidly basically mashes both meanings of quick and fast all together. :)
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Rapid and fast are both synonyms that simply mean something is moving or happening quickly. However, there is a minor difference. "Fast" can be used as an adverb and adjective, while "rapid" cannot. Ex. That horse is a fast runner. (Adjective) Ex. The horse is running too fast! (Adverb) Rapid is simply, an adjective. Hope this helped!
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@ulllliana Yes! An example would be: Ex. His heart began to race at a rapid speed. (Adjective) In this case, it's describing the rate of his heart beat. Another one would be: Ex. There was rapid growth of weeds in their garden. (Adjective once again) Like I stated previously, it can not be used as an adverb. Ex. The boy was running too rapid. (In this context, it does not work.) I hope this clarified things~
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“When used as adjectives, fast means firmly or securely fixed in place, whereas rapid means very swift or quick.”
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Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990.
Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition.
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Walker HK, Hall WD, Hurst JW, editors.
Boston: Butterworths; 1990.
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Chapter 191 Urinalysis
David M. Roxe.
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Definition
Urinalysis is the examination of urine for certain physical properties, solutes, cells, casts, crystals, organisms, or particulate matter. Because urinalysis is easy, cheap, and productive, it is recommended as part of the initial examination of all patients and should be repeated as clinically warranted. This chapter focuses on what the physician may do in a few minutes with a urine sample, reagent strips, a microscope, and an inquiring mind. Such analysis may lead to more sophisticated chemical, immunologic, or bacteriologic studies.
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Technique
The rationale and technique of urinalysis are straightforward. Nevertheless, various circumstances may alter the information obtained. For example, one should not be surprised if the urine analyzed did not come from the patient named, or that the protein and red cells were added to the urine after it reached the collection bottle. Such illicit treatment of the sample is not frequent, but may be used in an attempt to justify disability, military discharge, or need for hospitalization. Interestingly, if the temperature of fresh urine is checked, it may help to diagnose factitious oral or rectal fevers.
A sample should be obtained that is free of skin epithelium or bacteria, glandular secretions, skin salves, hair, lint, talcum, or other debris. In rare circumstances, it may be necessary to have the sample passed under observation to assure its source and freedom from exogenously introduced materials.
Ordinarily, a suitable sample may be obtained from a male simply by asking that the foreskin, if present, be pulled back, that the initial part of the stream be allowed to pass into the toilet, that the next ounce or two be collected, and that the last part of the stream be discarded with the first. In females, care must be taken to separate the labia, and the urine is collected similarly. During menstrual flow, a tampon helps keep menstrual fluid from mixing with voided urine. The utility of cotton balls, soaps, and scrubs to cleanse the meatal area is dubious, as the initial 40 to 50 ml of urine flow that is discarded is generally adequate to flush away meatal debris. The sample should be examined while fresh—indeed, while still warm—to give best results. On standing, particulate matter sediments out, bacteria proliferate and alter pH, casts may dissolve, and crystals may be lost. Refrigeration may cause precipitation of orange red crystals of uric acid, which can be redissolved by rewarming the urine.
Approximately 10 ml of well-mixed urine is taken for microscopic examination. If the urine is alkaline, 1 ml of dilute acetic acid will help dissolve phosphates that may obscure formed elements. There is no "gold standard" for how fast or how long one should centrifuge urine. I have found 3000 rpm for 3 minutes to be convenient. Others have favored 2000 rpm for 5 minutes. After centrifugation, the supernatant is discarded; the sediment is resuspended in the few drops that remain. A drop of this is placed on a slide, covered with a coverslip, and examined microscopically. No stain is routinely needed, nor is phase contrast ordinarily required. The microscope is adjusted so that relatively low light is used, and the slide is scanned under low power (100×), high power (400×), and, when protein is present, polarizing light. Low-power examination serves to identify areas of interest, high power permits identification and quantification, and polarizing light aids in identification of doubly refractile fat bodies and certain crystals. Criteria for "positive" findings on microscopic examination have been established and are useful guides for review (Table 191.1).
Table 191.1
Criteria for Classification of Positive Urinary Sediment Examinations Using Bright Field Microscopy.
While urine is being centrifuged, a macroscopic examination consisting of inspection and reagent strip testing should be done. The color and clarity of the urine are apparent on inspection. Causes of unusually colored urines are shown in Table 191.2. Turbidity of fresh urine suggests pus or red blood cells. When normal urine has cooled, crystals may precipitate and cause turbidity.
Table 191.2
Colored Urine.
Available reagent-impregnated strips provide information regarding renal function, carbohydrate metabolism, acid–base balance, liver function, and bacteriuria. It is necessary to follow instructions closely to obtain reliable results. In certain circumstances, misleading information may be suggested to the unwary observer.
The performance characteristics of one reagent strip are shown in Table 191.3.
Table 191.3
Reagent Strip Sensitivity for Specific Tests.
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Basic Science
The glomerular and tubular physiology involved in the formation and delivery of urine to the bladder and the mechanisms of bladder function are beyond the scope of this chapter.
The specific gravity of urine depends on a person's state of hydration, the integrity of the posterior pituitary, and the renal tubules. Normally, all urine leaving Henle's loop is dilute relative to plasma, and under forced hydration may contain as little as 50 mOsm/kg, roughly equivalent to a specific gravity of 1.001 or 1.002. Specific gravity of urine equals the weight of a given volume of urine divided by the weight of an equal volume of water:
When a urinometer is used, a correction must be made for temperature when very exact results are needed such that 0.001 is added or subtracted for each 3°C change above or below the calibration temperature recorded on the instrument. When protein is present in large amounts, all common methods are affected. There is no convenient correction factor for reagent strips. For refractometers or urinometers, it is necessary to subtract 0.003 for every 1 g/dl of protein in urine to be accurate. When glucose is present in large amounts, reagent strips should not be affected. It is necessary to subtract 0.004 from refractometer or urinometer readings for every 1 g/dl of glucose to correct the value. Values above 1.032 suggest the presence of exogenous solutes such as mannitol or iodinated contrast media.
The reagent strip method commonly available has three principal components: polymethylvinyl ether/maleic acid, bromothymol blue, and buffers. When specific gravity is high, the pKa of the polyelectrolyte is decreased and pH falls, resulting in a color change of the indicator. Highly buffered alkaline urine may, therefore, result in a factitiously low apparent specific gravity.
Urinary pH is an expression of the proton concentration in urine. Although the number of free protons excreted contributes only a trivial fraction of the approximately 80 mEq of acid an average person excretes daily, the free protons determine the efficacy of the titratable acid mechanism and the ammonium excretion mechanism, which together account for the bulk of excreted acid. When urinary pH is high, there are few proton acceptors in urine and the non-ionic diffusion of NH 3, into the tubular lumen is relatively impeded. When pH is low, phosphates and other solutes efficiently accept protons and diffusion of NH 3 into the tubular lumen is facilitated. Normally, urine can be acidified to a pH value of 5.2 or less. Failure to do so in the face of systemic acidosis may indicate partial or complete renal tubular acidosis. Bacterial infection with urea-splitting organisms may produce an elevated urinary pH, so if fresh urine has leukocytes, bacteria, and an elevated pH, Proteus species would be suspected as the offending organism. (Other bacteria can split urea but are not as commonly responsible for urinary tract infection.)
Protein enters urine either because of altered glomerular permeability or because of tubular damage. Glomerular proteinuria always includes a large component of albumin. Tubular proteinuria is of low molecular weight, such as (β 2-microglobulin. Reagent strip tests for protein are virtually (but not completely) specific for albumin, and depend upon the capacity of protein to change the color of an acid-base indicator at a constant pH maintained by a buffer in the strip section. Contrast media, tolbutamide, tolmetin, or penicillin, which may give false positive readings with heat and acetic acid, do not affect results with the reagent strip. Alkaline urine may give a false positive result with the reagent strip and a false negative result with acid precipitation techniques.
Glucose is normally present in human urine in small amounts. Glucose is usually not detectable because ketones, ascorbic acid, or other substances found in urine may cause false negative results by reagent strips even when urinary glucose approaches clearly abnormal values near 100 mg/dl. Copper reduction tests are not specific for glucose and may react with other hexoses, pentose, creatinine, uric acid, salicylates, and numerous other agents. When bacteria are present, glucose may be consumed, so a false negative could result from testing urine that is not fresh. When a hexokinase reagent strip is used, glucose concentrations below 2 mg/dl in morning urine from a fasting person correlate well with urinary tract infection. Large amounts of urinary glucose suggest diabetes mellitus, or, rarely, renal glycosuria.
Ketone bodies appear in urine as a consequence of accelerated fat metabolism. (β-Hydroxybutyric acid is quantitatively greatest, followed by acetoacetic acid and acetone. When large amounts are present, a fruity odor may be detectable. The commonly available tests for ketone bodies depend on the development of a purple compound in the presence of nitroprusside and alkali. Such tests will react with acetone or acetoacetic acid but not with β-hydroxybutyric acid. l-Dopa will give a false positive result with the nitroprusside-based tests. A ferric chloride method is available that gives false positive results for both l-dopa and salicylates. Ketone bodies are most likely to be present in the urine of an adult during diabetic ketoacidosis or when the patient has been fasting.
Bilirubin and urobilinogen appear in urine when there are abnormalities of bilirubin metabolism or liver function. Albumin-bound bilirubin (indirect bilirubin) is not water soluble and does not appear in the urine. Bilirubin conjugated with glucuronic or sulfuric acid is water soluble and appears in urine in amounts roughly correlated with the direct reacting serum bilirubin. The presence of conjugated bilirubin in detectable amounts (greater than 0.2 mg/dl) does not enable one confidently to distinguish between hepatocellular and obstructive jaundice, but does not commonly occur when hyperbilirubinemia is consequent to hemolysis. Reagent strips and tablets may give a false negative reaction when urine contains ascorbic acid. Phenothiazines may cause a false positive reaction in both cases.
When conjugated bilirubin reaches the bowel, bacterial action produces urobilinogen, which is reabsorbed into the portal circulation. Increased production of bilirubin or decreased hepatic clearance of urobilinogen from the portal circulation will increase the amount delivered to the kidney and excreted in the urine. Thus, hemolysis or hepatocellular dysfunction may increase urinary urobilinogen, while biliary obstruction will decrease delivery of conjugated bilirubin to the bowel, thereby reducing production of urobilinogen. Antibiotics, by altering bowel flora, may prevent production of urobilinogen.
These two tests, taken together, help to distinguish between hemolysis, hepatocellular disease, and biliary obstruction, as shown in Table 191.4.
Table 191.4
Profiles of Urine Urobilinogen and Bilirubin in Health and Disease.
Hemoglobin is not normally present in urine. It may appear if there is intravascular hemolysis so that hemoglobin is filtered into the urine, or if red cells break apart within the urinary tract, liberating hemoglobin. Myoglobin also causes reddish brown urine and reacts with the reagent strip for hemoglobin. Typically, if serum and urine are both red, hemolysis is more likely, as the molecular weight of hemoglobin impedes its filtration. If serum is normal in color and urine is red, myoglobin is more likely because its smaller size favors filtration. Alternatively, of course, clear serum and red urine may represent bleeding within the urinary tract itself.
False negative reagent strip results may occur if urine contains large amounts of ascorbic acid. False positive results may be seen if povidone iodine solutions are rinsed into urine before testing.
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Clinical Significance
The reduction of nitrate to nitrite by bacteria with consequent color change in a solution or on a reagent strip is a useful indicator of bacteriuria, and when positive, should be taken seriously. A negative result gives no assurance that significant infection is not present. Urine may not have been retained in the bladder long enough for bacteria to have reduced nitrate, and this will invariably be the case when a urinary catheter is in place. Moreover, certain pathogens, such as Streptococcus faecalis, do not reduce nitrate at all.
Patients who have hypokalemia, hypercalcemia, protein malnutrition, or polydipsia will not be able to concentrate urine appropriately. Diuretics similarly impair concentrating ability, especially if kaliopenia develops. These conditions must be considered when interpreting specific gravity in the two circumstances for which it has greatest use: evaluation of possible acute renal failure, and evaluation of renal dysfunction to distinguish glomerular from tubulointerstitial disease. It is often said that a specific gravity of more than 1.020 makes acute renal failure less likely, but this is true only if the value is not spuriously elevated by endogenous or exogenous solutes. It is also said that specific gravity tends to be elevated in primary glomerular disease and reduced in tubulointerstitial disease, but this is of value only when factors that might elevate or depress specific gravity are not present.
Reagent strips cannot be relied on to detect globulins or light chains. The combination of a "negative" strip result and a positive acid-precipitation test would be suggestive of multiple myeloma. Those with an aging population in their practice may find it productive to use both methods.
Similarly, most reagent strips cannot be depended on to detect sugars other than glucose in urine. This may be a drawback in pediatric screening circumstances when inborn errors of metabolism may cause other sugars to appear in urine.
Remember that the presence of excess amounts of ascorbic acid in urine may affect three areas of the commonly used reagent strip: those that detect glucose, bilirubin, and hemoglobin. As people commonly take large amounts of ascorbic acid for various reasons, it is prudent to inquire specifically about this before interpreting the reagent strip results.
Careful analysis of the sediment usually takes about 5 minutes of patient examination of the area under the coverslip. The principal elements to be identified and quantified are cells, casts, and crystals.
Cells are ordinarily divided into red blood cells, white blood cells, epithelial cells, and atypical cells.
Red blood cells are associated with primary parenchymal disease such as glomerulonephritis, diabetes mellitus, polycystic kidney disease, drug reactions (e.g., penicillin), or collagen vascular disease. They may also be found with renal calculi, tumors of the urinary tract, upper or lower tract infections (cystitis or prostatitis), and trauma. The differential diagnosis of hematuria is considered in some detail in Chapter 184.
White blood cells suggest inflammatory processes, with infection being the commonest of these. Other inflammatory stimuli, such as collagen vascular disease and allergic interstitial nephritis, also may cause pyuria. There is practical value in evaluating the pyuria by the company it keeps; for example, if white cells are found with red cells and red cell casts, then collagen vascular disease is more likely. If white cells are found by themselves in a patient with an acute onset of fever and dysuria, then infection is more likely.
Epithelial cells from any site in the urinary tract may be found in the sediment. Drugs, toxins, ischemia, instrumentation of the urinary tract, and tumors are among the causes for such cells to be sloughed into the sediment. Urine cytology studies may have particular utility when large numbers or unusual forms of epithelial cells are seen.
Casts are formed within the renal tubules and have a matrix of translucent protein that, by itself, forms the substance of the so-called hyaline cast. When there are inclusions in the cast such as red cells or white cells, they permit conclusion that the cells originated within the kidney. This is especially important when there is consideration of vasculitis or of pyelonephritis. Degenerating cellular debris may form granular casts. Hemoglobin, myoglobin, or bilirubin may be incorporated into pigmented casts.
Lipiduria is an important finding because of its association with the nephrotic syndrome. Fat may appear in urine as free fat, as inclusions within an oval fat body, or as the characteristically doubly birefringent cholesterol ester that is referred to as a "Maltese cross" after the cruciform emblem emblazoned on the shields and tunics of the Knights of Malta. Lipiduria is so abnormal that when found, it strongly suggests either fat embolization of the kidneys or the nephrotic syndrome, and the clinical context will readily differentiate in most cases. Finding lipiduria in a patient with proteinuria and edema permits one to anticipate the diagnosis of nephrotic syndrome while waiting for the laboratory to quantify cholesterol and serum albumin.
It is useful to look for bacteria. If the urine is truly fresh, the presence of even one bacterium per oil immersion field of unspun, gram-stained urine correlates reasonably well with a colony count of greater than 100,000 colonies per ml (Wilson, 1975).
There has been controversy in recent years as to whether microscopic urinalysis is always necessary, especially if the macroscopic urinalysis is entirely negative. In the majority of cases, it will prove that when specific gravity is over 1.020, and the macroscopic examination is completely normal, the microscopic examination will likewise be normal. There will also be false negative results, and it is estimated that these will vary from 3 to 37% of all cases (Schumann and Greenberg, 1979; Szwed and Schaust, 1982). Even when a special type of reagent strip is used, which is sensitive to leukocyte esterase, up to 3.3% of positive findings may be missed (Shaw, Poon, and Wong, 1985). With routine reagent strips, the same authors found a 13% false negative result rate. Certainly, one could not delete the microscopic examination in a patient with signs or symptoms of hypertension or renal disease. In other cases, if it is omitted, one must accept missing 10% or more of positive findings—a loophole too large for most physicians" comfort.
Certain circumstances may warrant special stains as an adjunct to routine urinalysis. It has been suggested that a Wright's stain of the sediment may help to distinguish glomerular from lower tract sources of hematuria (Chang, 1984). Technically, it is useful to add a drop of albumin solution to the sediment to obtain clear slides after staining. Glomerular lesions produce dysmorphic RBCs, while distal lesions produce cells similar to those seen in peripheral blood. Such stains would have to be done on fresh urine, as hypotonic urine would lead to cell lysis, and hypertonic urine to pyknotic cells if the urine stood for any length of time. The same staining technique may be used to search for eosinophiluria when allergic interstitial nephritis is suspected. This may be helpful when B-lactam antibiotics, non-steroidal inflammatory drugs, or other potentially toxic agents are used. Quantification of eosinophils as a percentage of total urinary white blood cells is helpful. When over 5% of urinary white cells are eosinophils, interstitial nephritis is more likely. When less than 5% are eosinophils, infection is commonly the cause (Corwin, Korbat, and Schwartz, 1985).
The urine sediment may also reveal crystals, parasites, foreign bodies, spermatozoa or other findings. Assessment of the meaning of such findings—indeed, assessment of all the findings of urinalysis—depends upon the inquiring mind that asks an appropriate question. The mind, therefore, is the most important element of informative urinalysis. Because of that, it is best for the physician rather than anyone else to review the macroscopic findings, inspect the sediment, and interpret results for that unique patient, even as the physician does the physical examination rather than depend upon a technician's report.
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References
Chang BS. Red cell morphology as a diagnostic aid in hematuria. JAMA. 1984;252:1747–49. [PubMed: 6471304]
Corwin HL, Korbat SM, Schwartz MM. Clinical correlates of eosinophiluria. Arch Intern Med. 1985;145:1097–99. [PubMed: 4004436]
Ellenbogen C, Nord BM. Freshly voided urine temperature: a test for factitial fever. JAMA. 1972;219:912. [PubMed: 5066682]
Package Insert Revised 5/84. Ames Reagent Strips for Clini-Tek. Ames Division, Miles Laboratories.
Schumann GB, Greenberg NF. Usefulness of macroscopic urinalysis as a screening procedure. Am J Clin Pathol. 1979;71:452–56. [PubMed: 443204]
Shaw ST Jr,, Poon SY, Wong ET. Routine urinalysis"—is the dipstick enough? JAMA. 1985;253:1596–1600. [PubMed: 3974041]
Szwed J, Schaust C. The importance of microscopic examination of the urinary sediment. Am J Med Technol. 1982;48:141–43. [PubMed: 7064990]
Wilson DM. Urinalysis and other tests of renal function. Minn Med. 1975;58:9–17. [PubMed: 1088975]
Definition
Technique
Basic Science
Clinical Significance
References
Copyright © 1990, Butterworth Publishers, a division of Reed Publishing.
Bookshelf ID: NBK302 PMID: 21250145
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Roxe DM. Urinalysis. In: Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990. Chapter 191. Available from:
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Problem
Find all functions such that for all positive real numbers .
Solution
We claim that is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let be the assertion to the Functional Equation.
Claim 1: is injective.
Proof: Assume for some . Now, from and we have:
Now comparing, we have as desired.
This gives us the power to compute . From we get and injectivity gives . Showing that is unbounded above is also easy as we can fix and let blow up to in the original Functional equation..
Claim 2: is surjective.
Proof: gives
This gives that . Putting , we get . By induction, surjectivity is proved as and we are essentially done.
Now, we have claim that if for some , then . This can be achieved by putting , , and . Let us calculate
Everything is fine, but we would have show that such numbers do exist. We will show this in the next lemma.
Lemma 1: There exits such that Proof: Notice that is surjective. So, we will solve for in the system of linear equation. For convinience, let for as is surjective. Solving we get: Now, you can chose such that and is positive. This finishes the proof to the lemma.
Claim 3: obeys Jensen's Functional Equation,i.e., Proof:
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Three complexity functions
Sébastien Ferenczi; Pascal Hubert
RAIRO - Theoretical Informatics and Applications
(2012)
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Ferenczi, Sébastien, and Hubert, Pascal. "Three complexity functions." RAIRO - Theoretical Informatics and Applications 46.1 (2012): 67-76.
@article{Ferenczi2012,
abstract = {For an extensive range of infinite words, and the associated symbolic dynamical systems, we compute, together with the usual language complexity function counting the finite words, the minimal and maximal complexity functions we get by replacing finite words by finite patterns, or words with holes. },
author = {Ferenczi, Sébastien, Hubert, Pascal},
journal = {RAIRO - Theoretical Informatics and Applications},
keywords = {Infinite words; symbolic dynamical systems; complexity; infinite words},
language = {eng},
month = {3},
number = {1},
pages = {67-76},
publisher = {EDP Sciences},
title = {Three complexity functions},
url = {
volume = {46},
year = {2012},
}
TY - JOUR
AU - Ferenczi, Sébastien
AU - Hubert, Pascal
TI - Three complexity functions
JO - RAIRO - Theoretical Informatics and Applications
DA - 2012/3//
PB - EDP Sciences
VL - 46
IS - 1
SP - 67
EP - 76
AB - For an extensive range of infinite words, and the associated symbolic dynamical systems, we compute, together with the usual language complexity function counting the finite words, the minimal and maximal complexity functions we get by replacing finite words by finite patterns, or words with holes.
LA - eng
KW - Infinite words; symbolic dynamical systems; complexity; infinite words
UR -
ER -
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How to remember in-order traversal and pre-order traversal in an easy and unforgettable way?
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Binary Tree traversal such as preorder traversal, inorder traversal, postorder traversal, and level order traversal is commonly interviewed by many IT companies.
It confuses me to remember the iterative implementation of pre-order traversal and inorder traversal.
Here are the problems from leetcode.
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asked Dec 23, 2020 at 3:04
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I remember it thinking about it with reference to the root node.
Inorder -> meaning the root is InBetween Left and Right,
PreOrder -> The Root is 'pre'/before Left and right
Post -> Root is after Left and Right.
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answered Mar 2, 2021 at 15:26
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preorder traversal:
We walk the graph, from top going counter-clockwise. Shout every time we pass the LEFT of a node.
inorder traversal:
We walk the graph, from top going counter-clockwise. Shout every time Shout when you cross the bottom.
post-order traversal
We walk the graph, from top going counter-clockwise. Shout every time Shout when you cross the right
If you want to see more recursive and iterative implementation details, please read the following post
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answered Jan 4, 2021 at 8:31
Jerry AnJerry An
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1061 | https://math.stackexchange.com/questions/1998722/is-log-1n-frac1-log-n | logarithms - Is $\log^{-1}n = \frac{1}{\log n}$? - Mathematics Stack Exchange
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Is log−1 n=1 log n log−1 n=1 log n?
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Is log−1 n=1 log n log−1 n=1 log n? I've encountered a solution to a problem where the above step was used. I can't determine the validity of the above line. I have thought a lot about it, but can't prove whether it is right or wrong. Please help?
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asked Nov 4, 2016 at 5:31
Somenath SinhaSomenath Sinha
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Take n=e n=e, then if it was the inverse function you would have e e=1 e e=1.copper.hat –copper.hat 2016-11-04 05:36:51 +00:00 Commented Nov 4, 2016 at 5:36
You could edit the question to show more context, such as the step that included log−1 n log−1 n and the step before it. The notation seems unusual, and maybe so is the interpretation of the notation in that "solution".David K –David K 2016-11-04 05:43:02 +00:00 Commented Nov 4, 2016 at 5:43
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Never log−1 n log−1 n means antilogarithm(n)antilogarithm(n) i.e e n e n because log(e n)=n log(e n)=n and 1 n 1 n means inverse of n n .
So they are not the same
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answered Nov 4, 2016 at 5:35
LearnmoreLearnmore
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If we'd need to remove the denominator, what could we rewrite 1 log n 1 log n as? I want an expression which has a numerator only. Is it possible?Somenath Sinha –Somenath Sinha 2016-11-04 05:41:06 +00:00 Commented Nov 4, 2016 at 5:41
1 If you really must, then you can use a change of base formula to say that log−1 e(n)=log n(e)log−1 e(n)=log n(e).ConMan –ConMan 2016-11-04 05:42:19 +00:00 Commented Nov 4, 2016 at 5:42
I am not getting it ;can you please rewrite using $$Learnmore –Learnmore 2016-11-04 05:42:31 +00:00 Commented Nov 4, 2016 at 5:42
@Amartya done as you asked.Somenath Sinha –Somenath Sinha 2016-11-04 05:44:14 +00:00 Commented Nov 4, 2016 at 5:44
1 @SomenathSinha Some people might write log log n log log n as log 2 n log 2 n but I don't think I've ever seen it. I think a notation like that would normally be explained in the same document before it was used, in order to avoid misunderstanding.David K –David K 2016-11-04 05:50:39 +00:00 Commented Nov 4, 2016 at 5:50
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Ouch!
This is very classic ambiguous notation. The notation f−1(x)f−1(x) can mean one of two entirely different meanings depending on context.
It could (but usually doesn't) mean (f(x))−1=1 f(x)(f(x))−1=1 f(x). The is the reciprical of f f. It is simply the function as a value raised to the −1−1 power.
More frequently it means the functional inverse (if there is one). f−1(x)=y f−1(x)=y means that y y is the value so that f(y)=x f(y)=x. For example, if f(x)=x 3+2 f(x)=x 3+2 then f−1(x)=3√x−2 f−1(x)=x−2−−−−−√3 because if y=3√x−2 y=x−2−−−−−√3 then f(y)=x f(y)=x. And f(f−1(x))=x f(f−1(x))=x by definition.
These two interpretations have NOTHING to do with each other and if you do confuse them (as people often do) you get utterly confusing contradictory nonsense. Example if f(x)=x 3+2 f(x)=x 3+2 and we thought both f−1(x)=1 f(x)=1 x 3+2 f−1(x)=1 f(x)=1 x 3+2 and that f−1(y)=3√x−2 f−1(y)=x−2−−−−−√3 and THAT MAKES NO FREAKING SENSE WHATSOEVER. (You will not BELIEVE the number of students who get frustrated because that makes no sense at all to them, but are convinced it must be so and it must be their fault that they simply can't understand it.)
So if you are interpreting log−1 x log−1 x as 1 log x 1 log x (and I highly suggest you do NOT do so-- if you want notation for 1 log x 1 log x use (log x)−1(log x)−1 inside parenthesis instead) then, the answer to the question as ... yeah.
But if you are interpreting log−1 x log−1 x as the inverse function--- that is, the function so that if y=log x y=log x then log−1 y=log−1 log x=x log−1 y=log−1 log x=x (and this is how you SHOULD interpret it) then the answer is an emphatic !!!!NO!!!!!!. As it turns out, log−1 y=e y log−1 y=e y (assuming log log is the natural log, base e e-- if log log it that base 10 log then log−1 10 y=10 y log−1 10 y=10 y).
To confuse the two and think log−1(log y)=y log−1(log y)=y and log−1(log y)=1 log y=y log−1(log y)=1 log y=y will result in meaningless, impossible and unresolvable GARBAGE. So don't ever confuse the two.
And I STRONGLY urge you to NEVER use f−1(x)f−1(x) to mean the reciprical but to ALWAYS mean the functional inverse.
So log−1 e y=e y log−1 e y=e y and log−1 e y≠1 log e y log−1 e y≠1 log e y.
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answered Nov 4, 2016 at 6:14
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1062 | https://www.quora.com/What-is-the-relation-between-the-sum-of-the-odd-numbers-and-the-sum-of-the-even-numbers | Something went wrong. Wait a moment and try again.
Sum of Series
Positive Odd Integers
Relations (mathematics)
Sums of Mathematics
Maths in Relation
Number Theory
Odd and Even Multiples
5
What is the relation between the sum of the odd numbers and the sum of the even numbers?
Mark Gritter
recreational mathematician
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 5.7K answers and 11.7M answer views
·
4y
The sum of the odd integers from 1 to n, and the sum of the even numbers from 1 to n, add up to the sum of all integers from 1 to n.
For odd n=2k+1, we have
(k+1)2+k(k+1)=(2k+1)(2k+2)/2
odd sum + even sum = complete sum
For even n=2k, we have
k2+k(k+1)=(2k)(2k+1)/2
From this, we can see that the sum of odd integers is always a square. So, if we know the sum S of odd numbers, then the sum of the even numbers up to the same point is (√S)(√S+1)).
The other direction is not quite as straight forward. Let T be the sum of even numbers. Then assuming the stopping point of the su
The sum of the odd integers from 1 to n, and the sum of the even numbers from 1 to n, add up to the sum of all integers from 1 to n.
For odd n=2k+1, we have
(k+1)2+k(k+1)=(2k+1)(2k+2)/2
odd sum + even sum = complete sum
For even n=2k, we have
k2+k(k+1)=(2k)(2k+1)/2
From this, we can see that the sum of odd integers is always a square. So, if we know the sum S of odd numbers, then the sum of the even numbers up to the same point is (√S)(√S+1)).
The other direction is not quite as straight forward. Let T be the sum of even numbers. Then assuming the stopping point of the sum is the last even number, we have
k(k+1)=T
4k2+4k+1=4T+1
(2k+1)2=4T+1
2k+1=√4T+1
2k=√4T+1−1
4k2=4T+1−2√4T+1+1
So, the odd sum is k2=T−12√4T+1+12
For example, take the even numbers sum T=2+4+6+8=20
k2=20−12√4⋅20+1+12=20−92+12=16
and the corresponding odd sum is 1+3+5+7=16.
Note that this latter case is ambiguous, the stopping point could be either k2 or (k+1)2 and we’d get the same even sum.
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Enrico Gregorio
Associate professor in Algebra
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Author has 18K answers and 15.7M answer views
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4y
Let E(n) denote the sum of the first n even number, so
E(1)=2,E(2)=2+4,…,E(n)=2+4+⋯+(2n)
Let O(n) denote the sum of the first n odd numbers, so
O(1)=1,O(2)=1+3,…,O(n)=1+3+⋯+(2n−1)
If you add n to O(n), you get E(n), because you can distribute a single 1 to each summand.
Therefore
E(n)=O(n)+n
which you may want to rigorously prove by induction.
By the way, O(n)=n2, which is interesting of its own and admits a proof without words.
Sum of odd numbers always gives a perfect square.
Let E(n) denote the sum of the first n even number, so
E(1)=2,E(2)=2+4,…,E(n)=2+4+⋯+(2n)
Let O(n) denote the sum of the first n odd numbers, so
O(1)=1,O(2)=1+3,…,O(n)=1+3+⋯+(2n−1)
If you add n to O(n), you get E(n), because you can distribute a single 1 to each summand.
Therefore
E(n)=O(n)+n
which you may want to rigorously prove by induction.
By the way, O(n)=n2, which is interesting of its own and admits a proof without words.
Sum of odd numbers always gives a perfect square.
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Bot
·
1y
The relationship between the sum of odd numbers and the sum of even numbers can be explored in several ways. Here are some key points:
Sum of the First
n
Odd Numbers
:
The sum of the first
n
odd numbers is given by the formula:
S
odd
=
1
+
3
+
5
+
…
+
(
2
n
−
1
)
=
n
2
For example, the sum of the first 5 odd numbers (1, 3, 5, 7, 9) is
25
, which is
5
2
.
2. Sum of the First
n
Even Numbers
:
The sum of the first
n
even numbers is given by the formula:
S
even
=
2
+
4
+
6
+
…
+
2
n
=
n
(
n
+
1
)
For example, the sum of the first 5 even numbers (2, 4, 6, 8, 10) is
30
,
The relationship between the sum of odd numbers and the sum of even numbers can be explored in several ways. Here are some key points:
Sum of the First
n
Odd Numbers
:
The sum of the first
n
odd numbers is given by the formula:
S
odd
=
1
+
3
+
5
+
…
+
(
2
n
−
1
)
=
n
2
For example, the sum of the first 5 odd numbers (1, 3, 5, 7, 9) is
25
, which is
5
2
.
2. Sum of the First
n
Even Numbers
:
The sum of the first
n
even numbers is given by the formula:
S
even
=
2
+
4
+
6
+
…
+
2
n
=
n
(
n
+
1
)
For example, the sum of the first 5 even numbers (2, 4, 6, 8, 10) is
30
, which is
5
×
6
.
3. Comparison
:
If we compare the two sums for the same
n
:
- Sum of the first
n
odd numbers:
n
2
- Sum of the first
n
even numbers:
n
(
n
+
1
)
We can see that:
S
even
=
S
odd
+
n
This shows that the sum of the first
n
even numbers is always
n
greater than the sum of the first
n
odd numbers.
Conclusion
:
The sum of the first
n
odd numbers grows quadratically as
n
2
.
- The sum of the first
n
even numbers grows quadratically as
n
(
n
+
1
)
.
- For any positive integer
n
, the sum of the even numbers up to
2
n
is greater than the sum of the odd numbers up to
2
n
−
1
by exactly
n
.
This relationship highlights the predictable nature of these sums and their growth patterns.
Related questions
How would you show that the sum of odd and even numbers are always odd?
How would you show that the sum of even numbers and odd numbers is always odd?
What is the sum of odd numbers from 1 to 50? What is the sum of even numbers from 1 to 50?
What is the relation between the sum of odd integers and the sum of even integers?
What is the sum of odd numbers?
Anil Bapat
Have studied Mathematics up to pre-Degree Level
·
Author has 2.8K answers and 3.7M answer views
·
Updated 4y
Originally Answered: What is the relation between the sum of odd integers and the sum of even integers?
·
What is the relation between the sum of odd integers and the sum of even integers? [ ]
I am not sure as to what the exact question is but looks like this is around test of divisibility.
When we are checking certain number for test of divisibility with 11, this test comes into being.
Here, we add Od
What is the relation between the sum of odd integers and the sum of even integers? [ ]
I am not sure as to what the exact question is but looks like this is around test of divisibility.
When we are checking certain number for test of divisibility with 11, this test comes into being.
Here, we add Odd Digits separately as also the Even Digits. Let the sum of odd digits be x and sum of even digits be y. We then subtract x from y or vice versa and if the final number is 0 or a multiple of 11 then the original number is divisible by 11.
By extending this a bit further, we can check if the same number is divisible by 22, 33, 44, 55, 66 etc. as shown below.
22: If the number is divisible by 11 and the last digit is even
33: If the number is divisible by 11 and x+y is divisible by 3
44: If the number is divisible by 11 and 4 divides last 2 digits
55: If the number is divisible by 11 and the last digit is 5
66: If the number is divisible by 11 and x + y is divisible by 3 and the last digit is even.
Further tests could also be developed for the other multiples of 11
EDIT 1: I am writing the answer again; I somehow thought it was about adding odd digits and even digits, but it’s ...
James Gere
Author has 1.6K answers and 1.3M answer views
·
4y
Originally Answered: What is the relation between the sum of odd integers and the sum of even integers?
·
The sum of even integers will always be even.
The sum of odd integers will be even if the number of terms is even, and odd if the number of terms is odd.
The sum of mixed even and odd integers will only be odd if the number of odd terms is odd.
If S is the ( even ) sum of n even terms. Adding one to each term produces a sum of odd terms S’ = S + n . If n is odd then so is S’.
The next statements apply to all finite Arithmetic Series, not solely to Arithmetic Series of integers.
The sum of n consecutive integers in a sequence starting at an integer, a, and ending at, b = a + ( n - 1 ) , is, S =
The sum of even integers will always be even.
The sum of odd integers will be even if the number of terms is even, and odd if the number of terms is odd.
The sum of mixed even and odd integers will only be odd if the number of odd terms is odd.
If S is the ( even ) sum of n even terms. Adding one to each term produces a sum of odd terms S’ = S + n . If n is odd then so is S’.
The next statements apply to all finite Arithmetic Series, not solely to Arithmetic Series of integers.
The sum of n consecutive integers in a sequence starting at an integer, a, and ending at, b = a + ( n - 1 ) , is, S = n•( a + b )/ 2 .
The sum of n consecutive odd integers or n consecutive even integers is S = n• ( a + b )/ 2 , where a is the first integer in the sequence and b is the last such integer.
There are other forms of this identity, derived from the common difference of successive terms, but having just one formula that fits all cases is handy.
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Paul Zimmermann
Owner at PVZ Consulting, Inc. (2018–present)
·
Author has 386 answers and 271.3K answer views
·
4y
The sum of the odd numbers is infinite; the sum of the even numbers is infinite. By inspection we can see the sum of the even numbers must be greater than the sum of the odd numbers (by an infinite number, no less!).
However, all infinities are equal.
This leads to the conclusion that the sum of the odd numbers and the sum of the even numbers are equal (as they are both infinite).
Definitely not intuitive.
Related questions
What is the difference of the sum of odd numbers and sum of even numbers between 100 and 120?
What is the three odd number whose sum is even?
What is the sum of the first 95 odd numbers?
What is the sum of the 1st 6 odd numbers?
What is the sum of n even numbers and odd numbers separately?
Ellis Cave
40+ years as an Electrical Engineer
·
Author has 7.9K answers and 4.3M answer views
·
4y
Originally Answered: What is the relation between the sum of odd integers and the sum of even integers?
·
Sum of the integers from 1 to n:
sa=n(n+1)/2
Sum of odd integers from 1 to n:
so=n^2
Sum of even integers from 1 to n:
se=n(n+1)
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968)
·
Author has 16.4K answers and 29.1M answer views
·
Updated 4y
Originally Answered: What is the relation between the sum of odd integers and the sum of even integers?
·
What is the relation between the sum of odd integers and the sum of even integers?
You need to specify the number of terms in some way. Suppose you take the first N odd integers and the first N even integers. Each of the even integers is 1 more than the preceding odd integer. and there are N of them.
Therefore if ON is the sum of the first N odd integers and E_N is the sum of the first N even integers Then EN=ON+N.
The total of all of these integers is the sum of the first 2N integers which is 2N(2N+1)2=N(2N+1)=2N2+N. It follows that the sum of the first N odd integers is ON=N2, so E
What is the relation between the sum of odd integers and the sum of even integers?
You need to specify the number of terms in some way. Suppose you take the first N odd integers and the first N even integers. Each of the even integers is 1 more than the preceding odd integer. and there are N of them.
Therefore if ON is the sum of the first N odd integers and E_N is the sum of the first N even integers Then EN=ON+N.
The total of all of these integers is the sum of the first 2N integers which is 2N(2N+1)2=N(2N+1)=2N2+N. It follows that the sum of the first N odd integers is ON=N2, so EN=N2+N=N(N+1).
Alon Amit
PhD in Mathematics; Mathcircler.
·
Upvoted by
Ellen Sassani
, M.A. Mathematics, University of South Florida and
Richard Shearer
, MSc (Biochemistry) Biochemistry & Mathematics, University of the Witwatersrand · Author has 8.7K answers and 172.8M answer views
·
Updated 8y
Related
What is the sum of first
n
odd numbers?
Let's look and see. Anyone can at least observe the first few instances, right?
1=1
1+3=4
1+3+5=9
1+3+5+7=16
1+3+5+7+9=25
Now, do you recognize the numbers on the right?
1,4,9,16,25,…
Yes! Those are the perfect squares. 1×1,2×2,3×3,4×4 and so on.
We now have a conjecture. Let's put it to the test:
1+3+5+7+9+11=36
Yes! The six smallest odd numbers add up to 62, just as we had predicted. You can try a few more: it works.
If we are physicists, we stop here. We've made an observation, we formed a hypothesis, we tested our hypothesis experimentally once and twice and a hundred
Let's look and see. Anyone can at least observe the first few instances, right?
1=1
1+3=4
1+3+5=9
1+3+5+7=16
1+3+5+7+9=25
Now, do you recognize the numbers on the right?
1,4,9,16,25,…
Yes! Those are the perfect squares. 1×1,2×2,3×3,4×4 and so on.
We now have a conjecture. Let's put it to the test:
1+3+5+7+9+11=36
Yes! The six smallest odd numbers add up to 62, just as we had predicted. You can try a few more: it works.
If we are physicists, we stop here. We've made an observation, we formed a hypothesis, we tested our hypothesis experimentally once and twice and a hundred times, it always works, done. Our theory is correct until an experiment refutes it.
But we are mathematicians, aren't we. We require proof. And there are rigorous proofs aplenty of this nice little fact.
But there's also a crystal clear visual proof. Here it is:
EDIT: many people have asked for a rigorous proof. Here's a relatively simple one which can be derived from this visual proof.
We notice that the odd numbers are just the differences between consecutive squares, like so:
1=12−02
3=22−12
5=32−22
7=42−32
and so on. Therefore when we add them up, everything cancels out except the last square:
1+3+5+7=(12−02)+(22−12)+(32−22)+(42−32)=42
So now let's write this formally for any number of odd numbers being added up. For any k,
2k+1=(k+1)2−k2
and therefore the sum of the first n odd numbers, which is
n−1∑k=02k+1
is equal to
n−1∑k=0(k+1)2−k2=n∑k=1k2−n−1∑k=0k2=n2. QED
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Bob Chan
Lives in Canada
·
Author has 189 answers and 670.9K answer views
·
7y
Related
How would you show that the sum of even numbers and odd numbers is always odd?
How would you show that the sum of even numbers and odd numbers is always odd?
Other answers show you mathematically how it must be so.
Let me try what I told my daughter when she was in kindergarten.
Even number (of items) can be paired up without one left alone.
When you try to pair up odd number (of items), there will be an “odd” one left behind.
Now it is easy to see that even plus even is even. Neither group had an “odd” one to begin with.
Even plus odd is odd because the “odd” one still had nobody to pair with.
Odd plus odd gives even because now those “odd” fellows can pair with each other.
Hop
How would you show that the sum of even numbers and odd numbers is always odd?
Other answers show you mathematically how it must be so.
Let me try what I told my daughter when she was in kindergarten.
Even number (of items) can be paired up without one left alone.
When you try to pair up odd number (of items), there will be an “odd” one left behind.
Now it is easy to see that even plus even is even. Neither group had an “odd” one to begin with.
Even plus odd is odd because the “odd” one still had nobody to pair with.
Odd plus odd gives even because now those “odd” fellows can pair with each other.
Hope this helps.
Lucas Curtis
Science teacher (2001–present)
·
Author has 7.3K answers and 21.5M answer views
·
7y
Related
What is the sum of ten odd numbers and eleven even numbers?
Well, that greatly depends on what the numbers actually are. I can tell you this: no matter what the numbers are, their sum is even. Why?
The sum of an odd number and an even number is odd. The sum of two odd numbers is even.
So if you add one odd number and one even number, you get an odd number. If you add two odd numbers and two even numbers, that’s the same as adding two odd numbers, which means you get an even number.
Let 2j and 2m be even numbers. Let 2k + 1 and 2n + 1 be odd numbers.
(2j) + (2m) + (2k + 1) + (2n + 1)
= 2j + 2m + 2k + 2n + 2
= 2(j + m + k + n + 1)
Clearly the result is even.
So
Well, that greatly depends on what the numbers actually are. I can tell you this: no matter what the numbers are, their sum is even. Why?
The sum of an odd number and an even number is odd. The sum of two odd numbers is even.
So if you add one odd number and one even number, you get an odd number. If you add two odd numbers and two even numbers, that’s the same as adding two odd numbers, which means you get an even number.
Let 2j and 2m be even numbers. Let 2k + 1 and 2n + 1 be odd numbers.
(2j) + (2m) + (2k + 1) + (2n + 1)
= 2j + 2m + 2k + 2n + 2
= 2(j + m + k + n + 1)
Clearly the result is even.
So if you add an odd number of odd/even pairs, you get an odd sum, but if you add an even number of odd/even pairs, you get an even number.
Since 10 is an even number, you’re going to get an even sum if you add 10 odd numbers and 10 even numbers, regardless of what those numbers actually are.
But in this case we’re adding 11 even numbers to 10 odd numbers, which is like adding another even number to the 10 odd-even pairs we’ve already summed up. Adding an even number to an even number produces another even number; ergo, the sum in question must be even.
Mike Alexander
Studied A-Level Mathematics, Physics & Computer Science
·
Author has 2.3K answers and 2.8M answer views
·
4y
Related
What is the sum of the first 50 odd numbers?
The sum of the first n odd numbers is n2.
So it's 50×50 = 2500.
To see why, imagine we start with 1 square tile. To make a 2×2 tile arrangement, we add one tile above it, one tile to the right of it, and one tile to fill in the top right corner. That's 3 tiles we've added. Now, to extend this to a 3×3 arrangement, we must add 2 tiles above the top row, 2 tiles beside the right column, and 1 in the top right corner. That's 5 tiles we've added. Next round we'll add 3+3+1 = 7 tiles, etc. So the series 1+3+5+7+9+11… will always make a square.
The sum of the first n odd numbers is n2.
So it's 50×50 = 2500.
To see why, imagine we start with 1 square tile. To make a 2×2 tile arrangement, we add one tile above it, one tile to the right of it, and one tile to fill in the top right corner. That's 3 tiles we've added. Now, to extend this to a 3×3 arrangement, we must add 2 tiles above the top row, 2 tiles beside the right column, and 1 in the top right corner. That's 5 tiles we've added. Next round we'll add 3+3+1 = 7 tiles, etc. So the series 1+3+5+7+9+11… will always make a square.
Rana Faiz Ahmad
Science Enthusiast
·
Author has 193 answers and 446.8K answer views
·
8y
Related
What is the sum of first
n
odd numbers?
Sum of first n odd numbers is n2.
Let S be the sum of first n odd numbers.
S=1+3+⋯+(2n−1)
S=(2n−1)+(2(n−1)−1)+⋯+1
Adding both of these equations:
2S=(2n)n
S=n2
Related questions
How would you show that the sum of odd and even numbers are always odd?
How would you show that the sum of even numbers and odd numbers is always odd?
What is the sum of odd numbers from 1 to 50? What is the sum of even numbers from 1 to 50?
What is the relation between the sum of odd integers and the sum of even integers?
What is the sum of odd numbers?
What is the difference of the sum of odd numbers and sum of even numbers between 100 and 120?
What is the three odd number whose sum is even?
What is the sum of the first 95 odd numbers?
What is the sum of the 1st 6 odd numbers?
What is the sum of n even numbers and odd numbers separately?
Is the sum of the first 999 odd numbers even or odd?
How is it possible that the sum of a three odd number is even?
What is odd number x even number =?
Can the sum of 3 odd numbers be even?
Is there any even number which results from the sum of 3 odd numbers?
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1063 | https://www.convertunits.com/from/cm+H2O/to/mmHg | Convert cm H2O to mmHg - Conversion of Measurement Units
Convert centimeter water [4 °C] to millimeter of mercury [0 °C]
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How many cm H2O in 1 mmHg?
The answer is 1.3595100263597.
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The SI derived unit for pressure is the pascal.
1 pascal is equal to 0.010197162129779 cm H2O, or 0.0075006156130264 mmHg.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between centimeters water and millimeters of mercury.
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Quick conversion chart of cm H2O to mmHg
1 cm H2O to mmHg = 0.73556 mmHg
5 cm H2O to mmHg = 3.6778 mmHg
10 cm H2O to mmHg = 7.35559 mmHg
20 cm H2O to mmHg = 14.71118 mmHg
30 cm H2O to mmHg = 22.06677 mmHg
40 cm H2O to mmHg = 29.42236 mmHg
50 cm H2O to mmHg = 36.77796 mmHg
75 cm H2O to mmHg = 55.16693 mmHg
100 cm H2O to mmHg = 73.55591 mmHg
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Definition: Millimeter of mercury
The torr (symbol: Torr) or millimetre of mercury (mmHg) is a non-SI unit of pressure. It is the atmospheric pressure that supports a column of mercury 1 millimetre high. The unit is named after Evangelista Torricelli, Italian physicist and mathematician, for his discovery of the principle of the barometer in 1643.
Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! |
1064 | https://mpra.ub.uni-muenchen.de/12414/1/MPRA_paper_12414.pdf | Munich Personal RePEc Archive Derivation of the Hicks Elasticity of Substitution from the Input Distance Function Stern, David I.
n.a.
30 December 2008 Online at MPRA Paper No. 12414, posted 01 Jan 2009 09:22 UTC Derivation of the Hicks Elasticity of Substitution from the Input Distance Function December 2008 David I. Stern 19/30 Watson Street, Turner, ACT 2612, AUSTRALIA E-mail: sterndavidi@yahoo.com Abstract The Hicks or direct elasticity of substitution is traditionally derived from the production function. This paper exploits duality theory to present a more general derivation from the input distance function, which is exactly dual to the Shadow Elasticity of Substitution. The new elasticity is more general than the traditional one as it can handle situations of technical inefficiency, non-separability between inputs and outputs, and multiple outputs, but is equal to the traditional elasticity under the classical conditions. The new derivation is related to the Morishima and Antonelli Elasticities of Complementarity in the same way that the Shadow Elasticity of Substitution is related to the Morishima and Allen-Uzawa Elasticities of Substitution. Furthermore, distance (technical efficiency) is not constant for the Morishima and Antonelli Elasticities of Complementarity JEL Codes: B21, D12, D24 Keywords: Microeconomics, production, substitution Word Count (not including this page): 1823 2 1. Introduction The elasticity of substitution introduced by Hicks (1932) was reformulated by Lerner (1933) as the degree to which the marginal rate of substitution between two inputs varies as the ratio of the quantity of those inputs varies while output is held constant. Assuming integrability (Hicks and Allen, 1934b), Hicks and Allen (1934a) proposed the following formula: " ij = #ln( fi / f j) #ln(x j /xi) y (1) where output, y, which is held constant, is produced according to the production function y = f (x) whose derivatives are indicated by subscripts. x is the vector of inputs. This is now known as the Hicks or Direct elasticity of substitution, though it is usually defined as the reciprocal of (1). Traditionally, this elasticity of substitution is derived from the production function, which is natural given (1). This paper exploits duality theory to present an alternative, more general derivation from the input distance function. The new derivation is related to the Morishima Elasticity of Complementarity (Kim, 2000) in the same way that the Shadow Elasticity of Substitution (MacFadden, 1963) is related to the Morishima Elasticity of Substitution (Morishima, 1967; Blackorby and Russell, 1975). I also estimate a system of cost share equations derived from the input distance function using the classic Berndt and Wood (1975) data set and compute the elasticity for each factor combination. 3 2. Relationships between Elasticities of Substitution and Complementarity Pigou (1934) pointed out that, in general, unless further restrictions are made, the value of the two factor, two price class (see Mundlak, 1968) of elasticity of substitution: " ij = #ln( fi / f j) #ln(x j /xi) (2) or its inverse is not symmetric to which price or quantity in the denominator changes. In the case of the Hicks Elasticity of Substitution (HES), Pigou (1934) suggested holding the output quantity constant, as in (1). Given certain conditions, McFadden (1963) gave a formula for the reciprocal of (1): HESij = 1 fixi + 1 fj xj ! fii fi fi + 2 fij fi fj ! fjj f j fj (3) Mundlak (1968), Morishima (1967), and Blackorby and Russell (1975) introduced the asymmetric Morishima Elasticity of Substitution: MESij = "ln(Ci(y,p)/C j(y,p)) "ln(p j / pi) (4) where C(y, p1,..., pm) is the cost function in output quantity and input prices whose derivatives are indicated by subscripts. The rationale behind this formula is that according to Shephard’s Lemma the derivative of the cost function is equal to the optimal factor input. This elasticity is exactly Robinson’s (1933) notion of the elasticity of the change in the input ratio with respect to the price ratio holding output constant and letting all other inputs adjust optimally while holding 4 their prices constant. Equation (2), on the other hand, holds all other inputs constant and lets the shadow prices of the inputs adjust. Blackorby and Russell (1989) show that for a change in pi (4) can be found as: MESij = "ln x j(y,p) "ln pi # "ln xi(y,p) "ln pi (5) The two terms on the RHS of (5) are the cross-price and own-price demand elasticities. The MES is, therefore, asymmetric, so that MESij ≠ MESji. Mundlak (1968) indicates that we can make this elasticity symmetric by imposing a condition of constant cost yielding McFadden’s (1963) shadow elasticity of substitution: SESij = ! Cii CiCi + 2 Cij CiCj ! Cjj CjCj 1 Cipi + 1 Cj pj (6) Equations (3) and (6) look very similar but there is an asymmetry between them. While the HES holds all other inputs and output constant, the SES holds all other input prices, output, and also cost constant. Blackorby and Russell (1981) and Kim (2000) proposed the Morishima elasticity of complementarity based on the input distance function as the dual of the MES. The formula is: MECij = "ln Di(y,x)/D j(y,x) ( ) "ln(x j /xi) = "ln pi(y,x) "ln x j # "ln p j(y,x) "ln x j (7) 5 where D(y,x) is the input distance function (Färe and Primont, 1995). The MEC is, therefore, the difference between two elasticities of price with respect to input quantities based on the inverse demand functions derived from the input distance function. Like the MES, the MEC is asymmetric. The elasticity shows how the price ratio changes as one of the inputs in the input ratio changes under the assumption of cost-minimisation and holding all other inputs and output constant. But it is impossible for only one input to change and for output to be held constant, unless distance is changing. And this is exactly what must be the case for the MEC and the related Antonelli Elasticity of Complementarity (Kim, 2000). In other words, the free disposal assumption is violated. We can impose the restriction of constant distance on the MEC (7) in the same way that constant cost was imposed on the MES (4) to yield the SES (6). As an input is changed, another input will then be forced to change so that constant distance is maintained. The elasticity will then measure substitution along a curve of constant distance in the same way as the SES measures substitution along an isocost curve. Because output is an argument of the input distance function, output will also be held constant in the proposed elasticity and substitution also occurs along an isoquant. It stands to reason that this restricted elasticity will be symmetric in the same way that the SES is symmetric. 6 3. New Derivation of the Hicks Elasticity of Substitution The input distance function for a single output is given by (for more details see Färe and Primont, 1995): D(y,x) "1 (8) Based on Shephard’s Lemma, the derivatives with respect to inputs i and j are given by (Färe and Primont, 1995): pi C = Di and pj C = Dj (9) where the prices p can be either market or shadow prices. We wish to evaluate the elasticity of complementarity – i.e. where prices respond to changes in inputs as in (7), which is given by: " ij = d(pi(y,x)/ p j(y,x)) d(x j /xi) p jx j pixi (10) subject to the quantities of all other inputs and output being held constant: dxk = dy = 0 !k " i, j (11) and, additionally, distance is held constant: "D "xk dxk = dD = 0 k=1 m # (12) Substituting (9) into (12) conditions (11) and (12) imply that: dxi = !(pj / pi)dx j (13) 7 We evaluate the two total differentials in (10) by applying the quotient rule and substituting in the expression in (13) in place of dxi : d(pi(x,y)/ p j(x,y)) = " DiiD j " D jiDi D jD j p j pi dx j + DijD j " D jjDi D jD j dx j (14) d(xj / xi) = pj xj xi 2pi dxj + 1 xi dx j (15) Then we take the ratio of (14) and (15) and multiply by p jx j ( )/ pixi ( ) to obtain the elasticity: " ij = #DiiD jD j + 2DijDiD j # D jjDiDi Dixi + D jx j x jxi DiD j (16) which can be simplified to (17): " ij = # Dii DiDi + 2 Dij DiD j # D jj D jD j 1 Dixi + 1 D jx j (17) This expression is identical to the reciprocal of (3) except that each derivative of the production function is replaced with a derivative of the input distance function. It is also clear that the elasticity is symmetric. In the single output case, when D(y,x) =1 and inputs and outputs are separable, evaluating (10) subject to (11) should lead to the same result as evaluating (10) subject to (11) and (12). In this 8 classical case, (17) should equal the reciprocal of (3) and be the original concept proposed by Lerner (1933) and Hicks and Allen (1934a). However, (17), which is the exact dual of (6), is a more general indicator as it can also handle situations of technical inefficiency, non-separability between inputs and outputs, and multiple outputs. 4. Numerical Example This section provides a numerical illustration by estimating a series of cost share equations derived from the input distance function and deriving the elasticities of substitution from the results. Following Kim (2000), I use the classic Berndt and Wood (1975) data set and the translog function to allow comparability with previous studies. Linear, shares based regression is used, despite the existence of better nonlinear approaches (e.g. Kim, 1992). I estimate the model using the instrumental variables in the Berndt and Wood (1975) dataset. Technical change is modeled using the standard linear trends biased technical change approach despite its potential shortcomings (Lim and Shumway, 1997). Among studies using the same data set, Kim (2000) models technological change using linear time trends, while Berndt and Wood (1975) do not include trends in their cost share equations. The translog input distance function for a single output, y, in period t, is given by: "0 + "y ln yt + "i ln xit i=1 m # + "$$ t + 1 2 %$$ t 2 + 1 2 %yy ln yt 2 + %iy ln xit ln yt i=1 m # + 1 2 %ij ln xit ln x jt j=1 m # i=1 m # + %y$ ln yt$ t + %i$ ln xit$ t i=1 m # &1 (18) 9 where τ is a linear time trend and the α’s and β’s are coefficients to be estimated. The input distance function is homogenous of degree one, increasing, and concave with respect to the inputs and decreasing with respect to the output (Kim, 2000). Using (9) and imposing the condition that D = 1 (technical efficiency) results in the following cost share equations: Sit = "i + #iy ln yt + #ij ln x jt j=1 m $ + #i%% (19) where Sit is the cost share of input i in period t. I estimate Equation (19) for the capital, labor, and energy cost shares imposing homogeneity of degree zero in the inputs by normalizing on the quantity of materials and symmetry of the coefficients and deducting the means of the explanatory variables. I then compute the elasticities at the mean where all the variables are zero. Under these assumptions, the elasticity of substitution (18) is estimated as: " ij = #i $ %ii #i 2 + 2 %ij #i# j + # j $ % jj # j 2 1 #i + 1 # j i " j (20) which is reported in Table 1. As this elasticity is the inverse of the traditional one, greater values indicate less substitutability. The poorest substitutes are capital and energy and the best substitutes are labor and materials. Capital has restricted substitutability with all the other inputs. 10 References Allen, R. G. D., 1934, A comparison between different definitions of complementary and competitive goods, Econometrica 2, 168-175. Allen, R. G. D., 1938, Mathematical Analysis for Economists. (Macmillan, London). Berndt, E. R. and D. O. Wood, 1975, Technology, prices and the derived demand for energy, Review of Economics and Statistics 57, 259-268. Blackorby, C. and R. R. Russell, 1975, The Partial Elasticity of Substitution, Discussion Paper 75-1, Economics, University of California, San Diego. Blackorby, C. and R. R. Russell, 1981, The Morishima elasticity of substitution: symmetry, constancy, separability, and relationship to the Hicks and Allen elasticities, Review of Economic Studies 43, 147-158. Blackorby, C. and R. R. Russell, 1989, Will the real elasticity of substitution please stand up?, (A comparison of the Allen/Uzawa and Morishima elasticities). American Economic Review 79, 882-888. Färe R. and D. Primont, 1995, Multi-Output Production and Duality: Theory and Applications. (Kluwer, Boston). Hicks, J. R., 1932, Theory of Wages. (Macmillan, London). Hicks, J. R. and R. G. D. Allen, 1934a, A reconsideration of the theory of value, part I, Economica 1, 52-76. Hicks, J. R. and R. G. D. Allen 1934b, A reconsideration of the theory of value, part II, Economica 1, 196-219. Kim, H. Y., 1992, The translog production function and variable returns to scale, Review of Economics and Statistics 74, 546-552. 11 Kim, H. Y., 2000, The Antonelli versus Hicks elasticity of complementary and inverse input demand systems, Australian Economic Papers 39, 245-261. Lerner, A. P., 1933, Notes on the elasticity of substitution: II the diagrammatical representation, Review of Economic Studies 1, 68-70. Lim, H. and C. R. Shumway, 1997, Technical change and model specification: U.S. agricultural production, American Journal of Agricultural Economics 79, 543-554. McFadden, D., 1963, Constant elasticity of substitution production functions, Review of Economic Studies 31, 73-83. Morishima, M., 1967, A few suggestions on the theory of elasticity (in Japanese), Keizai Hyoron (Economic Review) 16, 144-150. Mundlak, Y., 1968, Elasticities of substitution and the theory of derived demand, Review of Economic Studies 35, 225-236. Pigou, A. C., 1934, The elasticity of substitution, Economic Journal 44, 232-241. Robinson, J. V., 1933, The Economics of Imperfect Competition. (Macmillan, London). Uzawa, H., 1962, Production functions with constant elasticities of substitution, Review of Economic Studies 30, 291-299. 12 Table 1. Hicks Elasticities of Substitution in Elasticity of Complementarity Form Capital Labor Energy Materials Capital Labor 1.36565 Energy 1.47382 0.72817 Materials 1.16291 0.34248 0.77318 |
1065 | https://www.log2base2.com/data-structures/linked-list/deleting-a-node-in-linked-list.html | Deleting a node in linked list
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Deleting a node in linked list
Delete a given node from the linked list.
Example
Linked List : 10 20 30 NULL
Input
20
Output
10 30 NULL
Input
x
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30
Output
10 20 NULL
Algorithm
If the head node has the given key,
make the head node points to the second node and free its memory.
Otherwise,
From the current node, check whether the next node has the given key
if yes, make the current->next = current->next->next and free the memory.
else, update the current node to the next and do the above process (from step 2) till the last node.
1. Head node has the given key
void deleteNode(struct node head, int key)
{
//temp is used to freeing the memory
struct node temp;
//key found on the head node. //move to head node to the next and free the head. if(head->data == key) { temp = head; //backup the head to free its memory head = (head)->next; free(temp); }
}
Visual Representation
Let's delete data 10 (head node).
Make the head points to the next node.
Free the head node's memory.
Finally, the new linked list.
2. For other nodes (Non-Head)
void deleteNode(struct node head, int key)
{
//temp is used to freeing the memory
struct node temp;
//key found on the head node.
//move to head node to the next and free the head.
if((head)->data == key)
{
temp = head; //backup to free the memory
head = (head)->next;
free(temp);
}
else { struct node current = head; while(current->next != NULL) { //if yes, we need to delete the current->next node if(current->next->data == key) { temp = current->next; //node will be disconnected from the linked list. current->next = current->next->next; free(temp); break; } //Otherwise, move the current node and proceed else current = current->next; } }
}
Visual Representation
Let's delete data 20.
Make the current node points to the head node. (current => data = 10).
current => next. (current=>next=>data = 20).
current => next => next. (current=>next=>next=>data = 30).
We have to remove the node 20. Since current => next = 20 we can remove the node by setting current => next = current =>next => next. And then free the node.
Finally, the new linked list.
Implementation of deleting a node in linked list
Example
/
Program: Deleting a node in the linked list
Language: C
/
include
include
struct node
{
int data;
struct node next;
};
void addLast(struct node head, int val)
{
//create a new node
struct node newNode = malloc(sizeof(struct node));
newNode->data = val;
newNode->next = NULL;
//if head is NULL, it is an empty list
if(head == NULL)
head = newNode;
//Otherwise, find the last node and add the newNode
else
{
struct node lastNode = head;
//last node's next address will be NULL.
while(lastNode->next != NULL)
{
lastNode = lastNode->next;
}
//add the newNode at the end of the linked list
lastNode->next = newNode;
}
}
void deleteNode(struct node head, int key) { //temp is used to freeing the memory struct node temp; //key found on the head node. //move to head node to the next and free the head. if((head)->data == key) { temp = head; //backup to free the memory head = (head)->next; free(temp); } else { struct node current = head; while(current->next != NULL) { //if yes, we need to delete the current->next node if(current->next->data == key) { temp = current->next; //node will be disconnected from the linked list. current->next = current->next->next; free(temp); break; } //Otherwise, move the current node and proceed else current = current->next; } } }
void printList(struct node head)
{
struct node temp = head;
//iterate the entire linked list and print the data
while(temp != NULL)
{
printf("%d ->", temp->data);
temp = temp->next;
}
printf("NULL\n");
}
int main()
{
struct node head = NULL;
addLast(&head,10);
addLast(&head,20);
addLast(&head,30);
printf("Linked List Elements:\n");
printList(head);
//delete first node
deleteNode(&head,10);
printf("Deleted 10. The New Linked List:\n");
printList(head);
//delete last node
deleteNode(&head,30);
printf("Deleted 30. The New Linked List:\n");
printList(head);
//delete 20
deleteNode(&head,20);
printf("Deleted 20. The New Linked List:\n");
printList(head);
return 0;
}
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1066 | https://www.youtube.com/watch?v=zkNBSyHh7vw | How to find the intersection point of two linear equations
Brian McLogan
1600000 subscribers
10578 likes
Description
1050374 views
Posted: 8 Dec 2011
In this video series I show you how to solve a system of equations by graphing. When solving a system of equations by graphing we graph each equation separately and then analyze where the two lines intersect or if at all. If the two lines intersect at a point then the values of the x and y coordinates are our solution of the system. If our graphs are the same line then we have infinite many solutions and if the graphs are parallel and never intersect then we have no solution.
x + y = 3 , 3x - 2y = 4
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Transcript:
okay um if i was gonna be solving a problem like this what it wants us to do is it wants us to find the intersection of the two points so what we're gonna need to do is see when will these two equations you know equal each other when will they cross so what we'll look at is you know if we said something like this okay where are their points going to be exactly the same so if you think of this here's your y and here's your x right so what i'm saying is at this intersection point the x value and the y value are going to equal each other for both the um for both equations so what i'm saying is what we're going to do is we have two different equations we want to find when does the x value equal the y value so mrs garvin would you please come to the main office mrs garvin please come to the main office and not go to r9 portal r9 also officer smith would you please come to the conference room would you please come to the principal's conference room officer smith okay um so then to go and figure this out what we're going to have to do video yeah just come by and say hi i want to see what you're doing right now okay we're doing intersections with two lines will you just sit down i'll be on with done with this video in two seconds thank you for showing me what you're doing um so then what we're gonna want to do is here let's solve for y so if i subtract the x on both sides i get y equals a negative x plus three okay so now if i want to set these equal to each other right i want to say well i know that my y values have to be the same for both of these equations my two y values have to be exactly the same so therefore what i'm going to do is i'm going to plug this equation into the other one so i'll obtain 3x minus 2 times negative x plus 3 equals 4. all right then what i can do is when the y values are the same meaning i plugged one y value to the other one i can find when will the x values be the same so i'll use distributive property to give me 3x plus a 2x negative 2 times negative x gives you positive 2x minus 6 equals 4. combine my like terms to give me 5x minus 6 equals 4 add 6 5x equals 10 divided by 5 x equals 2. so what that tells me is when my y values are equal to each other my x value is equal to 2. so now the next thing i need to do is now well if x equals 2 what is my y value for both those problems so let's go and take a look at that what i can do is i can plug 2 in for x and figure out what my y is going to be so i'm going to plug it into both equations so you guys can see that when x equals 2 my y values are going to be exactly the same and that also will figure out what the y value is so i plug in 2 plus y equals 3. subtract 2 y equals 1. and we'll say that value is 1. or you could have plugged it in this equation 3 times two minus two y equals four three times two is six minus two y equals four subtract six negative two y equals a negative two divided by negative 2 y equals 1. so again remember when you're dealing with intersection you want to set your equations equal to each other plug one equation in for the other so y equals x minus 3. so therefore i'm going to plug that in solve for my one variable x then once i know the value of x where the two equations meet i can plug that into either one of the equations and i'll find the y value for when the two equations mean and that's how you find the intersection of two points |
1067 | https://liorpachter.wordpress.com/tag/mordell-weil-theorem/ | Mordell-Weil theorem | Bits of DNA
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I was wrong (part 3)
November 5, 2018 in academia, sexual harassment | Tags: #protestNIPS, Cox-Zucker machine, elliptic curves, Mordell-Weil theorem | by Lior Pachter | 23 comments
Last year I wrote a blog post on being wrong. I also wrote a blog post about being wrong three years ago. It’s not fun to admit being wrong, but sometimes it’s necessary. I have to admit to being wrong again.
To place this admission in context I need to start with Mordell’s finite basis theorem, which has been on my mind this past week. The theorem, proved in 1922, states the rational points on an elliptic curve defined over the rational numbers form a finitely generated abelian group. There is quite a bit of math jargon in this statement that makes it seem somewhat esoteric, but it’s actually a beautiful, fundamental, and accessible result at the crossroads of number theory and algebraic geometry.
First, the phrase elliptic curve is just a fancy name for a polynomial equation of the form y² = x³ + ax + b(subject to some technical conditions). “Defined over the rationals” just means that a and b are rational numbers. For example a=-36, b=0 or a=0, b=-26 would each produce an elliptic curve. A “rational point on the curve” refers to a solution to the equation whose coordinates are rational numbers. For example, if we’re looking at the case where a=0 and b=-26 then the elliptic curve is y² = x³ – 26 and one rational solution would be the point (35,-207). This solution also happens to be an integer solution; try to find some others! Elliptic curves are pretty and one can easily explore them in WolframAlpha. For example, the curve y² = x³ – 36x looks like this:
WolframAlpha does more than just provide a picture. It finds integer solutions to the equation. In this case just typing the equation for the elliptic curve into the WolframAlpha box produces:
One of the cool things about elliptic curves is that the points on them form the structure of an abelian group. That is to say, there is a way to “add” points on the curves. I’m not going to go through how this works here but there is a very good introduction to this connection between elliptic curves and groups in an exposition by Tanuj Nayak, an undergrad at Carnegie Mellon University.
Interestingly, even just the rational points on an elliptic curve form a group, and Mordell’s theorem says that for an elliptic curve defined over the rational numbers this group is finitely generated. That means that for such an elliptic curve one can describe all rational points on the curve as finite combinations of some finite set of points. In other words, we (humankind) has been interested in studying Diophantine equations since the time of Diophantus (3rd century). Trying to solve arbitrary polynomial equations is very difficult, so we restrict our attention to easier problems (elliptic curves). Working with integers is difficult, so we relax that requirement a bit and work with rational numbers. And here is a theorem that gives us hope, namely the hope that we can find all solutions to such problems because at least the description of the solutions can be finite.
The idea of looking for all solutions to a problem, and not just one solution, is fundamental to mathematics. I recently had the pleasure of attending a lesson for 1st and 2nd graders by Oleg Gleizer, an exceptional mathematician who takes time not only to teach children mathematics, but to develop mathematics ( not arithmetic!)curriculum that is accessible to them. The first thing Oleg asks young children is what they see when looking at this picture:
Children are quick to find the answer and reply either “rabbit” or “duck”. But the lesson they learn is that the answer to his question is that there is no single answer! Saying “rabbit” or “duck” is not a complete answer. In mathematics we seek all solutions to a problem. From this point of view, WolframAlpha’s “integer solutions” section is not satisfactory (it omits x=6, y=0), but while in principle one might worry that one would have to search forever, Mordell’s finite basis theorem provides some peace of mind for an important class of questions in number theory. It also guides mathematicians: if interested in a specific elliptic curve, think about how to find the (finite) generators for the associated group. Now the proof of Mordell’s theorem, or its natural generalization, the Mordell-Weil theorem, is not simple and requires some knowledge of algebraic geometry, but the statement of Mordell’s theorem and its meaning can be explained to kids via simple examples.
I don’t recall exactly when I learned Mordell’s theorem but I think it was while preparing for my qualifying exam in graduate school, when I studied Silverman’s book on elliptic curves for the cryptography section on my qualifying exam- yes, this math is even related to some very powerful schemes for cryptography! But I do remember when a few years later a (mathematician) friend mentioned to me “the coolest paper ever”, a paper related to generalizations of Mordell’s theorem, the very theorem that I had studied for my exam. The paper was by two mathematicians, Steven Zucker and David Cox, and it was titled Intersection Number of Sections of Elliptic Surfaces. The paper described an algorithm for determining whether some sections form a basis for the Mordell-Weil group for certain elliptic surfaces. The content was not why my friend thought this paper was cool, and in fact I don’t think he ever read it. The excitement was because of the juxtaposition of author names. Apparently David Cox had realized that if he could coauthor a paper with his colleague Steven Zucker, they could publish a theorem, which when named after the authors, would produce a misogynistic and homophobic slur. Cox sought out Zucker for this purpose, and their mission was a “success”. Another mathematician, Charles Schwartz, wrote a paper in which he built on this “joke”. From his paper:
So now, in the mathematics literature, in an interesting part of number theory, you have the Cox-Zucker machine. Many mathematicians think this is hilarious. I thought this was hilarious. In fact, when I was younger I frequently boasted about this “joke”, and how cool mathematicians are for coming up with clever stuff like this.
I was wrong.
I first started to wonder about the Zucker and Cox stunt when a friend pointed out to me, after I had used the term C-S to demean someone, that I had just spouted a misogynistic and homophobic slur. I started to notice the use of the C-S phrase all around me and it made me increasingly uncomfortable. I stopped using it. I stopped thinking that the Zucker-Cox stunt was funny (while noticing the irony that the sexual innuendo they constructed was much more cited than their math), and I started to think about the implications of this sort of thing for my profession. How would one explain the Zucker-Cox result to kids? How would undergraduates write a term paper about it without sexual innuendo distracting from the math? How would one discuss the result, the actual math, with colleagues? What kind of environment emerges when misogynistic and homophobic language is not only tolerated in a field, but is a source of pride by the men who dominate it?
These questions have been on my mind this past week as I’ve considered the result of the NIPS conference naming deliberation. This conference was named in 1987 by founders who, as far as I understand, did not consider the sexual connotations (they dismissed the fact that the abbreviation is a racial slur since they considered it all but extinct). Regardless of original intentions I write this post to lend my voice to those who are insisting that the conference change its name. I do so for many reasons. I hear from many of my colleagues that they are deeply offended by the name. That is already reason enough. I do so because the phrase NIPS has been weaponized and is being used to demean and degrade women at one of the main annual machine learning conferences. I don’t make this claim lightly. Consider, for example, TITS 2017 (the (un)official sister event to NIPS). I’ve thought about this specific aggression a lot because in mathematics there is a mathematician by the name of Tits who has many important objects named after him (e.g. Tits buildings). So I have worked through the thought experiment of trying to understand why I think it’s wrong to name a conference NIPS but I’m fine talking about the mathematician Tits. I remember when I first learned of Tits buildings I was taken aback for a moment. But I learned to understand the name Tits as French and I pronounce it as such in my mind and with my voice when I use it. There is no problem there, nor is there a problem with many names that clash across cultures and languages. TITS 2017 is something completely different. It is a deliberate use of NIPS and TITS in a way that can and will make many women uncomfortable. As for NIPS itself perhaps there is a “solution” to interpreting the name that doesn’t involve a racial slur or sexual innuendo (Neural Information Processing Systems). Maybe some people see a rabbit. But others see a duck. All the “solutions” matter. The fact is many women are uncomfortable because instead of being respected as scientists, their bodies and looks have become a subtext for the science that is being discussed. This is a longstanding problem at NIPS (see e.g., Lenna). Furthermore, it’s not only women who are uncomfortable. I am uncomfortable with the NIPS name for the reasons I gave above, and I know many other men are as well. I’m not at ease at conferences where racial slurs and sexual innuendo are featured prominently, and if there are men who are (cf. NIPS poll data) then they should be ignored.
I think this is an extremely important issue not only for computer science, but for all of science. It’s about much more than a name of some conference. This is about recognizing centuries of discriminatory and exclusionary practices against women and minorities, and about eliminating such practices when they occur now rather than encouraging them. The NIPS conference must change their name.#protestNIPS
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1068 | https://learn.genetics.utah.edu/content/basics/inheritance/ | What is Inheritance?
Sources of variation
Genetic variation is important because when conditions change (food becomes scarce, environment changes), some
individuals in a population will be more likely to have variations that will allow them to survive. Those who
reproduce pass their genes to the next generation. Variation helps species survive, and it's the reason species
change over time.
In asexual reproduction, variation comes mainly from mutation. Mutation is a natural process that introduces
permanent changes in a DNA sequence. However, microbes also acquire genetic variation through transformation,
transduction, and conjugation (gene transfer). These mechanisms often come into play when conditions are harsh.
In sexual reproduction, variation comes from both mutation and recombination. Mutation creates the different
versions (or alleles) of the same gene. Parental alleles are then shuffledâor recombinedâduring meiosis. Because
of recombination, sexual reproduction produces more variation than asexual does.
Living things have many ways of reproducing, each with its own advantages & disadvantages.
See Sexual vs. Asexual Reproduction
Patterns of inheritance
Most commonly, traits are influenced by multiple genes and the environment. So for most traits, patterns of
inheritance are complex and unpredictable. For instance, when a sexually reproducing organism inherits a defective
or "broken" allele that codes for a non-functional protein, the second copy can often make up for the loss. This
type of interplay between alleles, especially when traits are influenced by multiple genes, makes predicting
inheritance difficult, if not impossible.
A small number of traits are influenced primarily by a single gene. Traits influenced mainly by one gene usually
have predictable patterns of inheritance. However, most of the time even supposed "single-gene" traits are
influenced by multiple genes, as in the case of eye color. And often times, a single gene can influence multiple
traits. For example, some genes involved in eye color also influence hair and skin color.
Why do so many textbooks pretend that eyes come in just two colors, brown and blue? Eye color is influenced by two
major genes and several minor genes. Variations in these genes determine whether your eyes are brown, blue, green,
gold, hazel, greyish-blue, brown with green flecks, blue with a gold ring...you get the idea.
In rare cases, changes in DNA sequences can cause disease. The pattern of disease inheritance can be predictable or
unpredictable, depending on whether the disease is influenced mainly by one gene or by several genes.
See Genetic Disorders
Sex chromosomes
It's not quite accurate to say that everyone inherits two copies of every gene. In most mammals, the X and Y sex
chromosomes determine whether an individual is male or female. Females have two X chromosomes, and therefore two
copies of every gene. Males, however, have one X and one Y chromosome. For genes that appear only on the X chromosome
or only on the Y (some do appear on both), males inherit just one copy.
Other living things have sex chromosomes too. Birds and reptiles have Z and W sex chromosomes. Unlike with X and Y,
males have two Z chromosomes, and females have one Z and one W. Some insects and a few mammals have only an X
chromosome. Females have two copies and males have one. However, sex is not always determined by chromosomes. With
alligators, crocodiles, and most turtles, it's egg incubation temperature. And some fish can change gender in
response to cues from the environment. Sex determination happens in many different ways.
Traits that involve genes on the sex chromosomes have different inheritance patterns than non-sex-linked traits.
Color-blindness and Duchenne muscular dystrophy, for example, are linked to genes on the X chromosome. Both of
these disorders are more common in males because they have just one X chromosome. They have no second allele to
mask the effects of a defective gene. |
1069 | https://www.superteacherworksheets.com/generator-basic-addition.html | Free Addition Worksheet Generator
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1070 | https://forum.wordreference.com/threads/rebuke-vs-scold-vs-berate-vs-criticise-vs-reprimand-vs-tell-off-vs-lambast.3419061/ | Rebuke vs scold vs berate vs criticise vs reprimand vs tell off vs lambast | WordReference Forums
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Rebuke vs scold vs berate vs criticise vs reprimand vs tell off vs lambast
Thread starterbradx
Start dateJan 30, 2018
bradx
Member
arabic
Jan 30, 2018
#1
I get really confused when I am trying to put one of these words in proper sentence,
as example : My girlfriend always rebukes me for not taking her out
The question here is if I can use them interchangeably or does each word have it's convenient context ?
( feel free to mention any mistakes in my writing )
T
Thomas Tompion
Member Emeritus
Southern England
English - England
Jan 30, 2018
#2
Some are stronger than others. Context makes a great deal of difference.
bradx
Member
arabic
Jan 30, 2018
#3
what about the context i gave above:
My girlfriend always rebukes me for not taking her out. what do you think is the better verb to be used? or which one could be used and which one does not?
C
cando
Senior Member
English - British
Jan 30, 2018
#4
The word "rebuke" sounds a bit formal for the boyfriend/girlfriend context. It is somewhat literary and usually used of a superior/inferior relationship. Using "scold" would work, although it could sound slightly old fashioned, but "tells me off" is the most natural in your sentence.
For completeness -
Berate: implies an extended or repeated verbal assault.
Reprimand: is formal and serious and always by an authority figure.
Criticise: can be mild but is often focused on details of behaviour.
Lambast (or lambaste): is very serious and angry.
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1071 | https://www.geogebra.org/m/jhFbxzC2 | Revolving the area between curves about the x-axis – GeoGebra
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Revolving the area between curves about the x-axis
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I created this applet because I couldn't find one that would allow me to actually REVOLVE the bounded region around the axis. There are lots of applets that will generate the resulting solid, but this one also allows you to revolve the region through any angle you choose. This should aid students in being able to visualize how these solids are created from the original pair of function. You can input any pair of functions you want to work with, as well as change the upper and lower bounds of the "integration". You can also turn on or off each function and the "face" of the region as it is revolved around the axis.
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1072 | https://www.accessdata.fda.gov/drugsatfda_docs/label/1999/21092lbl.pdf | Revised Package Insert Page 1 Ez-HBT™ Test Instructions I. Intended Use The Ez-HBT™ is intended for use in the qualitative detection of 13CO2 in whole blood specimens, collected after the ingestion of 13C-urea. Helicobacter pylori (H. pylori) organisms colonizing the lining of the human stomach, produce urease which converts 13C-urea into 13CO2 and ammonia (NH4 +).
The device is indicated as an aid in the diagnosis of H. pylori infection in symptomatic adult subjects, 18 years or older.
For use by health care professionals. Administer test under a physician’s supervision. Metabolic Solutions, Inc. or a qualified laboratory using Gas Isotope Ratio Mass Spectrometry or equivalent instrumentation must analyze the test samples.
II. Summary and Explanation In the last decade, the most significant advancement in gastrointestinal medicine has been the discovery of H. pylori’s role in causing chronic active gastritis and peptic ulcer disease (PUD).1,2 It has been shown that eradication of the organism prevents ulcer relapse.3,4 This evidence has led gastroenterologists to routinely test patients who exhibit symptoms of peptic ulcers for H. pylori infection.
Several tests for the presence of H. pylori include invasive biopsy collection of gastric samples by means of esophagogastroduodenoscopy (EGD). Biopsy samples can be analyzed for the presence of H. pylori by microbiological culture, histological examination or direct detection of urease activity in the tissue. All of these biopsy-based methods are expensive, subject the patient to discomfort and risk and may give false results due to sampling errors when colonization of the gastric mucosa is patchy.5 Several serology-based methods are commercially available which detect the presence of antibodies to H. pylori organisms. These tests are unable to distinguish between active H. pylori infection and previous exposures to infection and thus can not conclusively demonstrate current H. pylori infection.
Several breath-based tests are also available commercially. These include the use of radioisotopes as well as stable isotopes of carbon for the detection of active H. pylori infection. The Metabolic Solutions’ Ez-HBT™ Helicobacter Blood Test (described in the next section) uses a stable or non-radioactive isotope for detection of active H. pylori infection and uses a single blood sample collected by standard well-established venipuncture techniques 30 minutes after ingestion of the drug component.
III. Principle of the Ez-HBT™ The patient ingests an oral dose of 125 mg of reconstituted 13C-urea. In the presence of the enzyme urease associated with gastric H. pylori, 13C-urea is converted into 13CO2 and ammonia (NH4 +) according to the following reaction: Hp Urease (NH2)2 13CO + H2O + 2H+ 13CO2 + 2NH4 + The 13CO2 is absorbed into the bloodstream. This results in an increase in the ratio of 13CO2 to 12CO2 in blood if H. pylori is present in the stomach.6,7 Analysis of the blood for increased levels of 13CO2 is performed at Metabolic Solutions, Inc. or a qualified laboratory using Gas Isotope Ratio Mass This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 2 Spectrometry or equivalent instrumentation. In the absence of gastric H. pylori, the ratio of 13CO2 to 12CO2 in the blood does not increase.
IV.
Description of Helicosol™ Diagnostic Drug Component The Helicosol™ diagnostic drug component of the kit is 13C-urea, a synthetic urea prepared as a lyophilized, white powder for reconstitution with sterile water (also provided in the kit) to produce a clear, colorless solution for oral administration. Greater than or equal to 99% of the carbon molecules in the Helicosol drug component are in the form of 13C, a stable naturally occurring, non-radioactive isotope of carbon.
Helicosol™ is supplied in a 100 ml glass vial containing 125 mg 13C-urea lyophilized powder.
13C-urea has the following chemical formula : 13CH4N2O. The drug is the diamide of 13C-carbonic acid and is highly soluble in water (1 gram per mL at 25°C).
An average adult body normally produces about 30 grams per day of urea, which is a product of protein metabolism. Of this amount, about 9 grams is retained. Naturally occurring urea in the body is composed of 98.9% 12C-urea and 1.1% 13C-urea.
V.
Warnings and Precautions 1. For in vitro diagnostic use only. The Helicosol drug solution is taken orally as part of the diagnostic procedure.
2. Discard Helicosol solution if not used within 4 hours after reconstitution.
3. A negative result does not rule out the possibility of H. pylori infection. False negative results occur at a rate of approximately 5-10% with this procedure. If clinical signs are suggestive of H.
pylori infection, retest with a new sample or an alternate method.
4. A false positive test may occur due to urease associated with other gastric spiral organisms observed in humans such as Helicobacter hominis.
5. Antimicrobials, proton pump inhibitors, and bismuth preparations are known to suppress H .
pylori, and ingestion of these within two weeks prior to performing the Ez-HBT™ Helicobacter Blood Test may give false negative results.
6. Premature collection of the blood sample can lead to a false negative diagnosis for a patient with a marginally positive Ez-HBT™ Helicobacter Blood Test result.
7. A false positive test could occur in patients who have achlorhydria.8 8. If particulate matter is visible in the reconstituted Helicosol™ solution after thorough mixing, the solution should not be used.
VI.
Shelf Life and Storage The components of the Ez-HBT™ test kit should be stored at 25°C (77°F), excursions permitted to 15° - 30°C (59° - 86°F). See USP Controlled Room Temperature procedures for further details.
The following components of the Ez-HBT™ test kit have expiration dates: 13C-Urea (Helicosol™), Ensure™ nutrition drink, Vacutainer® blood tubes and sterile water. Do not use any of these components beyond the indicated expiration date on the respective labels.
This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 3 VII. Patient Preparation 1. The patient should have fasted at least four (4) hours prior to administering the Ez-HBT™ test.
2. The patient should not have taken antimicrobials, proton pump inhibitors or bismuth preparations within two weeks prior to administering the Ez-HBT™ test.
VIII. Procedure for Collection of Blood Samples Materials Provided in the Collection Kit • One (1) glass vial containing Helicosol™ powder (13C-urea, 125 mg) • One (1) bottle containing sterile water (75 ml) • Two (2) Drinking Straws • One (1) 8 oz. Ensure™ drink, vanilla flavor • One (1) Vacutainer® 3 ml tube containing sodium heparin • Vacutainer® Brand Blood Collection System including needle and adapter • Gloves • Alcohol wipe • Bandage • Tourniquet • Gauze Materials Needed But Not Provided in the Collection Kit • A timer capable of timing an interval up to 30 minutes • Test request forms (optionally provided by qualified laboratory) • Specimen labels (optionally provided by qualified laboratory) • Specimen return envelopes (optionally provided by qualified laboratory) Note: A Gas Isotope Ratio Mass Spectrometer and related analytical equipment are required for analysis of blood samples. Analyses are performed at Metabolic Solutions or a qualified laboratory.
IX. Step-Wise Procedure Time intervals listed in the following procedure are critical. The timer icon highlights these critical steps: ¿ ¿ 1. Verify that the patient has been prepared for the test as specified in Section VII.
2. Open the Ez-HBT™ Kit which should contain all the materials listed above. To avoid confusion, be sure to keep these items patient-specific.
3. Complete all the areas of the test request form.
4. Open the Ensure™ can.
5. Instruct the patient to drink the Ensure™ using one of the straws. (The drink acts to delay gastric emptying during the test procedure).
This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 4 ¿ ¿ 6.
Set the timer for five (5) minutes after the patient finishes drinking Ensure™ before administering the Helicosol™ solution.
¿ ¿7.
Prepare the Helicosol™ solution just prior (within 30 minutes) of use: • Snap open the bottle of sterile water and Helicosol™ bottle.
• Pour sterile water into the Helicosol™ bottle.
• Re-seal the Helicosol™ bottle and gently invert the bottle 10-15 times to dissolve the Helicosol™ powder (about 15 seconds). The solution should be clear and colorless with no particulate matter. Invert several times again if the solution is not clear. If the particulate matter is present after through mixing, the solution should not be used.
¿ ¿8.
Instruct the patient to drink all the Helicosol™ solution directly from the glass vial using second straw.
¿ ¿9. Set a timer to 30 minutes when the patient completes drinking the Helicosol solution. The patient should sit quietly for the 30 minute interval. The patient should not eat, drink, or smoke.
¿ ¿10.
Thirty (30) minutes after the Helicosol™ solution is consumed, collect a blood sample.
Apply a tourniquet on one arm of patient. Cleanse the area of venipuncture with alcohol and dry with gauze.
11. Insert needle into holder of the Vacutainer® Blood System. Using gloves and standard venipuncture techniques, collect a blood sample into the 3-ml Vacutainer tube.
12. After the blood collection is complete, remove the tourniquet from arm and then remove needle from vein. Apply pressure to the venipuncture site and then place a bandage over site.
13. Dispose of needle using standard universal precaution techniques.
14. Clearly mark the blood tube with regard to patient name, date and time of venipuncture.
Mark this information directly on the Vacutainer® tube or on the labels provided by the qualified laboratory.
15. Review the completed test form(s) and labels for accuracy and completeness. Retain a copy of the test form.
16. Prior to shipment store samples at 15° - 30°C (59° - 86°F).
¿ ¿ 17. Send to lab within 4 days of test administration. Blood samples must arrive at the laboratory before 7 days after collection, otherwise the samples are not suitable for analysis.
X. Quality Control Metabolic Solutions’ clinical laboratory and other qualified laboratories performing the Ez-HBT blood analysis follow written policies and procedures designed to monitor and evaluate the overall quality of the total testing process. Prior to becoming a qualified testing facility, the laboratory must meet specifications for precision, accuracy, memory carryover and linearity of 13CO2 measurements. In addition, proficiency samples provided by Metabolic Solutions are used to qualify laboratories.
As part of the Quality Assurance program, control and reference gases are used to detect persistent and sporadic errors. Persistent errors are detected using control and reference gases periodically placed throughout the analytical run to accept or reject whole runs or portion of runs. Sporadic errors, which This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 5 occur unpredictably on individual specimens, are detected by quality criteria applied to each blood tube measurement. These criteria include a certain minimum amount of CO2 in the blood tube for measurement and analytical result within an expected range (see Section XIII).
In the event of quality criteria failure detected by the laboratory, you will be notified as soon as possible about the nature of the failure and the recommended remedial action.
XI.
Test Results A.
Analytical Method The ratio of 13CO2/12CO2 in the blood sample is determined by Metabolic Solutions, Inc. or another qualified laboratory using Gas Isotope Ratio Mass Spectrometry or equivalent instrumentation.
B.
Calculation of Results The result of the Ez-HBT test is provided as the delta per mil. No calculations are required by the customer. Delta per mil is the relative difference between the ratio of 13CO2 to 12CO2 in the sample and a reference standard.
C. Determination of the Cutoff Point The cut-off point is the level used to discriminate between H. pylori infected and uninfected individuals. For the Ez-HBT™ Helicobacter Blood Test, the delta per mil cutoff point was found to be -17.5 in a controlled study of 338 patients (125 infected and 178 noninfected out of 303 with congruent results). The reference standards were histological examination of biopsy tissue and rapid urease testing. The cutoff point was evaluated by determining the Ez-HBT™ values at which histologically and urease activity negative and positive subjects were best distinguished. Figure 1 shows graphically the Ez-HBT cutoff point that distinguishes histologically positive and negative patients. An indeterminate zone of 1 per mil around the cutoff (-17.0 to -18.0) was established. No determination as to the presence or absence of H. pylori should be made for a subject whose Ez-HBT™ result falls within this zone. If a test result is indeterminate, a repeat test should be performed with a timed blood collection at 45 minutes, instead of the usual 30 minutes, in order to confirm or reject H. pylori infection.
Figure 1 HP Status Delta per Mil -20 0 20 40 Positive Negative Indeterminate Zone This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 6 D.
Interpretation of Results An Ez-HBT™ result greater than -17 delta per mil units is considered diagnostically positive, indicating the presence of urease associated with H. pylori. An Ez-HBT™ result between -17.0 and -18.0 delta per mil is considered indeterminate. An Ez-HBT™ results less than -18 delta per mil is interpreted as diagnostically negative, indicating the absence of urease associated with H. pylori.
XII. Limitations of the Test 1. The performance characteristics of the test have not been established for monitoring the efficacy of antimicrobial therapies for the treatment of H. pylori infection. 2. The performance characteristics of the test have not been established for persons under the age of 18.
3. The specimen integrity due to storage of blood samples under ambient conditions has not been determined beyond seven (7) days.
4. A correlation between the number of H. pylori organisms in the stomach and the Ez-HBT™ test has not been established.
5. The Ez-HBT™ Helicobacter Blood Test should be used only to evaluate patients with clinical signs and symptoms suggestive of active duodenal ulcer disease.
6. The performance of the assay was assessed using the Europa Scientific 20/20 ABCA system and the Finnegan MAT Plus. Any other systems should be validated using the manufacturers specifications.
XIII . Expected Values Expected values for the Ez-HBT test were determined in a controlled clinical study of 338 adult patients with dyspeptic symptoms (125 infected and 178 uninfected out of 303 with congruent results by histology and rapid urease test). The range of Ez-HBT values for the uninfected group was determined to be –26.4 to 1.29 delta per mil and 23.29 to 44.57 delta per mil for the infected group.
A histogram for the distribution of delta per mil values from the uninfected patients is shown in Figure 2 and from the infected patients is shown in Figure 3.
This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 7 Figure 2 Delta per Mil Frequency 0 20 40 60 80 -25 -18 -17 -10 0 10 25 45 Histogram of Ez-HBT Values for Uninfected Subjects in the Pivotal Study Figure 3 Delta per Mil Frequency 0 5 10 15 -25 -18 -17 -10 0 10 25 45 Histogram of Ez-HBT Values for Infected Subjects in the Pivotal Study XIV. Interlaboratory Comparison Study A study evaluated the variability of Ez-HBT blood measurements among three laboratories utilizing two different gas isotope ratio mass spectrometers. All laboratories met specifications for precision, accuracy, memory and linearity. A human blood pool was spiked with six different levels of sodium 13C-bicarbonate to simulate blood samples of negative, positive and indeterminate values. Samples were randomized and coded to blind each laboratory. Two different proficiency blood sets were used to compare laboratory #1 both with laboratory #2 and with laboratory #3.
The interlaboratory comparisons are shown graphically in Figures 4 and 5. The linear regression plots for each laboratory with the amount of 13C-bicarbonate added to blood versus observed delta per mil is displayed. The slope and intercept of the linear regression plots between laboratory #1 and laboratory #2 or between laboratory #1 and laboratory #3 were not statistically different (P > 0.05).
This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 8 Figure 4 Figure 5 In summary, the results of this interlaboratory comparison indicate that a qualified laboratory using the appropriate equipment can perform the Ez-HBT analysis routinely.
C o m p a r i s o n o f L a b o r a t o r y 1 v s . L a b o r a t o r y 2 - 2 5 - 2 0 - 1 5 - 1 0 - 5 0 5 1 0 0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 n M o l e s o f 1 3 C - B i c a r b o n a t e a d d e d Laboratory 1 Laboratory 2 Linear 2 (y = 0.087x - 20.45) Linear 1 (y = 0.094x - 20.32) C o m p a r i s o n o f L a b o r a t o r y 1 v s . L a b o r a t o r y 3 - 2 5 - 2 0 - 1 5 - 1 0 - 5 0 5 1 0 0 5 0 1 0 0 1 5 0 n M o l e s o f 1 3 C - B i c a r b o n a t e a d d e d Laboratory 1 Laboratory 3 Linear 1 ( y = 0.107x -19.95) Linear 3 ( y = 0.114x -20.44) This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 9 XV. Performance Characteristics Method Comparisons in Clinical Trials Experimental Design The method comparison data presented here were collected from a double blind clinical field trial. The study had 338 adult patients with gastrointestinal symptoms who were enrolled in the study at 7 clinical sites around the United States and evaluated by three diagnostic methods: 1. Histopathology Biopsy specimens, fixed with 10% buffered formalin, were cut into 4-mm sections, stained with hematoxylin and eosin and Warthin-Starry stains and examined by an experienced pathologist.
2. PyloriTek (Serim Research Corp.) Biopsy specimens were tested for urease activity with the PyloriTek test method according to the instructions on the package insert.
3. Ez-HBT™ Helicobacter Blood Test (Metabolic Solutions, Inc.) The Ez-HBT™ Helicobacter blood test was performed in accordance with the procedures described in this package insert.
Results: Method comparison results are presented in two-way contingency tables. Tables 1 and 2 compare the Ez-HBT™ to histological examination and the PyloriTek® test method respectively. In Table 3, the Ez-HBT™ is compared to a congruent result from the two biopsy-based methods (histology and PyloriTek®).
Table 1: Comparison to Histological Examination Ez-HBT™ Helicobacter Blood Test Histology Positive Negative Indeterminate Total Positive 119 12 8 139 Negative 11 177 7 195 Total 130 189 15 334 SENSITIVITY: 90.8% [95 % CI ( 85.9 : 95.8)] SPECIFICITY: 94.1% [95 % CI ( 90.8 : 97.5)] This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 10 Table 2: Comparison to PyloriTek® Ez-HBT™ Helicobacter Blood Test PyloriTek® Positive Negative Indeterminate Total Positive 117 16 8 141 Negative 12 171 6 189 Total 129 187 14 330 SENSITIVITY: 88.0% [95 % CI ( 82.4 : 93.5)] SPECIFICITY: 93.4% [95 % CI ( 89.9 : 97.0)] Table 3: Comparison to Congruent Endoscopic Methods (Histlogy, PyloriTek®) Ez-HBT™ Helicobacter Blood Test Congruent Endoscopy Positive Negative Indeterminate Total Positive 115 10 7 132 Negative 9 169 5 183 Total 124 179 12 315 SENSITIVITY: 92.0% [95 % CI ( 87.2 : 96.8 )] SPECIFICITY: 94.9% [95 % CI ( 91.7 : 98.2)] XVI. References: 1) Marshall, B.J., Warren, J.R., Unidentified curved bacilli on gastric epithelium in active chronic gastritis, Lancet, June 4: 1273-1275; 1983.
2) NIH Consensus Development Panel, H. pylori in peptic Ulcer Disease, JAMA, July 6, 1994- Vol. 272, No.1, 65-69.
3) Hentschel E, Brandstatter G, Dragosics B, et al. Effect of Ranitidine and Amoxicillin plus Metronidazole on the eradication of Helicobacter pylori and the recurrence of duodenal ulcers, (1993) New Eng. J. Med., 328: 308-312.
4) Hopkins RJ, Girardi LS, Turney EA. Relationship between H. pylori eradication and reduced duodenal and gastric ulcer recurrence: A review. Gastroenterology 1996;110:1244-52.
5) Northfield TC, Mendall M, Goggin PM (Eds), Helicobacter pylori Infection. Pathophysiology, Epidemiology, and Management, Kluwer Academic Publisher (1993).
6) Moulton-Barrett, R., Triadafilopoulos, G., Michener, R., Gologorsky, D., Serum 13C-bicarbonate in the assessment of gastric Helicobacter pylori urease activity, Am. J. Gastroenterology, Vol. 88, No. 3, 1993.
7) Kim M.J., Michener, R., Triadafilopoulos, G., Serum 13C-bicarbonate assay for the diagnosis of gastric Helicobacter pylori infection and response to treatment, Gastroenterology 1997;113:31-37.
8) Borriello SP, Reed PJ, Dolby JM, Barclay FE, Webster ADB. Microbial and metabolic profile of achlorhydric stomach: comparison of pernicious anaemia and hypogammaglobulinaemia. J Clin Pathol 1985;38:946-953.
This label may not be the latest approved by FDA. For current labeling information, please visit Revised Package Insert Page 11 The Ez-HBT™ Helicobacter Blood Test is manufactured for Metabolic Solutions, Inc., Nashua, NH 03063 by Medikmark, Inc., Buffalo Grove, IL 60089. Helicosol™ is manufactured by Lyophilization Services of New England, Inc., Manchester, NH 03103.
This label may not be the latest approved by FDA. For current labeling information, please visit |
1073 | https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-89/issue-4/Universal-scaling-behaviour-for-iterated-maps-in-the-complex-plane/cmp/1103922930.pdf | Communications in Common. Math. Phys. 89, 555-570 (1983) Mathematical Physics © Springer-Verlag 1983 Universal Scaling Behaviour for Iterated Maps in the Complex Plane N. S. Manton and M. Nauenberg Institute for Theoretical Physics, University of California, Santa Barbara, CA93106, USA Abstract. According to the theory of Schroder and Siegel, certain complex analytic maps possess a family of closed invariant curves in the complex plane.
We have made a numerical study of these curves by iterating the map, and have found that the largest curve is a fractal. When the winding number of the map is the golden mean, the fractal curve has universal scaling properties, and the scaling parameter differs from those found for other types of maps. Also, for this winding number, there are universal scaling functions which describe the behaviour as n-> oo of the Qn th iterates of the map, where Qn is the nth Fibonacci number.
I. Introduction During the last few years, considerable progress has been made towards under-standing the onset of turbulence or chaos of dynamical systems by studying the properties of one- and two-dimensional maps of real variables. Scaling and renormalization group ideas have elucidated the appearance of universal quan-tities associated with these maps which are directly relevant to physical systems. In particular, the period doubling transitions have now been found experimentally in various systems with exponents in good agreement with the predictions of the scaling theory . Also progress has been made in applying scaling and renormalization group concepts to one-dimensional maps of the unit circle, which are believed to be generic for transitions which exhibit mode-locking on the route towards chaos , and also to the area-preserving twist map relevant to the study of critical Kolmogorov-Arnold-Moser (KAM) trajectories . Other routes to turbulence, such as the scenario described by Ruelle and Takens , are understood only qualitatively.
In this paper, we study the scaling properties of a map z—»/(z), where z is a complex variable and /(z) is complex analytic, and usually a polynomial. Apart from the fact that such a map is a natural extension of the well-studied quadratic map on the unit interval , we were motivated by finding numerically that under Permanent address: Physics Board, University of California, Santa Cruz, CA95060, USA 556 N. S. Manton and M. Nauenberg certain conditions, repeated iteration of the map generates sequences of points lying on closed curves in the complex plane. In fact, the existence of such curves was recognized by Schroder in the last century, although it was only relatively recently that Siegel made Schroder's argument rigorous . We review this theory in Sect. II. Fundamental studies of iterates of rational complex maps were initiated by Fatou and Julia , and have continued up to the present time [8-11].
Note that the Schroder theory for complex maps corresponds to Moser's twist theorem, which states sufficient conditions for the existence of a KAM trajectory for area-preserving maps in the plane , .
Our main results concern the limiting curve beyond which the Schroder theory breaks down. This curve is not smooth, but rather has fractal properties. In Sect. Ill, we consider the class of maps whose winding number (the average angle of rotation about the origin per iterate) is the golden mean. We present evidence which supports the conjecture that the fractal curves associated with almost all of these maps have universal scaling behaviour. The relationship between this scaling behaviour and the Schroder theory is discussed in Sect. IV. In Sect. V we study the generalization of the period doubling bifurcations which occur for iterated maps in the complex plane.
Finally, a note on our style of presentation: we have direct numerical evidence for almost all the statements in this paper which are not proved. We have presented as Conjectures those of our results which are of greatest generality, and for which it would be most valuable to have an analytic proof.
II. Schroder Theory and the Fractal Let z-»/(z) be a complex analytic map with a fixed point at z = 0, and a convergent power series (2.1) j=2 We denote the n ih iterate of z by / n(z), defined inductively by j n(z)=j(f n~ ί(z))9 /°(z) = z. We shall usually take / to be a polynomial, which guarantees the existence of all these iterates. The fixed point is stable if \a\<ί.
In this section, we assume that \a\ = l and ω^(2π)" 1argα is irrational.
Numerical iteration of the map then leads to a striking result. For any point z sufficiently close to the origin, all iterates of z appear to lie on a smooth curve passing through z, which is filled more and more densely as the iteration proceeds.
Topologically the curve is a circle surrounding the origin. If the initial point is far from the origin, then the iterates may escape to infinity or there may be a transient behaviour until some iterate lands on one of the invariant curves. Thereafter, all iterates stay on this curve.
The invariant curves very close to the origin are approximately true circles, for then the nonlinearity of / is small, but those further out are less circular, and, remarkably, there is a limiting invariant curve with fractal properties. We shall refer to this curve as the fractal. It seems to be continuous but not differentiable, and the density of points on it, after a fixed number of iterations, is highly non-Universal Scaling Behaviour for Iterated Complex Maps 557 uniform. A key observation is that there is a point on the fractal, denoted by z0, which is a stationary point of the map, i.e., /'(z0) = 0. An example of such a fractal curve, and one of the nearby smooth curves, is shown in Fig. 1. Here /(z) = αz — z 2, and the winding number is the golden mean ω0=-|(]/5 — 1).
-0.2 -0.2 0.4 Fig. 1. Fractal and nearby smooth invariant curves for the map z~+az— z 2, where fl = exp2πiω0 and Some of the above results follow from the classical Schroder theory .
Suppose there is an invertible mapping φ~ l from the z-plane to a complex w-plane such that the effect of / in the z-plane is reduced to a simple rotation about the origin in the w-plane. Then all iterates of z lie on the image under φ of a circle in the w-plane, hence on a curve. The functional equation for the Schroder function φ is / / i / \ / \ //^ f\ In the w-plane the rotation per iterate is ω, so the average rotation per iterate, or winding number, of the map / is also exactly ω.
A formal series solution of (2.2) can be obtained. Let cφO.
k=2 Then (2.1) and (2.2) imply that OO OO Σ ck(0 k-α)w k= Σ//CW + (2.3) (2.4) 7 = 2 k=2 The coefficient of w k on the left hand side is ck(a k — a), whereas the coefficient on the right hand side involves only {cί :2^ί^k— 1}. The coefficients ck can therefore be computed sequentially in terms of the coefficients of / and the arbitrary scale parameter c. Since ω is irrational, the denominator a k — a is never zero.
The Schroder series may have zero radius of convergence if ω is too close to an infinite set of rationals, but Siegel proved the following theorem : 558 N. S. Manton and M. Nauenberg Theorem. // there exist positive numbers λ, μ (depending on ω) such that for all integers m, n(n>l) \ω-m/n>λn~ μ, (2.5) then the Schroder series has finite radius of convergence for all functions f (whose power series (2.1) has finite radius of convergence).
The winding numbers satisfying (2.5) form a set of measure one on the unit interval.
The best way to study rational approximations to the irrational ω is via the continued fraction representation —< , (2.6) 1 >"2+—-where {rj is a set of positive integers. The rational approximant pn/qn is obtained by truncating the continued fraction at rn, and can be computed inductively by Pn = rnPn-l+Pn-2> 4n = rn ^ The condition (2.5) fails only if r increases too rapidly with i. In particular, if rt0). These image and preimage points w are dense, so the fractal is not even piecewise smooth. We shall conjecture below that the fractal is nowhere differentiable. This argument could fail if/' had poles on the curve, but such poles are certainly absent for polynomial maps.
Because the map / is conformal, the neighbourhood of any point on the fractal, other than the stationary point, is geometrically similar to its image under Universal Scaling Behaviour for Iterated Complex Maps 559 /. This statement has content even if the curve has neither left nor right derivatives, and hence no corners. For example, if the part of the fractal near z is bounded by lines passing through z, then the same is true at /(z), with the same angles between the lines.
Although it appears difficult to explicitly compute the Schroder series (2.3) by the method described earlier, even for the simplest map j(z) = az — z 2, we have been able to compute the coefficients numerically. For convenience, we fix the scale parameter c so that 0(1) = z0. The radius of convergence is now unity, assuming that Conjecture 1 is true. Since the fractal is the image of the unit circle φ(e ίθ) = ce iθ+ f cke ίkθ, (2.9) k = 2 the coefficients can be obtained from its Fourier transform. In practice, we need values of φ(e iθ) for a uniformly spaced set of angles θ. There is no exact way of getting these, but if we take a rational approximant p/q of ω, and replace 0 = exp2πzω by a = exp2πip/q, then we can regard z0 and the q—1 iterates of z0 as approximately the set φ(exp2πinp/q),Q^n^q — 1. The angles 2πnp/q have uniform spacing 2π/q. Alternatively, one can leave a unchanged, take q—1 iterates, and interpolate.
We have good numerical evidence in just one special case (see Sect. IV), but are prepared to make the following general conjectures about the coefficients.
00 Conjecture 2. (i) £ \ck\ converges.
k=2 (ii) lim \kck\ does not exist in particular, it is not zero.
The first of these conjectures has the important consequence that the Schroder series (2.9) converges absolutely and uniformly on the unit circle, so the fractal is continuous. Since φ(Qxpiθ) is bounded, the interior of the fractal has finite area.
The second conjecture prevents term by term differentation of the series and suggests that the fractal is nowhere differentiable.
A general property of φ, which follows from the Schroder equation and the single-valuedness of/, is that φ(expiθ) can equal φ(Qxpίθ') for some θ + θ' only if φ(expiθ) is periodic in θ with period 2π/JV, N>1. But φ is 1 — 1 on circles close to the origin, so by continuity of the winding number, N φ 1 is impossible. Therefore φ is continuous and 1 — 1 on the unit circle, and its image, the fractal, is a Jordan curve. It follows that its interior is a domain D, and φ~ 1 is the Riemann map from D to the unit disc.
A final conjecture, based on numerical evidence for one particular map discussed in Sect. Ill, is that the fractal has infinite length, and a dimension d>l.
III. Scaling Properties In this section we describe results obtained when the winding number is the golden mean ω0==^(]/5 —1) = 0.618034... The golden mean has the continued fraction 560 N. S. Manton and M. Nauenberg representation ω0= ^ , (3.1) l + 1+^L and its rational approximants are of the form βπ/βπ+1, where Qn is the n th Fibonacci number. SiegeΓs theorem guarantees that the Schroder series has finite radius of convergence for this winding number.
We have worked mainly with maps /(z) = a0z — z 2 -f Cz 3, where a0 = exp2πzω0, and in particular with the quadratic map (C = 0). The fractal curves associated with them exhibit scaling behaviour which is independent of C (except for some special values to be discussed), and we believe that this same behaviour is universal to a much larger class of maps, including almost all polynomials whose constant term vanishes and whose linear term is α0z, and for which /'(z) has just simple roots.
When z lies inside the fractal curve, the Schroder function is differentiable, so the sequence of Fibonacci iterates of z, f Qn(z) approaches z along a smooth curve at a rate determined by the winding number ω0. We have Qna>Q = Qn-1 — ( — ω0) n.
Hence, for large n, if z = φ(w), then so / On/ \ •/ \n ( \ i 2n ί \ i'i τ\ I Z I — Z /" >"'1\ — CO i C (Z) ~τ~ CO C ( Z ) I .j .j i where dφ Cl(z)=-2πw-, (3.4) dφ _M\ 2W a\ Note, in particular, that " +1'"' " -ω0. (3.5) When z lies on the fractal, the derivatives of φ(w) do not exist and we expect a different behaviour. Consider first the stationary point of the map z0, which lies on this curve. In contrast with (3.3), for large Qn (we have gone up to n = 2S) we find J Q"(z ) —z ^c + αV (~ ) n e +, (3.6) where α = 0 741932..., 2Θ+ = 107-27°..., and c+ is a complex constant. A more precise way of expressing this result is ]im /°"'(θ)-Zθ = g e T2» t - (17) Universal Scaling Behaviour for Iterated Complex Maps 561 Because the map / is conformal everywhere on the fractal except at z0, Eq. (3.7) is also valid if z0 is replaced by any preimage of z0. However, assuming that f"(z0) φ 0 (which excludes some values of the cubic coefficient C), at the image of z0 we have (3.8) and this scaling behaviour is propagated to all succeeding iterates of z0.
Since both the images and preimages of z0 are dense on the fractal, this type of scaling behaviour excludes the possibility that the fractal is even piecewise differentiable. It is also not surprising that the scaling behaviour (3.5) on the smooth curves is intermediate between that of (3.7) and (3.8), i.e. that α 2<ω 0<α, 2θ+<π<4θ+. (3.9) The Fibonacci preimages of z0 [denoted/~ QM(ZO)] are obtained by iterating the inverse map /~ 1(z). When /(z) = α0z —z 2, 1/2 (3.10) with the branch cut directed outside the fractal, and / -1(0) set to zero. Defined this way, / ~ 1 has the same invariant curves, including the fractal, as /. We find that /-^(z0)-z0^c_Λ-ί(-^-, (3.11) where 2Θ_ =119.6°... and c_ is a new constant. This scaling behaviour is similar to that of (3.6).
These results are illustrated in Fig. 2a, which shows part of the fractal close to z0 for the quadratic map. The points where the fractal and the lines with opening angle 2Θ + intersect are Fibonacci iterates of z0 (the number of iterates is indicated).
The Fibonacci preimages of z0 are where the fractal intersects the lines with opening angle 29 _.
These observations concerning Fibonacci iterates are special cases of a more general approximate scaling property of invariant curves near z0, including the fractal. We find that the fractal is self-similar after a uniform expansion about z0 by the scale factor α" 1, and a reflection in the symmetry axis shown in Fig. 2a, which bisects the opening angle between the two lines generated by the Fibonacci iterates of z0. Fig. 2b shows part of the curve in Fig. 2a, but expanded by α" 1. In Fig. 2c, this is further reflected in the axis of symmetry and brought into coincidence with Fig. 2a. (The small offset is for clarity.) Furthermore, a segment of a smooth curve through a point z0 + η on the symmetry axis coincides, after the same rescaling and reflection, with a segment of the curve through the point z0 + ^/α, provided \η\<ζl.
The approximate self-similarity of the fractal near z0 enables us to define a new non-closed curve. Thus z0 + δ lies on this new curve if there is a sequence of points {z0 + δn} on the fractal such that lim δn/a 2n exists and equals δ. Heuristically, the resulting curve is like Fig. 2a, bounded between the lines shown, but stretching to 562 N. S. Manton and M. Nauenberg -0.20 -0.25 -0.30 -0.35 -0.40 -0.2355 -0.3839 -0.3919 -0.40 -0.35 -0.30 -0.25 -0.20 -0.2435 Fig. 2a. Part of the fractal of Fig. 1, close to z0, and some Fibonacci iterates of z0. Positive numbers denote forward iterates and negative numbers denote iterates of the inverse map. The dashed line is the symmetry axis, b Same as a, but rescaled by α~ 1. c Superposition of b, after a reflection in the symmetry axis, and a. This shows the self-similarity of the fractal near z0 Universal Scaling Behaviour for Iterated Complex Maps 563 infinity. It is exactly self-similar. This curve is important because we have evidence that up to a linear transformation, it is universal, occurring for almost all maps with winding number ω0. We shall therefore refer to it as the universal fractal.
We now present a generalized scaling result involving Fibonacci iterates of any point z lying on an invariant curve and close to z0. This has been verified numerically to high accuracy in a certain domain. To start it is convenient to introduce a new variable ξ = (z— z0)e~ ιχ, where χ is the angle between the symmetry axis and the real axis. Under a reflection in this axis, ξ->ξ. Now define F n(ξ) = U"(zo + e iχξ)-z0~]e-ίχ. Our result is that for large Qn and small \ξ\9 ί(aξ)). (3.12) When £ = 0, this relation leads to Eq. (3.6).
The general form of the solution to Eq. (3.12) is rig2(ξ/x n)-] , (3.13) where gί and g2 are real analytic functions which will be referred to as scaling functions. More precisely, we define the scaling functions as a limit (which we believe exists), namely F Qn(tt nξ} lim —^L=gι(ξ)±ig2(ξ). (3.14) ϋSS"}— α It is clear that they are defined on the universal fractal and on the half-space to one side of it corresponding to the fractal's interior. We conjecture that the scaling functions themselves are universal.
To recover Eq. (3.7), it is necessary that gl(0) = c+cosθ+, g2(ty = c+sinθ+. (3.15) On the other hand, suppose that \ξ\ is large, but that zQ + a ne iχξ is interior to the fractal for all sufficiently large n. Now we have made a crossover from the scaling associated with the fractal to the scaling associated with the smooth invariant curves, and Eq. (3.3) is satisfied provided that as |£|->αo, 2, (3.16) with Pl=2p2 + l, p2=^- 1=0.612.... (3.17) Note that for z near z0, Eqs. (3.16) and (3.17) also give the z dependence of the functions cx and c2 occuring in (3.3) as a power law with coefficients determined by ω0 and α.
Because of the defining property of Fibonacci numbers Qn + Qn+l =Qn+2^ there are two composition rules, namely F Q» +1(ί lQn(ξ)) = f βn + 2(ξ), (3.18) 564 N. S. Manton and M. Nauenberg We can use these to obtain two functional equations for the scaling functions. Let us first change notation slightly, and define g(ζ) = g 1(ξ) + ig2(ζ\ Then g^(ξ^) = g 1(ξ) - ig2(ξ)9 because g1 and g2 are real analytic. From (3.18) and (3.14) we deduce (for n either even or odd) that (3.20) and from (3.19) and (3.14), that (3.21) We do not know if these equations are sufficient to essentially determine g.
However, they correspond to Shenker's equations for maps on the unit circle when ξ and g(ξ) are real . Asymptotically, the form of g given by (3.16) satisfies (3.20) and (3.21) provided the exponents satisfy (3.17) and alsoβl=^p2βl.
Let us now summarize the main results of this section as the following conjecture : Conjecture 3. When the winding number of the map f is the golden mean, the shape of the invariant fractal near z0, the map's stationary point, is characterized by a universal self-similar curve. The Fibonacci iterates of points dose to z0 have a scaling behaviour which enables one to define a scaling junction g on one of the half-spaces bounded by the universal curve, and there are a pair of functional equations which may essentially determine g.
Another scaling property concerns the length of the fractal curve. We have found numerically that the length increases as the number of points increases, in such a way that we can associate with the fractal a definite dimension d>l. For d to have an unambiguous value, we must take the Qn(+ί) points {/ I(z0): — D^βJ^J^CiδJ}, i.e. an equal number of forward and backward iterates from the stationary point of the map. These points have a well-defined order given by the order of their images on the unit circle under the inverse Schroder map. Joined up in this order they form a polygonal curve, whose length Ln defines the length of the fractal for Qn points. We find that .. (3.22) -« Ln A simple definition of the dimension d is suggested by this procedure. Let &n = Ln/Qn denote the average distance between adjacent points, and let us require that lira Qnε d n exist. Since Qn/Qn+ 1 ~ω0, we obtain 1.028.... (3.23) ιnω0 -f-lnr To conclude this section, we note that the quadratic map has two associated inverse maps corresponding to the two branches of the square root in Eq. (3.10).
The region generated from the fractal and its interior by iterating both these Universal Scaling Behaviour for Iterated Complex Maps 565 inverses is a domain whose boundary is the Julia set for the quadratic map. This Julia set is shown in Fig. 3, indicating that it, too, is a fractal. (For reference we have included some points of the original fractal.) r ' I Λ r ' ' t -y Fig. 3. Julia set for the map whose invariant curves appear in Fig. 1 IV. Scaling and the Schroder Series The simplest map z->/(z) exhibiting the scaling properties described in the last section is j(z) = aQz — z 2, where α0 = exp2τπω0 and ω0 is the golden mean. Figure 4 is a plot of the absolute values of the coefficients {cn} of the Schroder series (2.3) for 0.10 0.08 Fig. 4. Absolute values of the Schroder series coefficients for the map whose invariant curves appear in Fig. 1. The data is not reliable for l 566 N. S. Manton and M. Nauenberg this map. This data was obtained as described in Sect. II, by Fourier transforming the fractal curve shown in Fig. 1. The coefficient cn is particularly large when n— 1 is a Fibonacci number, but this is not surprising, since the denominator a n Q — aQ is then very small.
The apparent periodicity exhibited in Fig. 4, modified by the increasing density of sample points, suggests the following conjecture: Conjecture 4.
[ωo(ι +T)] Σ CB (4.1) W / n = [ω0r] Ύ jor all integers r and T satisfying — > r + T > r > l .
ω0 Here α is the scaling parameter, and χ is the angle introduced in Sect. III.
[x] denotes the nearest integer to x. We require T>1, so that [ω0T]/T is approximately ω0, but (r+ T)/Γ need not be much greater than 1.
We have checked the conjecture numerically for a number of values of r and T, for the simple quadratic map. However, we believe it is true for all maps whose fractal curve has the universal scaling properties described in the last section.
In what follows we shall argue in the opposite direction. Assuming the conjecture is true, we deduce a fundamental scaling property for the Schroder function φ, from which most of the results of Sect. Ill follow.
The scaling property of the Schroder function is: Theorem. For \θ\<ζ] [_φ(e ίβ}~φ(l)Y = ae2' X + 0WV> (^ where σ>0.
This theorem will imply scaling behaviour of the fractal (θ real), and of the nearby smooth curves (Imθ>0), close to the stationary point of the map (θ = 0).
The following is an outline of a proof of (4.2). The argument is formally correct, but the analysis of the error terms is left as an exercise. To get some control over the errors we must assume that φ(e iθ) -φ(l) = 0(\θ\ τ), (4.3) where 0 < τ < 1. This is reasonable, since the theorem itself implies that τ = Inα/lnω0 = 0.62... We can now choose a Fibonacci number QN such that, for small |0|, -.-.- ίΓ ( 4 4) Then φ~ "»<•-!), (4.5) Universal Scaling Behaviour for Iterated Complex Maps 567 since the terms in the sum with n<QN approximately cancel. By splitting up the summation into intervals of length T, and assuming that the exponential is approximately constant over these intervals, we obtain φ(e~ <»<) -P=0 QN+PT Now, use of the conjecture (4.1) and a reorganization of the exponential gives oo Qzv-ι+(P+l)[ω 0Γ]-l ae 2iχ Σ(e-i(QN-1 + p[ω°™--l) Σ c. (4.7) Then, reversing the steps leading to (4.6), but with QN and T replaced by QN_ ί and [ω0T], we obtain φ(e ~ ίωoθ) -φ(l] ^ ae 2i(φ(e iθ) - 0(1)) , (4.8) which proves the theorem.
The self-similarity of the fractal is an immediate consequence of (4.2). Recall that z0 = 0(1) is the stationary point of the map, and suppose z0 + e iχξ = φ(e iθ) is on the fractal, with |θ| ^ 1 and θ real. Then z0 + ae iχξ = φ(e~ iωoθ), and so this point is also on the fractal.
The scaling properties of the Fibonacci iterates also follow from (4.2). If z = φ(et θ)9 then j<»(z) = φ(e 2πi(0oQn + ίθ). Using the same formula for ω0Qn as in (3.2), and assuming that \θ\ <| 1, we obtain _7 ) ZQ _ 2iχ ~ for |£|<^1. This result is equivalent to Eq. (3.12).
V. Bifurcations Most of this paper has been concerned with maps of the form (2.1) whose stability parameter a has modulus one. For \a\ < 1, z — 0 is a stable, period one, fixed point.
Here we show that when a is close to a primitive n th root of unity, and \a\ is just greater than one, there is a bifurcation to a stable cycle of period n.
Note that for maps on the real line, only period doubling bifurcations are possible, since a is real. Moreover, such a bifurcation can fail to occur if the equation determining the stable cycle has complex roots. One consequence of working with complex analytic maps is to make these bifurcations more robust.
Suppose first that a = A, with A a primitive n th root of unity (A n=l, A mή=l for 1 < m < n).
Lemma. /"(z) = z + 0(z" +1) (5.1) for z in some neighbourhood of zero.
Proof. A modification of the Schroder series construction shows that there is a polynomial ψ(w) of the form k=2 568 N. S. Manton and M. Nauenberg for which /(ψ(w)) = ψ(Aw) + O(w n+ί) . (5.3) As before, the coefficients ck can be computed sequentially. No higher degree polynomial can be used to reduce the error term in (5.3) because A n+ 1 — A = 0, so the Schroder series construction would break down. Equation (5.3) implies /"(V?(w)) - φ(w) + 0(w n+ ) . (5.4) Since ψ is invertible near the origin, the lemma follows. It implies that /"(z) — z has an (ft+l)-fold zero at z = 0.
The lemma can be checked explicitly for small values of n. The coefficients of z 2, z 3, etc. involve a in the combinations l + α + α 2-f ... +α"~ 1, H-α 2 + α 4+ ... +α 2("~ 1}, etc. These vanish when a = A.
Now suppose α = ,4(l+ε), where |ε| is small and Reε>0. By continuity (more precisely, by Rouche's theorem) the number of zeros of /"(z) — z near z = 0 is still n+1, and one of them stays at the origin. The remaining n zeros form a closed cycle under iteration by / (since f n(z) = z=>f(f n(z))=f(z)=>f n(f(z))=f(z) etc.).
Let us estimate the locations of the new zeros. Using the lemma, we have j»(z} = (ljrεγz + 0(ε)z+ ...+κz n+1 + .... (5.5) Assume that κ;Φθ. To leading order in ε,/"(z) — z = 0 when nεz + κz n+l=Q, i.e., when z = ( — nε/κ) lln or z = 0. Provided |ε| is sufficiently small, z is small and the approximations are valid. In particular, the trajectory of each non-trivial zero z(ε) has the same tangent vector at ε = 0 as that given by the estimate.
To check stability we need to estimate \df n/dz\ at z = ( — nε/κ) 1/n. The leading approximation gives z n = l-n 2ε (5.6) dz at each of the zeros. For suitable ε, \df n/dz\0, nor whether there are gaps between the values of ε for which there is a stable cycle. However, some aspects of this problem are discussed by Mandelbrot .
VI. Conclusions We have seen that iterated maps in the complex plane have interesting properties which complement those already known for real one-dimensional maps, and for area-preserving and dissipative maps in the plane. One finds a family of invariant curves enclosing the origin when the origin itself is a marginally stable fixed point Universal Scaling Behaviour for Iterated Complex Maps 569 and the winding number of the map is irrational. This family has a limiting curve which is fractal. When the winding number is the golden mean, the fractal has universal scaling behaviour, but the scaling parameter in this case, α = 0.741932 ..., is different from those associated with the real maps mentioned above.
This difference can be understood if one considers the complex map as a real map in the plane. In polar coordinates z = rexpi(0 + 2πω0), the quadratic map /(z) = (exp2πiω0)z — z 2 becomes θ' = θ + 2πω0 - arctan ΓSm — -I , \l-rcosθ/ (6-1) For small r this reduces to the "standard" return map on the circle. The Jacobian J of the transformation is j = |/'(z)| = (1 _ 4r cos θ + 4r 2) 1/2 , (6.2) so J>1 if r>cos$, and J^Ξl otherwise. Hence, from the viewpoint of dynamical systems, this map is neither conservative nor dissipative. Also, the smooth invariant curves of the complex map, while being starlike close to the origin, cease to be so as they approach the fractal. "Starlike" here means that any ray from the origin intersects the curve exactly once. Therefore, for the fractal and the curves nearby, it is not possible to define an effective one-dimensional map from the circle to itself by ignoring the radial coordinate, i.e. by a radial projection onto the unit circle.
We expect that scaling behaviour occurs for complex maps with any winding number whose continued fraction is periodic, and not just for the golden mean. We have some preliminary evidence for this, but further work is needed.
Acknowledgement. We would like to thank G. Culler, S. Ostlund, J. Sethna, and E. Siggia for helpful conversations. We enjoyed also talking with D. Ruelle and would like to thank him for his enlightening comments.
This work is supported by the National Science Foundation, Grant PHY 77-27084.
References 1. Eckmann, J.-P. : Roads to turbulence in dissipative dynamical systems. Rev. Mod. Phys. 53, 643-654 (1981); Ott, E. : Strange attractors and chaotic motions of dynamical systems. Rev. Mod.
Phys. 53, 655-671 (1981); Helleman, R.H.G. : In: Fundamental problems in statistical mechanics, Cohen, E.G.D. (ed.). Amsterdam: North Holland 1980; These are three recent review articles which include references to earlier work 2. Shenker, S.J. : Scaling behaviour in a map of a circle into itself: empirical results. Physica 5D, 405-411 (1982); Ostlund, S., Rand, D., Sethna, J., Siggia, E. : A universal transition from quasi-periodicity to chaos in dissipative systems. Phys. Rev. Lett. 49, 132-135 (1982) 3. Kadanoff, L.P. : Scaling for a critical Kolmogorov-Arnold-Moser trajectory. Phys. Rev. Lett. 47, 1641-1643 (1981); Shenker, S.J, Kadanoff, L.P. : Critical behaviour of a KAM surface: 1. empirical results. J. Stat. Phys. 27, 631-656 (1982); Widom, M., Kadanoff, L.P.: Renormalization group analysis of bifurcations in area-preserving maps. Physica 5D, 287-292 (1982) 4. Ruelle, D, Takens, F.: On the nature of turbulence. Commun. Math. Phys. 20, 167-192 (1971); Newhouse, S, Ruelle, D, Takens, F. : Occurence of strange axiom A attractors near quasi periodic flows on 7™, m;>3. Commun. Math. Phys. 64, 35-40 (1978) 570 N. S. Manton and M. Nauenberg 5. Schroder, E.: Uber iterirte Funktionen. Math. Ann. 3, 296-322 (1871); Siegel, C.L.: Iteration of analytic functions. Ann. Math. 43, 607-612 (1942) For a modern treatment, see Moser, J.K., Siegel, C.L.: Lectures on celestial mechanics. New York: Springer 1971 6. Fatou, P.: Sur les equations fonctionelles. Bull. Soc. Math. France 47, 161-271 (1919), 48, 33-94 and 208-314 (1920) 7. Julia, G.: Memoire sur Γiteration des fonctions rationelles. J. Math. Pures Appl. 4, 47-245 (1918) 8. Brolin, H.: Invariant sets under iteration of rational functions. Ark. Mat. Astron. Pys. 6, 103-144 (1965) 9. Mandelbrot, B.: Fractal aspects of the iteration of z-λz(l — z) for complex λ and z. Ann. N. Y.
Acad. Sci. 357, 249-259 (1980) 10. Ruelle, D.: Repellers for real analytic maps. IHES preprint 81/52 11. Douady, A., Hubbard, J.M.: Iteration des polynόmes quadratiques complexes. C. R. Acad. Sci.
(Paris) Ser. 1, 294, 123-126 (1982) 12. Moser, J.: On invariant curves of area-preserving maps of an annulus. Nachr. Akad. Wiss., Gδttingen, Math. Phys. Kl. 1-20 (1962) Communicated by O. E. Lanford Received June 9, 1982; in revised form January 31, 1983 |
1074 | https://arxiv.org/pdf/1508.00319 | Published Time: Fri, 20 Jan 2023 19:57:58 GMT
A Study on the Modular Sumset Labeling of Graphs
Sudev Naduvath
Centre for Studies in Discrete Mathematics Vidya Academy of Science & Technology Thrissur, Kerala, India. E-mail: sudevnk@gmail.com
Abstract
For a positive integer n, let Zn be the set of all non-negative integers modulo
n and P(Zn) be its power set. A modular sumset valuation or a modular sumset labeling of a given graph G is an injective function f : V (G) → P(Zn) such that the induced function f + : E(G) → P(Zn) defined by f +(uv ) = f (u) + f (v). A modular sumset indexer of a graph G is an injective modular sumset valued function f : V (G) → P(Zn) such that the induced function f + : E(G) → P(Zn)is also injective. In this paper, some properties and characteristics of this type of modular sumset labeling of graphs are being studied.
Key Words : Modular sumset graphs; weak modular sumset graphs; strong modular sumset graphs, maximal modular sumset graphs exquisite modular sumset graphs; modular sumset number of a graph.
Mathematics Subject Classification : 05C78.
1 Introduction
For all terms and definitions, not defined specifically in this paper, we refer to [ 4], and . For graph classes, we further refer to , and and for notions and results in number theory, we refer to [ 3] and [ 13 ]. Unless mentioned otherwise, all graphs considered here are simple, finite and have no isolated vertices. Let A and B be two sets integers. The sumset of A and B is denoted by A + B
and is defined as A + B = {a + b : a ∈ A, b ∈ B}. If either A or B is countably infinite, then A + B is also countably infinite. We denote the cardinality of a set A
by |A|. Then, we have the following theorem on the cardinality of the sumset of two sets.
Theorem 1.1. [13 ] For two non-empty sets A and B, |A| + |B| − 1 ≤ | A + B| ≤ |A| | B|.
Another theorem on sumsets of two sets of integers proved in [ 13 ] is given below. 1
arXiv:1508.00319v2 [math.GM] 31 Jan 2017
2 On the Modular Sumset Labeling of Graphs
Theorem 1.2. Let A and B be two non-empty sets of integers. Then, |A + B| =
|A| + |B| − 1 if and only if A and B are arithmetic progressions with the same common difference.
Using the concepts of the sumset of two sets, the notions of an integer additive set-labeling and an integer additive set-indexer of a given graph G was introduced as follows.
Definition 1.3. Let N0 be the set of all non-negative integers and let P(N0) be its power set. An integer additive set-labeling of a graph G is an injective function
f : V (G) → P(N0), where the induced function f +(uv ) : E(G) → P(N0) is defined by f +(uv ) = f (u) + f (v). A graph G which admits an integer additive set-labeling is called an integer additive set-labeled graph or integer additive set-valued graph .
Definition 1.4. [ 8] An integer additive set-indexer (IASI) is as an injective function
f : V (G) → P(N0) such that the induced function f + : E(G) → P(N0) defined by
f +(uv ) = f (u) + f (v) = {a + b : a ∈ f (u), b ∈ f (v)} is also injective. A graph G
which admits an IASI is called an integer additive set-indexed graph .Certain studies about integer additive set-indexers of graphs have been initiated in [ 8], [ 9], [ 14 ] and [ 15 ]. A series of studies about different types of integer additive set-labeled graphs followed thereafter. These papers about the characteristics and properties of integer additive set-valued graphs are the main motivations behind this paper.
2 Modular Sumset Labeling of Graphs
Let n be a positive integer. We denote the set of all non-negative integers modulo n by
Zn and its power set by P(Zn). The modular sumset of the two subsets A and B of Zn,denoted by A+B, is the set defined by A+B = {x : a+b ≡ x (mod n ), a ∈ A, b ∈ B}.Through out our discussion, A + B is the sumset of A and B. It can also be noted that A, B ⊆ Zn =⇒ A + B ⊆ Zn.Then, using the concepts of modular sumsets of sets and analogous to the definition of integer additive set-labelings of graphs, let us now define the following notions.
Definition 2.1. A function f : V (G) → P(Zn), whose induced function f +(uv ) :
E(G) → P(Zn) is defined by f +(uv ) = f (u) + f (v), is said to be a modular sumset labeling if f is injective. A graph G which admits a modular sumset labeling is called an modular sumset graph .
Definition 2.2. A modular sumset indexer is an injective function f : V (G) → P(Zn)such that the induced function f + : E(G) → P(Zn) defined by f +(uv ) = f (u) +
f (v) = {x : a + b ≡ x (mod n ); a ∈ f (u), b ∈ f (v)} is also injective. In View of Theorem 1.1, the bounds for the cardinality of an edge of a modular sumset graph G is given in the following theorem. Sudev Naduvath 3
Theorem 2.3. Let f : V (G) → P(Zn) be a modular sumset labeling of a given graph G. Then, for any edge uv ∈ E(G), we have |f (u)| + |f (v)| − 1 ≤ | f +(uv )| =
|f (u) + f (v)| ≤ | f (u)| | f (v)| ≤ n.
We use the following terms and definitions analogous to those for IASL-graphs. The cardinality of the set-label of an element of G is said to be the set-labeling number
of that element. If all the vertices of a graph G have the same set-labeling number, say l, then we say that V (G) is l-uniformly set-labeled . A modular sumset labeling of G is said to be a k-uniform modular sumset labeling if the set-labeling number of all edges of G have the same set-labeling number k.The first and most important problem in this context is to verify whether a given graph admits a modular sumset labeling. The existence of a modular sumset labeling (and a modular sumset indexer) for any given graph G, is established in the following theorem.
Theorem 2.4. Every finite graph admits a modular sumset labeling (or a modular sumset indexer). Proof. Let G be a graph with vertex set V (G) = {v1, v 2, . . . , v p}.. Let A1, A 2 . . . , A p
be p distinct non-empty subsets of Zn, n being a positive integer. Let f : V (G) →
P(Zn) defined by f (vi) = Ai. Clearly f is injective. Define f + : E(G) → P(Zn) by
f +(vivj ) = {x : ai + bj ≡ x (mod n ); ai ∈ Ai; bj ∈ Aj }. Clearly, f +(vivj ) is also a set of non-negative integers modulo n and hence is a subset of Zn. Therefore, for suitable choices of n, f is a modular sumset labeling of G.For taking n sufficiently large, we can make f + also an injective function, which establishes the existence of a modular sumset indexer for G.Figure 1a depicts the existence of a modular sumset labeling for the Petersen Graph, with respect to the ground set Z4 and Figure 1b illustrates the existence of a modular sumset indexer for the Petersen Graph, with respect to the ground set Z5.
(a) Petersen Graph together with a sum-set valuation defined on it
(b) Petersen Graph together with a mod-ular sumset indexer defined on it
Figure 1 4 On the Modular Sumset Labeling of Graphs
As in the case of all other set-valuations of graphs, the problem of finding the minimum cardinality required for the ground set Zn with respect to which a given graph admits a modular sumset labeling is relevant and arouses much interest. In view of this, let us introduce the following notion.
Definition 2.5. The smallest value of n such that f : V (G) → P(Zn) is a modular sumset labeling of a given graph G is called the modular sumset number of G. The modular sumset number of a graph G is denoted by σ(G). Similarly, the minimum value of n such that f : V (G) → P(Zn) is a modular sumset indexer of G is called the sumset index of G and is denoted by ¯ σ(G). The following theorem establishes the modular sumset number of a given graph on
m vertices.
Theorem 2.6. The modular sumset number of a graph G is 1 + blog 2 mc.Proof. Let f : V (G) → P(Zn) be a modular sumset labeling of a given graph G on
m vertices. Then each vertex of G has a non-empty subset of Zn as its set-label. Therefore, the ground set Zn must have at least m non-empty subsets. That is, 2n − 1 ≥ m. Hence, n ≥ 1 + log 2 m. Therefore, σ(G) = 1 + blog 2 mc.
3 Certain Types Modular Sumset Graphs
In this section, we study certain types of modular sumset labelings of given graphs according to the nature of the set-labels of the elements of those graphs. Analogous to the studies of integer additive set-labeled graphs, let us now proceed to study different types of modular sumset labelings of G in terms of the cardinality of the set-labels of vertices and edges of G.
3.1 Weak Modular Sumset Graphs
As in the study of weak IASL graphs, our first aim is to check the existence of edges in modular sumset graph which has the same set-labeling number as that of one or both of its end vertices. The following result establishes the condition for a sumset to have the same cardinality of one or both of its summands.
Proposition 3.1. Let A and B be two non-empty subsets of Zn. Then, |A+B| = |A|
(or |A + B| = |B|) if and only if either |A| = |B| = Zn or |B| = 1 (or |A| = 1 ). More over, |A + B| = |A| = |B| if and only if |A| = |B| = Zn or |A| = |B| = Zn.
In view of the above proposition, it can be noted that the set-labeling number of an edge of G is equal to that of an end vertex of that edge if and only if the set-labeling number of the other end vertex is 1. Hence, analogous to the corresponding notion of IASL-graphs (see [9, 14]), we have the following definition.
Definition 3.2. A modular sumset labeling f of a graph G is said to be a weak modular sumset labeling of G if the set-labeling number of every edge of G is equal to the set-labeling number of at least one of its end vertices. A graph G which admits a weak modular sumset labeling is called a weak modular sumset graph .Sudev Naduvath 5It is to be noted that for a weak modular sumset graph, no two adjacent vertices can have non-singleton set-labels.
Remark 3.3. The elements of G having the set-labeling number 1 are called the
monocardinal elements of G. An edge can be monocardinal if and only if its end vertices are monocardinal. The set-labeling number of an edge of a given graph G is equal to the set-labeling number of both of its end vertices if and only if the edge and its end vertices are monocardinal Hence, analogous to the corresponding of weak IASL-graphs the following results are valid for weak modular sumset graphs.
Theorem 3.4. A graph G admits a weak modular sumset labeling if and only if G
is bipartite or contains monocardinal edges.
Theorem 3.5. A graph G admits a weakly uniform modular sumset labeling if and only if G is bipartite.
Figure 2 illustrates a weak modular sumset graph with respect to the ground set Z6.
Figure 2: An example to a weak modular sumset graph. It can be noticed that the study about weak modular sumset graphs is very much close to the study of weak IASL-graphs and hence offers no novelty in this context.
3.2 Weak Modular Sumset Number of Graphs
As a special case of modular sumset number of graph we introduce the following notion.
Definition 3.6. The weak modular sumset number of a graph G is defined to be the minimum value of n such that a modular sumset labeling f : V (G) → P(Zn) is a weak modular sumset labeling of G.In this section, we initiate a study about the weak modular sumset number of different graph classes. First we observe the following proposition. In the following theorem, we determine the weak sumset number of an arbitrary graph G in terms of its covering and independence numbers. 6 On the Modular Sumset Labeling of Graphs
Theorem 3.7. Let G be a modular sumset graph and α and β be its covering number and independence number respectively. Then, the weak modular sumset number of G
is max {α, r }, where r is the smallest positive integer such that 2r − r − 1 ≥ β.Proof. First recall the result α(G) + β(G) = |V (G)|. Since G is a modular sumset graph, no two adjacent vertices can have non-singleton set-labels simultaneously. Therefore, the maximum number of vertices that have non-singleton set-labels is
β. Let V ′ be the set of these independent vertices in G. Therefore, the minimum number of vertices that are monocardinal is |V (G)| − β = α. Since all these vertices in V − V ′ must have distinct singleton set-labels, the ground set must have at least
α elements. Also, the number non-empty, non-singleton subsets of the ground set must be greater than or equal to α, since, otherwise all the vertices in V ′ can not be labeled by non-singleton subsets of this ground set. Note that the number of non-empty, non-singleton subsets of a set A is 2 |A| − | A| − 1. Therefore, the weak modular sumset number G is α if 2 α − α − 1 ≥ β. Otherwise, the ground set must have at least r elements such that 2 r − r − 1 ≥ β. Therefore, in this case, the weak modular sumset number of G is r, where r is the smallest positive integer such that 2 r − r − 1 ≥ β. Hence, σ∗(G) = max {α, r }.We shall now discuss the weak modular sumset number of certain standard graph classes. First consider a path graph Pm on m vertices.
Proposition 3.8. The weak modular sumset number of a path Pm on m vertices is given by
σ∗(Pm) =
{
m if m ≤ 2
bm
2
c if m > 2
Proof. Let Pm denote be a path on m vertices. Then, α = bm
2
c and β = dm
2
e. It is obvious that we need a singleton set to label the single vertex of Pm if m = 1 and a two element set to label the two vertices of Pm when m = 2. Hence, let m ≥ 3. For any positive integer m ≥ 3, we have 2 b m
2c
Therefore, 2 b m
2c
bm
2
c − 1 > dm
2
e.That is, 2 α − α − 1 > β . Hence, by Theorem 3.7, σ∗(Pm) = α = bm
2
c.Next graph we consider is a cycle Cm on m vertices. The weak modular sumset number a cycle Cm is given in the following result.
Proposition 3.9. The weak modular sumset number of a cycle Cm, on m vertices is
σ∗(Cm) =
{
m − 1 if n = 3 , 4
dm
2
e if n ≥ 5 .
Proof. For C3, α = 2 and β = 1. Therefore, 2 α − α − 1 = 1 = β. Therefore, by Theorem 3.7, σ∗(C3) = 2. If C4, α = 2 and β = 2. Now, 2 α − α − 1 = 1 < β . Also, for r = 3, 2 r − r − 1 = 2 3 − 3 − 1 = 4 > β . Therefore, by Theorem 3.7, σ∗(C4) = 3 For a cycle Cm; m ≥ 5, we have α = dm
2
e and β = bm
2
c. Since m ≥ 3, 2 d m
2e
≥ 4. Therefore, 2 d m
2e
dm
2
e − 1 > bm
2
c. That is, 2 α − α − 1 > β . Hence, by Theorem 3.7,
σ∗(Cm) = α = dm
2
e.Sudev Naduvath 7Another graph that can be generated from a cycle is a wheel graph , denoted by Wn+1 , which is the graph obtained by joining every vertex of a cycle Cn to an external vertex. That is, Wn+1 = Cn + K1. The following result provides the weak modular sumset number of a wheel graph.
Proposition 3.10. The weak modular sumset number of a wheel graph Wm+1 is
1 + dm
2
e.Proof. For a wheel graph Wm+1 = Cm + K1, we have α = 1 + dm
2
e and β = bm
2
c.Since m ≥ 3, as explained in the previous theorems, 2 α − α − 1 > β . Hence, by Theorem 3.7, σ∗(Wm+1 ) = 1 + dm
2
e.Another graph that is close to cycles and wheel graphs is a helm graph , denoted by Hm, which is the graph obtained by adjoining a pendant edge to each vertex of the outer cycle Cm of a wheel graph Wm+1 . It has 2 m + 1 vertices and 3 m edges. The next result establishes the weak modular sumset number of a helm graph.
Proposition 3.11. The weak modular sumset number of a helm graph Hm is m.Proof. For Hm, we have α = m and β = m + 1. It is to be noted that 2 m−1 ≥ (m + 1) for all positive integer m ≥ 3. Therefore, 2 m −m−1 > m +1. That is, 2 α −α−1 > β .Hence by Theorem 3.7, σ∗(Hm) = α = m.Another graph which are related to paths are ladder graph, denoted by Lm and is defined as the Cartesian product of a path on m vertices and a path on two vertices. That is, Lm = pmP2 (or Lm = PmK2), where m > 2. The weak modular sumset number of a ladder graph is determined in the following proposition.
Proposition 3.12. The weak modular sumset number of a ladder graph Lm is m.Proof. Let Lm = PmP2, where m > 2. Irrespective of whether m is odd or even,
α = m and β = m. For any positive integer m > 2, we have 2 m − m − 1 > m . That is, 2 α − α − 1 > β . Then, by Theorem 3.7, σ∗(Ln) = m.Next we proceed to determine the weak modular sumset number of a complete graph Kn. is determined in the following result. We already mentioned the weak modular sumset number of K1, K2 and K3 when we discussed the weak modular sumset number of paths and cycles. Hence, need consider complete graphs having 4 or more vertices. The weak modular sumset number of a complete graph Km, where
m ≥ 4, is determined in the following result.
Proposition 3.13. For a positive integer m ≥ 4, the weak modular sumset number of a complete graph Km is m − 1.Proof. For a complete graph Km, we have α = m − 1 and β = 1. Therefore, for any positive integer m, we have 2 m−1 − m − 2 > 1. That is, 2 α − α − 1 > β . Therefore, the weak modular sumset number of Km is m − 1. The minimum cardinality of the ground set when the given graph G admits a weakly uniform modular sumset labeling arouses much interest in this occasion. Hence, we have the following result. 8 On the Modular Sumset Labeling of Graphs
Theorem 3.14. Let G be a graph with covering number α and independence number
β and let G admits a weakly k-uniform modular sumset labeling, where k < α being a positive integer. Then, the minimum cardinality of the ground set Zn is max {α, r },where r is the smallest positive integer such that (rk
) ≥ β.Proof. Let G be a graph which admits a weakly k-uniform sum set labeling f A = Zn.Then, by Theorem 3.5, G is bipartite. Let X, Y be the bipartition of the vertex set
V (G). If |X| ≤ | Y |, then α = |X| and β = |Y |. Then, distinct elements of X must have distinct singleton set-labels. Therefore, n ≥ α.On the other hand, since f is k-uniform, all the elements in Y must have distinct
k-element set-labels. The number of k-element subsets of a set A (obviously, with more than k elements) is (|A|
k
). The ground set A has α elements only if (αk
) ≥ β.Otherwise, the ground set A must contain at least r elements, where r > α is the smallest positive integer such that (rk
) ≥ β. Therefore, n = max {α, r }.In the above theorems we considered the value of k which is less than or equal to α. What is the case when k ≥ α? the following theorem provides a solution to this problem.
Corollary 3.15. Let G be a weakly k-uniform modular sumset graph, where k ≥ α,where α is the covering number of G. Then, the minimum cardinality of the ground set is the smallest positive integer n such that (nk
) ≥ β, where β is the independence number of G.Proof. If k ≥ α, then (αk
) does not exist. Then, the proof is immediate from Theorem 3.14. Now, we proceed to verify the properties of certain graphs in which the set-labeling number of edges are the product of the set-labeling numbers of their end vertices.
3.3 Strong Modular Sumset Graphs
Analogous to strong IASL-graphs, we define the following notion.
Definition 3.16. Let f : V (G) → P(Zn) be a modular sumset labeling defined on a given graph G. Then, f is said to be a strong modular sumset labeling if for the associated function f + : E(G) → P(Zn), |f +(uv )| = |f (u)| | f (v)| ∀ uv ∈ E(G). A graph which admits a strong modular sumset labeling is called a strong modular sumset graph .Let DA be the difference set of a given set A defined by DA = {| ai − aj | : ai, a j ∈
A}. Then, the necessary and sufficient condition for a graph to admit a strong modular sumset labeling is as given below.
Theorem 3.17. A modular sumset labeling f : V (G) → P(Zn) of a given graph G is a strong modular sumset labeling of G if and only if Df (u) ∩ Df (v) = ∅, ∀ uv ∈ E(G),where |f (u)| | f (v)| ≤ n.Sudev Naduvath 9
Proof. Let f : V (G) → P(Zn) be a modular sumset labeling on a given graph G.For any vertex u ∈ V (G), define Df (u) = {ai − aj : ai, a j ∈ f (u)}.Let uv be an arbitrary edge in E(G). Assume that f is a strong modular sumset labeling of G. Then, by definition |f +(uv )| = |f (u)| | f (v)|. Therefore, for any elements ai, a j ∈ f (u) and br, b s ∈ f (v), we have ai + br 6 = aj + bs in
f +(uv ) ∀ uv ∈ E(G). That is, |ai − aj | 6 = |bs − br| for any ai, a j ∈ f (u) and
br, b s ∈ f (v). That is, Df (u) ∩ Df (v) = ∅. Therefore, the difference sets of the set-label of any two adjacent vertices are disjoint. Conversely, assume that the difference Df (u) ∩ Df (v) = ∅ for any edge uv in
G. That is, |ai − aj | 6 = |bs − br| for any ai, a j ∈ f (u) and br, b s ∈ f (v). Then,
ai − aj 6 = bs − br. That is, ai + br 6 = aj + bs. Therefore, all elements in f (u) + f (v)are distinct. That is, |f +(uv )| = |f (u)| | f (v)| for any edge uv ∈ E(G). Hence, f is a strong modular sumset labeling of G.Also, note that the maximum possible cardinality in the set-label of any element of G is n, the product |f (u)| | f (v)| can not exceed the number n. This completes the proof. Figure 3 illustrates a strong modular sumset labeling of a graph with respect to the ground set Z6.
Figure 3: An example to a strong modular sumset graph. Analogous to the weak modular sumset number of a graph G, we can define the
strong modular sumset number of G as the minimum cardinality required for the ground set Zn so that G admits a strong modular sumset labeling. The choice of ground set Zn is very important in this context because n should be sufficiently large so that the vertices of the given graph can be labeled in such a way that the difference sets of these set-labels of all adjacent vertices must be pairwise disjoint. Analogous to the corresponding results on strong IASL-graphs and strongly uniform IASL-graphs proved in [ 15 ], the following results are valid for strong modular sumset graphs also.
Theorem 3.18. For a positive integer k ≤ n, a modular sumset labeling f : V (G) →
P(Zn) of a given connected graph G is a a strongly k-uniform modular sumset labeling of G if and only if either k is a perfect square or G is bipartite. 10 On the Modular Sumset Labeling of Graphs
Proof. If k is a perfect square, say k = l2, then we can label all the vertices of a graph by distinct l-element sets in such a way that the difference sets of the set-labels of every pair of adjacent vertices are disjoint. Hence, assume that k is not a perfect square. Let G be a bipartite graph with bipartition ( X, Y ). Let r, s be two divisors of
k. Label all vertices of X by distinct r-element sets all of whose difference sets are the same, say DX . Similarly, label all vertices of Y by distinct s-element sets all of whose difference sets the same, say DY , such that DX ∩ DY = ∅. Then, all the edges of G has the set-labeling number k = rs . Therefore, G is a strongly k-uniform modular sumset graph. Conversely, assume that G admits a strongly k-uniform modular sumset labeling, say f . Then, f +(uv ) = k ∀ uv ∈ E(G). Since, f is a strong modular sumset labeling, the set-labeling number of every vertex of G is a divisor of the set-labeling numbers of the edges incident on that vertex. Let v be a vertex of G with the set-labeling number r, where r is a divisor of k, but r2 6 = k. Since f is k-uniform, all the vertices in N (v), must have the set-labeling number s, where rs = k. Again, all vertices, which are adjacent to the vertices of N (v), must have the set-labeling number r.Since G is a connected graph, all vertices of G have the set-labeling number r or
s. Let X be the set of all vertices f G having the set-labeling number r and Y be the set of all vertices of G having the set-labeling number s. Since r2 6 = k, no two elements in X (and no elements in Y also) can be adjacent to each other. Therefore,
G is bipartite.
Theorem 3.19. For a positive non-square integer k ≤ n, a modular sumset labeling
f : V (G) → P(Zn) of an arbitrary graph G is a a strongly k-uniform modular sumset labeling of G if and only if either G is bipartite or a disjoint union of bipartite components.
For a positive integer k ≤ n, the maximum number of components in a strongly
k-uniform modular sumset graph is as follows.
Proposition 3.20. Let f be a strongly k-uniform modular sumset labeling of a graph
G with respect to the ground set Zn. Then, the maximum number of components in
G is the number of distinct pairs of divisors r and s of k such that rs = k.
Remark 3.21. It can be observed that a strongly k-uniform modular sumset graph can have a non-bipartite component if and only if k is a perfect square. More over, a strongly k-uniform modular sumset graph G can have at most one non-bipartite component.
3.4 Maximal Modular Sumset Graph
It can be observed that the maximum value of the set-labeling number of an edge of a modular sumset graph is n, the cardinality of the ground set Zn. Hence, we introduce the following notion.
Definition 3.22. Let f : V (G) → P(Zn) be a modular sumset labeling of a given graph G. Then, f is said to be a maximal modular sumset labeling of G if and only if f +(E(G)) = {Zn}.Sudev Naduvath 11 In other words, a modular sumset labeling f : V (G) → P(Zn) of a given graph
G is a maximal modular sumset labeling of G if the set-label of every edge of G is the ground set Zn itself. What are the conditions required for a graph to admit a maximal modular sumset labeling? Let us proceed to find out the solutions to this problem.
Proposition 3.23. The modular sumset labeling f : V (G) → P(Zn) of a given graph G is a maximal modular sumset labeling of G if and only if for every pair of adjacent vertices u and v of G some or all of the following conditions hold. (i) |f (u)| + |f (v)| ≥ n if Df (u) ∩ Df (v) 6 = ∅. The strict inequality hold when Df (u)
and Df (v) are arithmetic progressions containing the same elements. (ii) |f (u)| | f (v)| ≥ n if Df (u) ∩ Df (v) = ∅.Proof. For two adjacent vertices u and v in G, let Df (u) = Df (v) = {d} are arithmetic progressions containing the same elements. Then, the elements in f (u) and f (v)are also in arithmetic progression, with the same common difference d. Then, by Theorem 1.2, |f (u) + f (v)) | = |f (u)| + |f (v)| − 1. Therefore, the set-labeling number of the edge uv is n if and only if |f (u)| + |f (v)| > n .Now, let Df (u) ∩Df (v) 6 = ∅ such that Df (u) 6 = Df (v). Then, clearly |f (u)+ f (v)) | ≥ |f (u)| + |f (v)|. Therefore, we have |f +(uv )| = n if and only if |f (u)| + |f (v)| ≥ n.Next assume that Df (u) ∩ Df (v) = ∅. Then, |f (u) + f (v)) | = |f (u)| | f (v)|.Therefore, we have |f +(uv )| = n if and only if |f (u)| | f (v)| ≥ n.Figure 4 illustrates a maximal modular sumset labeling of a graph with respect to the ground set Z4.
Figure 4: An example to a maximal modular sumset graph. The following result explains a necessary and sufficient condition for a weak modular sumset labeling of a given graph G to be a maximal modular sumset labeling of G.
Proposition 3.24. A weak modular sumset labeling of a graph G is a maximal labeling of G if and only if G is a star graph. 12 On the Modular Sumset Labeling of Graphs
Proof. Let f be a weak modular sumset labeling of given graph G. First, assume that f is a maximal modular sumset labeling of G. Then, the set-labeling number of one end vertex of every edge of G is 1 and the set-labeling number of the other end vertex is n. Therefore, Zn be the set-label of one end vertex of every edge of G,which is possible only if G is a star graph with the central vertex has the set-label
Zn and the pendant vertices of G have distinct singleton set-labels. Conversely, assume that G is a star graph. Label the central vertex of G by the ground set Zn and label other vertices of G by distinct singleton subsets of Zn.Then, all the edges of G has the set-indexing number n. That is, this labeling is a maximal modular sumset labeling of G.A necessary and sufficient condition for a strong modular sumset labeling of a graph G to be a maximal modular sumset labeling of G.
Theorem 3.25. Let f be a strong sumset-labeling of a given graph G. Then, f is a maximal sumset-labeling of G if and only if n is a perfect square or G is bipartite or a disjoint union of bipartite components. Proof. The proof is an immediate consequence of Theorem 3.18, when k = n.In the coming discussion, we check whether the sumset of two sets can contain both sets and according to that property we define a particular type of modular sumset graphs.
3.5 Exquisite Modular Sumset Graphs
Analogous to the exquisite integer additive set-labeling of a graph G defined in [ 16 ], let us define the following.
Definition 3.26. For a positive integer n, let Zn be the set of all non-negative integers modulo n and P(Zn) be its power set. An exquisite modular sumset labeling
(EMSL) is a modular sumset labeling f : V (G) → P(Zn) with the induced function
f + : E(G) → P(Zn) defined by f +(uv ) = f (u) + f (v), uv ∈ E(G), such that
f (u), f (v) ⊆ f +(uv ) for all adjacent vertices u, v ∈ V (G). A graph which admits an exquisite modular sumset labeling is called an exquisite modular sumset graph .What is the condition required for a given graph to admit an exquisite modular sumset labeling? The following proposition leads us to a solution to this question.
Proposition 3.27. Let A and B be two subsets of the set Zn. Then, A (or B) is a subset of their sumset A + B if and only if every element ai of A is the sum (modulo
n) of an element aj (not equal to ai) in A and an element bl in B.Proof. Let A, B ⊆ Zn. Assume that every element of A is the sum (modulo n)of an element in A and an element B. That is, ai ∈ A =⇒ ∃ aj ∈ A, b l ∈
B such that ai = aj + bl. Hence ai ∈ A + B. Therefore, A ⊆ A + B.Conversely, assume that A ⊆ A + B. Then, ai ∈ A =⇒ ai ∈ A + B =⇒ ai =
aj + bl for some aj ∈ A, b l ∈ B.In view of the Proposition 3.27, we have the following theorem. Sudev Naduvath 13
Theorem 3.28. A graph G admits an exquisite modular sumset labeling if and only if every element in the set-label of any vertex of G is the sum (modulo n) of an element in that set-label and an element in the set-label of its adjacent vertex.
Analogous to the corresponding result of exquisite IASL-graphs, we have the following result for an exquisite modular sumset graphs.
Proposition 3.29. A graph admits an exquisite modular sumset labeling f if the set-label of every vertex of G consists of the element 0.
Proposition 3.30. Every maximal modular sumset labeling of a graph G is also an exquisite modular sumset labeling of G.Proof. Let f be a maximal sumset-labeling of G. Then, f +(uv ) = Zn for all edge
uv ∈ E(G). Then, the set-labels f (u) and f (v) are the subsets of f +(uv ). Hence, f
is an exquisite modular sumset labeling of G.Invoking the above results, a necessary and sufficient condition for a modular sumset labeling of a graph G to be an exquisite modular sumset labeling of G.
Theorem 3.31. A modular sumset labeling f : V (G) → P(Zn) of a graph G is an exquisite modular sumset labeling of G if and only if either the set-label of every vertex consists of the element 0 or f is a maximal modular sumset labeling of G.
Figure 5 illustrates an exquisite modular sumset labeling of a graph, the set-labels of all whose vertices containing 0.
Figure 5: An example to a maximal modular sumset graph. Can a weak modular sumset labeling of a graph G be an exquisite modular sumset labeling of G? Let us now proceed to find the solution to this problem.
Theorem 3.32. Let f : V (G) → P(Zn) be a weak modular sumset labeling of a graph G. Then, f is an exquisite modular sumset labeling of G if and only if G is a star. 14 On the Modular Sumset Labeling of Graphs
Proof. Since f is a weak modular sumset labeling of G, then at least one end vertex of every edge of G must have singleton set-labels. First, assume that G is a star graph. Label the central vertex by {0} and label all pendant vertices by the subsets of Zn containing 0. Then, this labeling is an exquisite modular sumset labeling. (Or label the central vertex of G by Zn and label the pendant vertices of G by distinct singleton subsets of Zn. This labeling is also an exquisite modular sumset labeling of G.) Conversely, assume that f is an exquisite modular sumset labeling of G. Then by Theorem 3.31, the set-labels of all vertices of G consist of 0 or f is a maximal modular sumset labeling. If the set-labels of all vertices of G contain 0, then {0} is the only singleton set that can be the set-label of a vertex of G. Therefore, G must be a star in which the set-label of the central vertex is {0} and the pendant vertices of G have distinct non-singleton subsets of Zn as their set-labels. If 0 is not an element of the set-label of every vertex of G, then f is a maximal modular sumset labeling of G. Then, by Theorem 3.24, G is a star graph in which the set-label of the central vertex is Zn and the pendant vertices of G have distinct singleton subsets of Zn as their set-labels. This completes the proof.
4 Conclusion
In this paper, we have introduced the notion of modular sumset labeling for given graphs and discussed certain properties and characteristics of modular sumset graphs. More properties and characteristics of various modular sumset graphs, both uniform and non-uniform, are yet to be investigated. Some promising problems in this area are the following.
Problem 1. Find the minimum cardinality of the ground set Zn so that the modular sumset labeling f : V (G) → P(Zn) of a given graph G is a uniform modular sumset labeling of G.
Problem 2. Characterise the graphs admitting the modular sumset labelings which are exquisite as well as weak.
Problem 3. Find the minimum cardinality of the ground set Zn so that the modular sumset labeling f : V (G) → P(Zn) defined on a given graph G is a modular sumset indexer of G.
Problem 4. Verify whether the existence of induced modular sumset labelings for certain graphs associated to the given modular sumset graphs like line graphs, total graphs etc.
Problem 5. Determine the strong modular sumset number of different graph classes. The problems of establishing the necessary and sufficient conditions for various graphs and graph classes to have certain other types of sumset valuations are also open. Studies on those sumset valuations which assign different sets having specific properties, to the elements of a given graph are also noteworthy. Sudev Naduvath 15
Acknowledgement
The author dedicates this paper to his research supervisor Prof. (Sr.) K. A. Germina, honouring her outstanding research career of over two decades.
References
B. D. Acharya, Set valuations and their applications , MRI Lecture Notes in Applied Mathematics, No.2, The Mehta Research Institute of Mathematics and Mathematical Physics, Allahabad, 1983. B. D. Acharya, Set-indexers of a graph and set-graceful graphs, Bull. Allahabad Math. Soc., 16 (2001), 1-23. T. M. Apostol, Introduction to analytic number theory , Springer-Verlag, New York, 1989. J. A. Bondy and U. S. R. Murty, Graph theory , Springer, 2008. A. Brandst¨ adt, V. B. Le and J. P. Spinrad, Graph classes: A survey , SIAM, Philadelphia, 1999. G. Chartrand and P. Zhang, Introduction to graph theory , McGraw-Hill Inc., 2005. J. A. Gallian, A dynamic survey of graph labeling , Elec. J. Combin., DS-6, 2015. K. A. Germina and T. M. K. Anandavally, Integer additive set-indexers of a graph: Sum square graphs , J. Combin. Inform. System Sci., 37 (2-4)(2012), 345-358. K. A. Germina and N. K. Sudev, On weakly uniform integer additive set-indexers of graphs , Int. Math. Forum, 8(37-40)(2013), 1827-1834, DOI: 10.12988/imf.2013.310188. R. Hammack, W. Imrich and S. Klavzar, Handbook of product graphs ,CRC Press, Boca Raton, 2011. F. Harary, Graph theory , Narosa Publications, Delhi, 2001. K. D. Joshi, Applied discrete structures , New Age International, New Delhi, 2003. M. B. Nathanson, Additive number theory: Inverse problems & geom-etry of sumsets , Springer, 1996. N. K. Sudev and K. A. Germina, On integer additive set-indexers of graphs ,Int. J. Math. Sci. Eng. Appl., 8(2)(2014), 11-22. 16 On the Modular Sumset Labeling of Graphs
N K Sudev and K A Germina, Some new results on strong integer additive set-indexers of graphs , Discrete Math. Algorithm. Appl., 7(1)(2015), 1-11, DOI: 10.1142/S1793830914500657. N. K. Sudev and K. A. Germina, The exquisite integer additive set-labeling of graphs , Int. J. Sci. Res., 4(4)(2015), 2858-2862. E. W. Weisstein, CRC concise encyclopaedia of mathematics , CRC press, 2011. D. B. West, Introduction to graph theory , Pearson Education Inc., 2001. Information system on graph classes and their inclusions, graphclasses.org/smallgraphs . |
1075 | https://www.youtube.com/watch?v=zSWRm0zb8tw | √ Mastering Geometric Series: A Logarithmic Approach
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📐 Are you ready to take your understanding of geometric series to the next level? Welcome to an enlightening journey into mathematics, where we'll explore how logarithms can be powerful tools for evaluating geometric series.
🌟 Unveiling the Geometric Series: We'll start by revisiting the basics of geometric series, understanding their structure, and appreciating their significance in various mathematical and real-world contexts.
🔍 The Power of Logarithms: Logarithms are more than just mathematical operations—they're essential problem-solving tools. We'll unravel the magic behind logarithms, exploring their properties and applications.
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okay now let's look at geometric series using logarithms okay so even with the geometric series questions you need to know your log rules very well because the questions required to use them so when you're doing a question related to geometric series so basically when you're using the some formula you need to apply some log rule when you have something that looks like this so this is the sum rule isn't it a are n minus 1 over R minus 1 if that is equal to some value P then I'm just going to multiply this side by R minus 1 so I have P times R minus 1 and then divide it by a so I have P times R minus 1 over a and then decided we're just left with R to the power of n minus 1 now simply I'm just gonna move negative 1 over to the other side so I have P R minus 1 into a plus 1 and this side I'm just left with R to the power of n now this is a stage when we're using the log rule remember we want to solve for n then n becomes the subject and that's equal to log the augur is down is the base and then this whole thing goes after the log ok sir basically guys again we're using this rule here ax equals to be then x equals to log a B ok sorry in this case this B is this big bunch here that the whole thing is based that's why we put the whole thing after the log ok and AH just goes down as a little base okay basically we're using this rule again so hope you're familiarized with it even if you haven't known very well about it before okay so this is our log rule I'm gonna try to use this into our questions case they have a look at that okay and now I might go into some questions using some log rules sir let's have a look at question 14 find the number of terms if nine plus 18 plus 36 plus ba ba-ba equals two 1143 okay see this time they actually give us what the value of the summation is is going to be 1143 our job is to tell i'm find the number of terms in this series that make a sum of this okay so guys have a look at the series first of all decide what your egg is and what your r is okay what your first term is and what your common ratio is so I hope you can see that the first term is going to be nine okay the first one in the list so a is nine and what's our our is simply two because 9 times 2 is 18 18 times 2 is 36 so therefore the common ratio must be 2 okay so we know what a is we know what R is we don't know what n is that's what we're gonna find cakes it says find the number of terms Sara what I'm gonna do put all these values into our sum formula just like this so a which is 9 times R which is 2 so 2 to the power of n okay so we're gonna put in terms of n because we need to solve for n it's going to be 2 to the power of n minus 1 over R minus 1 R is 2 so 2 minus 1 and this whole thing equals to 1143 because this is the sum and they give us what the sum is okay now let's solve it solve for n so I'm just gonna go like this so 2 n minus 1 is basically 1143 divided by 9 because look 2 minus 1 is simply 1 so we don't really need to worry about that so 1143 times 1 is just 1143 and then divide it by 9 so on this side we're left with 2 n minus 1 okay and then now just move one over to the other side so we have one thousand one hundred and forty three over nine plus one simplify that guys you should get 128 alright just simplify it use your calculator if you need to it should be 128 okay now guys 128 is 2 to the power of seven isn't it right I don't know put it sometimes if you don't really know just do some trial and error with your calculator ok just plug some numbers try 2 to the power of 6 to the power of 8 or to the power 7 you should see that 2 to the power of 7 is 128 and the reason why I wanted it to be 2 to the power of something is because the base of this is true okay so that's why I want this to be 2 to the power of something stir if the base is the same that means the indices must also be the same so therefore and is 7 you can see how they're both two so therefore n must be equal to 7 okay the indices must be equal so N equals to 7 so therefore there are seven terms okay guys basically that's the answer there are seven terms there are seven times in this series that make a sum of 1143 and you're probably wondering why we're going to use any logs well this one was pretty simple because we can change this into two two five seven but say for example if you just want to do 2n equals 228 okay this is this one to N equals 228 now you can use your locals if you want to if you don't want to change it around like this you can just use your log rules like this so n is going to be log - it goes down and 128 so this will be your Valley for EM okay so all the way is fine guys you can go ahead and use your log rules if you want to okay if you don't really want to if you don't think you can find out what to ^ what is makes 128 okay but I really think in this case this would be a poor simple away because see how we just got the X through in a few steps cuz this one you'll have to have a bit more steps because look we can't find what log 2 is right so we'll have to go ahead and do that fraction thing again pull off your log I'm gonna make a base of 10 log 10 hundred and twenty eight over log 10 - and you'll have to put this into your calculator okay but still this way is perfectly fine guys you can use your log rules if you and but I prefer to using this one if you can't make this 2 to the power of something okay if you can't with this if this is 2 to the power of something nothing then I'd go for the log rules like this okay either way you look at the same answer okay so if you did use your look logs this would be your case okay this will be your working out so have a look at that guy's that's question 14 basically there are seven terms that make a sum of 1143 |
1076 | https://mathoverflow.net/questions/433779/elementary-convexity-example | real analysis - Elementary convexity example - MathOverflow
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Elementary convexity example
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I'm trying to check that certain examples of Young functions in the harmonic analysis literature are actually Young functions, and in doing so need to prove the following convexity-like inequality for p>1,δ>0 p>1,δ>0 and 0≤a≤1<b 0≤a≤1<b:
(a+b 2)p[1+ln(a+b 2)]p−1+δ≤1 2 a p+1 2 b p[1+ln b]p−1+δ(a+b 2)p[1+ln(a+b 2)]p−1+δ≤1 2 a p+1 2 b p[1+lnb]p−1+δ
.
Clearly for fixed b>1 b>1 this is true if a=0 a=0 and a=1 a=1 (the latter by the convexity of x p[1+ln x]p−1+δ x p[1+lnx]p−1+δ for x≥1 x≥1). Also, this is easy for all 0≤a≤1<b 0≤a≤1<b assuming p−2+δ≥0 p−2+δ≥0:
For fixed 0≤a≤1 0≤a≤1 let L(b)L(b) and R(b)R(b) denote the left and right hand sides. Then then by convexity and the fact that 1+ln(a+1 2)∈(1−ln 2,1]1+ln(a+1 2)∈(1−ln2,1] we have
L(1)≤(1 2 a p+1 2)[1+ln(a+1 2)]p−1+δ≤1 2 a p+1 2=R(1).L(1)≤(1 2 a p+1 2)[1+ln(a+1 2)]p−1+δ≤1 2 a p+1 2=R(1).
using basic calculus and recalling that a<b a<b, we see that
L′(b)=1 2(a+b 2)p−1[p(1+ln(a+b 2))p−1+δ+(p−1+δ)(1+ln(a+b 2))p−2+δ]≤1 2 b p−1[p(1+ln b)p−1+δ+(p−1+δ)(1+ln b)p−2+δ]=R′(b),L′(b)=1 2(a+b 2)p−1[p(1+ln(a+b 2))p−1+δ+(p−1+δ)(1+ln(a+b 2))p−2+δ]≤1 2 b p−1[p(1+lnb)p−1+δ+(p−1+δ)(1+lnb)p−2+δ]=R′(b),
as desired. Here, we clearly need p−2+δ≥0 p−2+δ≥0 to use the fact that ln ln is increasing.
Any thoughts on how to remove this assumption? Does the inequality I seek follow from anything out in the literature?
P.S. For anyone curious, the Young functions in question are Φ(x)=x p[1+ln+x]p−1+δ Φ(x)=x p[1+ln+x]p−1+δ for p>1 p>1 and δ>0 δ>0, which induce Orlicz spaces that appear very naturally in modern harmonic analysis (see C. Perez "On sufficient conditions for...", Proc. London Math. Soc., (1995), and other papers that cite this).
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edited Nov 3, 2022 at 16:43
Joshua IsralowitzJoshua Isralowitz
asked Nov 2, 2022 at 21:27
Joshua IsralowitzJoshua Isralowitz
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Here is how to remove the assumption that p−2+δ≥0 p−2+δ≥0.
Let
s:=p−1+δ.s:=p−1+δ.
The conditions p>1 p>1 and δ>0 δ>0 imply s≥0 s≥0. No other conditions on p p and s s will be used or needed in what follows.
The inequality in question will follow from the inequality
L≤R(s)L≤R(s)
for
s∈[0,∞),p∈[1,∞),0<a≤1≤b,s∈[0,∞),p∈[1,∞),0<a≤1≤b,
where
L:=2 1−p(a+b)p,R(s):=R(s,p):=a p(1+ln a+b 2)−s+b p(1+ln b 1+ln a+b 2)s.L:=2 1−p(a+b)p,R(s):=R(s,p):=a p(1+lna+b 2)−s+b p(1+lnb 1+lna+b 2)s.
Clearly, R(s)R(s) is convex in s s.
Lemma 1:R′(0)≥0 R′(0)≥0, and hence R(s)R(s) is increasing in s s.
By Lemma 1 and the power-means inequality,
R(s)≥R(0)=a p+b p≥2 1−p(a+b)p=L,R(s)≥R(0)=a p+b p≥2 1−p(a+b)p=L,
as desired.
It remains to provide
Proof of Lemma 1: Let
R 1(p,a,b):=R′(0)/a p=(b a)p ln 1+ln b 1+ln a+b 2−ln(1+ln a+b 2),(0)(0)R 1(p,a,b):=R′(0)/a p=(b a)p ln1+lnb 1+lna+b 2−ln(1+lna+b 2),
which is clearly nondecreasing in p p. So,
a R 1(p,a,b)≥a R 1(1,a,b)=b ln(1+ln b)−(a+b)ln(1+ln a+b 2),a R 1(p,a,b)≥a R 1(1,a,b)=b ln(1+lnb)−(a+b)ln(1+lna+b 2),
which latter is clearly decreasing in a a. So,
a R 1(p,a,b)≥a R 1(1,a,b)≥R 1(1,1,b)=b g(b),(1)(1)a R 1(p,a,b)≥a R 1(1,a,b)≥R 1(1,1,b)=b g(b),
where
g(b):=ln(1+ln b)−1+b b ln(1+ln 1+b 2).g(b):=ln(1+lnb)−1+b b ln(1+ln1+b 2).
So, in view of (0)(0), it remains to prove that
g(b)≥(?)0(2)(2)g(b)≥(?)0
(for real b>1 b>1).
To begin the proof of (2)(2), let
g 1(b):=g′(b)b 2=b(1 1+ln b−1 1+ln 1+b 2)+ln(1+ln 1+b 2).(3)(3)g 1(b):=g′(b)b 2=b(1 1+lnb−1 1+ln1+b 2)+ln(1+ln1+b 2).
Now use the substitution b=2 e t−1 b=2 e t−1, so that t>0 t>0 and ln 1+b 2=t ln1+b 2=t. Let
G 2(t):=d g 1(2 e t−1)d t(1+t)2(1+ln(2 e t−1))2(2 e t−1)t=2(t+e t(1+t 2))ln(2 e t−1)(2 e t−1)t−1−ln 2(2 e t−1).(4)(4)G 2(t):=d g 1(2 e t−1)d t(1+t)2(1+ln(2 e t−1))2(2 e t−1)t=2(t+e t(1+t 2))ln(2 e t−1)(2 e t−1)t−1−ln 2(2 e t−1).
Next, note that
h(t):=t 3+t 2+t+2 e t(t 2+1)−1>0 h(t):=t 3+t 2+t+2 e t(t 2+1)−1>0
(for t>0 t>0) and let
G 3(t):=G′2(t)t 2(2 e t−1)2 2 e t h(t)=2 t(t+e t(1+t 2))h(t)−ln(2 e t−1),(5)(5)G 3(t):=G 2′(t)t 2(2 e t−1)2 2 e t h(t)=2 t(t+e t(1+t 2))h(t)−ln(2 e t−1),
G 4(t):=G′3(t)(2 e t−1)h(t)2 2 t=2−t+t 3+2 e 2 t(1−7 t−4 t 2−2 t 3−t 4+t 5)−e t(5−8 t+6 t 3+3 t 4+2 t 5),(5.5)(5.5)G 4(t):=G 3′(t)(2 e t−1)h(t)2 2 t=2−t+t 3+2 e 2 t(1−7 t−4 t 2−2 t 3−t 4+t 5)−e t(5−8 t+6 t 3+3 t 4+2 t 5),
G 5:=G′4,G 6:=G′5,G 7:=G′6,(6)(6)G 5:=G 4′,G 6:=G 5′,G 7:=G 6′,
G 8(t):=G′7(t)e−t P(t)=16 e t Q(t)P(t)−1,(7)(7)G 8(t):=G 7′(t)e−t P(t)=16 e t Q(t)P(t)−1,
where
P(t):=189+736 t+768 t 2+294 t 3+43 t 4+2 t 5,Q(t):=−65−91 t−8 t 2+40 t 3+18 t 4+2 t 5.P(t):=189+736 t+768 t 2+294 t 3+43 t 4+2 t 5,Q(t):=−65−91 t−8 t 2+40 t 3+18 t 4+2 t 5.
Next,
G′8(t)=16 e t S(t)P(t)2>0,G 8′(t)=16 e t S(t)P(t)2>0,
where
S(t):=18356+31777 t+25602 t 2+49850 t 3+84244 t 4+72903 t 5+34758 t 6+9548 t 7+1492 t 8+122 t 9+4 t 10.S(t):=18356+31777 t+25602 t 2+49850 t 3+84244 t 4+72903 t 5+34758 t 6+9548 t 7+1492 t 8+122 t 9+4 t 10.
So, G 8(t)G 8(t) is increasing (in t>0 t>0). Also, G 8(0)<0 G 8(0)<0 and G 8(∞−):=lim t→∞G 8(t)=∞>0 G 8(∞−):=lim t→∞G 8(t)=∞>0. So, G 8 G 8 is −+−+ (on (0,∞)(0,∞)) -- that is, for some real c>0 c>0 we have G 8<0 G 8<0 on (0,c)(0,c) and G 8>0 G 8>0 on (c,∞)(c,∞).
So, by (7)(7), G 7 G 7 is down-up (on (0,∞)(0,∞)) -- that is, for some real c>0 c>0 we have that G 7 G 7 is decreasing on (0,c)(0,c) and increasing on (c,∞)(c,∞). Also, G 7(0)<0 G 7(0)<0 and G 7(∞−)=∞>0 G 7(∞−)=∞>0. So, G 7 G 7 is −+−+.
So, by (6)(6), G 6 G 6 is down-up. Also, G 6(0)<0 G 6(0)<0 and G 6(∞−)=∞>0 G 6(∞−)=∞>0. So, G 6 G 6 is −+−+.
So, by (6)(6), G 5 G 5 is down-up. Also, G 5(0)<0 G 5(0)<0 and G 5(∞−)=∞>0 G 5(∞−)=∞>0. So, G 5 G 5 is −+−+.
So, by (6)(6), G 4 G 4 is down-up. Also, G 4(0)<0 G 4(0)<0 and G 4(∞−)=∞>0 G 4(∞−)=∞>0. So, G 4 G 4 is −+−+.
So, by (5.5)(5.5), G 3 G 3 is down-up. Also, G 3(0)=0 G 3(0)=0 and G 3(∞−)=−ln 2<0 G 3(∞−)=−ln2<0. So, G 3<0 G 3<0.
So, by (5)(5), G 2 G 2 is decreasing. Also, G 2(0+):=lim t↓0 G 2(t)=3>0 G 2(0+):=lim t↓0 G 2(t)=3>0 and G 2(∞−)=−∞<0 G 2(∞−)=−∞<0. So, G 2 G 2 is +−+− (on (0,∞)(0,∞)) -- that is, for some real c>0 c>0 we have G 2>0 G 2>0 on (0,c)(0,c) and G 2<0 G 2<0 on (c,∞)(c,∞).
So, by (4)(4), g 1 g 1 is up-down -- that is, for some real c>1 c>1 we have that g 1 g 1 is increasing on (1,c)(1,c) and decreasing on (c,∞)(c,∞). Also, g 1(1)=0 g 1(1)=0 and g 1(∞−)=−∞<0 g 1(∞−)=−∞<0. So, g 1 g 1 is +−+− (on (1,∞)(1,∞)).
So, by (3)(3), g g is up-down. Also, g(1+)=0 g(1+)=0 and g(∞−)=0 g(∞−)=0. Thus, (2)(2) follows. □◻
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edited Nov 3, 2022 at 16:40
answered Nov 3, 2022 at 15:51
Iosif PinelisIosif Pinelis
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Wow, thanks for the quick response! I'll try to read this today or tomorrow and then count this as answered once I do. :)Joshua Isralowitz –Joshua Isralowitz 2022-11-03 16:25:18 +00:00 Commented Nov 3, 2022 at 16:25
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Here is a much simpler proof, actually of the more general fact that
f(x):=x p(1+ln+x)s f(x):=x p(1+ln+x)s
is convex in x≥0 x≥0 for any real p≥1 p≥1 and s≥0 s≥0, where ln+x:=ln max(1,x)ln+x:=lnmax(1,x).
For the left and right derivatives f−f− and f+f+ of f f one has f+(1)=p+s≥p=f−(1)f+(1)=p+s≥p=f−(1).
So, it remains to show that f′′≥0 f″≥0 on (0,1)(0,1) and on (1,∞)(1,∞).
We have f′′(x)=p(p−1)x p−2≥0 f″(x)=p(p−1)x p−2≥0 for x∈(0,1)x∈(0,1).
Consider finally the case x∈(1,∞)x∈(1,∞). Then
f′′(x)=x p−2(1+ln x)s−2 f 2(x),f″(x)=x p−2(1+lnx)s−2 f 2(x),
where
f 2(x):=F 2(p,s,ln x)f 2(x):=F 2(p,s,lnx)
and
F 2(p,s,u):=(p−1)p u 2+u(2 p 2+2 p(s−1)−s)+p 2+p(2 s−1)+(s−2)s.F 2(p,s,u):=(p−1)p u 2+u(2 p 2+2 p(s−1)−s)+p 2+p(2 s−1)+(s−2)s.
It remains to show that
F 2(p,s,u)≥0 F 2(p,s,u)≥0
for p≥1 p≥1, s≥0 s≥0, and u≥0 u≥0. Note that the partial derivative of F 2(p,s,u)F 2(p,s,u) in u u at u=0 u=0 is 2(p−1)p+(2 p−1)s≥0 2(p−1)p+(2 p−1)s≥0.
It remains to note that
F 2(p,s,0)=(p−1)p+2(p−1)s+s 2≥0.□F 2(p,s,0)=(p−1)p+2(p−1)s+s 2≥0.◻
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answered Nov 25, 2022 at 17:53
Iosif PinelisIosif Pinelis
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1Is the space of bounded ψ∞ψ∞ Orlicz norm random variables equal to L∞L∞?
8Prove ∑m j=1 j 2 n∑m j=1(j−1 2)2 n>(m+1 2 m)2 n+1∑j=1 m j 2 n∑j=1 m(j−1 2)2 n>(m+1 2 m)2 n+1
0Reflexivity of the Orlicz space associated with the Huber function
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1077 | https://en.wikipedia.org/wiki/Household_debt | Household debt - Wikipedia
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1 Overview
2 Historical perspective
3 Global economic impact
4 StatisticsToggle Statistics subsection
4.1 United States
5 In United StatesToggle In United States subsection
5.1 Contribution to the 2008 financial crisis
5.2 Effects on economic growth
5.3 Effects on employment
5.4 Effects on economic recovery
5.5 Charts of U.S. household debt variables
6 Reducing household debtToggle Reducing household debt subsection
6.1 Paying off debt from income or savings
6.2 Debt restructuring or write-down
6.3 Inflation
7 Household over-indebtedness
8 See also
9 References
10 Further reading
Household debt
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From Wikipedia, the free encyclopedia
Combined debt of all people in a household
The examples and perspective in this article deal primarily with the United States and do not represent a worldwide view of the subject. You may improve this article, discuss the issue on the talk page, or create a new article, as appropriate.(December 2013) (Learn how and when to remove this message)
Household debt in Great Britain 2008-10
Household debt is the combined debt of all people in a household, including consumer debt and mortgage loans. A significant rise in the level of this debt coincides historically with many severe economic crises and was a cause of the U.S. and subsequent euro area crisis. Several economists have argued that lowering this debt is essential to economic recovery in the U.S. and selected Eurozone countries.
Overview
[edit]
Household debt can be defined in several ways, based on what types of debt are included. Common debt types include home mortgages, home equity loans, auto loans, student loans, and credit cards. Household debt can also be measured across an economy, to measure how indebted households are relative to various measures of income (e.g., pre-tax and disposable income) or relative to the size of the economy (GDP).
The burden of debt can also be measured in terms of the amount of interest it generates relative to the income of the borrower. For example, the U.S. Federal Reserve measures the "household debt service ratio" (DSR), an estimate of the ratio of debt payments to disposable personal income. Debt payments consist of the estimated required payments on outstanding mortgage and consumer debt. The Fed also measures the "financial obligations ratio" (FOR), which adds automobile lease payments, rental payments on tenant-occupied property, homeowners' insurance, and property tax payments to the debt service ratio. Homeowner and renter FORs are calculated by applying homeowner and renter shares of payments and income derived from the Survey of Consumer Finances and Current Population Survey to the numerator and denominator of the FOR. The homeowner mortgage FOR includes payments on mortgage debt, homeowners' insurance, and property taxes, while the homeowner consumer FOR includes payments on consumer debt and automobile leases.
Historical perspective
[edit]
In the 20th century, spending on consumer durables rose significantly. Household debt rose as living standards rose, and consumers demanded an array of durable goods. These included major durables like high-end electronics, vehicles, and appliances, that were purchased with credit. Easy credit encouraged a shift from saving to spending. This trend has continued into the 2020s where swiping a credit card has become a standard in day-to-day life
Households in developed countries significantly increased their household debt relative to their disposable income and GDP from 1980 to 2007 — one of the many factors that led to the subprime mortgage crisis and the euro area crisis. Research indicates that U.S. household debt increased from 43% to 62% of GDP from 1982 to 2000. Looking at the early years of the 21st century, many industrialized countries, with a notable exception of Germany, experienced a major spike of household debt versus GDP around 2007–8, with the United States leading up to 2007; by 2017, the American ratio was second only to that of the United Kingdom.
U.S. households made significant progress in deleveraging (reducing debt) post-crisis, much of it due to foreclosures and financial institution debt write-downs. By some measures, consumers began to add certain types of debt again in 2012, a sign that the economy may be improving as this borrowing supports consumption. However according to a business insider report Where the average American is almost 105,000 dollars in debt and in total Americans have towered over 17.6 trillion dollars of debt which is over double what the U.S government spent in 2022.
Global economic impact
[edit]
The International Monetary Fund (IMF) reported in April 2012:
"Household debt soared in the years leading up to the Great Recession. In advanced economies, during the five years preceding 2007, the ratio of household debt to income rose by an average of 39 percentage points, to 138 percent. In Denmark, Iceland, Ireland, the Netherlands, and Norway, debt peaked at more than 200 percent of household income. A surge in household debt to historic highs also occurred in emerging economies such as Estonia, Hungary, Latvia, and Lithuania. This occurred largely because the central banks implemented a prolonged period of artificially low policy interest rates, temporarily increasing the amount of debt that could be serviced with a given income. The leveraging up fueled a consumption boom that, ironically, boosted GDP in the countries in question, but represented not a sustainable 'boost to aggregate demand' but instead a mere pulling forward of consumption, as people took on new 30-40 year debt to pay for current year expenditures.
The predictable fallout of this policy is the slow growth that these countries are experiencing today. At the time, the concurrent boom in both house prices and the stock market - as the prices of financed and financial assets were bid up by virtue of the same low rates - meant that household debt relative to assets held broadly stable, which masked households' growing exposure to the eventual sharp fall in asset prices. House prices in particular were vulnerable to sharp movements in policy rates in countries in which most mortgages are adjustable-rate, as 30-40 year loans are highly rate sensitive, house purchases are financed at 80-95% of the price, and only 4-5% of homes change owners in a year.
Despite this low volume, the appraisal rules in most countries limit the number of times that a sale can serve as a comp only in terms of time (12 months), not the number of times that a given sale can be used as a comp - because of this, a "cash out refi boom" followed the low-rate-driven-price-increase - this was the primary vehicle through which developed countries' households leveraged up. On the supply side, home builders are also financed at short rates (in the US, Prime). Their cost of capital was effectively cut in half by virtue of the central bank rate cuts, enabling them to overbuild in response to the rise in prices - with all the new supply making the eventual price collapse worse than it would otherwise have been.
The entire episode was straight out of Ludwig von Mises' 'Human Action', Chapter 20, with the new twist that most of the debt was incurred on the consumer side. When rising consumer prices forced the central banks to allow policy rates to rise back up toward market rates, the wind propping up house prices was taken away. House prices immediately declined, leading to the 2008 financial crisis. Many households, who had reduced their savings out of a belief in the "wealth effect," saw their wealth shrink relative to their debt, and, with less income and more unemployment, especially in the previously booming real estate sector, found it harder to meet mortgage payments. "Strategic defaults" became common, as homeowners with significant negative equity simply abandoned the home and the debt.
By the end of 2011, real house prices had fallen from their peak by about 41% in Ireland, 29% in Iceland, 23% in Spain and the United States, and 21% in Denmark. Household defaults, underwater mortgages (where the loan balance exceeds the house value), foreclosures, and fire sales became endemic to a number of economies. Household deleveraging by paying off debts or defaulting on them has begun in some countries. It has been most pronounced in the United States, where about two-thirds of the debt reduction reflects defaults.
Of note, Germany, in which almost all mortgages carry 10 year pricing, and in which the 10 year interest rate did not decline during the 2000s, no housing bubble occurred at all. Also, in Canada, in which most mortgages are ARMs but 3-5 year, and which experienced a muted drop in intermediate-term rates, the housing bubble was more muted than in neighboring US.
Statistics
[edit]
Main article: List of countries by household debt
Among OECD countries, the larger share of all household debt is represented by mortgages, the incidence of which varies across countries, between 60% and 90% of total household debt.
2022 Household debt| No. | Country | % of GDP |
---
| 1 | Switzerland | 128.3% |
| 2 | Australia | 111.75% |
| 3 | South Korea | 105.09% |
| 4 | Canada | 102.39% |
| 5 | Hong Kong | 95.75% |
| 6 | Netherlands | 94.85% |
| 7 | New Zealand | 94.45% |
| 8 | Sweden | 88.23% |
| 9 | Thailand | 86.9% |
| 10 | Denmark | 86.22% |
United States
[edit]
Total US household debt timeline. Top to bottom:
Other
Student loans
Credit cards
Auto loans
Home equity revolving line of credit
Mortgage
Household debt of people in the US in the bottom 25% by wealth.
U.S. household (HH) debt (measured by the FRED variable "CMDEBT") rose relative to both GDP and disposable income over the 1980 to 2011 period.
Household debt as a% disposable income rose from 68% in 1980 to a peak of 128% in 2007, prior to dropping to 112% by 2011.
Household debt as a% nominal GDP rose from 47% in 1980 to a peak of 94% in 2009, prior to dropping to 77% in 2012.
U.S. household debt rose from nearly zero in the 1950s to $13.8 trillion in 2008, before declining to $12.9 trillion by Q2 2012.
Consumer credit outstanding includes credit cards, auto loans, student loans, and other types of household debt, but excludes mortgages. It rose from 14.0% GDP in January 1990 to 18.0% GDP by January 2009. It fell to a trough of 16.4% GDP in July 2010 and was back up to 17.5% GDP by January 2013.
In United States
[edit]
Main article: Subprime mortgage crisis
Contribution to the 2008 financial crisis
[edit]
This increase in debt from 1980 to 2007 enabled spending that stimulated the economy, arguably "papering over" or hiding the sustainable economic growth rate excluding this increase in leverage. This debt overhang then began holding back the economy as consumers paid down debt (which reduces economic activity) rather than spending.
Paul Krugman wrote in December 2010: "The root of our current troubles lies in the debt American families ran up during the Bush-era housing bubble. Twenty years ago, the average American household’s debt was 83 percent of its income; by a decade ago, that had crept up to 92 percent; but by late 2007, debts were 130 percent of income. All this borrowing took place both because banks had abandoned any notion of sound lending and because everyone assumed that house prices would never fall. And then the bubble burst. What we’ve been dealing with ever since is a painful process of 'deleveraging': highly indebted Americans not only can't spend the way they used to, they're having to pay down the debts they ran up in the bubble years. This would be fine if someone else were taking up the slack. But what's actually happening is that some people are spending much less while nobody is spending more—and this translates into a depressed economy and high unemployment. What the government should be doing in this situation is spending more while the private sector is spending less, supporting employment while those debts are paid down. And this government spending needs to be sustained: we're not talking about a brief burst of aid; we’re talking about spending that lasts long enough for households to get their debts back under control. The original Obama stimulus wasn’t just too small; it was also much too short-lived, with much of the positive effect already gone."
In April 2009, U.S. Federal Reserve Vice Chair Janet Yellen discussed the situation: "Once this massive credit crunch hit, it didn’t take long before we were in a recession. The recession, in turn, deepened the credit crunch as demand and employment fell, and credit losses of financial institutions surged. Indeed, we have been in the grips of precisely this adverse feedback loop for more than a year. A process of balance sheet deleveraging [paying down debt] has spread to nearly every corner of the economy. Consumers are pulling back on purchases, especially on durable goods, to build their savings. Businesses are cancelling planned investments and laying off workers to preserve cash. And, financial institutions are shrinking assets to bolster capital and improve their chances of weathering the current storm. Once again, Minsky understood this dynamic. He spoke of the paradox of deleveraging, in which precautions that may be smart for individuals and firms—and indeed essential to return the economy to a normal state—nevertheless magnify the distress of the economy as a whole."
The policy prescription of Ms Yellen's predecessor Mr Bernanke was to increase the money supply and artificially reduce interest rates. This stoked another debt and asset bubble. Mr Krugman's policy was to ensure that such borrowing took place at the federal government level, to be repaid via taxes on the individuals who he admitted were already overburdened with their own debts. These policies were arguably a mere return to the policies that inflated the debt bubble in the first place.
Economists Atif Mian and Amir Sufi wrote in 2014 that:
Historically, severe economic downturns are almost always preceded by a sharp increase in household debt.
U.S. household spending declines were largest in geographic areas with a combination of higher household debt and larger price declines.
When housing prices fall, poorer homeowners (with a larger proportion of their net worth in their home) are hit the hardest financially and reduce their consumption relatively more than wealthier households.
Declines in residential investment preceded the recession and were followed by reductions in household spending and then non-residential business investment as the recession worsened.
Effects on economic growth
[edit]
Ezra Klein wrote in August 2011: "[W]hat distinguishes crises like this one from typical recessions is household debt. When the financial markets collapsed, household debt was nearly 100 percent of GDP. It’s now down to 90 percent. In 1982, which was the last time we had a big recession, the household-debt-to-GDP ratio was about 45 percent. That means that in this crisis, indebted households can’t spend, which means businesses can’t spend, which means that unless government steps into the breach in a massive way or until households work through their debt burden, we can’t recover. In the 1982 recession, households could spend, and so when the Federal Reserve lowered interest rates and made spending attractive, we accelerated out of the recession. The utility of calling this downturn a “household-debt crisis” is it tells you where to put your focus: you either need to make consumers better able to pay their debts, which you can do through conventional stimulus policy like tax cuts and jobs programs, or you need to make their debts smaller so they're better able to pay them, which you can do by forgiving some of their debt through policies like [mortgage principal reduction] or eroding the value of their debt by increasing inflation. I’ve heard various economist make various smart points about why we should prefer one approach or the other, and it also happens to be the case that the two policies support each other and so we don't actually need to choose between them. All of these solutions, of course, have drawbacks: if you put the government deeper into debt in order to help households now, you increase the risk of a public-debt crisis later. That's why it's wise to pair further short-term stimulus with a large amount of long-term deficit reduction. If you force banks to swallow losses or face inflation now, you need to worry about whether they'll be able to keep lending at a pace that will support recovery over the next few years. But as we’re seeing, not doing enough isn't a safe strategy, either."
Economist Amir Sufi at the University of Chicago argued in July 2011 that a high level of household debt was holding back the U.S. economy. Households focused on paying down private debt are not able to consume at historical levels. He advocates mortgage write-downs and other debt-related solutions to re-invigorate the economy when household debt levels are exceptionally high. Several European countries also had high household debt levels relative to historical averages leading up to the European sovereign-debt crisis. Recent research also supports the view that excessive household leverage has contributed to the weakness in consumption.
Effects on employment
[edit]
Rana Foroohar wrote in July 2012: "[R]esearch shows that the majority of job losses in the U.S. since the Great Recession were due to lower consumer spending because of household debt, a decline that resulted in layoffs at U.S. firms. Parting back debt is the precursor to greater spending and greater growth."
Effects on economic recovery
[edit]
Neil Irwin explained the economic effects of rising consumer credit outstanding (i.e., all types of household debt other than mortgages) in July 2013: "Americans are finally feeling more confident about the economy and thus willing to take on debt. Lenders, meanwhile, are growing more comfortable extending loans. The spending enabled by this rising consumer debt can help create a virtuous cycle in which more demand for goods and services creates more jobs, which creates rising income. Indeed, more borrowing by households (and the spending that results) is likely offsetting some of the pain caused by federal spending cuts and deficit reduction."
Charts of U.S. household debt variables
[edit]
FRED – Household Debt to GDP Ratio
FRED - Mortgage and Consumer Credit Debt Relative to GDP
FRED - Household Debt Dollars (CMDEBT)
FRED - Change in Household Debt Dollars by Quarter (CMDEBT)
FRED – Household Debt Service Payments as a Percent of Disposable Personal Income (TDSP)
FRED – Household Financial Obligations as a Percent of Disposable Personal Income (FODSP)
Reducing household debt
[edit]
Household debt can be challenging to reduce. Major approaches include:
Paying down debt over time from income or accumulated savings, if available
Debt write-down or refinancing via negotiation, bankruptcy or government bailout
Inflation, although if wage increases do not exceed inflation, this approach is mostly academic
Paying off debt from income or savings
[edit]
Since most middle income households have the majority of their wealth invested in their home, paying down debt from income can take a decade or longer when a housing bubble bursts, as it did for much of the developed world in 2007–2012. For example, this debt accumulated over a 30-year period (1980–2007 peak) in the U.S. and much of the increase was mortgage-related.
Debt restructuring or write-down
[edit]
Debt can sometimes be reduced by negotiation with creditors or by a legal bankruptcy process, which can result in dismissal of certain types of debt such as credit cards. Some lenders may agree to write down mortgage values (reducing the homeowner's obligation) rather than taking even larger losses in foreclosure. Economist Joseph Stiglitz argued for a rapid bankruptcy process for homeowners, to allow debts to be written down.
Economists Joseph Stiglitz and Mark Zandi both advocated significant mortgage refinancing or write-downs in August 2012. They explained that more than four million Americans lost their homes since the housing bubble began bursting in 2006. An additional 3.5 million homeowners are in the foreclosure process or are so delinquent on payments that they will be soon. Roughly 13.5 million homeowners are underwater (in negative equity), meaning they owe more than their home is now worth, increasing the odds that millions more will lose their homes. The household debt and foreclosures are significantly holding back the economy.
Economists Kenneth S. Rogoff and Carmen M. Reinhart wrote in April 2013: "In the United States, we support reducing mortgage principal on homes that are underwater (where the mortgage is higher than the value of the home). We have also written about plausible solutions that involve moderately higher inflation and “financial repression” — pushing down inflation-adjusted interest rates, which effectively amounts to a tax on bondholders. This strategy contributed to the significant debt reductions that followed World War II.
Professor Luigi Zingales (University of Chicago) advocated a mortgage debt for equity swap in July 2009, where the mortgage debt would be written down in exchange for the bank taking an interest in future appreciation of the home upon sale (a debt-for-equity swap). Fund Manager John Hussman also advocated debt-for-equity swaps for households and banks during April 2009.
Journalist Steven Clemons wrote in July 2012: "The Great Recession of 2008–2009, from which there are still repercussions in the US and global economies, resulted from a massive surge in consumer loans—98% in just five years—and ... the combined total of US business and consumer loans is basically at the same level as at the moment this crisis hit." He argued for debt forgiveness or restructuring to help stimulate the economy.
During the Great Depression, the U.S. created the Home Owners' Loan Corporation (HOLC), which acquired and refinanced one million delinquent mortgages between 1933 and 1936.
Inflation
[edit]
If wages increase due to inflation, but debts remain fixed, the debts can be more easily repaid. Economists may advocate increasing inflation to help reduce the debt burden in highly leveraged economies. For example, economist Kenneth Rogoff has advocated both mortgage write-downs and inflation during August 2011: "I argued that the only practical way to shorten the coming period of painful deleveraging and slow growth would be a sustained burst of moderate inflation, say, 4–6% for several years. Of course, inflation is an unfair and arbitrary transfer of income from savers to debtors. But, at the end of the day, such a transfer is the most direct approach to faster recovery. Eventually, it will take place one way or another, anyway, as Europe is painfully learning." He also suggested that the government could take an equity interest in the homes in exchange for paying down the mortgages.
Household over-indebtedness
[edit]
While it is challenging to define over-indebtedness, definitions of over-indebtedness tend to have a few core elements in common. The first one refers to the capacity to meet the expenses associated with the contracted financial commitments. Over-indebtedness implies an inability to meet such recurring expenses. Second, this inability is structural. This is the time dimension, which holds that the definition must capture persistent and ongoing financial problems and exclude one-off occurrences that arise due to forgetfulness, for instance. The third core element refers to standard of living. The household must be unable to meet contracted commitments without reducing its minimum standard of living expenses. Fourth, comes illiquidity. The household is unable to remedy the situation by recourse to (financial and non-financial) assets and other financial sources such as credit. Lastly, all contracted financial commitments are included, among them mortgage and consumer credit commitments, utility and telephone bills as well as rent payments (recurring expenses).
Causes of over-indebtedness can be diverse. A 'risky life event' (for example, unemployment, relationship break-up, leaving the parental home, business failure, illness or unexpected home repairs. Such events can trigger income and expenditure shocks) can lie at the root, sometimes instantly turning indebtedness into over-indebtedness. Other households have unconsciously, or consciously, gradually become over-committed. They make use of available forms of credit, sometimes without realizing that they might not be able to repay in the future. Lack of financial management skills and aggressive marketing by lenders may both lie at the origin of this. Another potential cause includes escalating consumption habits. The third group of people are the least well-off. They need to obtain credit in order to attain a reasonable standard of living. They are frequently subject to relatively high interest rates. Potential consequences for the over-indebted household include financial hardship, depression, poor health, relationship breakdown, exclusion from basic financial services, a strain on social relations, absenteeism at work and lack of organizational commitment, a feeling of insecurity.
Responses to household over-indebtedness can be preventive, alleviative and rehabilitative. Preventive measures include financial education and regulation. An example of the last is the European Union's "Directive on Credit Agreements for Consumers Household debt advisory services". It stipulates, for example, that lenders should list the interest rates they change in a homogenized way (yearly rates) and that paper contracts should be signed for debts above a certain amounts, thus inhibiting for example impulsive borrowing through SMS-loans. Alleviative measures include debt advisory services, which aim to help households getting their finances back on track, mainly by means of information provision, budget planning and balancing, help with legal arrangements, negotiation with creditors, providing psychological support by having someone to talk to, and even by effectively, voluntarily taking over the managing of a household's finances. Rehabilitative measures include consumer bankruptcy and legal debt settlement procedures. While precise arrangements vary largely between countries, in general such procedures work as follows. During such procedures, the over-indebted household hands over all income above a minimum threshold to the creditors/state and is cleared of its debts after the period, varying in length from for example 1 year in the UK to 5 in Portugal and 12 in Ireland.
See also
[edit]
Net worth
References
[edit]
^NYT-Paul Krugman-Block those Metaphors-December 2010
^ Jump up to: abIMF-Report Extract Chapter 3 – April 2012
^ Jump up to: abMian, Atif and, Sufi, Amir (2014). House of Debt. University of Chicago. ISBN978-0-226-08194-6.{{cite book}}: CS1 maint: multiple names: authors list (link)
^U.S. Federal Reserve-Household Debt Service and Financial Obligations Ratios
^Federal Reserve Bank of Chicago, Welfare Implications of the Transition to High Household Debt, November 2006
^"10 Years after the Great Recession, New Threats Have Appeared". Analytics: Seeing Trends in the Data. Fortune (Paper periodical). Graphics by Nicolas Rapp: 11. 1 October 2018.{{cite journal}}: CS1 maint: others (link)
^NYT-Rise in Household Debt Might Be Sign of a Strengthening Recovery-Annie Lowrey-October 2012
^"Average American Debt in 2024: Household Debt Statistics". Business Insider.
^"McKinsey-Debt and deleveraging: The global credit bubble and its economic consequences-Updated-July 2011". Archived from the original on 2013-03-07. Retrieved 2012-07-13.
^Comelli, Martino (2021). "The impact of welfare on household debt". Sociological Spectrum. 41 (2): 154–176. doi:10.1080/02732173.2021.1875088. hdl:20.500.14018/13843.
^"Household debt, loans and debt securities". IMF.
^Fred Database-CMDEBT Variable-Retrieved July 2012
^FRED-Household Debt as Percent of GDP-Retrieved September 2013
^"U.S. Federal Reserve-FRED Database-Retrieved July 2012". Archived from the original on 2012-02-12. Retrieved 2012-07-12.
^Federal Reserve Database-Consumer Credit as% GDP-Retrieved July 2013
^Project Syndicate-Joseph Stiglitz-To Cure the Economy-October 2011
^NYT=-Paul Krugman-Block those Economic Metaphors-December 2010
^"Federal Reserve-Janet Yellen-A Minsky Meltdown-April 2009". Archived from the original on 2013-01-05. Retrieved 2012-07-12.
^Washington Post-Ezra Klein-It's the Household Debt, Stupid-August 2011
^Bloomberg-Household Debt is at Heart of Weak U.S. Economy-July 2011
^Brookings Institution, Is a Household Debt Overhang Holding Back Consumption?, Spring 2012Archived 2012-10-31 at the Wayback Machine
^Time Magazine-Foroohar-Digging out of the Debt Hole-July 2012
^Washington Post-Neil Irwin-Consumer Debt is Soaring-July 2013
^NYT-Joe Nocera-November 2011
^Justice for Some Project Syndicate-Joseph Stiglitz-Justice for Some-November 2010
^NYT-Stiglitz and Zandi-The One Housing Solution Left-Mass Mortgage Refinancing-August 2012
^NYT-Reinhart and Rogoff-Debt, Growth and the Austerity Debate-April 2013
^"Time For A New Mortgage Plan: Debt-For-Equity Swaps-July 2009". Archived from the original on 2013-01-18. Retrieved 2012-07-12.
^Hussman Funds Newsletter-John Hussman-April 2009
^The Atlantic-Economic Growth Idea: Forgive or Restructure Debt U.S. Citizens Hold-Clemons-July 2012
^The Atlantic-Clemons and Vague-How to Predict the Next Financial Crisis-July 2012
^Federal Reserve Bank of St. Louis-The Federal Response to Home Mortgage Distress: Lessons from the Great Depression-June 2008
^Project Syndicate-Kenneth Rogoff-The Second Great Contraction-August 2011
^European Commission 2008 Towards a common operational European definition of over-indebtedness
^Eurofound 2010 Managing household debts: Social service provision in the EU
^Eurofound 2012 Household debt advisory services in the European Union
Further reading
[edit]
Federal Reserve Bank of New York - Quarterly Report on Household Credit and Debt
McKinsey & Company – Debt and Deleveraging – 2012 Update
How to Predict the Next Financial Crisis-Clemons and Vague-July 2012
Mian and Sufi – What Explains High Unemployment? The Deleveraging – Aggregate Demand Hypothesis-October 2011
Krugman and Eggertsson – Debt, Deleveraging, and the Liquidity Trap: A Fisher‐Minsky‐Koo approach – Revised February 2011
Haliassos, Michael. Household Portfolios. Second Edition Ed. Steven Durlauf and Lawrence Blume. Palgrave Macmillan,2008 The New Palgrave Dictionary of EconomicsOnline..Palgrave Macmillan. 9 Sept. 2011. .
Bertola, Giuseppe, Richard Disney, and Charles Grant. The Economics of Consumer Credit. Cambridge: The MIT Press, 2006. N. pag. Print.
Agarwal, Sumit, and Brent W. Ambrose. Household Credit Usage. New York: Palgrave Macmillan, 2007. N. pag. Print.
Olney, Martha L. Buy Now Pay Later. Chapel Hill Ans London: The North Carolina Press, 1991. Print.
Schutz, Howard G., Pamela C. Baird, and Glenn R. Hawkes. Lifestyles and Consumer Behavior of Older Americans. New York: Praeger Publishing, 1979. Print.
Kasser, Tim, and Allen D. Kanner. Psychology and Consumer Culture. Washington, DC: American Psychological Association, 2004. Print.
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5: Discrete Probability Distributions
Mostly Harmless Statistics (Webb)
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5.5: Hypergeometric Distribution
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Contents
1. Home
2. Bookshelves
3. Introductory Statistics
4. Mostly Harmless Statistics (Webb)
5. 5: Discrete Probability Distributions
6. 5.5: Hypergeometric Distribution
Expand/collapse global location
Mostly Harmless Statistics (Webb)
Front Matter
1: Introduction to Data
2: Organizing Data
3: Descriptive Statistics
4: Probability
5: Discrete Probability Distributions
6: Continuous Probability Distributions
7: Confidence Intervals for One Population
8: Hypothesis Tests for One Population
9: Hypothesis Tests and Confidence Intervals for Two Populations
10: Chi-Square Tests
11: Analysis of Variance
12: Correlation and Regression
13: Nonparametric Tests
Back Matter
5.5: Hypergeometric Distribution
Last updated Mar 12, 2023
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5.6: Poisson Distribution
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Rachel Webb
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1.
1. Solution
Solution
Solution
Solution
The hypergeometric distribution is a discrete
probability distribution
used to find the probability of success when there are two
outcomes
to each trial, and there are a set number of dependent trials. A hypergeometric
probability distribution
results from a random
experiment
that meets all of the following requirements. 1 2
1. The procedure has a fixed sample size n taken from a population N.
2. The trials are taken without replacement (dependent).
3. Each trial has exactly two outcomes, the number of successes = a and the number of failures = b. Note a + b = N.
If a random
experiment
satisfies all of the above, the distribution of the
random variable
X, where X counts the number of successes, is called a hypergeometric distribution, we write X ~ H(n, a, N).
The hypergeometric distribution is P(X = x) = aC x⋅bC n−x NC n, x = 0, 1, 2, … , n or a, whichever is smaller. Where n is the
sample
size taken without replacement from the
population
size N, x is the number of successes out of n that you are trying to find the probability for, a is the number of a successes and b = N – a is the number of failures out of the
population
N.
A committee of 5 people is to be formed from 15 volunteers and 6 appointed positions. Compute the probability that the committee will consist of 4 randomly selected volunteers.
Solution
There are a total of N= 15 + 6 = 21 people. A
sample
of n = 5 people are selected without replacement. The success is choosing a volunteer, so a = 15. Of the 15 volunteers, we are asked to find the probability of selecting x = 4. Substitute in each value: P(X = 4) = 15C 4⋅6C 1 21C 5=0.4025.
There is a pattern to the numbers in the formula. The top two numbers on the left of the C’s add up to the bottom number on the left of C, and the top two numbers on the right of the C’s add up to the bottom number on the right of C.
The waiting room for jury selection has 40 people with a college degree and 30 people without. A jury of 12 is selected from this waiting room. What is the probability that exactly 5 of the jury members will have a college degree?
Solution
There are a total of N = 40 + 30 = 70 people. A
sample
of n = 12 people are selected without replacement. The success is choosing someone with a college degree, so a = 40. Of the 12 jurors we are asked to find the probability of selecting x= 5.
Substitute in each value: P(X = 5) = 40C 5⋅30C 7 70C 12=0.1259.
A wallet contains three 100billsandfive 1 bills. You randomly choose four bills. What is the probability that you will choose exactly two $100 bills?
Solution
There are a total of N = 3 + 5 = 8 bills. A
sample
of n= 4 bills are selected without replacement. The success is choosing a $100 bill, so a = 3. Of the 3 $100 bills we are asked to find the probability of selecting x = 2. Substitute in each value: P(X = 2) = 3C 2⋅5C 2 8C 4=0.4286.
Note: the calculator does not have the hypergeometric distribution shortcut. The following online calculator will calculate the probability: The M = a is the number of successes a from our example, or use Excel.
Figure 5-7 may help you decide when to use the “True” for cumulative in Excel. Figure 5-7
A bookstore owner examines 5 books from each lot of 25 to check for missing pages. If they find at least 2 books with missing pages, the entire lot is returned. If, indeed, there are 4 books with missing pages, find the probability that the lot will be returned.
###### Solution Sometimes the hardest part is determining which of the numbers goes in the right place in the formula. The population size is always the largest number, N = 25. A success is what you are trying to find the probability for, which is getting a book with a missing page. The number of successes out of the population is a = 4 (the 4 books with the missing pages). This means b = 25 – 4 = 21 (the number of books without missing pages). Note a + b = N. The sample size is the n = 5 is the number of books the owner examines. The owner will send the lot back if X is at least 2 missing pages. We can stop at X = 4 since there are only 4 books with missing pages. Your stopping point will be the smaller of n or a. P(X ≥ 2) = P(X = 2) + P ( X = 3) + P(X = 4) = 4C 2⋅21C 3 25C 5+4C 3⋅21C 2 25C 5+4C 4⋅21C 1 25C 5=0.1502+0.0158+0.0004=0.1664. In Excel use the formula: =1- HYPGEOM.DIST(1,5,4,25,TRUE).
Figure 5-8 shows the online calculator at
Figure 5-8
This page titled 5.5: Hypergeometric Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Rachel Webb via source content that was edited to the style and standards of the LibreTexts platform.
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Finding the equation of a circle given two points on the circle
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11. Find the equation of the circle which touches $x^{2} + y^{2} - 6x + 2y + 5 = 0$ at $(4, -3)$ and passes through $(0, 7)$.
My textbook has a worked example for obtaining the equation of a circle from three points on the circle. It also talks you through obtaining the equation of a circle if you're given two points on the circle and if it touches an axis (in this case you know a coordinate of the centre will be $\pm r$, where $r$ is your radius.)
I am reasonably well-practised at using these two techniques. I have also been practising finding the length of a tangent from a given point.
However, I have so far been unable to make the "jump" to this question, I suspect there's something I'm not seeing. So I was hoping for a hint that would help me work out how to approach this question.
circles
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asked Apr 27, 2015 at 21:48
user86389user86389
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Lets say the equation of the circle is given by:$$(x-a)^2+(y-b)^2=r^2$$Make use of the fact that the circle passes through $(4,-3)$ and $(0,7)$ to form two equations.
You are also told that it touches the circle:$$x^{2} + y^{2} - 6x + 2y + 5 = 0$$at $(4,-3)$ which mean the tangents of both circles at this point must be equal. This gives you a third equation.
You now have three equations and three unknowns which you should be able to solve.
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answered Apr 27, 2015 at 22:04
MufasaMufasa
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$\begingroup$ So we have $(4 - a)^{2} + (-3 - b)^{2} = r^{2}$, $a^{2} + (7 - b)^{2} = r^{2}$ and, unless I'm much mistaken, we also have the equation of the tangent to the circle at $(4, -3)$, which is $4y - 7x + 40 = 0$. But, I'm sorry, I'm not sure how to go from here to three valid equations pertaining to our circle in $a$, $b$ and $r$. $\endgroup$
user86389
– user86389
2015-04-27 22:24:43 +00:00
Commented Apr 27, 2015 at 22:24
$\begingroup$ Use implicit differentiation to work out the slope of the tangent line at $(4,-3)$ for both circles. These slopes must be equal - this gives you your third equation. $\endgroup$
Mufasa
– Mufasa
2015-04-27 22:30:59 +00:00
Commented Apr 27, 2015 at 22:30
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$\begingroup$
let the circle $$ (x-3)^2+ (y+1)^2 = 5$$ has a tangent at $(4, -3)$ with the circle $C.$ the line connecting the centers of the circle has slope $$\frac{-1-(-3)}{3-4} = -2$$.therefore the center of $C$ is at $(3+t, -1-2t)$ for some $t.$
since both $(4,-3), (0,7)$ are on $C,$ equating the radius squared, we have $$(3+t-4)^2 (-1-2t+3)^2 =(3+t-0)^2 +(-1-2t-7)^2 $$ that gives $$-8(3+t)+16 + -6(1+2t)+9=14(1+2t)+49\to 48t=-68, t = -17/7 $$
$$\text{ center of } C \text{ is }(4/7, 27/7), \text{ radius is } 7.666$$
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edited Apr 27, 2015 at 23:40
answered Apr 27, 2015 at 22:49
abelabel
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$\begingroup$ Sounds good, but - forgive me - why $(3 + t, -1 - 2t)$. I understand that the centre of $(x - 3)^{2} + (y + 1)^{2} = 5$ (the given circle) is $(3, -1)$, but why do we add $t$ to the $x$-coordinate and take $2t$ from the $y$-coordinate, I don't think I've followed. $\endgroup$
user86389
– user86389
2015-04-27 23:14:16 +00:00
Commented Apr 27, 2015 at 23:14
$\begingroup$ @Au101, the reason is the the centers and the common point of contact are collinear. please see my edit. $\endgroup$
abel
– abel
2015-04-27 23:41:48 +00:00
Commented Apr 27, 2015 at 23:41
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0
$\begingroup$
I've come back to this question after a while and have found a solution which agrees with that in the textbook. My method is based primarily on the tips given by @Mufasa.
Let the circle which touches $x^{2} + y^{2} 6x + 2y + 5 = 0$ at $(4,3)$ and passes through $(0,7)$ be $C_{1}$.
Let the centre of $C_{1}$ be $(p, q)$ and let the radius be $a$.
$(4 - p)^{2} + (-3 - q)^{2} = a^{2}$
$p^{2} + (7 - q)^{2} = a^{2}$
Let the circle $x^{2} + y^{2} 6x + 2y + 5 = 0$ be $C_{2}$.
The centre of $C_{2}$ is $(3, -1)$.
The gradient of the radius of $C_{2}$ to $(4, -3)$ is $-2$.
$\therefore$ The gradient of the tangent to $C_{2}$ at $(4, -3)$ is $\frac{1}{2}$.
$\therefore$ The gradient of the tangent to $C_{1}$ at $(4, -3)$ is $\frac{1}{2}$.
$\therefore$ The gradient of the radius of $C_{1}$ to $(4, -3)$ is $-2$.
$\therefore \dfrac{q - (-3)}{p - 4} = -2$
$2p + q - 5 = 0 \qquad (1)$
$(4 - p)^{2} + (-3 - q)^{2} = a^{2} \qquad (2)$
$p^{2} + (7 - q)^{2} = a^{2} \qquad (3)$
Equating (2) and (3), we ultimately get:
$p = \dfrac{5}{2}q - 3 \qquad (4)$
Substituting $p$ into (1), we get:
$q = \dfrac{11}{6}$
Substituting $q$ into (4), we get:
$p = \dfrac{19}{12}$
Substituting $p$ and $q$ into (3) we ultimately get:
$a^{2} = \dfrac{4,205}{144}$
This gives us our equation of $C_{1}$:
$$\left(x - \frac{19}{12}\right)^{2} + \left(y - \frac{11}{6}\right)^{2} = \frac{4,205}{144}$$
Expanding and simplifying, we get:
$$6x^{2} + 6y^{2} - 19x - 22y - 140 = 0.$$
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answered May 11, 2015 at 0:03
user86389user86389
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Squaring a binomial has a few steps. Square the first term, square the last term, then double the product of the first and last terms.
Explanations (3)
Hannah Bonville
Text
7
Squaring Binomials
A binomial is an expression composed of two monomials (hence the prefix "bi") that are connected by either a plus sign or a minus sign.
So, how do we square a binomial? Well, we've got a couple of options:
Expand and use FOIL
Use the formula a2±2ab+b2
Squaring Binomials Using FOIL
Image source: by Caroline Kulczycky
Suppose that we are tasked with squaring the binomial 3x−4. In order to do this, we must first enclose that binomial in parentheses and give it the appropriate exponent. We want to square the binomial, which is the same as raising it to the second power, thereby creating the following expression: (3x−4)2.
Once we've done that, it's time for us to go ahead and expand our new expression.
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Multiplying a Difference of Squares - Definition & Examples
Squaring Multiples of 5 - Examples & Practice
Multiply Binomials with FOIL Method - Examples & Practice
Multiplying Polynomials using the Distributive Property - Process & Examples
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Alex Federspiel
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(Videos) Squaring a Binomial
by mathman1024
This video by mathman1024 shows you how to square a binomial.
Summary
When you see a binomial squared, (a+b)2, your first thought might be to square each term inside, a2+b2, similar to what we do with radicals. This is incorrect. Remember that squaring something is a short hand way to represent that thing being multiplied by itself. So,
(a+b)2=(a+b)(a+b)
We can solve this using the FOIL method and combining like terms to see that the answer is in fact different from just squaring the inside terms.
(a+b)(a+b)F O I La2+ab+ba+b2a2+2ab+b2
So, as you see, there's actually an extra term in the middle, 2ab. We can actually use this as a formula now. Instead of FOILing we can square each term and then double the product of them. Also, when it's subtraction we just subtract the middle term.
(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2
Let's try this out with an example now: (x+5)2.
(x+5)2(x)2+2(x)(5)+(5)2x2+10x+25
Try these next ones on yourself!
(x+7)2
(x−6)2
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Kevin Chang
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Squaring Binomials With Distribution
How do we square binomials? One way is to multiply a binomial by itself using a table, but there are even easier ways! Let's look at an example problem that tasks us with finding the square of the binomial 3x+7:
(3x+7)2=?
When dealing with problems like these, we can expand the initial expression by using the distributive property:
(3x+7)2=(3x+7)(3x+7)
=3x(3x+7)+7(3x+7)
=9x2+21x+21x+49
=9x2+42x+49
Now, let's try another example.
Expand (4x−2y)2.
Again, we expand the expression by using distribution:
(4x−2y)2=(4x−2y)(4x−2y)
=4x(4x−2y)−2y(4x−2y)
=16x2−8xy−8xy+4y2
=16x2−16xy+4y2
At this point, you may be wondering whether or not we could be using a formula to square binomials instead. While such a formula does exist, I prefer not to commit it to memory. After a lot of practice, this whole procedure becomes automatic, and a formula is no longer needed.
In order to derive this formula, we'll be expanding the expression (a+b)2, where a and b stand for two arbitrary terms.
On your own, try using the distributive property to come up with the expanded version of (a+b)2.
Squared Binomial Formula
When we expand the expression (a+b)2, we get:
(a+b)2=(a+b)(a+b)
=a(a+b)+b(a+b)
=a2+ab+ab+b2
=a2+2ab+b2
This is our formula. Now, let's show that this formula works by testing it out on one of our previous example expressions: (4x−2y)2. First, we'll let a=4x (since 4x is our binomial's first term), and we'll let b=−y (since −y is our binomial's last term). Then, we plug those terms into our formula and simplify:
(4x−2y)2=(4x)2+2(4x)(−2y)+(−2y)2
=16x2+(−16xy)+4y2
=16x2−16xy+4y2
Notice that this is the same exact answer that we got when we previously used the distributive property to expand (4x−2y)2.
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Top Questions
What is an oxide?
How are oxides formed?
What are the different types of oxides?
What is the difference between a metal oxide and a non-metal oxide?
How do oxides react with water?
What are some common examples of oxides found in everyday life?
Why are oxides important in industrial processes?
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oxide, any of a large and important class of chemical compounds in which oxygen is combined with another element. With the exception of the lighter inert gases (helium [He], neon [Ne], argon [Ar], and krypton [Kr]), oxygen (O) forms at least one binary oxide with each of the elements.
Both metals and nonmetals can attain their highest oxidation states (i.e., donate their maximum number of available valence electrons) in compounds with oxygen. The alkali metals and the alkaline earth metals, as well as the transition metals and the posttransition metals (in their lower oxidation states), form ionic oxides—i.e., compounds that contain the O2− anion. Metals with high oxidation states form oxides whose bonds have a more covalent nature. Nonmetals also form covalent oxides, which are usually molecular in character. A smooth variation from ionic to covalent in the type of bonding in oxides is observed as the periodic table is traversed from the metals on the left to the nonmetals on the right. This same variation is observed in the reaction of oxides with water and the resulting acid-base character of the products. Ionic metal oxides react with water to give hydroxides (compounds containing the OH− ion) and resultant basic solutions, whereas most nonmetal oxides react with water to form acids and resultant acidic solutions (see the table).
Periodic variation of the properties of oxides of the elements of the third period
| | group 1 | group 2 | group 13 | group 14 | group 15 | group 16 | group 17 |
| Source: From W. Robinson, J. Odom, and H. Holtzclaw, Jr., Chemistry: Concepts and Models, D.C. Heath and Co., 1992. | | | | | | | |
| reaction of oxides with water and the acid-base character of hydroxides | Na2O gives NaOH (strong base) | MgO gives Mg(OH)2 (weak base) | Al2O3 nonreacting | SiO2 nonreacting | P4O10 gives H3PO4 (weak acid) | SO3 gives H2SO4 (strong acid) | Cl2O7 gives HClO4 (strong acid) |
| bonding in oxides | Na2O ionic | MgO ionic | Al2O3 ionic | SiO2 covalent | P4O10 covalent | SO3 covalent | Cl2O7 covalent |
Certain organic compounds react with oxygen or other oxidizing agents to produce substances called oxides. Thus, amines, phosphines, and sulfides form amine oxides, phosphine oxides, and sulfoxides, respectively, in which the oxygen atom is covalently bonded to the nitrogen, phosphorus, or sulfur atom. The so-called olefin oxides are cyclic ethers.
Metal oxides
Metal oxides are crystalline solids that contain a metal cation and an oxide anion. They typically react with water to form bases or with acids to form salts.
The alkali metals and alkaline earth metals form three different types of binary oxygen compounds: (1) oxides, containing oxide ions, O2−, (2) peroxides, containing peroxide ions, O22−, which contain oxygen-oxygen covalent single bonds, and (3) superoxides, containing superoxide ions, O2−, which also have oxygen-oxygen covalent bonds but with one fewer negative charge than peroxide ions. Alkali metals (which have a +1 oxidation state) form oxides, M2O, peroxides, M2O2, and superoxides, MO2. (M represents a metal atom.) The alkaline earth metals (with a +2 oxidation state) form only oxides, MO, and peroxides, MO2. All the alkali metal oxides can be prepared by heating the corresponding metal nitrate with the elemental metal.
2MNO3 + 10M + heat → 6M2O + N2
A general preparation of the alkaline earth oxides involves heating the metal carbonates.
MCO3 + heat → MO + CO2
Both alkali metal oxides and alkaline earth metal oxides are ionic and react with water to form basic solutions of the metal hydroxide.
M2O + H2O → 2MOH (where M = group 1 metal)
MO + H2O → M(OH)2 (where M = group 2 metal)
Thus, these compounds are often called basic oxides. In accord with their basic behaviour, they react with acids in typical acid-base reactions to produce salts and water; for example,
M2O + 2HCl → 2MCl + H2O (where M = group 1 metal).
These reactions are also often called neutralization reactions. The most important basic oxides are magnesium oxide (MgO), a good thermal conductor and electrical insulator that is used in firebrick and thermal insulation, and calcium oxide (CaO), also called quicklime or lime, used extensively in the steel industry and in water purification.
Periodic trends of the oxides have been thoroughly studied. In any given period, the bonding in oxides progresses from ionic to covalent, and their acid-base character goes from strongly basic through weakly basic, amphoteric, weakly acidic, and finally strongly acidic. In general, basicity increases down a group (e.g., in the alkaline earth oxides, BeO < MgO < CaO < SrO < BaO). Acidity increases with increasing oxidation number of the element. For example, of the five oxides of manganese, MnO (in which manganese has an oxidation state of +2) is the least acidic and Mn2O7 (which contains Mn7+) the most acidic. Oxides of the transition metals with oxidation numbers of +1, +2, and +3 are ionic compounds consisting of metal ions and oxide ions. Those transition metal oxides with oxidation numbers +4, +5, +6, and +7 behave as covalent compounds containing covalent metal-oxygen bonds. As a general rule, the ionic transition metal oxides are basic. That is, they will react with aqueous acids to form solutions of salts and water; for example,
CoO + 2H3O+ → Co2+ + 3H2O.
The oxides with oxidation numbers of +5, +6, and +7 are acidic and react with solutions of hydroxide to form salts and water; for example,
CrO3 + 2OH- → CrO42− + H2O.
Those oxides with +4 oxidation numbers are generally amphoteric (from Greek amphoteros, “in both ways”), meaning that these compounds can behave either as acids or as bases. Amphoteric oxides dissolve not only in acidic solutions but also in basic solutions. For example, vanadium oxide (VO2) is an amphoteric oxide, dissolving in acid to give the blue vanadyl ion, [VO]2+, and in base to yield the yellow-brown hypovanadate ion, [V4O9]2−. Amphoterism among the main group oxides is primarily found with the metalloidal elements or their close neighbours.
Nonmetal oxides
All nonmetals form covalent oxides with oxygen, which react with water to form acids or with bases to form salts. Most nonmetal oxides are acidic and form oxyacids, which in turn yield hydronium ions (H3O+) in aqueous solution. There are two general statements that describe the behaviour of acidic oxides. First, oxides such as sulfur trioxide (SO3) and dinitrogen pentoxide (N2O5), in which the nonmetal exhibits one of its common oxidation numbers, are known as acid anhydrides. These oxides react with water to form oxyacids, with no change in the oxidation number of the nonmetal; for example,
N2O5 + H2O → 2HNO3.
Second, those oxides in which the metal does not exhibit one of its common oxidation numbers, such as nitrogen dioxide (NO2) and chlorine dioxide (ClO2), also react with water. In these reactions, however, the nonmetal is both oxidized and reduced (i.e., its oxidation number is increased and decreased, respectively). A reaction in which the same element is both oxidized and reduced is called a disproportionation reaction. In the following disproportionation reaction, N4+ is reduced to N2+ (in NO) and oxidized to N5+ (in HNO3).
3NO2 + H2O → 2HNO3 + NO
Oxides of nitrogen
Click Here to see full-size tableNitrogen (N) forms oxides in which nitrogen exhibits each of its positive oxidation numbers from +1 to +5. Nitrous oxide (dinitrogen oxide), N2O, is formed when ammonium nitrate, NH4NO3, is heated. This oxide, which is a colourless gas with a mild, pleasant odour and a sweet taste, is used as an anesthetic for minor operations, especially in dentistry. It is called laughing gas because of its intoxicating effect. It is also widely used as a propellant in aerosol cans of whipped cream. Nitric oxide, NO, can be created in several ways. The lightning that occurs during thunderstorms brings about the direct union of nitrogen and oxygen in the air to produce small amounts of nitric oxide, as does heating the two elements together. Commercially, nitric oxide is produced by burning ammonia (NH3), whereas in the laboratory it can be produced by the reduction of dilute nitric acid (HNO3) with, for example, copper (Cu).
3Cu + 8HNO3 → 2NO + 3Cu(NO3)2 + 4H2O
Gaseous nitric oxide is the most thermally stable oxide of nitrogen and is also the simplest known thermally stable paramagnetic molecule—i.e., a molecule with an unpaired electron. It is one of the environmental pollutants generated by internal-combustion engines, resulting from the reaction of nitrogen and oxygen in the air during the combustion process. At room temperature nitric oxide is a colourless gas consisting of diatomic molecules. However, because of the unpaired electron, two molecules can combine to form a dimer by coupling their unpaired electrons.
2NO ⇌ N2O2
Thus, liquid nitric oxide is partially dimerized, and the solid consists solely of dimers.
When a mixture of equal parts of nitric oxide and nitrogen dioxide, NO2, is cooled to −21 °C (−6 °F), the gases form dinitrogen trioxide, a blue liquid consisting of N2O3 molecules. This molecule exists only in the liquid and solid states. When heated, it forms a mixture of NO and NO2. Nitrogen dioxide is prepared commercially by oxidizing NO with air, but it can be prepared in the laboratory by heating the nitrate of a heavy metal, as in the following equation,
2Pb(NO3)2 + heat → 2PbO + 4NO2 + O2,
or by adding copper metal to concentrated nitric acid. Like nitric oxide, the nitrogen dioxide molecule is paramagnetic. Its unpaired electron is responsible for its colour and its dimerization. At low pressures or at high temperatures, NO2 has a deep brown colour, but at low temperatures the colour almost completely disappears as NO2 dimerizes to form dinitrogen tetroxide, N2O4. At room temperature an equilibrium between the two molecules exists.
2NO2 ⇌ N2O4
Dinitrogen pentoxide, N2O5, is a white solid formed by the dehydration of nitric acid by phosphorus(V) oxide.
P4O10 + 4HNO3 → 4HPO3 + 2N2O5
Above room temperature N2O5 is unstable and decomposes to N2O4 and O2. Two oxides of nitrogen are acid anhydrides; that is, they react with water to form nitrogen-containing oxyacids. Dinitrogen trioxide is the anhydride of nitrous acid, HNO2, and dinitrogen pentoxide is the anhydride of nitric acid, HNO3.
N2O3 + H2O → 2HNO2
N2O5 + H2O → 2HNO3
There are no stable oxyacids containing nitrogen with an oxidation number of +4.
Nitrogen dioxide reacts with water in one of two ways. In cold water NO2 disproportionates to form a mixture of HNO2 and HNO3, whereas at higher temperatures HNO3 and NO are formed. In their chemical activity, the nitrogen oxides undergo extensive oxidation-reduction reactions. Nitrous oxide resembles oxygen in its behaviour when heated with combustible materials. It is a strong oxidizing agent that decomposes upon heating to form nitrogen and oxygen. Because one-third of the gas liberated is oxygen, nitrous oxide supports combustion better than air. All the nitrogen oxides are, in fact, good oxidizing agents. Dinitrogen pentoxide reacts violently with metals, nonmetals, and organic materials, as in the following reactions with potassium (K) and iodine gas (I2).
N2O5 + K → KNO3 + NO2
N2O5 + I2 → I2O5 + N2
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1082 | https://arxiv.org/pdf/1406.4012 | arXiv:1406.4012v2 [math.OC] 16 Jul 2014
ACCELERATING THE ALTERNATING PROJECTION ALGORITHM FOR THE CASE OF AFFINE SUBSPACES USING SUPPORTING HYPERPLANES
C.H. JEFFREY PANG
Abstract. The von Neumann-Halperin method of alternating projections con-verges strongly to the projection of a given point onto the intersection of finitely many closed affine subspaces. We propose acceleration schemes making use of two ideas: Firstly, each projection onto an affine subspace identifies a hyperplane of codimension 1 containing the intersection, and secondly, it is easy to project onto a finite intersection of such hyperplanes. We give conditions for which our ac-celerations converge strongly. Finally, we perform numerical experiments to show that these accelerations perform well for a matrix model updating problem.
Contents
Introduction 12. Algorithms 33. Strong convergence results 54. Performance of acceleration 12 5. Conclusion 16 References 16 1. Introduction
Let X be a (real) Hilbert space, and let M1, M2, . . . , Mk be a finite number of closed linear subspaces with M := ∩kl=1 Ml. For any closed subspace N of X, let
PN denote the orthogonal projection onto N . The von Neumann-Halperin method of alternating projections, or MAP for short, is an iterative algorithm for determining the best approximation PM x, the projection of x onto M . We recall their theorem on the strong convergence of the MAP below.
Theorem 1.1. (von Neumann [vN50] for k = 2 , Halperin [Hal62] for k ≥ 2) Let M1,
M2, . . . , Mk be closed subspaces in the Hilbert space X and let M := ∩kl=1 Ml. Then
lim
n→∞
‖(PMk PMk−1 · · · PM1 )nx − PM x‖ = 0 for all x ∈ X. (1.1) In the case where k = 2 , this result was rediscovered numerous times. The method of alternating projections, as suggested in the formula (1.1), guarantees convergence to the projection PM x, but the convergence is slow in practice. Various
Date : July 17, 2014. 2010 Mathematics Subject Classification. 11D04, 90C59, 47J25, 47A46, 47A50, 52A20, 41A50.
Key words and phrases. best approximation problem, alternating projections, supporting hyperplanes.
1PROJECTION ONTO INTERSECTION OF AFFINE SPACES 2
acceleration schemes have been studied in [GPR67, GK89, BDHP03]. An identity for the convergence of the method of alternating projections in the case of linear subspaces is presented in [XZ02]. We remark that the Boyle-Dykstra Theorem [BD85] generalizes the strong conver-gence to the projection in Theorem 1.1 to Dykstra’s algorithm [Dyk83], where the Ml
do not have to be linear subspaces. When the sets Ml are not linear subspaces, a simple example using a halfspace and a line in R2 shows that the method of alternating projections may not converge to the projection PM x. Nevertheless, the method of alternating projections is still useful for the SIP (Set Intersection Problem). When M1, M2, . . . , Mk is a finite number of closed (not necessarily convex) subsets of a Hilbert space X, the SIP is the problem of finding a point in M := ∩kl=1 Ml, i.e., (SIP): Find x ∈ M :=
k
⋂
l=1
Ml, where M 6 = ∅. (1.2) An acceleration of the method of alternating projections for the case where each Ml were closed convex sets (but not necessarily subspaces) was studied in [Pan14b] and improved in [Pan14a]. The idea there, named as the SHQP strategy (Supporting Halfspaces and Quadratic Programming) was to store each of the halfspace produced by the projection process, and use quadratic programming to project onto an intersection of a reasonable number of these halfspaces. In the particular case of affine spaces, the SHQP strategy is even easier to state and implement: Consider an affine space M1 of a Hilbert space X. First, the projection of a point x0 onto M1 identifies the hyperplane of codimension 1
{x : 〈x0 − PM1 x0, x 〉 = 〈x0 − PM1 x0, P M1 x0〉} (1.3) as a superset of M1. Next, it is easy to project any point onto the intersection of finitely many hyperplanes of the form (1.3). A problem with many similarities but separate considerations and techniques is that of [NT14]. In that paper, a randomized block Kaczmarz method is analyzed. 1.1. Contributions of this paper. The techniques of [Pan14b] gives additional as-sumptions so that the SHQP strategy converges weakly to PM (x). The question we ask in this paper is whether the SHQP strategy converges strongly to the projection
PM x in the case when M := ∩ki=1 Mi and M1, M2, . . . , Mk is a finite number of closed affine subspaces like in the von Neumann-Halperin Theorem. We propose Algorithm 2.1, which is based on a naive implementation of the SHQP strategy. Based on the additional structure of affine spaces, we propose Algorithm 2.5, which is effective when one of the affine subspaces is easy to project onto. We prove that Algorithms 2.1 and 2.5 converge strongly to PM (x) under some assumptions in Section 3. We also give reasons (Example 3.14) to explain why these additional conditions cannot be removed. Our proof is adapted from the proof of the Boyle-Dykstra Theorem [BD85] on the strong convergence of Dykstra’s Algorithm [Dyk83] in the manner presented in [ER11]. Next, we examine an implementation of our acceleration on a Matrix Model Updating Problem (MMUP) from [ER11, Section 6.2], who in turn cited [DS01, MDR09]. The numerical experiments show the effectiveness of our algorithms. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 3
1.2. Notation. We shall assume that X is a Hilbert space with the inner product 〈· , ·〉
and norm ‖ · ‖ .2. Algorithms
In this section, we propose Algorithms 2.1 and 2.5 that seek to find the projection of a point onto the intersection of a finite number of closed linear subspaces. It is clear to see that our algorithms apply for affine spaces with nonempty intersection as well (a fact we use in our experiments in Section 4), since a translation can reduce the problem to involving only linear subspaces. We begin with our first algorithm.
Algorithm 2.1. (Accelerated Projections) Let M1, M2, . . . , Mk be a finite number of closed linear subspaces in a Hilbert space X. For a starting point x0 ∈ X, this algorithm seeks to find PM (x0), where M := ∩kl=1 Ml.
Step 0: Set i = 0 .
Step 1: Project xi onto Mli , where li ∈ { 1, . . . , k }, to get ˜xi. This projection identifies a hyperplane Hi := {x : 〈ai, x 〉 = bi}, where ai = xi − PMli (xi) ∈ X and
bi = 〈ai, P Mli (xi)〉 ∈ R, such that M ⊂ Mli ⊂ Hi. (When ai = 0 , then Hi = X.)
Step 2: Choose Ji ⊂ { 1, . . . , i } such that i ∈ Ji, and project ˜xi onto ˜Hi := ∩j∈Ji Hj
to get xi+1 . In short:
xiPMli (·)
−−−−−→ ˜xiP ˜Hi (·)
−−−−→ xi+1 , with M ⊂ ˜Hi and M ⊂ Mli for all i. (2.1)
Step 3: The algorithm ends if some convergence criterion is met. Otherwise, set
i ← i + 1 and return to step 1. Remark 2.2 . (Limit points of {xi}) We can easily figure out that
xi − ˜xi ∈ M ⊥
li
⊂ M ⊥ and ˜xi − xi+1 ∈ ˜H⊥
i
⊂ M ⊥,
from which we can deduce that x0 − xi ∈ M ⊥. Suppose {xi}i converges (weakly or strongly) to ¯x. We can then deduce that x0 − ¯x ∈ [∑kl=1 M ⊥
l
] = M ⊥. Furthermore, if
¯x ∈ M , the KKT conditions imply that ¯x = PM (x0). For the rest of this paper, we will concentrate our efforts in showing that in our algorithms, the iterates {xi} converge strongly to PM (x0).The next two easy results are preparation for Algorithm 2.5, which is an improve-ment of Algorithm 2.1 when one of the linear subspaces, say M1, is easy to project onto. Something similar was done in [Pie84, BCK06], where analytic formulas for the projection onto an affine space and a halfspace were derived.
Proposition 2.3. (Projection onto intersection of affine spaces) Suppose M and ˜H
are linear subspaces of a Hilbert space X such that ˜H⊥ ⊂ M . Then PM∩ ˜H (·) =
P ˜H ◦ PM (·).Proof. For x ∈ X, let y := PM (x). Then y − P ˜H (y) ∈ ˜H⊥ ⊂ M . Since y ∈ M ,we have P ˜H ◦ PM (x) = P ˜H (y) ∈ M . It is also clear that P ˜H ◦ PM (x) ∈ ˜H, so
P ˜H ◦ PM (x) ∈ M ∩ ˜H.Next, since y − x ∈ M ⊥ ⊂ [ ˜H ∩ M ]⊥, we have
P ˜H∩M (x) = P ˜H∩M (y) = P ˜H (y) = P ˜H ◦ PM (x).
Since the above holds for all x ∈ X, we are done. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 4
x’
2
M1 x+
x x’’ M
Figure 2.1. This figure illustrates the two dimensional space spanned by x, x′ and x′′ in Proposition 2.4.
Proposition 2.4. (2 subspaces) Let M1 and M2 be two linear subspaces of a Hilbert space X. Suppose x ∈ M1 and x′ = PM2 (x) and x′′ = PM1 (x′). Then the hyperplane
H := {˜x : 〈x − x′′ , ˜x〉 = 〈x − x′′ , x +〉} , where x+ := x + ‖x−x′‖2
‖x−x′′ ‖2
(x′′ − x) (See Figure 2.1), is such that M1 ∩ M2 ⊂ H. Moreover, x − x′′ , the only vector in H⊥ up to a scalar multiple, satisfies x − x′′ ∈ M1.Proof. By the properties of projection, we have M2 ⊂ H′ := {˜x : 〈x − x′, ˜x − x′〉 = 0 }
and M1 ⊂ H′′ := {˜x : 〈x′ − x′′ , ˜x − x′′ 〉 = 0 }. By elementary geometry (See Figure 2.1), we can figure out that the projection of x′′ onto H′ ∩ H′′ is x+. Thus
M1 ∩ M2 ⊂ H′ ∩ H′′ ⊂ H, from which the first part follows. The last sentence of the result is clear.
From the preparations in Propositions 2.3 and 2.4, we propose the following algo-rithm.
Algorithm 2.5. (Accelerated Projections 2) Let M1, M2, . . . , Mk be a finite number of closed linear subspaces in a Hilbert space X. Suppose M1 is easy to project onto. For a starting point x0 ∈ M1, this algorithm seeks to find PM (x0), where M := ∩kl=1 Ml.
Step 0: Set i = 0 .
Step 1: Project xi onto Mli , where li ∈ { 2, . . . , k }, to get x′
i
. Project x′
i
onto
M1 to get x′′
i
. This projection identifies a hyperplane Hi := {x : 〈ai, x 〉 = bi}, where
ai = xi −x′′
i
∈ M1 and bi = 〈ai, x i + ‖xi−x′
i‖2
‖xi−x′′
i‖2
(x′′
i
−xi)〉 ∈ R, such that M1 ∩Mli ⊂ Hi.
Step 2: Choose Ji ⊂ { 1, . . . , i } such that i ∈ Ji, and set xi+1 = P ˜Hi (xi), which also equals P ˜Hi (xi) since xi − x′′
i
∈ H⊥
i
⊂ ˜H⊥
i
. One has
xiPMli (·)
−−−−−→ x′
iPM1(·)
−−−−→ x′′
iP˜Hi(·)= PM1∩˜Hi(·)
−−−−−−−−−−−−→ xi+1 , (2.2)
with M ⊂ ˜Hi and M ⊂ Mli for all i.
Step 3: The algorithm ends if some convergence criterion is met. Otherwise, set
i ← i + 1 and return to step 1.
If x0 /∈ M1 in Algorithm 2.5, we can start the algorithm with x0 ← PM1 (x0) instead. It is clear that PM (x0) = PM (PM1 (x0)) . So if the algorithm with the adjusted starting point converges to PM (PM1 (x0)) , then it converges to PM (x0).PROJECTION ONTO INTERSECTION OF AFFINE SPACES 5
We show that xi ∈ M1 for all i and also explain why P ˜Hi (x′′
i
) = PM1 ∩ ˜Hi (x′′
i
) in (2.2). The assumptions state that x0 ∈ M1. Suppose xi ∈ M1. Then
xi+1 = P ˜Hi (xi) = P ˜Hi ◦ PM1 (xi) = PM1 ∩ ˜Hi (xi).
The last equation comes from applying the fact that ˜H⊥
i
= ∑
j∈Ji
H⊥
j
and H⊥
j
⊂ M1
for all j from Proposition 2.4 onto Proposition 2.3. The formula (2.2) is a useful tool in the analysis of Algorithm 2.5. The linear subspace M1 ∩ ˜Hi in Algorithm 2.5 has a larger codimension than the ˜Hi
in Algorithm 2.1. Thus operations involving the projection PM1 ∩ ˜Hi (·) can be expected to get iterates closer to M than P ˜Hi (·). So when M1 is easy to project onto, we expect Algorithm 2.5 to converge in fewer iterations and less time than Algorithm 2.1. Such a condition is met for the example we present in Section 4, and we will see that Algorithm 2.5 is indeed better. Another factor that may play a role in the fast convergence observed is that M1 has a larger codimension than the other subspaces. 3. Strong convergence results
In this section, we prove the strong convergence results for Algorithms 2.1 and 2.5. We recall some easy results on the projection onto a closed linear subspace and Fejér monotonicity.
Theorem 3.1. (Orthogonal projection onto linear subspaces) Let X be a Hilbert space, and suppose T : X → X is a projection of a point x onto a closed linear subspace S.Then
‖x − T x ‖2 = ‖x‖2 − ‖ T x ‖2 for all x ∈ X.
Definition 3.2. (Fejér monotone sequence) Let X be a Hilbert space, C ⊂ X be a closed convex set, and {xi} be a sequence in X. We say that {xi} is Fejér monotone with respect to C if
‖xi+1 − c‖ ≤ ‖ xi − c‖ for all c ∈ C and i = 1 , 2, . . .
A tool for obtaining a Fejér monotone sequence is stated below.
Theorem 3.3. (Fejér attraction property) Let X be a Hilbert space. For a closed convex set C ⊂ X, x ∈ X, λ ∈ [0 , 2] , and the projection PC (x) of x onto C, let the
relaxation operator RC,λ : X → X [Agm83] be defined by
RC,λ (x) = x + λ(PC (x) − x).
Then
‖RC,λ (x) − c‖2 ≤ ‖ x − c‖2 − λ(2 − λ)d(x, C )2 for all y ∈ C.
(For this paper, we only consider the case λ = 1 , which corresponds to the projection.)
We need a few lemmas proven in [BD85] and a few classical results used in [BD85] for the proof of our result.
Theorem 3.4. (Uniform boundedness principle) Let {fn} be a sequence of continuous linear functionals on a Hilbert space X such that sup n |fn(x)| < ∞ for each x ∈ X.Then ‖fn‖ ≤ K < ∞.
Corollary 3.5. Let {fn} be a sequence of linear functionals on a Hilbert space X such that for each x ∈ X, {fn(x)} converges. Then there is a continuous linear functional
f such that f (x) = lim n fn(x) and ‖f ‖ ≤ lim inf n ‖fn‖.PROJECTION ONTO INTERSECTION OF AFFINE SPACES 6
Theorem 3.6. (Kadec-Klee property) In a Hilbert space, xn → x strongly if and only if xn → x weakly and ‖xn‖ → ‖ x‖.
Theorem 3.7. (Banach-Saks property) Let {xn} be a sequence in a Hilbert space that converges weakly to x. Then we can find a subsequence {xnk } such that the arithmetic mean 1
m
∑mk=1 xnk converges strongly to x.
Lemma 3.8. [BD85] (Sum of squares) Suppose a sequence of nonnegative numbers
{aj }j is such that ∑∞
j=1
a2
j
converges. Then there is a subsequence {jt}t such that the sequence {∑jt
s=1
asajt }t converges to zero.
We now prove a result that will be used in all the variants of our strong convergence results for Algorithms 2.1 and 2.5. This result is modified from that of [BD85], and we follow the treatment in [ER11].
Proposition 3.9. (Conditions for strong convergence) Let {Ml}kl=1 be linear subspaces of a Hilbert space X, and M := ∩kl=1 Ml. For a starting x0 ∈ X, suppose that the iterates {xi}i generated by an algorithm satisfy
(1) {xi}i is Fejér monotone with respect to M .
(2) There exists a subsequence {it}t such that
lim sup
t→∞
〈xit − x0, x it 〉 ≤ 0. (3.1) (3) For all l ∈ { 1, . . . , k } and i > 0, there is some pli > 0 such that xi+pli ∈ Ml
and lim i→∞ ‖xi − xi+pli ‖ = 0 .
(4) xi − x0 ∈ M ⊥ for all i.Then the sequence of iterates {xi} converges strongly to PM (x0).Proof. The proof of this result is modified from that of [BD85], following the presenta-tion in [ER11]. By property (2), we choose a subsequence satisfying (3.1). By property (1), {xit }t is a bounded sequence, so we can assume, by finding a subsequence if necessary, that the weak limit
h := w- lim
t→∞
xit
exists. Property (3) states that for each l ∈ { 1, . . . , k }, we can find a sequence
{pli}i ⊂ [0 , ∞) such that xi+pli ∈ Ml and lim i→∞ ‖xi − xi+pli ‖ = 0 . We therefore have w- lim
t→∞
xit +plit
= w- lim
t→∞
xit = h.
The Banach-Saks Property (Theorem 3.7) implies that we can further choose a subse-quence of {xit }t if necessary (we don’t relabel) so that
1
m
m
∑
t=1
xit +plit
converges strongly to h as m → ∞ . (3.2) The term on the left of (3.2) lies in Ml. Since l is arbitrary, we conclude that h ∈ M .Since {‖ xit ‖} t is bounded, we can choose a subsequence if necessary so that
u := lim
t→∞
‖xit ‖.
By applying the Uniform Boundedness Principle (Theorem 3.4 and Corollary 3.5), we have
‖h‖ ≤ lim inf
t→∞
‖xit ‖ = lim
t→∞
‖xit ‖ = u. (3.3) PROJECTION ONTO INTERSECTION OF AFFINE SPACES 7
Since xit − x0 ∈ M ⊥, we have 〈xit − x0, y 〉 = 0 for all y ∈ M . So for all y ∈ M ,
0 ≥ lim sup
t→∞
〈xit − x0, x it 〉
= lim sup
t→∞
〈xit − x0, x it − y〉
= u2 − 〈 h, y 〉 − 〈 x0, h 〉 + 〈x0, y 〉 (3.4)
≥ 〈h − x0, h − y〉 .
This means that h = PM x0. Next, we use (3.4) and substitute y = h to get 0 ≥
u2 − ‖ h‖2, which together with (3.3), gives u = ‖h‖. By the Kadec-Klee property (Theorem 3.6), we conclude that the subsequence {xit }t converges strongly to h.To see that {xi}i converges strongly to h, we make use of the Fejér monotonicity of the iterates with respect to M and h ∈ M .
Remark 3.10 . (Conditions (1) and (4) of Proposition 3.9) The sequence we apply Proposition 3.9 on for our next results on Algorithm 2.1 is actually x0, ˜x0, x 1, ˜x1,x2, ˜x2, . . . instead of {xi}. Similarly, the sequence we apply Proposition 3.9 on for our next results on Algorithm 2.1 is actually x0, x ′
0
, x ′′
0
, x 1, x ′
1
, x ′′
1
, x 2, x ′
2
, x ′′
2
, . . . .We remark that for Algorithm 2.1, condition (1) holds because of (2.1). Similarly, in Algorithm 2.5, condition (1) holds due to (2.2). Condition (4) holds for Algorithm 2.1 because (2.1) implies that
xi − ˜xi ∈ M ⊥
li
⊂ M ⊥
and ˜xi − xi+1 ∈ ˜H⊥
i
⊂ M ⊥,
from which we can easily deduce x0 − xi ∈ M ⊥ and x0 − ˜xi ∈ M ⊥ for all i as needed. The analysis for Algorithm 2.5 is similar. We now prove the convergence of Algorithm 2.1 for the easier case first.
Theorem 3.11. (Strong convergence of Algorithm 2.1: Version 1) Suppose that in Algorithm 2.1, the additional conditions are satisfied:
(A) There is a number ¯p such that for all l ∈ { 1, . . . , k } and i > 0, there is a
pli ∈ [0 , ¯p] such that ˜xi+pli = PMl (xi+pli ).
(B) The hyperplanes ˜Hi are chosen such that x0 − xi ∈ span ({aj : j ∈ Ji}) for all iterations i.Then the sequence of iterates {xi}i converges strongly to PM (x0).Proof. We apply Proposition 3.9. The sequence we apply Proposition 3.9 to is ac-tually x0, ˜x0, x 1, ˜x1, x 2, ˜x2, . . . instead of {xi}. By Remark 3.10, it suffices to check conditions (2) and (3) of Proposition 3.9.
Step 1: Condition (A) implies Condition (3) of Proposition 3.9.
By condition (A), for any i > 0 and l ∈ { 1, . . . , k }, there exists a pli ∈ [0 , ¯p] such that ˜xi+pli ∈ Ml. By using Theorem 3.1 repeatedly, we have
∞
∑
i=0
[‖xi − ˜xi‖2 + ‖˜xi − xi+1 ‖2] ≤ ‖ x0‖2 < ∞.
Therefore the sequence ‖x0 − ˜x0‖, ‖˜x0 − x1‖, ‖x1 − ˜x1‖, . . . converges to zero. Since
‖xi − ˜xi+pli ‖ ≤ ‖ xi+pli − ˜xi+pli ‖ +
¯p−1
∑
j=0
[‖xi+j − ˜xi+j ‖ + ‖˜xi+j − xi+j+1 ‖],PROJECTION ONTO INTERSECTION OF AFFINE SPACES 8
we see that {‖ xi − ˜xi+pli ‖} i is bounded by a finite sum of terms with limit zero. Hence
‖xi − ˜xi+pli ‖ → 0 as i → ∞ . Thus condition (3) holds.
Step 2: Condition (B) implies Condition (2) of Proposition 3.9.
We prove
〈x0 − xi, x i〉 = 0 for all i > 0, (3.5) which clearly implies Condition (2). We use standard induction. It is easy to check that formula (3.5) holds for i = 1 . Suppose it holds for i = i∗. We want to show that it holds for i = i∗ + 1 . We have xi∗ +1 = P ˜Hi∗ (˜ xi∗ ), or equivalently, xi∗ +1 ∈ ˜xi∗ + ˜H⊥
i∗
.Since i∗ ∈ Ji∗ , we have xi∗ − ˜xi∗ ∈ ˜H⊥
i∗
, so xi∗ ∈ xi∗ +1 + ˜H⊥
i∗
, or xi∗ +1 ∈ P ˜Hi∗ (xi∗ ).Since x0 − xi∗ ∈ span ({aj : j ∈ Ji∗ }), we have x0 − xi∗ ∈ ˜H⊥
i∗
, so xi∗ +1 ∈ P ˜Hi∗ (x0)
using a similar argument. Since 0 ∈ ˜Hi∗ , we can deduce (3.5), ending our proof by induction.
Note that condition (A) of Theorem 3.11 satisfied in the classical method of alter-nating projections, but condition (B) is not. We propose a second convergence result such that includes the classical method of alternating projections. For the iterates in Algorithm 2.1, i ≥ 0 and l ∈ { 1, . . . , k }, we define ˜vi,l ∈ X and vi,l ∈ X to be such that
xi − ˜xi =
k
∑
l=1
˜vi,l ,
and ˜xi − xi+1 =
k
∑
l=1
vi,l , where ˜vi,l , v i,l ∈ M ⊥
l
for all l ∈ { 1, . . . , k }.
Such a representation is not unique. This part of the proof is modified from the treatment in [ER11] of [BD85].
Theorem 3.12. (Strong convergence of Algorithm 2.1: Version 2) Suppose that in Algorithm 2.1, the additional conditions are satisfied:
(A) There is a number ¯p such that for all l ∈ { 1, . . . , k } and i > 0, there is a
pli ∈ [0 , ¯p] such that ˜xi+pli = PMl (xi+pli ).
(B ′) There is a number K such that
k
∑
l=1
[‖˜vj,l ‖2 + ‖vj,l ‖2] ≤ K[‖xi − ˜xi‖2 + ‖˜xi − xi+1 ‖2] for all i ≥ 0. (3.6)
Then the iterates {xi} converge strongly to PM (x0).Proof. Like in Theorem 3.11, we apply Proposition 3.9. The sequence we apply Propo-sition 3.9 on is actually x0, ˜x0, x 1, ˜x1, x 2, ˜x2, . . . instead of {xi}. The proof that con-dition (A) implies condition (3) of Proposition 3.9 is the same as that in Theorem 3.11. We proceed with the rest of the proof.
Step 1: Condition (2) of Proposition 3.9 holds.
By using Theorem 3.1 repeatedly, we have
∞
∑
i=0
[‖xi − ˜xi‖2 + ‖˜xi − xi+1 ‖2] ≤ ‖ x0‖2 < ∞. (3.7) PROJECTION ONTO INTERSECTION OF AFFINE SPACES 9
For j ∈ N0, define αj ∈ R to be
αj :=
k
∑
l=1
[‖˜vj,l ‖ + ‖vj,l ‖].
By repeatedly using the inequality 2cd ≤ c2 + d2 onto the expansion of α2
j
and (3.6), we have
α2
j
≤ 2k
k
∑
l=1
[‖˜vj,l ‖2 + ‖vj,l ‖2] ≤ 2kK [‖xi − ˜xi‖2 + ‖˜xi − xi+1 ‖2].
In view of (3.7), the sum ∑∞
j=0
α2
j
is finite. Next, we calculate the bounds on the inner product 〈xi − x0, x i〉. By Condition (A), for each l ∈ { 1, . . . , k } and i > 0, there is some pli ∈ [0 , ¯p] such that ˜xi+pli =
PMl (xi+pli ), from which we get ˜xi+pli ∈ Ml.Since xi − x0 = ∑is=0
∑kl=1 [˜ vs,l + vs,l ] and [˜ vs,l + vs,l ] ∈ M ⊥
l
, we have
〈xi − x0, x i〉 =
i−1
∑
s=0
k
∑
l=1
〈˜vs,l + vs,l , x i〉 =
i−1
∑
s=0
k
∑
l=1
〈˜vs,l + vs,l , x i − ˜xi+pli 〉.
Since
‖xi − ˜xi+pli ‖ ≤
¯p
∑
s=0
[‖xi+s − ˜xi+s‖ + ‖˜xi+s − xi+s+1 ‖]
≤
¯p
∑
s=0
k
∑
l=1
[‖˜vi+s‖ + ‖vi+s‖]=
¯p
∑
s=0
αi+s,
we continue the earlier calculations to get
〈xi − x0, x i〉 ≤
i
∑
s=0
k
∑
l=1
[‖˜vs,l ‖ + ‖vs,l ‖]
︸ ︷︷ ︸
=αs
‖xi − ˜xi+pli ‖ ≤
[ i∑
s=0
αs
] [ ¯p∑
s=0
αi+s
]
.
Define βj := ∑¯ps=0 αj[ ¯ p+1]+ s. The inequality above would imply
〈xj[ ¯ p+1] − x0, x j[ ¯ p+1] 〉 ≤
j
∑
s=0
βsβj .
Since β2
j
≤ [¯ p + 1] ∑¯ps=0 α2
j[ ¯ p+1]+ s
, we see that ∑∞
j=1
β2
j
≤ [¯ p + 1] ∑∞
j=1
α2
j
< ∞. By Lemma 3.8, we can find a subsequence {it} such that lim sup t→∞ 〈xit − x0, x it 〉 ≤ 0,which is exactly condition (2). Thus we are done.
In the case of alternating projections, it is clear to see that condition (B ′) is satisfied with K = 1 because ˜xi − xi+1 = 0 and vi,l = 0 for all i ≥ 0, and for each i ≥ 0, only one of the ˜vi,l among l ∈ { 1, . . . , k } equals to xi − ˜xi, and the rest of the ˜vi,l are zero. We remark that the condition (B ′) can be checked once we get the new iterate ˜x(i)
j
.The value K can be chosen to be any finite value. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 10
We now proceed to prove a strong convergence result of Algorithm 2.5. The proof is similar to that of Theorem 3.11, but we shall include the details for completeness.
Theorem 3.13. (Strong convergence of Algorithm 2.5) Suppose that in Algorithm 2.5, the additional conditions are satisfied:
(A) There is a number ¯p such that for all l ∈ { 2, . . . , k } and i > 0, there is a
pli ∈ [0 , ¯p] such that x′
i+pli
= PMl (xi+pli ).
(B) The hyperplanes ˜Hi are chosen such that x0 − xi ∈ span ({aj : j ∈ Ji}) for all iterations i.Then the sequence of iterates {xi}i converges strongly to PM (x0).Proof. We apply Proposition 3.9. The sequence we apply Proposition 3.9 to is actually
x0, x ′
0
, x ′′
0
, x 1, x ′
1
, x ′′
1
, x 2, x ′
2
, x ′′
2
, . . . . instead of {xi}. By Remark 3.10, it suffices to check conditions (2) and (3) of Proposition 3.9. The changes from the proof of Theorem 3.11 are minimal, but we still include details for completeness.
Step 1: Condition (A) implies Condition (3) of Proposition 3.9.
By condition (A), for any i > 0 and l ∈ { 2, . . . , k }, there exists a pli ∈ [0 , ¯p] such that x′
i+pli
∈ Ml. Note that xi+1 = PM1 ∩ ˜Hi (x′
i
) by Proposition 2.3. By using Theorem 3.1 repeatedly, we have
∞
∑
i=0
[‖xi − x′
i
‖2 + ‖x′
i
− xi+1 ‖2] ≤ ‖ x0‖2 < ∞.
Therefore the sequence ‖x0 − x′
0
‖, ‖x′
0
− x1‖, ‖x1 − x′
1
‖, . . . converges to zero. Since
‖xi − x′
i+pli
‖ ≤ ‖ xi+pli − x′
i+pli
‖ +
¯p
∑
j=0
[‖xi+j − x′
i+j
‖ + ‖x′
i+j
− xi+j+1 ‖],
it is clear that ‖xi − x′
i+pli
‖ → 0 as i → ∞ . Thus condition (3) holds.
Step 2: Condition (B) implies Condition (2) of Proposition 3.9.
We prove
〈x0 − xi, x i〉 = 0 for all i > 0, (3.8) which clearly implies Condition (2). We use standard induction. It is easy to check that formula (3.8) holds for i = 1 . Suppose it holds for i = i∗. We want to show that it holds for i = i∗ + 1 . We have xi∗ +1 = PM1 ∩ ˜Hi∗ (x′
i∗
), or equivalently,
xi∗ +1 ∈ x′
i∗
[ M1 ∩ ˜Hi∗ ]⊥.
Since x′′
i∗
= PM1 (x′
i∗
), we have x′
i∗
− x′′
i∗
∈ M ⊥
1
⊂ [M1 ∩ ˜Hi∗ ]⊥. Next, since i∗ ∈ Ji∗ ,we have xi∗ − x′′
i∗
∈ ˜H⊥
i∗
⊂ [M1 ∩ ˜Hi∗ ]⊥, Thus
xi∗ +1 ∈ xi∗ + [ M1 ∩ ˜Hi∗ ]⊥,
or xi∗ +1 = PM∩ ˜Hi∗ (xi∗ ). Since x0 − xi∗ ∈ span ({aj : j ∈ Ji∗ }), we have
x0 − xi∗ ∈ ˜H⊥
i∗
⊂ [M1 ∩ ˜Hi∗ ]⊥,
so xi∗ +1 = PM1 ∩ ˜Hi∗ (x0) using a similar argument. Since 0 ∈ M1 ∩ ˜Hi∗ , we can deduce (3.8), ending our proof by induction.
It is clear that some variant of condition (A) is necessary so that we project onto each set Ml infinitely often, otherwise we may converge to some point outside M . We now give our reasons to show that it will be hard to prove the result if conditions (A) and (B) were dropped. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 11
Example 3.14. (Difficulties in dropping conditions in strong convergence theorems) Consider the case when k = 2 . The linear operator PM2 PM1 (·) is nonexpansive. But
M ⊥
1
M ⊥
2
is a closed subspace if and only if ‖PM2 PM1 PM⊥ ‖ < 1 [BBL97]. We look at the case when
‖PM2 PM1 PM⊥ ‖ = 1 . (3.9) The hyperplanes ˜Hi considered in the algorithm satisfy 0 ∈ ˜Hi. Suppose that this is the condition imposed on the ˜Hi rather than ˜Hi being the intersection of hyperplanes found by previous iterations. We refer to Figure 3.1. The points xi1 and xi2 , where
i1 < i 2, are iterates of Algorithm 2.1, and xi2 is obtained after projecting consecutively onto four subspaces from xi1 . This arises when a third subspace M3 is the Hilbert space X and we project onto different hyperplanes passing through 0 after projecting onto M3. We now show that it is possible for the iterates xi1 and xi2 to be such that
‖xi2‖
‖xi1‖
is arbitrarily close to 1. Suppose the angle ∠xi1 0xi2 is θ. If xi2 is obtained by projecting consecutively onto k subspaces, where consecutive subspaces are at an angle of θ/k . We can use Theorem 3.1 and some elementary geometry to bound ‖xi2 ‖ by
‖xi1 ‖2
[
1 − k
[
sin θ
k
]2]
≤ ‖ xi2 ‖2 ≤ ‖ xi1 ‖2.
Some simple trigonometry gives us lim k→∞ k[sin θ
k
]2 = 0 . This would imply that ‖xi2 ‖
‖xi1‖
can be arbitrarily close to 1 if we allow for projections onto arbitrarily large number of subspaces containing M as claimed. Combining this fact together with (3.9), we cannot rule out that (by our method of proof at least) it is possible that the iterates
xi may not even converge to PM (x0).
i1
xi2
x
Figure 3.1. The iterate xi2 is obtained from xi1 after projecting onto 4 linear subspaces. Note that ‖xi2 ‖ is larger than the norm of the projection of xi1 onto the final subspace.
Remark 3.15 . (Connection to Dykstra’s algorithm) Dykstra’s algorithm [Dyk83] is an algorithm to find the projection of a point onto the intersection of finitely many closed convex sets (not necessarily affine subspaces). The difference between Dykstra’s algo-rithm and the method of alternating projections is the additional correction vectors in Dykstra’s algorithm. Readers familiar with Dykstra’s algorithm will know that in the case of finitely many affine subspaces, Dykstra’s algorithm reduces to the method of alternating projections. The Boyle-Dykstra Theorem [BD85] proves the correctness of Dykstra’s algorithm, and we have used ideas in [BD85] for our proof. A reason why we PROJECTION ONTO INTERSECTION OF AFFINE SPACES 12
only analyze the problem of accelerating alternating projections in the case of finitely many affine spaces and not the more general setting of accelerating Dykstra’s algorithm is that we feel that the idea of using supporting halfspaces and quadratic programming as explained in [Pan14b] will be more effective than Dykstra’s algorithm in general. Finally, we remark that a consequence of our strong convergence theorems is that strong convergence is guaranteed even when the projection order is not cyclic. These observations have already been made in [HD97] when they were analyzing the more general Dykstra’s algorithm. 4. Performance of acceleration
In this section, we consider a Matrix Model Updating Problem (MMUP) as presented in [ER11, Section 6.2], who in turn cited [DS01, MDR09], and show how one can use Algorithm 2.1 to solve the problem. We also show the numerical performance of our acceleration. The problem of interest is as follows. For M, D, K ∈ Rn×n, we want to solve
min
˜K, ˜D∈Rn×n
‖K − ˜K‖2
F
‖D − ˜D‖2
F
(4.1a) s.t. ˜K = ˜KT , ˜D = ˜DT , (4.1b)
M Y 1(Λ ∗
1
)2 + ˜DY 1(Λ ∗
1
) + ˜KY 1 = 0 , (4.1c) where Λ∗
1
∈ Cp×p and Λ∗
1
= diag (μ1, . . . , μ p) and Y1 ∈ Cn×p with columns y1, . . . , y p
are the matrices of the desired eigenvalues {μi}pi=1 and eigenvectors {yi}pi=1 . Problem (4.1) arises when we want to find minimal perturbations in K and D so that some undesirable eigenvalues are moved to more desirable values. We can transform (4.1) as follows. We start by writing (4.1c) as
A + ˜DB + ˜KC = 0 ,
where A, B, C ∈ Cn×p are
A = M Y 1(Λ ∗
1
)2, B = Y1(Λ ∗
1
) and C = Y1. (4.2) We can now write (4.1a) as a function of only one 2n×2n block matrix variable. Define the matrices X0 ∈ R2n×2n and ˜X ∈ R2n×2n by
X0 =
( K 00 D
)
and ˜X =
( ˜K 00 ˜D
)
.
We now write (4.1c) in terms of ˜X. Define the block matrices W and ˆI as
ˆI :=
( In×n
In×n
)
and W =
( CB
)
,
where In×n ∈ Rn×n is the identity matrix. Note that
A + ˆIT ˜XW = A + ( In×n In×n
) ( ˜K 00 ˜D
) ( CB
)
= A + ˜KC + ˜DB
= A + ˜X22 B + ˜X11 C. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 13
Then problem (4.1) is reduced to that of finding the matrix ˜X that solves the following optimization problem
min
˜X∈R2n×2n
‖X0 − ˜X‖2
F
(4.3a) s.t. ˜X = ˜XT , ˜X21 = ˜X12 = 0 (4.3b)
A + ˆIT ˜XW = 0 . (4.3c) The projection of a matrix X onto the set S of matrices satisfying the first constraint (4.3b) is given by
[PS (X)] 12 = [ PS (X)] 21 = 0 ,
[PS (X)] 11 = 1
2 [X11 + XT
11
],
and [PS (X)] 22 = 1
2 [X22 + XT
22
].
For the second constraint, we need to project onto the linear variety
V := {X ∈ R2n×2n : A + ˆIXW = 0 }. (4.4)
Theorem 4.1. [MDR09] If X ∈ R2n×2n is any given matrix, then the projection onto the linear variety V is given by
PV (X) = X + ZΣW T ,
where ΣT = − 1
2 [W T W ]−1(AT + W T XT Z).
4.1. Numerical experiments. We consider two algorithms for solving the MMUP problem (4.3). In the first algorithm, we make specific choices on step 1.
Algorithm 4.2. (MMUP algorithm 1) For a starting matrix X0, we wish to solve (4.3) .We apply Algorithm 2.1 by choosing the first affine space to be S and the second affine space to be V . We project onto S and V alternately, starting with S. Choose q to be a positive integer. The affine space ˜Hi is chosen to be the intersection of the last
q affine spaces identified, or all of the affine spaces if less than q affine spaces were identified.
It is clear that q = 1 corresponds to the alternating projection algorithm. We now describe a second algorithm for the MMUP.
Algorithm 4.3. (MMUP Algorithm 2) For a starting matrix X0 ∈ S, we wish to solve
(4.3) . We apply Algorithm 2.5 by choosing the first affine space to be S and the second affine space to be V . We choose S to play the role of M1 in Algorithm 2.5. Choose
q to be a positive integer. The affine space ˜Hi is chosen to be the intersection of the last q affine spaces identified, or all of the affine spaces if less than q affine spaces were identified.
One can see that the choice of ˜Hi in Algorithms 4.2 and 4.3 do not satisfy condition (B). Nevertheless, if the iterates do converge, we can show that limit points must be of the form [x0 + M ⊥] ∩ M , and the only point satisfying this property is PM (x0). We still obtain desirable numerical results in our experiments.
Remark 4.4 . (Sparsity in Algorithm 4.3) Note that the iterates Xi and ai, the normal vectors of the halfspaces produced, have to lie in the space S, which is sparse. Besides the ease of projection onto S and the large codimension of S, the sparsity of iterates and normals is another reason why Algorithm 4.3 performs better than Algorithm 4.2. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 14
Remark 4.5 . (The case of q = ∞) In our problem, the two affine spaces S and V
are both determined by finitely many equations. We can define both Algorithms 4.2 and 4.3 by setting the parameter q to be ∞. What this means is that we project onto the affine space produced by intersecting all previous hyperplanes generated in earlier iterations. We can converge in finitely many iterations for both algorithms once we identify all the equations defining the two subspaces, but the computational costs for solving the resulting system can be huge. (The reason why the alternating projection method is preferable is that the cost per iteration is small.) We now perform our experiments on two problems presented in [ER11, Section 6.2]. 4.1.1. Experiment 1. For our first experiment, we choose M, D, K ∈ R4×4 to be the symmetric positive definite matrices as described in [DS01]:
M =
1.4685 0.7177 0.4757 0.4311 0.7177 2.6938 1.2660 0.9676 0.4757 1.2660 2.7061 1.3948 0.4311 0.9676 1.3918 2.1876
,
D =
1.3525 1.2695 0.7967 0.8160 1.2695 1.3274 0.9144 0.7325 0.7967 0.9144 0.9456 0.8310 0.8160 0.7325 0.8310 1.1536
,
K =
1.7824 0.0076 −0.1359 −0.7290 0.0076 1.0287 −0.0101 −0.0493
−0.1359 −0.0101 2.8360 −0.2564
−0.7290 −0.0493 −0.2564 1.9130
.
The eigenvalues of P (λ) = λ2M + λD + K computed via MATLAB are −0.0861 ±
1.6242 i, −0.1022 ± 0.8876 i, −0.1748 ± 1.1922 i and −0.4480 ± 0.2465 i. We want to reassign only the most unstable pair of eigenvalues, namely −0.0861 ± 1.6242 i, to the locations −0.1 ± 1.6242 i. Let the matrix of eigenvectors to be assigned be
1.0000 1.0000 0.0535 + 0 .3834 i 0.0535 − 0.3834 i
0.5297 + 0 .0668 i 0.5297 − 0.0668 i
0.6711 + 0 .4175 i 0.6711 − 0.4175 i
.
The formulas for A, ˆI and W can work in principle, but we decide to use a differ-ent strategy when the targeted eigenvalues and eigenvectors are complex conjugates. Consider the targeted eigenvalue μ1 = −0.1 + 1 .6242 i and its targeted eigenvector
y1 = ( 1 .0000 0.0535 + 0 .3834 i 0.5297 + 0 .0668 i 0.6711 + 0 .4175 i )T .
Instead of choosing A, B and C in the manner of (4.2), we choose A, B, C ∈ R4×2 to be
A = M [ Re (y1μ21) Im (y1μ21) ] ,B = [ Re (y1μ1) Im (y1μ1) ] .
and C = [ Re (y1 ) Im (y1 ) ] .
We illustrate the results of this experiment in Figure 4.1. The experiments show that the effectiveness of Algorithm 4.2 and Algorithm 4.3. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 15
0510 15 20 25 30
10 −16
10 −14
10 −12
10 −10
10 −8
10 −6
10 −4
10 −2
10 0
Iteration count Feasbility test Rate of convergence of various algorithms
No Acceleration
General acceleration
2nd acceleration
Figure 4.1. This figure shows the results for the experiment de-scribed in Subsubsection 4.1.1. We plot ‖A + ˆIT ˜XW ‖, the distance of the iterates to V , against the number of times we project onto V
(4.4) for three different algorithms, namely the method of alternat-ing projections, Algorithm 4.2 (General acceleration) and Algorithm 4.3 (2nd acceleration). For Algorithm 4.2, we test with parameters
q = 2 , 3, 4, 5, 6, 7, 8. (Note that q = 1 corresponds to the alternating projection algorithm). For Algorithm 4.3, we test with parameters
q = 1 , 2, 3, 4, 5.4.1.2. Experiment 2. We repeat the experiment in [MDR09] for the case when M, D, K ∈
R30 ×30 are the matrices
M = D = 4 I30 ×30 =
4 0 0 · · · 0 00 4 0 · · · 0 00 0 4 · · · 0 0
... ... ... . . . ... ...
0 0 0 · · · 4 00 0 0 · · · 0 4
and K =
1 −1 0 · · · 0 0
−1 2 −1 · · · 0 00 −1 2 · · · 0 0
... ... . . . . . . . . . ...
0 0 · · · −1 2 −10 0 · · · 0 −1 1
.
The pencil P (λ) = λ2M + λD + K has 60 eigenvalues, but the eigenvalue that causes the instability is 0 with eigenvector 1√30 (1 , 1, . . . , 1) T , and the rest of the spectrum of P (λ) is below −0.0027 . We use Y1 = 1√30 (1 , 1, . . . , 1) T with targeted eigenvalue
−0.018 .Our experiments indicate that in one iteration of both Algorithms 4.2 and 4.3, the norm ‖A + ˆIT ˜XW ‖ goes down by a factor of 2.4 × 10 −14 , essentially reaching convergence within the numerical limits. For the alternating projection algorithm, the decrease is linear, and each iteration reduces the norm ‖A + ˆIT ˜XW ‖ by a factor of 0.5.This experiment once again illustrates the efficiency of the accelerations in Algorithms 4.2 and 4.3. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 16
Conclusion
In this paper, we propose acceleration methods for projecting onto the intersection of finitely many affine spaces. This strategy can be applied to general feasibility problems where not only affine spaces are involved, as long as there is more than one affine space.
Acknowledgement. The author acknowledges very helpful conversations with Lim Chuan Li which led to Propositions 2.3 and 2.4 and Algorithm 2.5. He is grateful to his employer, the National University of Singapore, for his startup grant.
References
[Agm83] S. Agmon, The relaxation method for linear inequalities , Canad. J. Math. 4 (1983), 479– 489. [BBL97] H.H. Bauschke, J.M. Borwein, and A.S. Lewis, The method of cyclic projections for closed convex sets in Hilbert space , Recent developments in optimization theory and nonlinear analysis (Jerusalem, 1995), Contemporary Mathematics 204, Amer. Math. Soc., Provi-dence, R.I., 1997, pp. 1–38. [BCK06] H.H. Bauschke, P.L. Combettes, and S.G. Kruk, Extrapolation algorithm for affine-convex feasibility problems , Numer. Algorithms 41 (2006), 239–274. [BD85] J.P. Boyle and R.L. Dykstra, A method for finding projections onto the intersection of convex sets in Hilbert spaces , Advances in Order Restricted Statistical Inference, Lecture notes in Statistics, Springer, New York, 1985, pp. 28–47. [BDHP03] H.H. Bauschke, F. Deutsch, H.S. Hundal, and S.-H. Park, Accelerating the convergence of the method of alternating projections , Trans. Amer. Math. Soc. 355 (2003), no. 9, 3433–3461. [DS01] B.N. Datta and D.R. Sarkissian, Theory and computations of some inverse eigenvalue problems for the quadratic pencil , Mathematics, Computer Science, and Engineering I, Contemporary Mathematics Volume 280, Structured Matrices, Amer. Math. Soc., New York, 2001, pp. 221–240. [Dyk83] R.L. Dykstra, An algorithm for restricted least-squares regression , J. Amer. Statist. Assoc.
78 (1983), 837–842. [ER11] R. Escalante and M. Raydan, Alternating projection methods , SIAM, 2011. [GK89] W.B. Gearhart and M. Koshy, Acceleration schemes for the method of alternating projec-tions , J. Comput. Appl. Math. 26 (1989), 235–249. [GPR67] L.G. Gubin, B.T. Polyak, and E.V. Raik, The method of projections for finding the common point of convex sets , USSR Comput. Math. Math. Phys. 7 (1967), no. 6, 1–24. [Hal62] I. Halperin, The product of projection operators , Acta. Sci. Math. (Szeged) 23 (1962), 96–99. [HD97] H.S. Hundal and F. Deutsch, Two generalizations of Dykstra’s cyclic projections algorithm ,Math. Programming 77 (1997), 335–355. [MDR09] J. Moreno, B.N. Datta, and M. Raydan, A symmetry perserving alternating projection method for matrix model updating , Mech. Syst. Signal Process 23 (2009), 1784–1791. [NT14] D. Needell and J. A. Tropp, Paved with good intentions: Analysis of a randomized block Kaczmarz method , Linear Algebra Appl. 441 (2014), 199–221. [Pan14a] C.H.J. Pang, Improved analysis of algorithms based on supporting halfspaces and quadratic programming for the convex intersection and feasibility problems , (preprint) (2014). [Pan14b] , Set intersection problems: Supporting hyperplanes and quadratic programming ,Math. Programming (Online first) (2014). [Pie84] G. Pierra, Decomposition through formalization in a product space , Math. Programming
28 (1984), 96–115. [vN50] J. von Neumann, Functional operators. II. The geometry of orthogonal spaces. , Annals of Mathematics Studies, no. 22., Princeton University Press, Princeton, NJ, 1950, [This is a reprint of mimeograghed lecture notes first distributed in 1933.]. [XZ02] Jinchao Xu and Ludmil Zikatanov, The method of alternating projections and the method of subspace corrections in Hilbert space , J. Amer. Math. Soc. 15 (2002), no. 3, 573–597. PROJECTION ONTO INTERSECTION OF AFFINE SPACES 17
Current address : Department of Mathematics, National University of Singapore, Block S17 08-11, 10 Lower Kent Ridge Road, Singapore 119076
E-mail address : matpchj@nus.edu.sg |
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SPIONs, or superparamagnetic nanoparticles, are defined as iron oxide core nanoparticles that possess a variable carbohydrate coating, which influences their cellular uptake and biological half-life. They are valued for their applications in novel therapeutic and diagnostic methods.
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Chapter
Epitaxy
2024, Encyclopedia of Condensed Matter Physics (Second Edition)D. Maryenko
Oxide compound
The growth of oxides constitutes one of the important fields in the material science, as the oxides show a diverse range of physical phenomena in bulk and thin films and therefore are considered as a promising material platform for realizing novel functionalities (Schlom et al., 2008; Hwang et al., 2012). High-quality structures are realized in an MBE chamber equipped with a suitable oxygen source, which can be molecular oxygen O2 (suitable for easy oxidizing compounds), oxygen plasma (activated by RF plasma source), or ozone, which is the most reactive and the cleanest oxygen source (Schlom et al., 1988; Eckstein et al., 1989; Berkley et al., 1988, 1989). By using a highly reactive oxidizer a long mean free path should be guaranteed to avoid gas-phase reactions. Usually the nozzle of oxygen line is positioned about 10 cm away from the substrate surface. A complete oxidation of epilayers is always a matter of concern in oxide MBE. Yet, it is possible to achieve high-quality oxide structures. ZnO is one of the examples. The heterostructures based on ZnO feature high electron mobility, giving rise to a variety of quantum phenomena (Tsukazaki et al., 2010; Falson et al., 2011, 2016, 2015; Maryenko et al., 2018, 2021). High-quality ternary oxides, such as SrTiO3 perovskite, are grown by employing metalorganic precursor (MO) as the source for Ti element, that is titanium(IV)isopropoxide (TTIP, Ti[OCH(CH3)2]4). A high vapor pressure of TTIP enables an adsorption controlled growth and therefore offers an excellent control over the epilayer stoichiometry (Endo et al., 1991; Jalan et al., 2009a, b, c; Brahlek et al., 2018). Once TTIP molecule reaches the hot substrate surface, it decomposes in nonvolatile TiO2 and by products consisting of volatile C3H6 and H2O. TiO2 reacts with SrO formed on the surface, leading to the formation of a stoichiometric SrTiO3 layer. The structures synthesized in this way allowed to achieve high electron mobility (Son et al., 2010) and to demonstrate the quantum Hall effect for the first time in oxide perovskite structures (Matsubara et al., 2016). Oxide MBE with MO sources is referred as hybrid MBE.
The approach to synthesize oxide perovskite with MOs has been extended on the growth on vanadium perovskite, using vanadium-triisopropoxide (VTIP) as a source of vanadium (Moyer et al., 2013; Brahlek et al., 2018) and high mobility BaSnO3, employing hexamethylditin (HMDT) as a precursors to supply tin (Chambers et al., 2016). The reactive environment of oxide growth requires the use of oxidation-resistive materials for components that are particularly exposed to high temperatures. Heating of substrate is also an important part of consideration for oxide MBE. The usage of laser heating system instead of resistive heaters allows not only to achieve even higher growth temperatures but also to reduce the heat load on the growth environment. As a result, the incorporation of undesired impurities collected on the chamber walls of the heating system is reduced. Due to the local heating by laser, the heat capacitance is reduced and the quick change of the growth temperature can be achieved, which can be favorable for growth protocols requiring the temperature changes. High temperature can even prepare in situ the substrate surfaces suitable for the growth (Jaeger et al., 2018; Braun et al., 2020).
Fig. 6 is the photograph of an oxide MBE machine installed at RIKEN. It is equipped with several retractable Knudsen cells, supply of metal-organic precursors, and ozone source. Laser heater system allows to achieve temperature well above 1000C. The pumping system, consisting of turbomolecular pump and cryopump, allows to remove the by products of metalorganic precursors effectively. Cryogenic panels cooled with liquid nitrogen minimize the thermal load of the growth chamber and allow to trap impurities from the gas phase during thin film deposition.
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Chapter
Chemistry of Modified Oxide and Phosphate Surfaces: Fundamentals and Applications
2023, Interface Science and TechnologyRobson Fernandes de Farias
Abstract
Oxides can be defined as binary compounds formed between elements, and oxygen and phosphates can be defined as salts based formally on phosphorus (V) oxoacids and in particular salts of phosphoric (V) acid, H3PO4. Both, oxides and phosphates, are among the most important classes of inorganic compounds. For example, silicon dioxide (SiO2) is the main compound on the earth crust constitution, and apatite, a complex mineral form of calcium phosphate, Ca5(PO4)3 (OH, F, Cl), is the main compound on the enamel of teeth.
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Review article
Advances in the growth and characterization of magnetic, ferroelectric, and multiferroic oxide thin films
2010, Materials Science and Engineering: R: ReportsL.W. Martin, ... R. Ramesh
At the heart of understanding oxide materials, is recognizing how the chemical structure of the material enables the evolution of the electronic structure. From chemical bonding to chemical doping, the interplay of physics and chemistry in oxide materials is a very rich subject. In the monoxides (MO) like the rock salt materials discussed above both the cation and the O-anion are in octahedral coordination and they possess largely ionic bonds and, therefore, typically have very large band gaps and insulating properties. Monoxides with the wurtzite structure, however, are hexagonal and possess tetrahedral coordination of the cations and anions (very similar to a diamond cubic structure of many traditional semiconductors like Si) and thus typically are more semiconducting in nature. This idea can be extended to a wide range of oxide materials, but is very elegantly described in the richness of properties found in the perovskite-derived oxides.
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Additive manufacturing of oxide-dispersion strengthened alloys: Materials, synthesis and manufacturing
2023, Progress in Materials ScienceMarkus B. Wilms, ... Bilal Gökce
2.3.2Alternative oxide compounds in ODS alloys
In addition to the widely used oxide compound of Y2O3 in ODS alloys, alternative oxide compounds exist, exhibiting high stability in high temperature regimes and therefore are considered candidates for dispersion phases in ODS alloys. Most of these oxides are based on different rare earth (RE) elements, such as cerium [180,186–189] and lanthanum [180,190–192]. These oxides show a behavior comparable to that of yttrium oxide, e.g. by the formation of pyrochlore-structured complex oxide compounds in titanium alloyed steels, e.g. La2Ti2O7 and Ce2Ti2O7 . Oxide compounds of ZrO2 [193–196], HfO2 and Al2O3 [198,199] are also feasible due to their high thermodynamic stability; however, the formation of complex oxide structures is not observed. (Table 2).
Table 2. Overview of alternative (non-yttrium-based) oxide compounds used in ODS alloys with the resulting oxide compound determined after consolidation by the used consolidation process. The references column lists the studies, where the respective oxide compounds were detected in ODS alloys.
| Oxide compound | Oxide load | Matrix alloy | Consolidation process | Resulting oxide compound | Reference |
--- --- --- |
| CeO2 | 0.3 wt% | Fe-14Cr-2 W-0.3Ti | Spark Plasma Sintering | Ce2Ti2O7 | |
| CeO2 | 0.7 wt% | Fe-12Cr-6.3Al-0.5Ti | Spark Plasma Sintering | Ce2O3, Al2O3 | |
| CeO2 | 0–4 wt% | Ni-16Cr-8Fe-4.5Si-3.3B | l-DED | Ce2O3 | |
| CeO2 | n.d. | Fe-22Cr-5Al | Laser Remelting | n.r. | |
| La2O3 | 0.3 wt% | Fe-14Cr-2 W-0.3Ti | Spark Plasma Sintering | La2Ti2O7 | |
| La2O3 | 0.5 wt% | Fe-14Cr-1Ti-0.3Mo | Spark Plasma Sintering | La-Ti-Cr-O | |
| La2O3 | 5 wt% | Al | Hot Pressing | La2O3 | |
| ZrO2 | 0.5 wt% | 316L | Arc Plasma Sintering | n.r. | |
| ZrO2 | 0.35 wt% | Fe-9Cr-2 W | Spark Plasma Sintering | n.r. | |
| ZrO2 | 0.3 wt% | Fe-21Cr-9Mn-6Ni-0.4Ti | Vacuum Hot Pressing | n.r. | |
| ZrO2 | 20–60 vol% | Cu | Spark Plasma Sintering | n.r. | |
| Al2O3 | 8 wt% | Fe-15Cr-1Mo-2.5C-1Si | l-DED | n.r. | |
| Al2O3 | 1–3 vol% | Ti-6Al-4 V | l-PBF | n.r. | |
| Al2O3 | 0.2–1 wt% | Co-27Cr-4.5Mo-2.3Ni | l-DED | CoAl2O4 | |
| HfO2 | 0.3 wt% | Fe-14Cr-2 W-0.35Ti | l-DED | n.r. | |
| Ta2O5 | 1; 3 vol% | Ti-6Al-4 V | l-PBF | n.r. | |
In conclusion, a large library of different oxide nanoparticle species has been applied in ODS materials. Oxide nanoparticles can be manipulated by alloying the matrix alloy with additional elements and thus tune the nanoparticle size typically toward refined structures. However, research in this field does not include the adaptation of nanoparticle and matrix alloys for each other. Machine learning approaches, exemplary shown by Martin et. al , may pose a powerful tool to identify suitable nanoparticle species for given matrix alloy, allowing the development of superior ODS materials.
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Chapter
Metal Oxides in Photovoltaics: All-Oxide, Ferroic, and Perovskite Solar Cells
2018, The Future of Semiconductor Oxides in Next-Generation Solar CellsAmador Pérez-Tomás, ... Mónica Lira-Cantú
8.1.1Introduction
An oxide is a chemical compound that contains at least one oxygen atom and one other element in its chemical formula. Simple oxides are also known as binary oxide but they can contain additional elements. A complex oxide is a chemical compound that contains oxygen and at least two other elements. Almost the entire Earth's crust is made of oxides as the individual crust elements are oxidized by the oxygen present in the atmosphere or in the water. So, in general, oxides can be regarded as naturally earth abundant and stable compounds. Oxides are then ubiquitous in nature due to the large oxygen electronegativity (i.e. the atom tendency to attract electrons and thus form bonds) that easily creates stable chemical bonds with almost all elements to give the corresponding oxides. Oxides are inextricably linked with human technological progress. Oxide splitting advances are currently used to divide ancient times in the Bronze and Iron Ages while noble, oxide-resistant metals such as platinum or gold have been used as jewelery from the Neolithic. As the industrial revolution evolved to global electrification by the beginning of the 20th century, the electronic property of materials has been a prime topic of research for entire generations of scientists and engineers. Here oxides cover the entire range of conductivity from insulators, through semiconductors and metallic conductors to superconductors . The implementation of the first solid-state transistor in 1948 paved the way for smaller and cheaper consumer electronic radios, calculators, and computers but the real technological breakthrough was the development of the complementary metal-oxide-semiconductor technology again inextricably linked to an engineered gate oxide; SiO2 in this case . Oxides are increasingly being incorporated into a myriad of solid-state electronic devices since then. It is the objective of this chapter to review the use of oxides within the field of photovoltaics (a device that converts photons into usable electric energy) . Oxides are used in solar cells as insulators, semiconductors, and conductors in general delivering lower cost and larger device stability. Although there are several photovoltaic technologies , the basic solar cell structure is a sandwich of a (depleted of free carriers) light harvester layer in between two carrier extraction layers with the final goal of maximizing the light into electricity power conversion efficiency. The focus of this chapter in particular is to summarize the oxide layer implementation into several kinds of solar cells (see Fig. 8.1), where we can find them as transparent conductive electrodes, as light absorbers, as transport layers, or bringing unique functionalities into the structure such as magnetism, ferroelectricity, pyroelectricity, etc.
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Chapter
Chemistry of Modified Oxide and Phosphate Surfaces: Fundamentals and Applications
2023, Interface Science and TechnologyRobson Fernandes de Farias
1Introduction
Oxides can be defined as binary compounds formed between elements, and oxygen and phosphates can be defined as salts based formally on phosphorus (V) oxoacids and in particular salts of phosphoric (V) acid, H3PO4. Both, oxides and phosphates, are among the most important classes of inorganic compounds. For example, silicon dioxide (SiO2) is the main compound on the earth crust constitution, and apatite, a complex mineral form of calcium phosphate, Ca5(PO4)3 (OH, F, Cl), is the main compound on the enamel of teeth [1–3].
Besides their uses as pigments, many oxides exhibit catalytic properties, making them a very important class of compounds, even for basic and technological research. Phosphates, on the other hand, have also a lot of practical uses, e.g., fertilizers based on ammonium phosphate.
It is worth noting that with the exception of the lighter inert gases (He, Ne, Ar, and Kr), oxygen (O) forms at least one binary oxide with each of the elements of the periodic table. Furthermore, both metals and nonmetals can attain their highest oxidation states in compounds with oxygen, showing how significant is the chemistry of oxides.
Also from a structural point of view oxides are very interesting and rich compounds. For example, vanadium oxides have an interesting behavior, as they can form polyanions, polyvanadates such as: V2O74−, V3O93−, V4O124−, V10O286−, etc., which can form tetrahedrons (VO4) in the case of anhydrous metavanadates, trigonal bipyramids (VO5) in the case of hydrated metavanadates, as well as octahedrons (VO6) in decavanadates. Nb and Ta oxides can also assume structures of several octahedrons (MO6) linked together.
Despite the fact that “pure” oxides and phosphates exhibit, themselves, a lot of possible chemical interactions, modified oxide and phosphates could, on the other hand,be considered as exhibiting “expanded” chemical interactions, with very interesting possible consequences to basic and applied chemistry.
Concerning the scope of this book, we are interested not in “pure” oxides and phosphates, but in modified ones, specifically those with chemical modifications that affected their superficial chemical composition (the bulk composition can or cannot be affected).
Due to the presence of Lewis and/or Bronsted acidic and basic sites on their surfaces, both, oxides and phosphates, can interact (with or without the occurrence of chemical reactions) with a lot of substances, especially with organic ones, as will be shown in this book.
If the chemical species that is modifying the oxide or phosphate surface is organic, then it would have produced a so-called inorganic–organic hybrid. A lot of modified oxides and phosphates are examples of that class of compounds.
In a very interesting connection between modified oxides and phosphates, it is possible to modify lamellar phosphates with inorganic oxides, a process called pillaring . This last example shows how rich is the chemistry of modified oxides and phosphates.
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An overview of biomedical applications of oxide materials
2023, Oxides for Medical ApplicationsGaneshlenin Kandasamy
1.1Introduction
Oxide materials are the chemical compounds composed of at least one oxygen atom in combination with other atom(s). Oxygen atoms associate/bond with metals (alkali, alkaline-earth, rare-earth, or transition) to form metal oxides mainly in the oxidation state of 2- (i.e., O2−-dianionic), where other oxidation states are also available. Various metal oxides exist naturally in the earth surface/crust and they have been used in different fields of applications like electronics and electrical (e.g., sensing, energy storage, etc.), mechanical and construction (e.g., thermal insulation), environmental (e.g., water purification), optical (e.g., display devices), optoelectronics (e.g., imaging detectors), magnetic (e.g., spintronics), chemical (e.g., catalysts/photocatalysts, adsorbents, etc.), and so on for many years. Nevertheless, the recent evolution of nanoscience/nanotechnology over the past 2 decades has widened the possibilities to prepare distinct metal oxide materials at the nanoscale dimensions (i.e., known as metal oxide nanoparticles or oxide nanoparticles) with tuneable properties. These properties majorly include morphology/structure, surface area-to-volume ratio, mechanical/chemical stability, magnetic behaviors, redox properties, biocompatibility, and so on. The tunability in the properties of the oxide nanoparticles has enabled their effective usage and also enhanced the efficacy in their above-mentioned applications.
Apart from above, oxide nanoparticles are exclusively utilized in biomedical applications, as there is a major need for the development of novel nanomaterials or nanoparticles in association with the state-of-the-art approaches to solve many of the existing problems—for instance, antibacterial activities, delivery of drugs/genes, therapeutics and diagnostics (theranostics) of different cancers, and so on (Jiang et al., 2018; Li et al., 2019; Maity & Kandasamy, 2019; Maity et al., 2019; McNeil, 2009; Vallabani & Singh, 2018). For this purpose, oxide nanoparticles such as iron oxide (Fe3O4/Fe2O3) nanoparticles, copper oxide (CuO) nanoparticles, manganese oxide (MnO) nanoparticles, magnesium oxide (MgO) nanoparticles, gadolinium oxide (GdO2) nanoparticles, titanium oxide (TiO2) nanoparticles, silica (SiO2) nanoparticles, cerium oxide (CeO2) nanoparticles, zirconia (ZrO2) nanoparticles, yttrium oxide (Y2O3) nanoparticles, zinc oxide (ZnO) nanoparticles, graphene oxide (GO) nanoparticles, and so on have been developed using distinct synthesis approaches and then consequently tested for their effectiveness in those biomedical applications including cancer theranostics (Cai et al., 2019; Kandasamy, 2019a, 2019b; Kandasamy & Kumar, 2020; Kandasamy & Maity, 2015; Sobańska et al., 2021; Wen et al., 2020). However, hydrophilicity, biocompatibility, cytotoxicity, accumulation inside human body, prolong exposure, nano–bio interactions at their interface, biological responses and so on are some of the major constrains that need to be resolved currently. Corresponding researches are ongoing worldwide in that directions, even the starting steps in the synthesis of these nanoparticles are critically getting considered.
Among above, the oxide nanoparticles like iron oxide (Fe3O4/Fe2O3) nanoparticles, zinc oxide (ZnO) nanoparticles, titanium oxide (TiO2) nanoparticles, silicon oxide (SiO2) nanoparticles, cerium oxide (CeO2) nanoparticles, and graphene oxide (GO) nanoparticles have attained specific advancements in the biomedical application due to their intrinsic properties. Hence, in this chapter, these nanoparticles and their corresponding biomedical applications, especially cancer theranostics, have been mainly reviewed.
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Nanotechnology
2011, Microfabrication for Industrial ApplicationsRegina Luttge
Natural versus Artificial Oxide Nanoarrays
Before discussing the various oxide materials that form nanoarrays, we must first define what is meant by the term “oxide” in this context. Turning to inorganic chemistry, an oxide is a chemical compound formed from an element and at least one oxygen atom. Materials can also be classified by their oxidation state (the number of valence electrons in the chemical compound). Nearly all elements can form oxides when they are exposed to oxygen. Noble metals such as gold or platinum form oxides with greater difficulty, but these do exist. Oxides can be also gaseous, but we will only consider solid oxides.
Oxides can be amorphous (for example window glass) or crystalline (quartz, sapphire). Natural silicon oxide formed on a silicon surface is also amorphous, and is normally removed from wafers prior to processing. In this context, there is no need to explain the dielectric properties of silica and its benefits or disadvantages in electronic applications. In other applications, amorphous oxides have been used as a platform for releasing molecules or electrons from its matrix. For example, in drug delivery, highly defined oxides have replaced the polymeric coatings which were previously used. Such a polymer-free platform is used for drug loading in cardiovascular stents . Additionally, biomedical devices for neural stimulation have been based on oxide thin-film technology. In this application, a Teflon-type substrate is covered with a gold/iridium oxide film. The gold layer acts as current collector, and the iridium oxide as an electrochemical actuator. The intrinsic material microstructure plays an important role in enabling a low-resistance interface . Another more recent example for an application of iridium oxides (IrOx) is given by Huang et al., who developed an IrOx micro pH sensor array .
Such miniaturized sensors could be used in chemical, food and bio-material devices. The authors present the design and fabrication process of an amorphous, IrOx, thin film, pH sensor array on flexible polymer substrates in their paper, so the utilization of amorphous oxides as micro- and mesoporous materials have therefore already found their way into applications. Most likely these materials will not be called nanoarrays, although their intrinsic microstructure means that they could be considered as such (e.g., see Figure 3.7, Chapter 3, Section 3.3.6), as could natural crystalline oxides. The properties and classifications of oxides is too broad a field to be discussed here, so this section will be restricted to some oxides that have been described in the literature as an artificially manufactured material. Examples are selected on the basis of their specific optical properties and the application of such oxide nanoarrays in the medical field. Figure 4.13 depicts the four different material types that can also be manipulated into nanoarrays: (A) crystalline, (B) polycrystalline, (C) amorphous and (D) layered composites.
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Types of ceramics
2017, Advances in Ceramic BiomaterialsChristophe Drouet, ... Anna Tampieri
2.1.1Introduction
Oxide materials are increasingly encountered in biomedical applications (Rey et al., 2011) such as tissue repair (e.g., bone regeneration), drug delivery, or medical diagnosis. Among these compounds, certain types of oxides not only play a “physical” role to fulfill a function (e.g., to fill bone defects and avoid invasion from surrounding soft tissues or to play an imaging role as in magnetic resonance imaging (MRI)), but can also take an active or “chemical” part in biological processes. Bioactivity, for example, consists of supplying constitutive building blocks (e.g., ions) for promoting tissue repair or of activating biological pathways (e.g., favoring or orienting cell activity). Such compounds are referred to as “bioactive oxides.” This activity is directly related to features such as chemical composition (and related thermodynamic stability in body fluids), structure, morphology, degree of stoichiometry, crystal dimensions, surface state/reactivity, etc.
Among bioactive oxides, calcium phosphates (CaPs) represent by far the main group of compounds, used either as is or in combination with other compounds, and will be treated in the following. The family of active oxides, however, also encompasses other systems such as calcium carbonates and sulfates, which will also be briefly commented on at the end of this section.
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Solid-State Materials, Including Ceramics and Minerals
2013, Comprehensive Inorganic Chemistry II (Second Edition)S.P.N. Nair, P. Murugavel
4.03.1Introduction
Oxides are a wide class of materials with different crystal structures, displaying amazingly diverse properties. These properties make them quite useful in a large range of applications. This chapter is an attempt to understand the development in the field of oxides in the past decade. Some of the important properties of interest are their dielectric, ferroelectric, piezoelectric, magnetic, dilute magnetic semiconductor properties, ionic conductivity, fuel cell characteristics, and optical properties among others. It is a difficult task to put together all the oxides in one review article. In this chapter, we begin our discussion on the recent developments on the perovskite oxide materials. Heterostructure materials based on these perovskite oxides have surprisingly thrown up a range of exotic properties. The rest of the chapter focuses on the crystal structures with a close relation to perovskite structures. These include Aurivillius phases, Ruddlesden–Popper (RP) phases, pyrochlores, brownmillerites, etc. Some of the results described here are of fundamental interest, such as the field of spin ice and the observation of magnetic monopoles in spin ice structures. The Aurivillius phase on the other hand has found importance in the field of nonvolatile random access memory devices because of their stability and high ferroelectric Curie temperature. RP and brownmillerite materials are useful because of their complex structural and magnetic phenomenon as well as their oxide ion conduction behavior that can be used in the field of solid oxide fuel cells (SOFCs). Finally, we discuss the recent development in the field of spinels and delafossite structures.
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1085 | https://home.strw.leidenuniv.nl/~keller/Teaching/ATI_2016/ATI_2016_L03_PhysicalOptics.pdf | Lecture 3: Physical Optics Outline 1 Interference 2 Coherence 3 Two-Element Interferometer 4 Fraunhofer and Fresnel Diffraction 5 Transfer Functions 6 Van Cittert-Zernike Theorem Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 1 Very Large Telescope Interferometer (VLTI) Image credit: ESO Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 2 T Leporis: a Mira-like Star with the VLTI in the NIR Image credit: ESO/J.-B. Le Bouquin et al.
Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 3 3 Ways to Understanding Optics 1 Geometrical Optics: Light can be described as rays 2 Physical Optics: Light can be described as waves 3 Quantum Optics: Light can be described as discrete particles Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 4 Plane-Wave Solutions Plane Vector Wave ansatz: ⃗ E = ⃗ E0ei(⃗ k·⃗ x−ωt) ⃗ k spatially and temporally constant wave vector ⃗ k normal to surfaces of constant phase |⃗ k| wave number ⃗ x spatial location ω angular frequency (2π× frequency) t time ⃗ E0 a (generally complex) vector independent of time and space real electric field vector given by real part of ⃗ E Scalar Wave electric field at position ⃗ r at time t is ˜ E(⃗ r, t) complex notation to easily express amplitude and phase real part of complex quantity is the physical quantity Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 5 Interference Young’s Double Slit Experiment Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 6 Young’s Double Slit Experiment monochromatic wave infinitely small holes (pinholes) source S generates fields ˜ E(⃗ r1, t) ≡˜ E1(t) at S1 and ˜ E(⃗ r2, t) ≡˜ E2(t) at S2 two spherical waves from pinholes interfere on screen electrical field at P (without propagators i/λ) ˜ EP(t) = ˜ E1(t −t1) + ˜ E2(t −t2) t1 = r1/c, t2 = r2/c r1, r2: path lengths from S1, S2 to P Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 7 no tilt tilt by 0.5 λ/d Change in Angle of Incoming Wave phase of fringe pattern changes, but not fringe spacing tilt of λ/d produces identical fringe pattern Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 8 long wavelength short wavelength wavelength average Change in Wavelength fringe spacing changes, central fringe broadens integral over 0.8 to 1.2 of central wavelength intergral over wavelength makes fringe envelope Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 9 First Fringes from VLT Interferometer Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 10 Visibility en.wikipedia.org/wiki/Interferometric_visibility “quality” of fringes described by Visibility function V = Imax −Imin Imax + Imin Imax, Imin are maximum and adjacent minimum in fringe pattern first introduced by Michelson Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 11 Coherence Why Not Fringes Everywhere?
extended sources are not phase locked sources are not monochromatic Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 12 Mutual Coherence total field in point P (neglecting propagators) ˜ EP(t) = ˜ E1(t −t1) + ˜ E2(t −t2) irradiance at P, averaged over time (expectation operator E) I = E|˜ EP(t)|2 = E n ˜ EP(t)˜ E∗ P(t) o writing out all the terms I = E n ˜ E1(t −t1)˜ E∗ 1(t −t1) o + E n ˜ E2(t −t2)˜ E∗ 2(t −t2) o +E n ˜ E1(t −t1)˜ E∗ 2(t −t2) o + E n ˜ E∗ 1(t −t1)˜ E2(t −t2) o Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 13 Mutual Coherence (continued) as before I = E n ˜ E1(t −t1)˜ E∗ 1(t −t1) o + E n ˜ E2(t −t2)˜ E∗ 2(t −t2) o +E n ˜ E1(t −t1)˜ E∗ 2(t −t2) o + E n ˜ E∗ 1(t −t1)˜ E2(t −t2) o stationary wave field, time average independent of absolute time IS1 = E n ˜ E1(t)˜ E∗ 1(t) o , IS2 = E n ˜ E2(t)˜ E∗ 2(t) o irradiance at P is now I = IS1 + IS2 +E n ˜ E1(t −t1)˜ E∗ 2(t −t2) o + E n ˜ E∗ 1(t −t1)˜ E2(t −t2) o Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 14 Mutual Coherence (continued) as before I = IS1 + IS2 +E n ˜ E1(t −t1)˜ E∗ 2(t −t2) o + E n ˜ E∗ 1(t −t1)˜ E2(t −t2) o time difference τ = t2 −t1 ⇒last two terms become E n ˜ E1(t + τ)˜ E∗ 2(t) o + E n ˜ E∗ 1(t + τ)˜ E2(t) o equivalent to 2 Re h E n ˜ E1(t + τ)˜ E∗ 2(t) oi cross-term becomes 2Re h E n ˜ E1(t + τ)˜ E∗ 2(t) oi Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 15 Mutual Coherence (continued) irradiance at P I = IS1 + IS2 +2Re h E n ˜ E1(t + τ)˜ E∗ 2(t) oi mutual coherence function of wave field at S1 and S2 ˜ Γ12(τ) = E n ˜ E1(t + τ)˜ E∗ 2(t) o therefore I = IS1 + IS2 + 2 Re ˜ Γ12(τ) I1 = IS1, I2 = IS2: irradiances at P from single aperture I = I1 + I2 + 2 Re ˜ Γ12(τ) Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 16 Self-Coherence S1 = S2 ⇒mutual coherence function = autocorrelation ˜ Γ11(τ) = ˜ R1(τ) = E n ˜ E1(t + τ)˜ E∗ 1(t) o ˜ Γ22(τ) = ˜ R2(τ) = E n ˜ E2(t + τ)˜ E∗ 2(t) o autocorrelation functions are also called self-coherence functions for τ = 0 IS1 = E n ˜ E1(t)˜ E∗ 1(t) o = Γ11(0) = E n |˜ E1(t)|2o IS2 = E n ˜ E2(t)˜ E∗ 2(t) o = Γ22(0) = E n |˜ E2(t)|2o autocorrelation function with zero lag (τ = 0) represent (average) irradiance (power) of wave field at S1, S2 Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 17 Complex Degree of Coherence normalized mutual coherence defines the complex degree of coherence ˜ γ12(τ) ≡ ˜ Γ12(τ) p Γ11(0)Γ22(0) = E n ˜ E1(t + τ)˜ E∗ 2(t) o r E n |˜ E1(t)|2 o E n |˜ E2(t)|2 o irradiance in point P as general interference law for a partially coherent radiation field I = I1 + I2 + 2 p I1I2 Re ˜ γ12(τ) Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 18 Spatial and Temporal Coherence complex degree of coherence ˜ γ12(τ) ≡ ˜ Γ12(τ) p Γ11(0)Γ22(0) = E n ˜ E1(t + τ)˜ E∗ 2(t) o r E n |˜ E1(t)|2 o E n |˜ E2(t)|2 o measures both spatial coherence at S1 and S2 temporal coherence through time lag τ ˜ γ12(τ) is a complex variable and can be written as: ˜ γ12(τ) = |˜ γ12(τ)|eiψ12(τ) 0 ≤|˜ γ12(τ)| ≤1 phase angle ψ12(τ) relates to phase angle between fields at S1 and S2 phase angle difference in P resulting in time lag τ Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 19 Fringes in Fiber-fed High-Resolution Spectrograph Figure 7 from Rains et al. 2016, SPIE 9908, 990876 Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 20 Detector Fringes in HyViSi Detector Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 21 Coherence of Quasi-Monochromatic Light quasi-monochromatic light, mean wavelength λ, frequency ν, phase difference φ due to optical path difference: φ = 2π λ (r2 −r1) = 2π λ c(t2 −t1) = 2πντ with phase angle α12(τ) between fields at pinholes S1, S2 ψ12(τ) = α12(τ) −φ and Re ˜ γ12(τ) = |˜ γ12(τ)| cos [α12(τ) −φ] intensity in P becomes I = I1 + I2 + 2 p I1I2 |˜ γ12(τ)| cos [α12(τ) −φ] Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 22 Visibility of Quasi-Monochromatic, Partially Coherent Light intensity in P I = I1 + I2 + 2 p I1I2 |˜ γ12(τ)| cos [α12(τ) −φ] maximum, minimum I for cos(...) = ±1 visibility V at position P V = 2√I1 √I2 I1 + I2 |˜ γ12(τ)| for I1 = I2 = I0 I = 2I0 {1 + |˜ γ12(τ)| cos [α12(τ) −φ]} V = |˜ γ12(τ)| Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 23 Interpretation of Visibility for I1 = I2 = I0 I = 2I0 {1 + |˜ γ12(τ)| cos [α12(τ) −φ]} V = |˜ γ12(τ)| modulus of complex degree of coherence = visibility of fringes modulus can therefore be measured shift in location of central fringe (no optical path length difference, φ = 0) is measure of α12(τ) measurements of visibility and fringe position yield amplitude and phase of complex degree of coherence Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 24 Two-Element Interferometer Fringe Pattern for I1 = I2 = I0 I = 2I0 {1 + |˜ γ12(τ)| cos [α12(τ) −φ]} V = |˜ γ12(τ)| source S on central axis, fully coherent waves from two holes I = 2I0(1 + cos φ) = 4I0 cos2 φ 2 Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 25 Fringe Pattern (continued) I = 4I0 cos2 φ 2 φ = 2π λ (r2 −r1) = 2πντ distance a between pinholes distance s to observation plane ΣO, s ≫a path difference (r2 −r1) in equation for φ in good approximation r2 −r1 = aθ = a s y and therefore I = 4I0 cos2 πay sλ Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 26 Interference Fringes from Monochromatic Point Source irradiance as a function of the y-coordinate of the fringes in observation plane ΣO irradiance vs. distance distribution is Point-Spread Function (PSF) of ideal two-element interferometer Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 27 Diffraction Huygens-Fresnel Principle en.wikipedia.org/wiki/File:Refraction_on_an_aperture_-_Huygens-Fresnel_principle.svg every unobstructed point of a wavefront at a given moment in time serves as a source of spherical, secondary waves with the same frequency as the primary wave the amplitude of the optical field at any point beyond is the coherent superposition of all these secondary, spherical waves Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 28 Diffraction www.smkbud4.edu.my/Data/sites/vschool/phy/wave/diffraction.htm if obstructing structures are small compared to the wavelength, waves will spread out ⇒diffraction really need to solve wave equation with boundary constraints ⇒ riguorous solution for only a few special cases various numerical ways to solve such problems (e.g. Rigorous Coupled Wave Analysis) Huygens-Fresnel is useful for most applications Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 29 Fraunhofer and Fresnel Diffraction en.wikipedia.org/wiki/File:Fraunhofer_diffraction_pattern_image.PNG wave shape changes as it moves away from obstruction Fresnel (or near-field) diffraction close to obstruction Fraunhofer (or far-field) diffraction far away from obstruction rule of thumb: Fraunhofer diffraction for R > a2/λ a greatest width of obscuration λ wavelength R greater of distance between source/detector and obscuration Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 30 Slit Diffraction hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html diffraction of a plane wave on a slit aperture Huygens-Fresnel: line of spherical wave sources Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 31 Slit Diffraction hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html with E0 the strength of each slit segment i at point P is Ei(P) = EL ri sin(kω −kri)∆yi i segment index (1 −M) EL source strength per unit length ri distance between segment and point P ∆yi small segment of slit D length of slit Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 32 Fraunhofer Diffraction at Single Slit integrate along slit E = EL Z D/2 −D/2 sin ωt −kr r dy express r as a function of y: r = R −y sin θ + y2 2R cos2 θ + ...
R distance between center of slit and point P substituting, integrating and squaring for intensity: I(θ) = I(0) sin β β 2 β = (kD/2) sin θ Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 33 Interpretation of Single Slit Diffraction hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html assume infinite distance from aperture for source and observation plane equivalent to plane waves coming from aperture into different directions first minimum when phase delay at edge is exactly one wave Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 34 Diffraction Effects in High-Resolution Spectrographs Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 35 Fraunhofer Diffraction at Circular Aperture integrate over circular aperture with radius a E = EAei(ωt−kR) R Z Z aperture eik(Yy+Zz)/RdS using polar coordinates in aperture and plane of observation and Bessel functions I(θ) = I(0) 2J1(ka sin θ) ka sin θ 2 J1 Bessel function of order 1 Airy function first dark ring at 1.22Rλ 2a images with perfect telescopes are convolution of Airy disk with actual image Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 36 Arbitrary Diffracting Aperture from before forgetting common phase term and 1/R amplitude drop-off E(Y, Z) = Z Z aperture A(y, z)eik(Yy+Zz)/RdS complex aperture function A(y, z) describing non-uniform absorption and phase delays finite aperture ⇒change integration boundaries to infinity with ky = kY/R and kz = kZ/R we obtain E(ky, kz) = Z Z aperture A(y, z)ei(kyy+kzz)dy dz field distribution in Fraunhofer diffraction pattern is Fourier transform of field distribution across aperture Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 37 Transfer Functions Introduction linear black box system measure response to delta function input (transfer function) express output as convolution between input signal and transfer function Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 38 Point-Spread Function intensity is modulus squared of field distribution ⇒point-spread function image of a point source: Point Spread Function (PSF) image of arbitrary object is a convolution of object with PSF i = o ∗s i observed image o true object, constant in time s point spread function ∗convolution Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 39 Optical Transfer Function after Fourier transformation: I = O · S Fourier transformed I Fourier transform of image O Fourier transform of object S Optical Transfer Function (OTF) OTF is Fourier transform of PSF and vice versa Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 40 Modulation Transfer Function (MTF) is the absolute value of the optical transfer function describes the amplitude reduction of a sinusoidal source is the autocorrelation of the aperture function A OTF = FT−1(PSF) = FT−1(|FT(A)|2) = FT−1(FT(A)·FT(A)∗) = A∗A MTF of Circular Telescope with Central Obscuration ϵ Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 41 2 pinholes 2 small holes 2 large holes Finite Hole Diameter fringe spacing only depends on separation of holes and wavelength the smaller the hole, the larger the ’illuminated’ area fringe envelope is Airy pattern (diffraction pattern of a single hole) Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 42 2-d Brightness Distribution PSF of single circular aperture PSF of two-element interferometer, aperture diameter d = 25 m, length of baseline vector |⃗ s| = 144 m double beam interference fringes showing modulation effect of diffraction by aperture of a single pinhole Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 43 Modulation Effect of Aperture typical one-dimensional cross-section through central part of interferogram visibilities are equal to one, because Imin = 0 |˜ γ12(τ)| = 1 for all values of τ and any pair of spatial points, if and only if the radiation field is strictly monochromatic Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 44 Van Cittert-Zernike Theorem The Problem relates brightness distribution of extended source and phase correlation between two points in radiation field extended source S incoherent, quasi-monochromatic positions P1 and P2 in observers plane Σ ˜ E1(t)˜ E∗ 2(t) = E{˜ E1(t)˜ E∗ 2(t)} = ˜ Γ12(0) Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 45 The Solution I(⃗ Ω) is intensity distribution of extended source as function of unit direction vector ⃗ Ωas seen from observation plane Σ ˜ Γ(⃗ r) is coherence function in Σ-plane vector ⃗ r represents arbitrary baseline van Cittert-Zernike theorem ˜ Γ(⃗ r) = Z Z source I(⃗ Ω)e 2πi⃗ Ω.
⃗ r λ d⃗ Ω I(⃗ Ω) = λ−2 Z Z Σ-plane ˜ Γ(⃗ r)e−2πi⃗ Ω.
⃗ r λ d⃗ r ˜ Γ(⃗ r) and I(⃗ Ω) are linked through Fourier transform, except for scaling with wavelength λ "true" Fourier transform with conjugate variables ⃗ Ω, ⃗ r/λ, Fourier pair: I(⃗ Ω) ⇔ ˜ Γ(⃗ r/λ) Christoph U. Keller, Leiden University, keller@strw.leidenuniv.nl Lecture 3: Physical Optics 46 |
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Converting numbers to percentages is a fundamental skill in data analysis, and Google Sheets provides a powerful tool to streamline this process. Whether you're working on financial analysis, grading students, or evaluating performance metrics, understanding how to convert numbers to percentages in Google Sheets is essential. In this guide, we'll walk you through the simple steps to master this skill and enhance your data analysis capabilities.
Introduction to Converting Numbers to Percentages
A percentage is a way to express a number as a fraction of 100, often denoted using the percent sign (%). Converting numbers to percentages is a crucial skill in data analysis, as it allows you to compare and interpret data more effectively. Here are some scenarios where converting numbers to percentages in Google Sheets can be particularly helpful:
Bardeen can help you gather data from various sources directly into your Google Sheet, saving you time and effort. Try these playbooks to automate data import:
By converting numbers to percentages, you can gain valuable insights from your data and make more informed decisions. Google Sheets provides a simple yet powerful tool to perform these calculations efficiently. To further improve your workflow, consider using GPT for Google Sheets.
Setting Up Your Google Sheet for Percentage Calculations
Before you start calculating percentages in Google Sheets, it's essential to format your cells properly. Here's how to set up your sheet for smooth percentage calculations:
Proper cell formatting ensures that your percentage calculations are displayed correctly and are easy to read. Another crucial aspect of setting up your sheet is using cell referencing in your formulas. Instead of typing in static numbers, reference the cells containing the values you want to calculate. This approach has several benefits:
By formatting your cells as percentages and using cell referencing in your formulas, you'll be well-prepared to perform efficient and accurate percentage calculations in Google Sheets.
Once your cells are formatted, you can use Bardeen to automate importing data from various sources directly into your Google Sheet. This saves time on manual data entry, allowing you to focus on analyzing the percentage calculations. Try these playbooks to bring data into Google Sheets:
Basic Percentage Calculation Formula in Google Sheets
The fundamental formula for calculating percentages in Google Sheets is (Part/Total) 100. This formula calculates what percentage a part is of a total. To input this formula in Google Sheets:
Bardeen can help automate tasks that happen before or after your percentage calculations in Google Sheets. For example, you can use these playbooks to qualify leads from various sources and save that data to Google Sheets, setting you up to then calculate conversion percentages or other metrics:
For example, if cell A2 contains the part value and cell B2 contains the total value, the formula would be:
=A2/B2100
Remember to format the result as a percentage by selecting the cell and clicking Format > Number > Percent in the menu. This basic formula makes it easy to quickly calculate percentages using cell references in Google Sheets. For more advanced tasks, you can connect Google Sheets with Excel using Bardeen.
Once you've calculated key percentages, Bardeen can also help you generate reports from that data to share with stakeholders. Try these playbooks:
Advanced Percentage Calculations: Conditional Percentages
In some cases, you may need to calculate percentages based on specific conditions. This is where the IF function in Google Sheets comes in handy. The IF function allows you to create a formula that returns different values based on whether a condition is true or false. When combined with percentage formulas, you can create powerful conditional calculations.
For example, let's say you want to calculate employee bonuses based on their performance rating:
To set up this conditional percentage calculation in Google Sheets:
The final formula would look like this:
=IF(A2>=0.9,0.1,IF(A2>=0.8,0.05,0))
For more advanced uses, you can automate enrichment and qualification tasks in Google Sheets. You can apply similar conditional percentage calculations to various scenarios, such as calculating discounts based on purchase volume or determining grade distributions based on test scores. By combining the power of the IF function with percentage formulas, you can create dynamic, condition-based calculations in Google Sheets.
Bardeen can help you save even more time by automating the process of enriching and qualifying data in Google Sheets. With just a few clicks, you can bring in additional data points and filter rows based on specific criteria. Try these playbooks to streamline your workflow:
Utilizing Google Sheets Functions for Percentage Operations
Google Sheets offers several built-in functions that can simplify complex percentage calculations, especially when working with large datasets. These functions automate the process and save time by eliminating the need for manual calculations.
Some useful functions for percentage operations include:
Bardeen can further automate percentage-related tasks in Google Sheets, saving you even more time from manual calculations. Try these playbooks to enrich your existing data with new percentage metrics:
To use these functions, simply input the range of cells containing your data and the value or percentile you want to calculate. For example, to find the 75th percentile of values in cells A1:A100, you would use the formula:
=PERCENTILE(A1:A100, 0.75)
These functions can be particularly helpful when analyzing large datasets, such as sales figures or survey responses. By utilizing them, you can quickly identify key percentiles, ranks, and trends without the need for time-consuming manual calculations.
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1087 | https://www.osmosis.org/learn/Muscles_of_the_back | Fall in love with Osmosis at 20% off! Save now until September 30 at 11:59 PM PT.
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Muscles of the back
Foundational SciencesAnatomyBackGross anatomy
6,341views
Muscles of the back
Back & Vertebral Column
Playlist
Anatomical terminology
Anatomy of the vertebral canal
Bones of the vertebral column
Joints of the vertebral column
Muscles of the back
Vessels and nerves of the vertebral column
Anatomy of the suboccipital region
Anatomy clinical correlates: Vertebral canal
Anatomy clinical correlates: Spinal cord pathways
Anatomy clinical correlates: Bones, joints and muscles of the back
Spinal fractures: Clinical sciences
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Notes
Anatomy of Muscles of the Back
Figure 1: Superficial Extrinsic back muscles. A. Superficial subgroup. B. Deep subgroup (trapezius and latissimus dorsi removed).
Figure 2: A. Superficial intrinsic back muscles. B.Intermediate intrinsic back muscles.
Figure 3: A. Intermediate intrinsic back muscles. B.Deep intrinsic back muscles (transversospinales).
Figure 4: Semispinalis (deep intrinsic back muscles).
Figure 5: Dorsal scapular artery and nerve. Dorsal scapular artery as: A. the deep branch of the transverse cervical artery or B. a direct branch from the subclavian artery
MUSCLE TABLE
| | | | | |
--- ---
| Muscle | Origin | Insertion | Innervation | Action |
| Superficial Extrinsic Back Muscles | | | | |
| Trapezius | Medial third of superior nuchal line External occipital protuberance Nuchal ligament Spinous processes of C7-T12 | Lateral 1/3 of clavicle Acromion & spine of the scapula | Motor: Spinal accessory nerve (CN XI) Sensory: Anterior Rami C3-C4 | Desc. part: elevation of scapula Middle part: retraction of scapula Asc. part: depression of scapula (lowering shoulder) Asc. + desc.: upward rotation of scapula |
| Latissimus Dorsi | Iliac crest Thoracolumbar fascia Inferior six thoracic vertebrae Inferior three or four ribs | Floor of intertubercular groove of the humerus | Thoracodorsal nerve (C6, C7, C8) | Extends, adducts and medially rotates the humerus |
| Rhomboid Major | Spinous processes of T2-T5 vertebrae | Medial border of scapula inferior to spine of scapula | Dorsal Scapular nerve (C4, C5) | Retraction and downward rotation of scapula Assist the serratus anterior in holding the scapula against the thoracic wall |
| Rhomboid Minor | Nuchal ligament Spinous processes of C7 & T1 vertebrae | Medial border of scapula at spine of scapula |
| Levator Scapulae | Posterior tubercles of the transverse processes C1-C4 | Medial border of the scapula (superior to the spine of the scapula) | Dorsal Scapular nerve (C4-C5) Anterior rami C3-C4 | Elevates scapula Fixes scapula (resists downward force) Downward rotation of scapula |
| | | | | |
--- ---
| Muscle | Origin | Insertion | Innervation | Action |
| Intermediate Extrinsic Back Muscles | | | | |
| Serratus Posterior Superior | Nuchal ligament Spinous processes C7-T3 | Superior borders of 2nd-5th ribs | Anterior rami T2-T5 | Elevate superior ribs during inspiration |
| Serratus Posterior Inferior | Spinous processes T11-L2 | Inferior borders of 9th-12th ribs | Anterior rami T9-T12 | Proprioception Depress(es) inferior ribs (controversial) |
| | | | | |
--- ---
| Muscle | Origin | Insertion | Innervation | Action |
| Superficial Extrinsic Back Muscles | | | | |
| Splenius Capitis | Nuchal ligament Spinous processes of C7-T4 | Mastoid process of the temporal bone Lateral 1/3 of superior nuchal line of the occipital bone | Posterior rami of the middlecervical spinal nerves | Unilaterally: Laterally flex neck, ipsilateral rotation Bilaterally: Extend head and neck |
| Splenius Cervicis | Spinous processes of T3-T6 | Posterior tubercles of the transverse processes of C1-C3 | Posterior rami of the lowercervical spinal nerves |
| | | | | |
--- ---
| Muscle | Origin | Insertion | Innervation | Action |
| Intermediate Extrinsic Back Muscles (Erector Spinae) | | | | |
| Iliocostalis | Broad tendon that attaches to: + Posterior part of the iliac crest + Posterior surface of the sacrum Sacroiliac ligaments Sacral and inferior lumbar spinous processes Supraspinous ligament | Iliocostalis cervicis Transverse processes of cervical vertebrae Iliocostalis thoracis: Angles of lower ribs Iliocostalis lumborum: Transverse processes of lumbar vertebrae | Posterior rami of spinal nerves | Unilaterally: Laterally flex vertebral column Bilaterally: Extend vertebral column and head When back is flexed: Control movement via eccentric contraction |
| Longissimus | Longissimus Capitis: Mastoid process of temporal bone Longissimus Cervicis: Transverse processes of cervical vertebrae Longissimus Thoracis: Ribs between tubercles and angles Transverse processes of thoracic vertebrae |
| Spinalis | Spinalis capitis: Occipital bone Spinalis Cervicis: Spinous processes in cervical region Spinalis Thoracis: Spinous processes in upper thoracic region variable, may be absent or fused with other muscles |
| | | | | |
--- ---
| Muscle | Origin | Insertion | Innervation | Action |
| Deep Intrinsic back muscles (transversospinales) | | | | |
| Semispinalis | Transverse processes of C4-T12 | Occipital bone Semispinalis: thoracis, cervicis, capitis: fibers run superomedially to occipital bone and spinous processes in thoracic and cervical regions, spanning 4–6 segments | Posterior rami of spinal nerves | Bilaterally: Extend head and cervical and thoracic regions Unilaterally: Contralateral rotation |
| Multifidus | Posterior sacrum Posterior superior iliac spine of the ilium Aponeurosis of the erector spinae Mammillary processes of the lumbar vertebrae, Transverse processes of the thoracic vertebrae Articular processes of C4 to C7 | Spinous processes located 2 to 4 segments superior to their origin | Stabilize vertebrae during local movements of the vertebral column |
| Rotatores (Brevis and longus) | Transverse processes | Lamina/spinous process of process 1 (brevis) to 2 (longus) segments above origin | Contralateral rotation of the vertebral column Stabilize vertebrae Assist with extension Proprioception |
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Illustrator: Patricia Nguyen, MScBMC
Editor: Andrew Horne, MSc., BSc.
Editor: Leah Labranche, PhD, MSc, BSc(Hons)
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Patricia Nguyen, MScBMC
Here at Osmosis, we support each other - we’ve got each other’s backs. And it turns out, our backs support all of us! Let’s look at the muscles of the back that help maintain our posture and also provide movements of the trunk and upper limb.
The muscles of the back are divided into two major groups: the extrinsic back muscles and the intrinsic back muscles. The extrinsic back muscles include the superficial back muscles, which produce and control upper limb movements, and the intermediate back muscles, which are thought to be involved in respiratory movements.
The superficial extrinsic back muscles include the trapezius, latissimus dorsi, levator scapulae, and the two rhomboids - rhomboid major and rhomboid minor.
All these muscles are posterior axioappendicular muscles, which connect the axial skeleton, primarily the vertebral column, to the superior appendicular skeleton, specifically the pectoral girdle and the humerus.
The posterior axioappendicular group is divided into two subgroups. The first subgroup consists of two superficial muscles: the trapezius and latissimus dorsi.
The trapezius is a large triangular muscle that covers the posterior aspect of the neck and the superior half of the back. There are two trapezius muscles in the back, which when seen together, look like a trapezium.
Proximally, the trapezius originates on the medial third of the superior nuchal line, the external occipital protuberance, the nuchal ligament, and the spinous processes of the C7 to T12 vertebrae.
Distally, the trapezius inserts on the lateral third of the clavicle, the acromion, and the spine of the scapula. Based on the orientation of its fibers, the trapezius is divided into the descending - or superior - part, the middle part, and the ascending - or inferior - part.
Contraction of the descending part results in elevation of the scapula, like when shrugging the shoulders. Contraction of the middle part retracts the scapula, and contraction of the ascending part depresses the scapula, lowering the shoulder.
And finally, when the ascending and descending fibers work together, it results in upward rotation of the scapula which causes the glenoid cavity to tilt superiorly.
The trapezius is innervated by a plexus of nerves formed by the spinal accessory nerve - or cranial nerve 11 - which provides motor innervation to the muscle, as well as by branches from the anterior rami of the C3 and C4 spinal nerves which carry sensory information - specifically pain and proprioception - from the muscle.
The arterial supply for the trapezius is primarily from the superficial branch of the transverse cervical artery, which arises from the thyrocervical trunk. These neurovascular structures travel through the posterior triangle and then along the deep surface of the trapezius muscle.
The second superficial muscle is latissimus dorsi, which extends from the trunk to the humerus, acting directly on the glenohumeral joint and indirectly on the pectoral girdle. The latissimus dorsi originates on the iliac crest, thoracolumbar fascia, spinous processes of the inferior six thoracic vertebrae, and the inferior three or four ribs.
Distally, the latissimus dorsi inserts on the floor of the intertubercular groove of the humerus. The latissimus dorsi is innervated by the thoracodorsal nerve, which consists of fibers from the anterior rami of C6, C7, and C8.
The contraction of latissimus dorsi extends, adducts and medially rotates the humerus, like when trying to scratch that annoying itch in the middle of your back. In fact, because it is such a powerful extensor and adductor, this muscle is used when pulling yourself up during rock climbing, chin ups, or movements such as swimming or chopping wood.
Okay, next up, we have the deep subgroup of the posterior axioappendicular muscles, consisting of the levator scapulae and the two rhomboids.
The levator scapulae is found in the neck, deep to the sternocleidomastoid and trapezius muscles. Proximally, it originates on the posterior tubercles of the transverse processes of the C1 to C4 vertebrae and then descends to insert on the medial border of the scapula, superior to the spine of the scapula.
“Levator” roughly translates to “lifter”, so when this muscle contracts, it elevates - or ‘lifts’ - the scapula. It also fixes the scapula and holds it in place to resist downward forces like when carrying a heavy piece of luggage up the stairs. It also produces downward rotation of the scapula, causing the glenoid cavity to tilt inferiorly.
Now, let’s have a look at the rhomboids, which are located immediately deep to trapezius. There are two muscles: rhomboid minor and rhomboid major, where the rhomboid major muscle is found immediately below the rhomboid minor.
Rhomboid minor originates on the nuchal ligament and spinous processes of the C7 and T1 vertebrae and inserts on the medial border of the scapula at the level of the spine of the scapula. The much larger rhomboid major originates on the spinous processes of the T2 to T5 vertebrae and inserts on the medial border of the scapula inferior to the spine of the scapula.
Rhomboid minor and major always contract together to produce retraction and downward rotation of the scapula. They also assist the serratus anterior in holding the scapula against the thoracic wall. These movements enable us to forcibly lower the arm, like when driving a stake... but not, like, literally driving a steak.
The rhomboids and the levator scapulae are all innervated by the dorsal scapular nerve - which arises from the anterior rami of C4 and C5, with the levator scapulae also receiving branches directly from the anterior rami of the C3 and C4 spinal nerves.
Now, the dorsal scapular nerve is accompanied by the dorsal scapular artery, which may arise directly from the subclavian artery, or as a branch from the transverse cervical artery, in which case it is also referred to as the deep branch of the transverse cervical artery.
Regardless of its origin, the artery buddies up with the dorsal scapular nerve, where they travel deep to the levator scapulae, and continue deep to the rhomboids, descending parallel to the medial border of the scapula, supplying all three muscles throughout their course.
Now, deep to the posterior axioappendicular muscles, are the intermediate extrinsic back muscles which are the two thin serratus posterior muscles. On each side, there’s a serratus posterior superior muscle and serratus posterior inferior muscle.
Serratus posterior superior originates from the nuchal ligament and the spinous processes of the C7 through T3 vertebrae and inserts on the superior borders of 2nd through 5th ribs. Based on these attachments, it is thought that this muscle acts to elevate these superior ribs during inspiration, and it is innervated by the anterior rami of the T2 to T5 spinal nerves.
The serratus posterior inferior originates from the spinous processes of the T11 through L2 vertebrae and inserts at the inferior borders of the 9th through 12th ribs near their angles. Based on these attachments, it is thought to depress the inferior ribs during inspiration , and it is innervated by the anterior rami of the T9 to T12 spinal nerves.
Although these muscles are often designated as respiratory muscles that cause movement of the ribs, these actions are controversial with many sources suggesting they may function primarily in proprioception by providing sensory information about the position of the vertebral column and ribs.
Let’s take a quick break and try to identify all the extrinsic back muscles.
Now, the intrinsic back muscles - also sometimes called the deep back muscles - are actually innervated by the posterior rami of spinal nerves, and function to maintain posture and control movements of the vertebral column.
They extend from the pelvis to the cranium and are enclosed by deep fascia which attaches medially to the nuchal ligament, the tips of the spinous processes of the vertebrae, the supraspinous ligament, and the median crest of the sacrum. Laterally, this fascia attaches to the cervical and lumbar transverse processes and the angles of the ribs.
The portions of the deep fascia in the thoracic and lumbar regions are collectively referred to as the thoracolumbar fascia, with the fascia forming a thin covering over the intrinsic back muscles in the thoracic region and a tough, thick covering in the lumbar region.
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1088 | https://www.geogebra.org/m/ugaQEBGc | Constructing Trapezoids – GeoGebra
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1089 | https://www.12000.org/my_courses/univ_wisconsin_madison/spring_2017/math_320/HWs/HW12/HW12.pdf | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 HW 12, Math 320, Spring 2017 Nasser M. Abbasi (Discussion section 383, 8:50 AM - 9:40 AM Monday) December 30, 2019 Contents 0.1 Section 5.1 problem 52 (page 299) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 0.2 Section 5.1 problem 54 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 0.3 Section 5.2 problem 40 (page 311) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 0.4 Section 5.5 problem 9 (page 351) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 0.5 Section 5.5 problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 0.6 Section 5.5 problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 0.7 Section 5.5 problem 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 0.8 Section 5.5 problem 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 0.9 Section 5.5 problem 37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 0.10 Section 5.5 problem 49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 0.11 Section 5.5 problem 50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 0.12 Section 5.5 problem 53 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 0.13 Section 5.5 problem 61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 0.14 Section 5.5 problem 62 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 2 0.1 Section 5.1 problem 52 (page 299) Problem Make the substitution 𝑣= ln 𝑥to find general solution for 𝑥> 0 of the Euler equation 𝑥2𝑦′′ + 𝑥𝑦′ −𝑦= 0 solution Let 𝑣= ln 𝑥. Hence 𝑑𝑦 𝑑𝑥= 𝑑𝑦 𝑑𝑣 𝑑𝑣 𝑑𝑥= 𝑑𝑦 𝑑𝑣 1 𝑥and 𝑑2𝑦 𝑑𝑥2 = 𝑑 𝑑𝑥൭ 𝑑𝑦 𝑑𝑣 1 𝑥൰ = 𝑑2𝑦 𝑑𝑣2 𝑑𝑣 𝑑𝑥 1 𝑥+ 𝑑𝑦 𝑑𝑣 𝑑 𝑑𝑥൭ 1 𝑥൰ = 𝑑2𝑦 𝑑𝑣2 1 𝑥2 −𝑑𝑦 𝑑𝑣 1 𝑥2 Hence the ODE becomes 𝑥2𝑦′′ + 𝑥𝑦′ −𝑦= 0 𝑥2 ൭ 𝑑2𝑦 𝑑𝑣2 1 𝑥2 −𝑑𝑦 𝑑𝑣 1 𝑥2 ൰+ 𝑥൭ 𝑑𝑦 𝑑𝑣 1 𝑥൰−𝑦(𝑣) = 0 𝑑2𝑦 𝑑𝑣2 −𝑑𝑦 𝑑𝑣+ 𝑑𝑦 𝑑𝑣−𝑦(𝑣) = 0 𝑑2𝑦 𝑑𝑣2 −𝑦(𝑣) = 0 This can now be solved using characteristic equation. 𝑟2 −1 = 0 or 𝑟2 = 1 or 𝑟= ±1. Hence the solution is 𝑦(𝑣) = 𝑐1𝑒𝑣+ 𝑐2𝑒−𝑣 But 𝑣= ln 𝑥, hence 𝑦(𝑥) = 𝑐1𝑒ln 𝑥+ 𝑐2𝑒−ln 𝑥 = 𝑐1𝑥+ 𝑐1 1 𝑥 The above is the solution.
But an easier method is the following. Let 𝑦= 𝑥𝑟. Hence 𝑦′ = 𝑟𝑥𝑟−1, 𝑦′′ = 𝑟(𝑟−1) 𝑥𝑟−2. Substituting this into the ODE gives 𝑟(𝑟−1) 𝑥𝑟+ 𝑟𝑥𝑟−𝑥𝑟= 0 𝑥𝑟(𝑟(𝑟−1) + 𝑟−1) = 0 Since 𝑥𝑟≠0, we simplify the above and obtain the characteristic equation 𝑟(𝑟−1) + 𝑟−1 = 0 𝑟2 −1 = 0 𝑟2 = 1 𝑟= ±1 Hence 𝑦(𝑥) = 𝑐1𝑥𝑟1 + 𝑐2𝑥𝑟2 = 𝑐1𝑥+ 𝑐2𝑥−1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 3 For 𝑥> 0.
0.2 Section 5.1 problem 54 Problem Make the substitution 𝑣= ln 𝑥to find general solution for 𝑥> 0 of the Euler equation 4𝑥2𝑦′′ + 8𝑥𝑦′ −3𝑦= 0 solution Let 𝑣= ln 𝑥. Hence 𝑑𝑦 𝑑𝑥= 𝑑𝑦 𝑑𝑣 𝑑𝑣 𝑑𝑥= 𝑑𝑦 𝑑𝑣 1 𝑥and 𝑑2𝑦 𝑑𝑥2 = 𝑑 𝑑𝑥൭ 𝑑𝑦 𝑑𝑣 1 𝑥൰ = 𝑑2𝑦 𝑑𝑣2 𝑑𝑣 𝑑𝑥 1 𝑥+ 𝑑𝑦 𝑑𝑣 𝑑 𝑑𝑥൭ 1 𝑥൰ = 𝑑2𝑦 𝑑𝑣2 1 𝑥2 −𝑑𝑦 𝑑𝑣 1 𝑥2 Hence the ODE becomes 𝑥2𝑦′′ + 𝑥𝑦′ −𝑦= 0 4𝑥2 ൭ 𝑑2𝑦 𝑑𝑣2 1 𝑥2 −𝑑𝑦 𝑑𝑣 1 𝑥2 ൰+ 8𝑥൭ 𝑑𝑦 𝑑𝑣 1 𝑥൰−3𝑦(𝑣) = 0 4𝑑2𝑦 𝑑𝑣2 −4𝑑𝑦 𝑑𝑣+ 8𝑑𝑦 𝑑𝑣−3𝑦(𝑣) = 0 4𝑑2𝑦 𝑑𝑣2 + 4𝑑𝑦 𝑑𝑣−3𝑦(𝑣) = 0 This can now be solved using characteristic equation. 4𝑟2 + 4𝑟−3 = 0, whose roots are 𝑟1 = −3 2 , 𝑟2 = 1 2 Hence the solution is 𝑦(𝑣) = 𝑐1𝑒 −3 2 𝑣+ 𝑐2𝑒 1 2 𝑣 But 𝑣= ln 𝑥, hence 𝑦(𝑥) = 𝑐1𝑒 −3 2 ln 𝑥+ 𝑐2𝑒 1 2 ln 𝑥 = 𝑐1𝑥 −3 2 + 𝑐1𝑥 1 2 0.3 Section 5.2 problem 40 (page 311) Problem Use reduction of order to find second L.I. solution 𝑦2. 𝑥2𝑦′′ −𝑥(𝑥+ 2) 𝑦′ + (𝑥+ 2) 𝑦= 0 with 𝑦1 = 𝑥and 𝑥> 0 solution Let 𝑦= 𝑣𝑦1, hence 𝑦′ = 𝑣′𝑦1 + 𝑣𝑦′ 1 𝑦′′ = 𝑣′′𝑦1 + 𝑣′𝑦′ 1 + 𝑣′𝑦′ 1 + 𝑣𝑦′′ 1 = 𝑣′′𝑦1 + 2𝑣′𝑦′ 1 + 𝑣𝑦′′ 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 4 Therefore the original ODE becomes 𝑥2𝑦′′ −𝑥(𝑥+ 2) 𝑦′ + (𝑥+ 2) 𝑦= 0 𝑥2 ൫𝑣′′𝑦1 + 2𝑣′𝑦′ 1 + 𝑣𝑦′′ 1 ൮−𝑥(𝑥+ 2) ൫𝑣′𝑦1 + 𝑣𝑦′ 1൮+ (𝑥+ 2) ൫𝑣𝑦1൮= 0 𝑣′′ ൫𝑥2𝑦1൮+ 𝑣′ ൫2𝑥2𝑦′ 1 −𝑥(𝑥+ 2) 𝑦1൮+ 𝑣 0 ാുുുുുുുുുുുുുുുുുുുുുിുുുുുുുുുുുുുുുുുുുുുീ ൫𝑥2𝑦′′ 1 −𝑥(𝑥+ 2) 𝑦′ 1 + (𝑥+ 2) 𝑦1൮= 0 Hence 𝑣′′ ൫𝑥2𝑦1൮+ 𝑣′ ൫2𝑥2𝑦′ 1 −𝑥(𝑥+ 2) 𝑦1൮= 0 But 𝑦1 = 𝑥, hence the above becomes 𝑥3𝑣′′ + 𝑣′ ൫2𝑥2 −𝑥(𝑥+ 2) 𝑥൮= 0 𝑥3𝑣′′ −𝑥3𝑣′ = 0 Since we are told 𝑥> 0 when we can divide by 𝑥3 and obtain 𝑣′′ −𝑣′ = 0 To solve the above, let 𝑧= 𝑣′ Therefore 𝑧′ −𝑧= 0 or 𝑑 𝑑𝑥(𝑧𝑒𝑥) = 0 or 𝑧𝑒𝑥= 𝑐1 or 𝑧= 𝑐1𝑒−𝑥. Therefore the above becomes 𝑣′ = 𝑐1𝑒−𝑥 Integrating 𝑣= 𝑐2 −𝑐1𝑒−𝑥 Since 𝑦= 𝑣𝑦1 therefore 𝑦= 𝑦1 (𝑐2 −𝑐1𝑒−𝑥) But 𝑦1 = 𝑥, hence the complete solution is 𝑦= 𝑐2𝑥−𝑐1𝑥𝑒−𝑥 Therefore, we see now that the two basis solutions are 𝑦1 = 𝑥 𝑦2 = 𝑥𝑒𝑥 These can be shown to be L.I. using the Wronskian as follows 𝑊(𝑥) = ૬ 𝑦1 𝑦2 𝑦′ 1 𝑦′ 2 ૬ = ૬ 𝑥 𝑥𝑒𝑥 1 𝑒𝑥+ 𝑥𝑒𝑥૬ = 𝑥𝑒𝑥+ 𝑥2𝑒𝑥−𝑥𝑒𝑥 = 𝑥2𝑒𝑥 Which is not zero since we are told 𝑥> 0. Hence indeed the second basis solution 𝑦2 found is L.I. to 𝑦1.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 5 0.4 Section 5.5 problem 9 (page 351) Problem Find the particular solution for 𝑦′′ + 2𝑦′ −3𝑦= 1 + 𝑥𝑒𝑥 solution First we find the homogenous solution. This will tell us if 𝑒𝑥is one of the basis solutions of not, so we know what to guess. The characteristic equation is 𝑟2 + 2𝑟−3 = 0 (𝑟−1) (𝑟+ 3) = 0 Hence 𝑦1 = 𝑒𝑥, 𝑦2 = 𝑒−3𝑥. 𝑒𝑥is a solution to the homogeneous ODE. The guess is therefore 𝑦𝑝= 𝐴+ (𝐵+ 𝐶𝑥) 𝑥𝑒𝑥 = 𝐴+ ൫𝐵𝑥+ 𝐶𝑥2൮𝑒𝑥 (1) Hence 𝑦′ 𝑝= (𝐵+ 2𝐶𝑥) 𝑒𝑥+ ൫𝐵𝑥+ 𝐶𝑥2൮𝑒𝑥 = 𝑒𝑥൫𝐵+ 2𝐶𝑥+ 𝐵𝑥+ 𝐶𝑥2൮ 𝑦′′ 𝑝= (2𝐶+ 𝐵+ 2𝐶𝑥) 𝑒𝑥+ 𝑒𝑥൫𝐵+ 2𝐶𝑥+ 𝐵𝑥+ 𝐶𝑥2൮ = 𝑒𝑥൫2𝐶+ 𝐵+ 2𝐶𝑥+ 𝐵+ 2𝐶𝑥+ 𝐵𝑥+ 𝐶𝑥2൮ = 𝑒𝑥൫2𝐶+ 2𝐵+ 4𝐶𝑥+ 𝐵𝑥+ 𝐶𝑥2൮ Plugging into the ODE 𝑒𝑥൫2𝐶+ 2𝐵+ 4𝐶𝑥+ 𝐵𝑥+ 𝐶𝑥2൮+ 2𝑒𝑥൫𝐵+ 2𝐶𝑥+ 𝐵𝑥+ 𝐶𝑥2൮−3 ൫𝐴+ ൫𝐵𝑥+ 𝐶𝑥2൮𝑒𝑥൮= 1 + 𝑥𝑒𝑥 𝑒𝑥(2𝐶+ 2𝐵+ 2𝐵) + 𝑥𝑒𝑥(4𝐶+ 𝐵+ 4𝐶+ 2𝐵−3𝐵) + 𝑥2𝑒𝑥(𝐶+ 2𝐶−3𝐶) −3𝐴= 1 + 𝑥𝑒𝑥 𝑒𝑥(2𝐶+ 4𝐵) + 𝑥𝑒𝑥(8𝐶) −3𝐴= 1 + 𝑥𝑒𝑥 Hence −3𝐴= 1 or 𝐴= − 1 3 and 8𝐶= 1 Or 𝐶= 1 8 And 2𝐶+ 4𝐵= 0 Or 𝐵= −1 16 Hence particular solution becomes, from (1) 𝑦𝑝= −1 3 + ൭−1 16𝑥+ 1 8𝑥2൰𝑒𝑥 = −1 3 + 1 16 ൫2𝑥2 −𝑥൮𝑒𝑥 0.5 Section 5.5 problem 10 Problem Find the particular solution for 𝑦′′ + 9𝑦= 2 cos 3𝑥+ 3 sin 3𝑥 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 6 solution First we find the homogenous solution. The characteristic equation is 𝑟2 + 9 = 0 𝑟2 = −9 𝑟= ±3𝑖 Hence 𝑦1 = 𝑒3𝑖𝑥, 𝑦2 = 𝑒−3𝑖𝑥or 𝑦ℎ= 𝑐1 cos 3𝑥+ 𝑐2 sin 3𝑥. We see that cos 3𝑥and sin 3𝑥are already in the homogeneous solution. Therefore the guess is 𝑦𝑝= 𝐴𝑥cos 3𝑥+ 𝐵𝑥sin 3𝑥 Hence 𝑦′ 𝑝= 𝐴cos 3𝑥−3𝐴𝑥sin 3𝑥+ 𝐵sin 3𝑥+ 3𝐵𝑥cos 3𝑥 𝑦′′ 𝑝= −3𝐴sin 3𝑥−3𝐴sin 3𝑥−9𝐴𝑥cos 3𝑥+ 3𝐵cos 3𝑥+ 3𝐵cos 3𝑥−9𝐵𝑥sin 3𝑥 Substitution into the ODE gives (−3𝐴sin 3𝑥−3𝐴sin 3𝑥−9𝐴𝑥cos 3𝑥+ 3𝐵cos 3𝑥+ 3𝐵cos 3𝑥−9𝐵𝑥sin 3𝑥) + 9 (𝐴𝑥cos 3𝑥+ 𝐵𝑥sin 3𝑥) = 2 cos 3𝑥+ 3 sin 3𝑥 Or −6𝐴sin 3𝑥−9𝐴𝑥cos 3𝑥+ 6𝐵cos 3𝑥−9𝐵𝑥sin 3𝑥+ 9𝐴𝑥cos 3𝑥+ 9𝐵𝑥sin 3𝑥= 2 cos 3𝑥+ 3 sin 3𝑥 sin 3𝑥(−6𝐴) + cos 3𝑥(6𝐵) + 𝑥sin 3𝑥(−9𝐵+ 9𝐵) + 𝑥cos 3𝑥(−9𝐴+ 9𝐴) = 2 cos 3𝑥+ 3 sin 3𝑥 −6𝐴sin 3𝑥+ 6𝐵cos 3𝑥= 2 cos 3𝑥+ 3 sin 3𝑥 Hence −6𝐴= 3 or 𝐴= −1 2 and 6𝐵= 2 or 𝐵= 1 3, therefore the particular solution is 𝑦𝑝= −1 2 𝑥cos 3𝑥+ 1 3𝑥sin 3𝑥 = 1 6 (2𝑥sin 3𝑥−3𝑥cos 3𝑥) 0.6 Section 5.5 problem 16 Problem Find the particular solution for 𝑦′′ + 9𝑦= 2𝑥2𝑒3𝑥+ 5 solution From the above problem, we found 𝑦ℎ= 𝑐1 cos 3𝑥+ 𝑐2 sin 3𝑥. Therefore there are no basis solutions in the RHS which are in the homogenous solution. The guess for the constant term is 𝐴.
The guess for 2𝑥2𝑒3𝑥is ൫𝐵0 + 𝐵1𝑥+ 𝐵2𝑥2൮𝑒3𝑥, hence 𝑦𝑝= 𝐴+ ൫𝐵0 + 𝐵1𝑥+ 𝐵2𝑥2൮𝑒3𝑥 𝑦′ 𝑝= (𝐵1 + 2𝐵2𝑥) 𝑒3𝑥+ 3 ൫𝐵0 + 𝐵1𝑥+ 𝐵2𝑥2൮𝑒3𝑥 𝑦′′ 𝑝= 2𝐵2𝑒3𝑥+ 3 (𝐵1 + 2𝐵2𝑥) 𝑒3𝑥+ 3 (𝐵1 + 2𝐵2𝑥) 𝑒3𝑥+ 9 ൫𝐵0 + 𝐵1𝑥+ 𝐵2𝑥2൮𝑒3𝑥 Simplifying 𝑦′′ 𝑝= 𝑒3𝑥(2𝐵2 + 3𝐵1 + 3𝐵1 + 9𝐵0) + 𝑥𝑒𝑥(6𝐵2 + 6𝐵2 + 9𝐵1) + 𝑥2𝑒3𝑥(9𝐵2) = 𝑒3𝑥(2𝐵2 + 6𝐵1 + 9𝐵0) + 𝑥𝑒𝑥(12𝐵2 + 9𝐵1) + 𝑥2𝑒3𝑥(9𝐵2) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 7 Substitution into the ODE gives 𝑒3𝑥(2𝐵2 + 6𝐵1 + 9𝐵0) + 𝑥𝑒𝑥(12𝐵2 + 9𝐵1) + 𝑥2𝑒3𝑥(9𝐵2) + 9 ൫𝐴+ ൫𝐵0 + 𝐵1𝑥+ 𝐵2𝑥2൮𝑒3𝑥൮= 2𝑥2𝑒3𝑥+ 5 𝑒3𝑥(2𝐵2 + 6𝐵1 + 9𝐵0) + 𝑥𝑒𝑥(12𝐵2 + 9𝐵1) + 𝑥2𝑒3𝑥(9𝐵2) + 9𝐴+ ൫9𝐵0 + 9𝐵1𝑥+ 9𝐵2𝑥2൮𝑒3𝑥= 2𝑥2𝑒3𝑥+ 5 𝑒3𝑥(2𝐵2 + 6𝐵1 + 18𝐵0) + 𝑥𝑒𝑥(12𝐵2 + 18𝐵1) + 𝑥2𝑒3𝑥(18𝐵2) + 9𝐴= 2𝑥2𝑒3𝑥+ 5 Comparing coefcients gives 9𝐴= 5 2𝐵2 + 6𝐵1 + 18𝐵0 = 0 12𝐵2 + 18𝐵1 = 0 19𝐵2 = 2 From last equation 𝐵2 = 1 9. Hence from third equation 18𝐵1 = − 12 9 , or 𝐵1 = − 2 27. And from second equation 2𝐵2 + 6𝐵1 + 18𝐵0 = 0 2 ൭ 1 9൰+ 6 ൭−2 27൰+ 18𝐵0 = 0 𝐵0 = 1 81 And 𝐴= 5 9. Therefore 𝑦𝑝= 𝐴+ ൫𝐵0 + 𝐵1𝑥+ 𝐵2𝑥2൮𝑒3𝑥 = 5 9 + ൭ 1 81 −2 27𝑥+ 1 9𝑥2൰𝑒3𝑥 = 5 9 + ൭ 1 81 −6 81𝑥+ 9 81𝑥2൰𝑒3𝑥 = 45 81 + ൭ 1 81 −6 81𝑥+ 9 81𝑥2൰𝑒3𝑥 = 1 81 ൫45 + 𝑒3𝑥−6𝑥𝑒𝑥+ 9𝑥2𝑒3𝑥൮ 0.7 Section 5.5 problem 25 Problem Setup the form for the particular solution but do not determine the values of the coefcients.
𝑦′′ + 3𝑦′ + 2𝑦= 𝑥𝑒−𝑥−𝑥𝑒−2𝑥 solution First we find the homogenous solution. The characteristic equation is 𝑟2 + 3𝑟+ 2 = 0 (𝑟+ 1) (𝑟+ 2) = 0 Hence 𝑦1 = 𝑒−𝑥, 𝑦2 = 𝑒−2𝑥. We see that the basis solutions are part of the RHS. Therefore the guess solution is 𝑦𝑝= 𝑥(𝐴1 + 𝐴2𝑥) 𝑒−𝑥+ 𝑥(𝐴3 + 𝐴4𝑥) 𝑒−2𝑥 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 8 0.8 Section 5.5 problem 26 Problem Setup the form for the particular solution but do not determine the values of the coefcients.
𝑦′′ −6𝑦′ + 13𝑦= 𝑥𝑒3𝑥sin 2𝑥 solution First we find the homogenous solution. The characteristic equation is 𝑟2 −6𝑟+ 13 = 0 The roots are 3 ± 2𝑖. Hence the homogenous solution is 𝑦ℎ= 𝑐1𝑒3𝑥cos 2𝑥+ 𝑐2𝑒3𝑥sin 2𝑥. We see that 𝑒3𝑥sin 2𝑥is already in the homogenous solution. Hence the guess is 𝑦𝑝= 𝑥guess ാുുുുുിുുുുുീ (𝐴1 + 𝐴2𝑥)𝑥 sin 2𝑥𝑒3𝑥guess ാുുുുുുുുുുുുുുുുിുുുുുുുുുുുുുുുുീ (𝐴3 sin 2𝑥+ 𝐴4 cos 2𝑥) 𝑒3𝑥 = ൫𝐴1𝑥+ 𝐴2𝑥2൮𝑒3𝑥cos 2𝑥+ ൫𝐴3𝑥+ 𝐴4𝑥2൮𝑒3𝑥sin 2𝑥 0.9 Section 5.5 problem 37 Problem Solve the initial value problem 𝑦′′′ −2𝑦′′ + 𝑦′ = 1 + 𝑥𝑒𝑥with 𝑦(0) = 0, 𝑦′ (0) = 0, 𝑦′′ (0) = 1 solution First we find the homogenous solution. The characteristic equation is 𝑟3 −2𝑟2 + 𝑟= 0 𝑟൫𝑟2 −2𝑟+ 1൮= 0 For 𝑟2 −2𝑟+ 1 = 0, it factors into (𝑟−1) (𝑟−1), hence roots are 𝑟1 = 0, 𝑟2 = 1, 𝑟3 = 1. Since double roots, the homogenous solution is 𝑦ℎ= 𝑐1 + 𝑐2𝑒𝑥+ 𝑐3𝑥𝑒𝑥 We notice that both 𝑒𝑥and 𝑥𝑒𝑥is in the RHS. Therefore we need to multiply by 𝑥2. The guess is therefore 𝑦𝑝= 𝐴𝑥+ 𝑥2 (𝐵+ 𝐶𝑥) 𝑒𝑥 = 𝐴𝑥+ ൫𝐵𝑥2 + 𝐶𝑥3൮𝑒𝑥 Therefore 𝑦′ 𝑝= 𝐴+ ൫2𝐵𝑥+ 3𝐶𝑥2൮𝑒𝑥+ ൫𝐵𝑥2 + 𝐶𝑥3൮𝑒𝑥 𝑦′′ 𝑝= (2𝐵+ 6𝐶𝑥) 𝑒𝑥+ ൫2𝐵𝑥+ 3𝐶𝑥2൮𝑒𝑥+ ൫2𝐵𝑥+ 3𝐶𝑥2൮𝑒𝑥+ ൫𝐵𝑥2 + 𝐶𝑥3൮𝑒𝑥 Simplifying gives 𝑦′ 𝑝= 𝐴+ 𝑥𝑒𝑥(2𝐵) + 𝑥2𝑒𝑥(3𝐶+ 𝐵) + 𝑥3𝑒𝑥(𝐶) 𝑦′′ 𝑝= 𝑒𝑥(2𝐵) + 𝑥𝑒𝑥(6𝐶+ 4𝐵) + 𝑥2𝑒𝑥(6𝐶+ 𝐵) + 𝑥3𝑒𝑥(𝐶) 𝑦′′′ 𝑝= 𝑒𝑥(2𝐵) + 𝑒𝑥(6𝐶+ 4𝐵) + 𝑥𝑒𝑥(6𝐶+ 4𝐵) + 2𝑥𝑒𝑥(6𝐶+ 𝐵) + 𝑥2𝑒𝑥(6𝐶+ 𝐵) + 3𝑥2𝑒𝑥(𝐶) + 𝑥3𝑒𝑥(𝐶) = 𝑒𝑥(6𝐵+ 6𝐶) + 𝑥𝑒𝑥(6𝐶+ 4𝐵+ 12𝐶+ 2𝐵) + 𝑥2𝑒𝑥(6𝐶+ 𝐵+ 3𝐶) + 𝐶𝑥3𝑒𝑥 = 𝑒𝑥(6𝐵+ 6𝐶) + 𝑥𝑒𝑥(18𝐶+ 6𝐵) + 𝑥2𝑒𝑥(9𝐶+ 𝐵) + 𝐶𝑥3𝑒𝑥 Substitution into the ODE gives 𝑦′′′ 𝑝−2𝑦′′ 𝑝+ 𝑦′ 𝑝= 1 + 𝑥𝑒𝑥 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 9 Hence 𝑒𝑥(6𝐵+ 6𝐶) + 𝑥𝑒𝑥(18𝐶+ 6𝐵) + 𝑥2𝑒𝑥(9𝐶+ 𝐵) + 𝐶𝑥3𝑒𝑥 −2 ൫𝑒𝑥(2𝐵) + 𝑥𝑒𝑥(6𝐶+ 4𝐵) + 𝑥2𝑒𝑥(6𝐶+ 𝐵) + 𝑥3𝑒𝑥(𝐶)൮+ 𝐴+ 𝑥𝑒𝑥(2𝐵) + 𝑥2𝑒𝑥(3𝐶+ 𝐵) + 𝑥3𝑒𝑥(𝐶) = 1 + 𝑥𝑒𝑥 Or 𝑒𝑥(6𝐵+ 6𝐶) + 𝑥𝑒𝑥(18𝐶+ 6𝐵) + 𝑥2𝑒𝑥(9𝐶+ 𝐵) + 𝐶𝑥3𝑒𝑥 −𝑒𝑥(4𝐵) −𝑥𝑒𝑥(12𝐶+ 8𝐵) −𝑥2𝑒𝑥(12𝐶+ 2𝐵) −𝑥3𝑒𝑥(2𝐶) + 𝐴+ 𝑥𝑒𝑥(2𝐵) + 𝑥2𝑒𝑥(3𝐶+ 𝐵) + 𝑥3𝑒𝑥(𝐶) = 1 + 𝑥𝑒𝑥 Or 𝑒𝑥(6𝐵+ 6𝐶−4𝐵) + 𝑥𝑒𝑥(18𝐶+ 6𝐵−12𝐶−8𝐵+ 2𝐵) + 𝑥2𝑒𝑥(9𝐶+ 𝐵−12𝐶−2𝐵+ 3𝐶+ 𝐵) + 𝑥3𝑒𝑥(𝐶−2𝐶+ 𝐶) + 𝐴= 1 + 𝑥𝑒𝑥 Or 𝑒𝑥(2𝐵+ 6𝐶) + 𝑥𝑒𝑥(6𝐶) + 𝐴= 1 + 𝑥𝑒𝑥 Hence 6𝐶= 1 2𝐵+ 6𝐶= 0 𝐴= 1 Therefore, 𝐶= 1 6, 𝐵= − 1 2, and the particular solution is 𝑦𝑝= 𝐴𝑥+ 𝑥2 (𝐵+ 𝐶𝑥) 𝑒𝑥 = 𝑥+ ൭−1 2𝑥2 + 1 6𝑥3൰𝑒𝑥 Hence the complete solution is 𝑦= 𝑦ℎ+ 𝑦𝑝 = 𝑐1 + 𝑐2𝑒𝑥+ 𝑐3𝑥𝑒𝑥+ 𝑥+ ൭−1 2𝑥2 + 1 6𝑥3൰𝑒𝑥 Applying initial conditions. 𝑦(0) = 0 gives 0 = 𝑐1 + 𝑐2 (1) And 𝑦′ = 𝑐2𝑒𝑥+ 𝑐3𝑒𝑥+ 𝑐3𝑥𝑒𝑥+ 1 + ൭−𝑥+ 1 2𝑥2൰𝑒𝑥+ ൭−1 2𝑥2 + 1 6𝑥3൰𝑒𝑥 Applying second initial conditions 𝑦′ (0) = 0 gives 0 = 𝑐2 + 𝑐3 + 1 (2) And 𝑦′′ = 𝑐2𝑒𝑥+ 𝑐3𝑒𝑥+ 𝑐3𝑒𝑥+ 𝑐3𝑥𝑒𝑥+ (−1 + 𝑥) 𝑒𝑥+ ൭−𝑥+ 1 2𝑥2൰𝑒𝑥+ ൭−𝑥+ 1 2𝑥2൰𝑒𝑥+ ൭−1 2𝑥2 + 1 6𝑥3൰𝑒𝑥 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 10 Applying initial conditions 𝑦′′ (0) = 1 gives 1 = 𝑐2 + 2𝑐3 −1 2 = 𝑐2 + 2𝑐3 The solution is 𝑐1 = 4, 𝑐2 = −4, 𝑐3 = 3, hence the general solution is 𝑦= 𝑐1 + 𝑐2𝑒𝑥+ 𝑐3𝑥𝑒𝑥+ 𝑥+ ൭−1 2𝑥2 + 1 6𝑥3൰𝑒𝑥 = 4 −4𝑒𝑥+ 3𝑥𝑒𝑥+ 𝑥−1 2𝑥2𝑒𝑥+ 1 6𝑥3𝑒𝑥 0.10 Section 5.5 problem 49 Problem Use method of variation of parameters to find particular solution 𝑦′′ −4𝑦′ + 4𝑦= 2𝑒2𝑥 solution We need to first find the homogenous solution. The characteristic equation is 𝑟2 −4𝑟+ 4 = 0 (𝑟−2) (𝑟−2) = 0 Hence 𝑟1 = 2, double root. Therefore 𝑦1 (𝑥) = 𝑒2𝑥 𝑦2 (𝑥) = 𝑥𝑒2𝑥 Let 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 Where 𝑢1 = −ഐ𝑦2 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 𝑢2 = ഐ𝑦1 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 Where 𝑓(𝑥) = 2𝑒2𝑥and 𝑊(𝑥) = ૬ 𝑦1 𝑦2 𝑦′ 1 𝑦′ 2 ૬= ૬ 𝑒2𝑥 𝑥𝑒2𝑥 2𝑒2𝑥 𝑒2𝑥+ 2𝑥𝑒2𝑥૬ = 𝑒2𝑥൫𝑒2𝑥+ 2𝑥𝑒2𝑥൮−2𝑥𝑒4𝑥 = 𝑒4𝑥+ 2𝑥𝑒4𝑥−2𝑥𝑒4𝑥 = 𝑒4𝑥 Hence 𝑢1 = −ഐ 𝑥𝑒2𝑥൫2𝑒2𝑥൮ 𝑒4𝑥 𝑑𝑥= −ഐ2𝑥𝑑𝑥= −𝑥2 And 𝑢2 = ഐ 𝑒2𝑥൫2𝑒2𝑥൮ 𝑒4𝑥 𝑑𝑥= 2𝑥 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 11 Therefore 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 = −𝑥2𝑒2𝑥+ 2𝑥2𝑒2𝑥 = 𝑥2𝑒2𝑥 0.11 Section 5.5 problem 50 Problem Use method of variation of parameters to find particular solution 𝑦′′ −4𝑦= sinh 2𝑥 solution We need to first find the homogenous solution. The characteristic equation is 𝑟2 −4 = 0 𝑟= ±2 Therefore 𝑦1 (𝑥) = 𝑒2𝑥 𝑦2 (𝑥) = 𝑒−2𝑥 Let 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 Where 𝑢1 = −ഐ𝑦2 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 𝑢2 = ഐ𝑦1 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 Where 𝑓(𝑥) = sinh 2𝑥= 𝑒2𝑥−𝑒−2𝑥 2 and 𝑊(𝑥) = ૬ 𝑦1 𝑦2 𝑦′ 1 𝑦′ 2 ૬= ૬ 𝑒2𝑥 𝑒−2𝑥 2𝑒2𝑥 −2𝑒−2𝑥૬ = −2 −2 = −4 Hence 𝑢1 = −ഐ 𝑒−2𝑥൬𝑒2𝑥−𝑒−2𝑥 2 ൯ −4 𝑑𝑥 = 1 4 ഐ𝑒−2𝑥൭ 𝑒2𝑥−𝑒−2𝑥 2 ൰𝑑𝑥 = 1 8 ഐ൫1 −𝑒−4𝑥൮𝑑𝑥 = 1 8 ൭𝑥+ 𝑒−4𝑥 4 ൰ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 12 And 𝑢2 = ഐ 𝑒2𝑥൬𝑒2𝑥−𝑒−2𝑥 2 ൯ −4 𝑑𝑥 = −1 8 ഐ𝑒2𝑥൫𝑒2𝑥−𝑒−2𝑥൮𝑑𝑥 = −1 8 ഐ൫𝑒4𝑥−1൮𝑑𝑥 = −1 8 ൭ 𝑒4𝑥 4 −𝑥൰ Therefore 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 = 1 8 ൭𝑥+ 𝑒−4𝑥 4 ൰𝑒2𝑥−1 8 ൭ 𝑒4𝑥 4 −𝑥൰𝑒−2𝑥 = ൭ 1 8𝑥𝑒2𝑥+ 𝑒−2𝑥 32 ൰−1 8 ൭ 𝑒2𝑥 32 −𝑥𝑒−2𝑥 8 ൰ = 1 8𝑥𝑒2𝑥+ 𝑒−2𝑥 32 −𝑒2𝑥 32 + 𝑥𝑒−2𝑥 8 = 1 4𝑥൭ 𝑒2𝑥+ 𝑒−2𝑥 2 ൰+ 1 16 ൭ 𝑒−2𝑥−𝑒2𝑥 2 ൰ = 1 4𝑥൭ 𝑒2𝑥+ 𝑒−2𝑥 2 ൰−1 16 ൭ 𝑒2𝑥−𝑒−2𝑥 2 ൰ = 1 4𝑥cosh 2𝑥−1 16 sinh 2𝑥 = 1 16 (4𝑥cosh 2𝑥−sinh 2𝑥) 0.12 Section 5.5 problem 53 Problem Use method of variation of parameters to find particular solution 𝑦′′ + 9𝑦= 2 sec 3𝑥 solution We need to first find the homogenous solution. The characteristic equation is 𝑟2 + 9 = 0 𝑟= ±3𝑖 Therefore 𝑦1 (𝑥) = sin 3𝑥 𝑦2 (𝑥) = cos 3𝑥 Let 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 Where 𝑢1 = −ഐ𝑦2 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 𝑢2 = ഐ𝑦1 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 13 Where 𝑓(𝑥) = 2 sec 3𝑥= 2 cos 3𝑥and 𝑊(𝑥) = ૬ 𝑦1 𝑦2 𝑦′ 1 𝑦′ 2 ૬= ૬sin 3𝑥 cos 3𝑥 3 cos 3𝑥 −3 sin 3𝑥૬ = −3 sin2 3𝑥−3 cos2 𝑥 = −3 Hence 𝑢1 = −ഐ cos 3𝑥൬ 2 cos 3𝑥൯ −3 𝑑𝑥 = 1 3 ഐ2𝑑𝑥 = 2 3𝑥 And 𝑢2 = ഐ sin 3𝑥൬ 2 cos 3𝑥൯ −3 𝑑𝑥 = −2 3 ഐtan 3𝑥𝑑𝑥 = −2 3 ൭ 1 6 ln 1 cos2 (3𝑥)൰ Therefore 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 = 2 3𝑥(sin 3𝑥) + −2 3 ൭ 1 6 ln 1 cos2 (3𝑥)൰cos 3𝑥 = 2 3𝑥(sin 3𝑥) −1 9 cos (3𝑥) ln ൭ 1 cos2 (3𝑥)൰ = 2 3𝑥(sin 3𝑥) + 1 9 cos (3𝑥) ln ൫cos2 (3𝑥)൮ = 2 3𝑥(sin 3𝑥) + 2 9 cos (3𝑥) ln |cos (3𝑥)| 0.13 Section 5.5 problem 61 Problem Find a particular solution to the Euler ODE 𝑥2𝑦′′ + 𝑥𝑦′ + 𝑦= ln 𝑥with homogenous solution 𝑦ℎ= 𝑐1 cos (ln 𝑥) + 𝑐2 sin (ln 𝑥) solution We see that 𝑦1 = cos (ln 𝑥) 𝑦2 = sin (ln 𝑥) Using variation of parameters on the ODE 𝑦′′ + 1 𝑥𝑦′ + 1 𝑥2 𝑦= ln 𝑥 𝑥2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 14 Where now we use 𝑓(𝑥) = ln 𝑥 𝑥2 . Let 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 Where 𝑢1 = −ഐ𝑦2 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 𝑢2 = ഐ𝑦1 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 And 𝑊(𝑥) = ૬ 𝑦1 𝑦2 𝑦′ 1 𝑦′ 2 ૬= ૬ cos (ln 𝑥) sin (ln 𝑥) − 1 𝑥sin (ln 𝑥) 1 𝑥cos (ln 𝑥)૬ = 1 𝑥cos2 (ln 𝑥) + 1 𝑥sin2 (ln 𝑥) = 1 𝑥 Hence 𝑢1 = −ഐ sin (ln 𝑥) ൬ln 𝑥 𝑥2 ൯ 1 𝑥 𝑑𝑥 = −ഐln 𝑥sin (ln 𝑥) 𝑥 𝑑𝑥 = ln (𝑥) cos (ln 𝑥) −sin (ln 𝑥) And 𝑢2 = ഐ cos (ln 𝑥) ൬ln 𝑥 𝑥2 ൯ 1 𝑥 𝑑𝑥 = ഐcos (ln 𝑥) (ln 𝑥) 𝑥 𝑑𝑥 = ln (𝑥) sin (ln 𝑥) + cos (ln 𝑥) Therefore 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 = (ln (𝑥) cos (ln 𝑥) −sin (ln 𝑥)) cos (ln 𝑥) + (ln (𝑥) sin (ln 𝑥) + cos (ln 𝑥)) sin (ln 𝑥) = ln (𝑥) cos2 (ln 𝑥) −sin (ln 𝑥) cos (ln 𝑥) + ln (𝑥) sin2 (ln 𝑥) + sin (ln) cos (ln 𝑥) = ln (𝑥) cos2 (ln 𝑥) + ln (𝑥) sin2 (ln 𝑥) = ln 𝑥 0.14 Section 5.5 problem 62 Problem Find a particular solution to the Euler ODE ൫𝑥2 −1൮𝑦′′ −2𝑥𝑦′ + 2𝑦= 𝑥2 −1 with homogenous solution 𝑦ℎ= 𝑐1𝑥+ 𝑐2 ൫1 + 𝑥2൮ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 15 solution We see that 𝑦1 = 𝑥 𝑦2 = 1 + 𝑥2 Using variation of parameters on the ODE 𝑦′′ −2 𝑥 ൫𝑥2 −1൮ 𝑦′ + 2 ൫𝑥2 −1൮ 𝑦= 1 Where now we use 𝑓(𝑥) = 1. Let 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 Where 𝑢1 = −ഐ𝑦2 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 𝑢2 = ഐ𝑦1 (𝑥) 𝑓(𝑥) 𝑊(𝑥) 𝑑𝑥 And 𝑊(𝑥) = ૬ 𝑦1 𝑦2 𝑦′ 1 𝑦′ 2 ૬= ૬ 𝑥 1 + 𝑥2 1 2𝑥૬ = 2𝑥2 −൫1 + 𝑥2൮ = 𝑥2 −1 Hence 𝑢1 = −ഐ ൫1 + 𝑥2൮(1) 𝑥2 −1 𝑑𝑥 = −𝑥−ln (𝑥−1) + ln (𝑥+ 1) And 𝑢2 = ഐ 𝑥 𝑥2 −1𝑑𝑥 = 1 2 ln (𝑥−1) + 1 2 ln (𝑥+ 1) Therefore 𝑦𝑝= 𝑢1𝑦1 + 𝑢2𝑦2 = (−𝑥−ln (𝑥−1) + ln (𝑥+ 1)) 𝑥+ ൭ 1 2 ln (𝑥−1) + 1 2 ln (𝑥+ 1)൰൫1 + 𝑥2൮ = −𝑥2 + 𝑥ln 𝑥+ 1 𝑥−1+ 1 2 ൫1 + 𝑥2൮ln |(𝑥−1) (𝑥+ 1)| = −𝑥2 + 𝑥ln 𝑥+ 1 𝑥−1+ 1 2 ൫1 + 𝑥2൮ln ്𝑥2 −1് |
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Simple and clean way of comparing three numbers
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I've got some code that has a sequence of ifs that work, but just feel messy. Basically, I want to choose the largest of three integers and set a status flag to say which was chosen. My current code looks like this:
```
a = countAs();
b = countBs();
c = countCs();
if (a > b && a > c)
status = MOSTLY_A;
else if (b > a && b > c)
status = MOSTLY_B;
else if (c > a && c > b)
status = MOSTLY_C;
else
status = DONT_KNOW;
```
This pattern occurs a few times, and with long variable names it gets a little difficult to visually confirm that each if is correct. I feel there might be a better and clearer way to do this; can anyone suggest anything?
There are some potential duplicates, but they don't quite align with this question.
In the suggested duplicate: Approaches to checking multiple conditions? all of the suggested solutions seem equally clumsy as the original code, so they do not provide a better solution.
And this post Elegant ways to handle if(if else) else deals only with nesting levels and asymmetry, which is not the problem here.
algorithms
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edited Apr 12, 2017 at 7:31
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asked Jun 22, 2015 at 4:41
Ken Y-NKen Y-N
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To me the only problem is the code repetition. What you have here is extremely clear to read, why change that? Just break it out into a function that takes in a,b,c and returns status. If you it makes you feel better, drop an "inline" on it. No macros, no complexity, just good old function extraction. I will thank you for your clear code if I need to work with it later.
– J Trana
Commented
Jun 22, 2015 at 5:23
possible duplicate of Approaches to checking multiple conditions?
– gnat
Commented
Jun 22, 2015 at 5:51
see also: Elegant ways to handle if(if else) else
– gnat
Commented
Jun 22, 2015 at 5:52
Note that your #defines are named poorly. Consider a = 40, b = 30, c = 30. The result is MOSTLY_A. But most of the things are actually not A, but B or C. You might want to try MORE_A (although still ambiguous - more A than B and more A than C, or more A than B and C combined?), or A_LARGEST, MAX_IS_A, ...? (which also suggests max(a, max(b, c)) as an answer to the question).
– tony
Commented
Jun 23, 2015 at 16:13
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Factorize logic, return early
As suggested in comments, it would be sufficient to simply wrap your logic in a function and exit early with return's in order to simplify things a lot. Also, you could factorize a bit of functionnality by delegating tests to another function. More concretely:
```
bool mostly(max,u,v) {
return max > u && max > v;
}
status_t strictly_max_3(a,b,c)
{
if mostly(a,b,c) return MOSTLY_A;
if mostly(b,a,c) return MOSTLY_B;
if mostly(c,a,b) return MOSTLY_C;
return DONT_KNOW;
}
```
This is shorter than my previous attempt:
```
status_t index_of_max_3(a,b,c)
{
if (a > b) {
if (a == c)
return DONT_KNOW;
if (a > c)
return MOSTLY_A;
else
return MOSTLY_C;
} else {
if (a == b)
return DONT_KNOW;
if (b > c)
return MOSTLY_B;
else
return MOSTLY_C;
}
}
```
The above is a little more verbose, but is easy to read IMHO and does not recompute comparisons multiple times.
Visual confirmation
In your answer you say:
my issue was mostly visual confirmation that all comparisons used the
same variables
... also, in your question, you say:
This pattern occurs a few times, and with long variable names it gets
a little difficult to visually confirm that each if is correct.
I may not understand what you are trying to achieve: do you want to copy-paste the pattern everywhere you need it? With a function like the one above, you capture the pattern once, and you check once for all that all comparisons use a, b and c as required. Then, you don't need to worry anymore when you call the function. Of course, maybe in practice your problem is a little more complex than the one you described: if so, please add some details if possible.
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answered Jun 22, 2015 at 9:46
coredumpcoredump
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I don't understand your comment about DONT_KNOW , what if c is the smallest and a and b are the same? The algorithm would return that b is the biggest, while a is the same as b.
– Pieter B
Commented
Jun 22, 2015 at 9:57
@PieterB I was wronlgy assuming that it would not matter if we returned either MOSTLY_A or MOSTLY_C in case a == c and a > b. This is fixed. Thanks.
– coredump
Commented
Jun 22, 2015 at 12:10
@DocBrown Granted, most of the benefits come from the early exit behavior.
– coredump
Commented
Jun 22, 2015 at 14:13
1
+1, now it is indeed an improvement over the OPs original code.
– Doc Brown
Commented
Jun 23, 2015 at 5:23
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TL:DR; Your code is already correct and "clean".
I see a lot of people waffling around the answer but everyone is missing the forest through the trees. Let's do the full computer science and mathematical analysis to completely understand this question.
First, we note that we have 3 variables, each with 3 states: <, =, or >. The total number of permutations is 3^3 = 27 states, which I'll assign an unique number, denoted P#, for each state. This P# number is a factorial number system.
Enumerating all the permutations we have:
```
a ? b | a ? c | b ? c |P#| State
------+-------+-------+--+------------
a < b | a < c | b < c | 0| C
a = b | a < c | b < c | 1| C
a > b | a < c | b < c | 2| C
a < b | a = c | b < c | 3| impossible a<b b<a
a = b | a = c | b < c | 4| impossible a b | a = c | b < c | 5| A=C > B
a < b | a > c | b < c | 6| impossible ac
a = b | a > c | b < c | 7| impossible ac
a > b | a > c | b < c | 8| A
a < b | a < c | b = c | 9| B=C > A
a = b | a < c | b = c |10| impossible a b | a < c | b = c |11| impossible ac
a < b | a = c | b = c |12| impossible a b | a = c | b = c |14| impossible a>a
a < b | a > c | b = c |15| impossible ac
a = b | a > c | b = c |16| impossible a>a
a > b | a > c | b = c |17| A
a < b | a < c | b > c |18| B
a = b | a < c | b > c |19| impossible bc
a > b | a < c | b > c |20| impossible ac
a < b | a = c | b > c |21| B
a = b | a = c | b > c |22| impossible a>a
a > b | a = c | b > c |23| impossible c>b b>c
a < b | a > c | b > c |24| B
a = b | a > c | b > c |25| A=B > C
a > b | a > c | b > c |26| A
```
By inspection we see we have:
3 states where A is the max,
3 states where B is the max,
3 states where C is the max, and
4 states where either A=B, or B=C.
Let's write a program (see the footnote) to enumerate all these permutations with values for A, B, and C. Stable sorting by P#:
```
a ?? b | a ?? c | b ?? c |P#| State
1 < 2 | 1 < 3 | 2 < 3 | 0| C
1 == 1 | 1 < 2 | 1 < 2 | 1| C
1 == 1 | 1 < 3 | 1 < 3 | 1| C
2 == 2 | 2 < 3 | 2 < 3 | 1| C
2 > 1 | 2 < 3 | 1 < 3 | 2| C
2 > 1 | 2 == 2 | 1 < 2 | 5| ??
3 > 1 | 3 == 3 | 1 < 3 | 5| ??
3 > 2 | 3 == 3 | 2 < 3 | 5| ??
3 > 1 | 3 > 2 | 1 < 2 | 8| A
1 < 2 | 1 < 2 | 2 == 2 | 9| ??
1 < 3 | 1 < 3 | 3 == 3 | 9| ??
2 < 3 | 2 < 3 | 3 == 3 | 9| ??
1 == 1 | 1 == 1 | 1 == 1 |13| ??
2 == 2 | 2 == 2 | 2 == 2 |13| ??
3 == 3 | 3 == 3 | 3 == 3 |13| ??
2 > 1 | 2 > 1 | 1 == 1 |17| A
3 > 1 | 3 > 1 | 1 == 1 |17| A
3 > 2 | 3 > 2 | 2 == 2 |17| A
1 < 3 | 1 < 2 | 3 > 2 |18| B
1 < 2 | 1 == 1 | 2 > 1 |21| B
1 < 3 | 1 == 1 | 3 > 1 |21| B
2 < 3 | 2 == 2 | 3 > 2 |21| B
2 < 3 | 2 > 1 | 3 > 1 |24| B
2 == 2 | 2 > 1 | 2 > 1 |25| ??
3 == 3 | 3 > 1 | 3 > 1 |25| ??
3 == 3 | 3 > 2 | 3 > 2 |25| ??
3 > 2 | 3 > 1 | 2 > 1 |26| A
```
In case you were wondering how I knew which P# states were impossible, now you know. :-)
The minimum number of comparisons to determine the order is:
Log2(27) = Log(27)/Log(2) = ~4.75 = 5 comparisons
i.e. coredump gave the correct 5 minimal number of comparisons. I would format his code as:
```
status_t index_of_max_3(a,b,c)
{
if (a > b) {
if (a == c) return DONT_KNOW; // max a or c
if (a > c) return MOSTLY_A ;
else return MOSTLY_C ;
} else {
if (a == b) return DONT_KNOW; // max a or b
if (b > c) return MOSTLY_B ;
else return MOSTLY_C ;
}
}
```
For your problem we don't care about testing for equality so we can omit 2 tests.
It doesn't matter how clean/bad the code is if it gets the wrong answer so this is a good sign that you are handling all the cases correctly!
Next, as for simplicity, people keep trying to "improve" the answer,
where they think improving means "optimizing" the number of comparisons,
but that isn't strictly what you are asking. You confused everyone where you asked
"I feel there might be a better" but didn't define what 'better' means. Less comparisons? Less code? Optimal comparisons?
Now since you are asking about code readability (given correctness) I would only make one change to your code for readability: Align the first test with the others.
```
if (a > b && a > c)
status = MOSTLY_A;
else if (b > a && b > c)
status = MOSTLY_B;
else if (c > a && c > b)
status = MOSTLY_C;
else
status = DONT_KNOW; // a=b or b=c, we don't care
```
Personally I would write it the following way but this may be too unorthodox for your coding standards:
```
if (a > b && a > c) status = MOSTLY_A ;
else if (b > a && b > c) status = MOSTLY_B ;
else if (c > a && c > b) status = MOSTLY_C ;
else / a==b || b ==c/status = DONT_KNOW; // a=b or b=c, we don't care
```
Footnote: Here is the C++ code to generate the permutations:
```
include
char txt[] = "< == > ";
enum cmp { LESS, EQUAL, GREATER };
int val = { 1, 2, 3 };
enum state { DONT_KNOW, MOSTLY_A, MOSTLY_B, MOSTLY_C };
char descr[]= "??A B C ";
cmp Compare( int x, int y ) {
if( x < y ) return LESS;
if( x > y ) return GREATER;
/ x==y / return EQUAL;
}
int main() {
int i, j, k;
int a, b, c;
printf( "a ?? b | a ?? c | b ?? c |P#| State\n" );
for( i = 0; i < 3; i++ ) {
a = val[ i ];
for( j = 0; j < 3; j++ ) {
b = val[ j ];
for( k = 0; k < 3; k++ ) {
c = val[ k ];
int cmpAB = Compare( a, b );
int cmpAC = Compare( a, c );
int cmpBC = Compare( b, c );
int n = (cmpBC 9) + (cmpAC 3) + cmpAB; // Reconstruct unique P#
printf( "%d %c%c %d | %d %c%c %d | %d %c%c %d |%2d| "
, a, txt[cmpAB2+0], txt[cmpAB2+1], b
, a, txt[cmpAC2+0], txt[cmpAC2+1], c
, b, txt[cmpBC2+0], txt[cmpBC2+1], c
, n
);
int status;
if (a > b && a > c) status = MOSTLY_A;
else if (b > a && b > c) status = MOSTLY_B;
else if (c > a && c > b) status = MOSTLY_C;
else / a ==b || b== c/status = DONT_KNOW; // a=b, or b=c
printf( "%c%c\n", descr[status2+0], descr[status2+1] );
}
}
}
return 0;
}
```
Edits: Based on feedback, moved TL:DR to top, removed unsorted table, clarified 27, cleaned up code, described impossible states.
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edited Jun 24, 2015 at 16:36
answered Jun 23, 2015 at 13:05
Michaelangel007Michaelangel007
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-1: wouldn't reducing the number of decisions lead to simpler code paths and more readable code? You argument is not clear: first, you say everybody is wrong; then, you put not one or two but three tables; I hoped they would lead to a simpler way to compute the result but instead you confirmed what everybody else already knew (OP's code does the right thing). Sure, the question is about readability, but readability is not achieved only by modifying code layout (you admit your changes would hardly fit existing code standards). It makes sense to simplify logic when optimizing for readability.
– coredump
Commented
Jun 24, 2015 at 8:03
More constructively: I'd suggest to simplify your answer by leaving out some details and thinking about the structure of your answer. I appreciate that you took time to write and post the C++ code generating permutations, but maybe you could give the main result and only one table: as it is now, it looks like you dumped all your work as-is. I almost failed to spot the TL;DR thing (you could start with that). Hope it helps.
– coredump
Commented
Jun 24, 2015 at 8:09
2
Thanks for the constructive feedback coredump. I've removed the middle unsorted table since it is easily verified.
– Michaelangel007
Commented
Jun 24, 2015 at 16:16
2
Jesus Christ! Who would say comparing three numbers is almost as complex as rocket science?
– Mandrill
Commented
Jun 24, 2015 at 17:49
@Mandrill As computer scientists it is our job to understand a problem thoroughly. Only by enumerating all 27 possible permutations for a 3-way comparison can we verify that our solution works in ALL cases. The last thing we want as programmers are hidden bugs and unaccounted edge cases. Verbosity is the price one pays for correctness.
– Michaelangel007
Commented
Jul 9, 2015 at 17:15
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@msw told you to use an array instead of a,b,c, and @Basile told you refactor the "max" logic into a function. Combining these two ideas leads to
```
val = countAs(); // in the real code, one should probably define
val = countBs(); // some enum for the indexes 0,1,2 here
val = countCs();
int result[]={DONT_KNOW, MOSTLY_A, MOSTLY_B, MOSTLY_C};
```
then provide a function which calculates the max index of an arbitrary array:
```
// returns the index of the strict maximum, and -1 when the maximum is not strict
int FindStrictMaxIndex(int values,int arraysize)
{
int maxVal=INT_MIN;
int maxIndex=-1;
for(int i=0;imaxVal)
{
maxVal=values[i];
maxIndex=i;
}
else if (values[i]==maxVal)
{
maxIndex=-1;
}
}
return maxIndex;
}
```
and call it like
```
return result[FindStrictMaxIndex(val,3)+1];
```
The total number of LOC seems to have increased over the original one, but now you have the core logic in a reusable function, and if you can reuse the function multiple times, it starts to pay off. Moreover, the FindStrictMaxIndex function is not interwoven with your "business requirements" any more (separation of concerns), thus the risk you will have to modify it later is much lower than in your original version (open-closed principle). For example, that function will not have to be changed even if the number of arguments changes, or need to use other return values than MOSTLY_ABC, or you are processing other variables than a,b,c. Moreover, the use of an array instead of 3 different values a, b, c might simplify your code in other places as well.
Of course, if in your whole program there are only one or two places for calling this function, and you do not have any further applications for holding the values in an array, then I would probably leave the original code as it is (or use @coredump's improvement).
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edited Jun 24, 2015 at 15:17
answered Jun 22, 2015 at 5:15
Doc BrownDoc Brown
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I like that - the guts of FindStrictMaxIndex() may not be too clean, but from the caller point of view it is reasonably obvious what is trying to be achieved.
– Ken Y-N
Commented
Jun 22, 2015 at 5:24
Or instead of holding two arrays, hold one array of key-value-pairs: { MOSTLY_A, countAs() }, take the first element ordered by value and read off the key.
– Julia Hayward
Commented
Jun 22, 2015 at 8:36
@JuliaHayward: the main reason I did not suggest such a solution was the "C" tag of the question - in C, one will need some more boilerplate code to deal with key-value pairs, and creating a function typed in terms of KVPs will probably be not as reusable in different contexts than a simple int typed function. But I agree to your comment if one uses a different language like Python or Perl.
– Doc Brown
Commented
Jun 22, 2015 at 11:11
1
@gnasher729: it depends on how many "duplicates" in the orginal code are, how similar they really are, and how often the FindStrictMaxIndex function can be reused. For one or just two times of reusage, this won't pay off, of course, but that is what I wrote already in my answer. Note also the other advantages I mentioned above concerning future changes.
– Doc Brown
Commented
Jun 23, 2015 at 9:26
1
... and note that the original 8 lines can be replaced by a simple one-liner return result[FindStrictMaxIndex(val,3)]; at the point in code where the original 8 lines were placed. The other parts, especially FindStrictMaxIndex itself are completely separated from the "business logic", which moves them out of the focus of changing business requirements.
– Doc Brown
Commented
Jun 23, 2015 at 9:38
|
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You probably should use a macro or a function MAX giving the maximum of two numbers.
Then you just want:
```
status = MAX(a,MAX(b,c));
```
You might have defined
```
#define MAX(X,Y) (((X)>(Y))?(X):(Y))
```
but be cautious -notably about side effects- when using macros (since MAX(i++,j--) would behave strangely)
So better define a function
```
static inline int max2ints(int x, int y) {
return (x>y)?x:y;
}
```
and use it (or at least #define MAX(X,Y) max2ints((X),(Y)) ....)
If you need to understand the origin of the MAX you might have a long macro like
#define COMPUTE_MAX_WITH_CAUSE(Status,Result,X,Y,Z) which is a long do{ ... }while(0) macro, perhaps
```
#define COMPUTE_MAX_WITH_CAUSE(Status,Result,X,Y,Z) do { \
int x= (X), y= (Y), z=(Z); \
if (x > y && y > z) \
{ Status = MOSTLY_FIRST; Result = x; } \
else if (y > x && y > z) \
{ Status = MOSTLY_SECOND; Result = y; } \
else if (z > x && z > y) \
{ Status = MOSTLY_THIRD; Result = z; } \
/ etc / \
else { Status = UNKNOWN; Result = ... } \
} while(0)
```
Then you could invoke COMPUTE_MAX_WITH_CAUSE(status,res,a,b,c) at several places. It is a bit ugly. I defined local variables x, y, z to lower bad side effects....
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edited Jun 22, 2015 at 5:20
answered Jun 22, 2015 at 4:58
Basile StarynkevitchBasile Starynkevitch
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2
2
Refactoring the common logic into a function is the right approach, but I would actually avoid two things here: 1. I would not "invent" new requirements (the OP did not ask for calculating the maximum). And 2nd: even if the resulting code may become more DRY, it is very debatable if this justifies a complicated macro.
– Doc Brown
Commented
Jun 22, 2015 at 7:33
1
Macros should be a tool of last resort. Definitely out of bounds for this problem.
– kevin cline
Commented
Jun 23, 2015 at 7:31
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I've had more of a think about this, so since my issue was mostly visual confirmation that all comparisons used the same variables, I think this might be a useful approach:
```
a = countAs();
b = countBs();
c = countCs();
if (FIRST_IS_LARGEST(a, b, c))
status = MOSTLY_A;
else if (SECOND_IS_LARGEST(a, b, c))
status = MOSTLY_B;
else if (THIRD_IS_LARGEST(a, b, c))
status = MOSTLY_C;
else
status = DONT_KNOW; / NO_SINGLE_LARGEST is a better name? /
```
That each macro takes a, b and c in the same order is easy to confirm, and the macro name saves me having to work out what all the comparisons and ANDs are doing.
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answered Jun 23, 2015 at 5:49
Ken Y-NKen Y-N
32911 gold badge33 silver badges99 bronze badges
1
1
(1) why auxiliary macros instead of functions? (2) why do you need visual confirmation here? is it really your core problem or is the need for visual confirmation a consequence of code duplication? Your best option is to factorize your code into a single, simple function that you check once for all.
– coredump
Commented
Jun 24, 2015 at 13:44
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1091 | https://www.sohu.com/a/701148275_121692140 | 小学数学：正方形的面积怎么算
正方形的面积=边长×边长。正方形是特殊的平行四边形之一。即有一组邻边相等，并且有一个角是直角的平行四边形称为正方形，又称正四边形。
正方形的面积怎么算
正方形的面积公式是：面积=边长的平方，用字母表示就是：S=a²(S指正方形面积，a指正方形边长)。
正方形是特殊的矩形，特殊的长方形，长方形面积=矩形面积=长×宽;用字母表示就是：S=ab(S表示长方形面积，a表示长方形的长，b表示长方形的宽)。
正方形的面积=边长×边长=a×a(其中a为正方形的边长)，正方形的面积=对角线×对角线÷2，正方形是特殊的平行四边形，也是特殊的长方形。
在同一平面内：四条边都相等且一个角是直角的四边形是正方形。有一组邻边相等的矩形是正方形。有一个角为直角的菱形是正方形。四边形对角线相等且互相垂直平分。
常见面积定理：
一个图形的面积等于它的各部分面积的和。
两个全等图形的面积相等。
等底等高的三角形、平行四边形、梯形(梯形等底应理解为两底的和相等)的面积相等。
等底(或等高)的三角形、平行四边形、梯形的面积比等于其所对应的高(或底)的比。
相似三角形的面积比等于相似比的平方。
等角或补角的三角形面积的比，等于夹等角或补角的两边的乘积的比;等角的平行四边形面积比等于夹等角的两边乘积的比。
任何一条曲线都可以用一个函数y=f(x)来表示，那么，这条曲线所围成的面积就是对X求积分。
面积公式大全：
长方形的周长=(长+宽)×2
正方形的周长=边长×4
长方形的面积=长×宽
正方形的面积=边长×边长
三角形的面积=底×高÷2
平行四边形的面积=底×高
梯形的面积=(上底+下底)×高÷2
直径=半径×2半径=直径÷2
圆的周长=圆周率×直径=圆周率×半径×2
圆的面积=圆周率×半径×半径
长方体的表面积=(长×宽+长×高+宽×高)×2返回搜狐，查看更多 |
1092 | https://www.sciencedirect.com/topics/computer-science/inequality-constraint | Inequality Constraint - an overview | ScienceDirect Topics
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Inequality Constraint
In subject area:Computer Science
An inequality constraint, in the context of optimization problems, refers to a restriction that imposes a minimum or maximum value on a variable or a combination of variables. It is used to ensure that the solution satisfies certain conditions or limitations. In the given source, an example of an inequality constraint is the constraint on energy dissipation due to viscous friction, which is directly related to pressure drops.
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3.3 Equality and inequality constraints
The optimization problem is subjected to different constraints. The most obvious ones are given by the process model equations (Eqs.(1-2)). The other constraints consist of :
-
An iso-volume constraint introduced in order to guarantee the same residence time between initial and optimized shapes (Eq.(4)).
-
An inequality constraint on the energy dissipation by the fluid due to viscous friction and is given by Eq. and (5). Such a constraint is relevant since the energy dissipation and the pressure drops are directly correlated.
(4) and (5)C 1=V Ω−V Ω 0=0 and C 2=2 v∫Ω ε U 2 dx−2 v∫Ω 0 ε U 2 dx<0
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Safety violations are in many cases defined by the value of one variable or its derivatives going over a limit: low level in a drum, high temperature in a reactor, etc. The safety condition is simply defined mathematically as an inequality constraint on state variables: T r<T max, h d>h min, where T max and h min are constants. More complex constraints can be specified depending on more than one variable or varying with time: composition of a mixture outside the flammability region, temperature increase rate during start up, etc.
All those safety specifications can be formulated as inequality path constraints. They must be satisfied along the whole trajectory in every mode or only in a particular mode, what we do not consider here for simplicity: h(x˙,x,u,p,t)≤0. Inequality point constraints, like the ones that could be imposed on the initial conditions, can be considered as a special case of the path constraints.
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3 Handling Equality and Inequality Constraints
Many optimization problems in chemical engineering involve both equality and inequality constraints. Of these, equality constraints are more challenging to handle. Constraint handling techniques in the stochastic techniques have received much attention in the last decade. In this study, we propose a new constraint handling scheme for both equality and inequality constraints. It incorporates adaptive relaxation of constrains and the feasibility approach for selection between trial and target individuals in SaDETL.
In the initialization, values of the objective function and constraints of the population are calculated. The median (—) of total absolute violation (TAV) of all individuals in the population is recorded. If TAV of an individual is less than —, then it is temporarily treated as a feasible solution; else, it is taken as an infeasible solution. Thus, constraints are relaxed. During the generations, — value is gradually reduced according to the percentage of feasible individuals (P F) in the latest population. After considering several possibilities, we propose the following equation for reducing — for the next generation.
(1)μ(G+1)=μ(G)-P F×μ(G)NP
Here, G is generation number. The relaxation of constraints in the initial generations helps greater exploration of the search space for locating the global optimum region.
The selection between target and trial individuals in the DE algorithm is based on the feasibility approach of Deb (2000). According to this, (a) a feasible solution is preferred over an infeasible solution; (b) among two feasible solutions, the one with better objective function value is preferred; and (c) among two infeasible solutions, the one with smaller TAV is chosen. The adaptive relaxation and feasibility approach will force the population towards the feasible region but more gradually than the feasibility approach alone without constraint relaxation. So, it can enhance the global search, especially for problems with equality constraints, which have very small feasible region.
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Discrete-Time Control System Implementation Techniques
1995, Control and Dynamic SystemsRobert R. Bless, Dewey H. Hodges
III TREATMENT OF CONTROL CONSTRAINTS
This section is concerned with the treatment of optimal control problems that are subject to inequality constraints which are explicit functions of the controls.
A GENERAL DEVELOPMENT
Consider the same system as defined in Eqs.(1) – (3). For simplicity, it is assumed that the problem has only one phase. Suppose now that the p constraints are written as
(67)g i x u≤0 f o r i=1,2,…,p
One way of handling inequality constraints is to use a “slack” variable . The idea is that if g i ≤ 0 then g i plus some nonnegative number, the i th slack variable k i 2 is equal to zero. Thus we form p × 1 column matrices for K and δK
(68)K=k 1 2 k 2 2…k p 2 T
(69)δ K=2 k 1 δ k 1 2 k 2 δ k 2…2 k p δ k p T
and a p × 1 column matrix g(x, u), the elements of which are g i(x, u). Now, from Eq.(67)
(70)g x u+K=0
Eq.(70) will be adjoined to the augmented cost functionJ a in Eq.(6) by using pLagrange multiplier functions μ(t). After dropping the superscripted phase number on H, we obtain
(71)J a=Φ+∫t 0 t 1 H−λ T x˙+μ T g+K d t
By redefining the Hamiltonian in Eq.(5) to be
(72)H=L+λ T f+μ T g+K
then, after the usual integration by parts and making use of Eqs.(8) – (9), we obtain the first variation of J a as
(73)δ J a=∫t 0 t 1 δ λ T f−x˙+δ x T∂H∂x T+λ˙+δ u T∂H∂u T+δ μ T g+K+δ K T μ d t+∂Φ∂t 1+H t 1 d t 1+d ν T ψ+∂Φ∂x t 1−λ T t 1 d x t 1+∂Φ∂x t 0+λ T t 0 d x t 0
Setting δJ a = 0 in Eq.(73) results in the same necessary conditions (for a single-phase problem) as are listed in Eqs.(12) – (19). There are also the additional conditions that
(74)g i+k i 2=0 μ i k i=0 f o r i=1,2,…,p
These conditions, when satisfied, ensure that the constraint will not be violated at element midpoints and nodes and that either the constraint or the multiplier will be zero. For example, if the constraint g i equals zero, then k i must be zero, and μ i need not be zero. On the other hand, if g i is less than zero (inactive), then k i must be nonzero, which in turn forces μ i to zero. An additional requirement for a minimizing problem is that the multipliers μ must be nonnegative for all time. This is enforced computationally, as will be shown in the example, by squaring the value of μ that appears in the costate equations and optimality conditions.
There may be numerical problems associated with the introduction of slack variables. As pointed out in , there can be poor conditioning or even singularities in the Jacobian occurring during the solution procedure if both μ i and k i were zero at the same time. The advantage of using a slack variable, however, is that control-constrained problems may be treated as single-phase problems. Since the control is allowed to come on and off the constraint as required, there is no need to define a switching structure (i.e., the sequence of constrained and unconstrained arcs) for the problem.
To derive the weak formulation, Eq.(73) is simplified by using the boundary conditions in Eqs.(12) – (16). Then an integration by parts to eliminate any time derivatives of x or λ from appearing in the formulation yields
(75)0=∫t 0 t 1 δ λ T f+δ λ˙T x+δ x T∂H∂x T−δ x˙T λ+δ u T∂H∂u T+δ μ T g+K+δ K T μ d t+δ x T t 1 λ t 1−δ λ T t 1 x t 1+δ λ T t 0 x t 0−δ x T t 0 λ t 0
Equations(12) – (16) and Eq.(75) define the weak formulation for optimal control problems with control constraints. Equation(75) can be generalized in a straightforward manner to multiple phases, similar to Eq.(21). The discretization of the weak form is similar to what is defined in Section II, B.
B FINITE ELEMENT DISCRETIZATION
To the discretization defined in Section II, B, we define shape functions for δK and δμ of the form (with δD defined as the Dirac delta function)
(76)δ K=2 k^i+δ k^i+δ D τ+2 k¯i δ k¯i+2 k^i+1−δ k^i+1−δ D τ−1
(77)δ μ=δ μ^i+δ D τ+δ μ¯i+δ μ^i+1−δ D τ−1
where the subscript i now refers to the element number. The Dirac delta functions appearing in the discretization of δK and δμ have the effect that the associated parts of the integrand of Eq.(73) are forced to zero pointwise wherever the Delta functions have nonzero values.
The shape functions for K and μ are of the form
(78)K=k^i 2+if τ=0;k¯i 2 if 0<τ<1;k^i+1 2−if τ=1
and
(79)μ=μ^i+if τ=0;μ¯i if 0<τ<1;μ^i+1−if τ=1
The superscripted “–” and “+” signs in Eqs.(76) – (79) signify values just before and after nodal times.
By substituting the shape and test functions defined in Eqs.(76) – (79) and the shape and test functions defined in Eqs.(24) – (29), and carrying out the element quadrature over τ from 0 to 1, the general algebraic form of the weak principle is obtained. The equation is similar to Eq.(44) with the addition of the appropriate δμ and δK equations. These equations will be written out explicitly for the example problem.
Note that if the time t does not appear explicitly in the problem formulation then all integration is exact and can be done by inspection. If t does appear explicitly, then t may be approximated by its midpoint value over each element and the integration may still be done by inspection. This is equivalent to redefining t to be a new state. The latter case occurs in the example problem presented next.
In addition to the unknowns and equations listed in Section II, C, there are an additional 2 p(M 1 + 2) equations corresponding to the coefficients of δμ and δk and an additional 2 p(M 1 + 2) unknowns to solve for the k¯, k^, μ¯, and μ^ quantities. Thus, there are still the same number of equations as unknowns.
Again, once the nodal values for the states and costates are found, then one may use the optimality condition (∂H/∂u = 0), the constraint equations (g + K = 0), and the condition that either k or μ be zero (kμ = 0) to solve for u, K, and μ at a node. This procedure is used in the following example problem.
C EXAMPLE
This example is taken from , section 3.8. The problem is to minimize
(80)J=1 2 x T 2+1 2∫0 T u 2 d t
where T = 8, x and u are scalars, and the initial condition is x(0) = – (5 + 8/π). The state equation is
x˙=h t u with h t=2 sin π t 4
The following two control inequality constraints are imposed.
(81)g 1=u−1≤0
(82)g 2=−u+1≤0
The exact solution is found to be x(T) = –1 and for the control
(83)u t=−x T h t f o r 0≤t≤1 1 f o r 1≤t≤3−x T h t f o r 3≤t≤5−1 f o r 5≤t≤7−x T h t f o r 7≤t≤8
The algebraic equations which come from the weak principle can be verified to be
(84)x^1+5+8/π=0
(85)t 1−8=0
for the state and time boundary conditions (in ψ),
(86)λ^1+ν 1=0
(87)λ^1−λ¯1=0
(88)λ¯i−λ¯i+1=0 f o r i=1,2,…,M 1−1
(89)λ¯M 1−λ^M 1+1=0
(90)λ¯M 1+1−x^M 1+1=0
for the costate equations (obtained from the coefficients of δx set equal to zero) and boundary conditions,
(91)u¯i+λ¯i h t¯i+μ¯1 i−μ¯2 i=0 k¯1 i μ¯2 i=0 k¯2 i μ¯2 i=0 u¯i−1+k¯1 i 2=0−u¯i−1+k¯2 i 2=0 f o r i=1,2,…,M 1
for the coefficients of δū, δ k¯, and δ μ¯, (note that the equations formed from setting the coefficients of δû, δ k^, and δ μ^ at the first and last node of the phase equal to zero are similar to these) and
(92)x¯1−Δ t 2 h t¯1 u¯1=x^1
(93)x¯i+1−x¯i−Δ t 2 h t¯i u¯i+h t¯i+1 u¯i+1=0 f o r i=1,2,…,M 1−1
(94)x¯M 1−Δ t 2 h t¯M 1 u¯M 1+x^M 1+1=0
for the δ λ coefficients. Note that t¯i is an average time value for the i th element and (if Δ t=Δ t i=t 1/M 1 for all i) can be expressed as
(95)t¯i=2 i−1 2 Δ t f o r i=1,2,…,M 1
Recall from that one of the additional necessary conditions for problems with control constraints is that the multipliers be nonnegative for a minimizing problem. Therefore, in practice, the multipliers μ appearing in the first of Eqs.(91) are squared to ensure that they remain nonnegative. Further recall that if the constraint is not violated, then μ = 0. This condition is satisfied by the second and third equation in Eq.(91) which implies that either k or μ is zero for each element.
It is readily apparent from Eqs.(86) – (90) that all the costate variables are equal to x^M 1+1=x T. Therefore, these equations were eliminated and all costates that occurred in the remaining equations were replaced with x^M 1+1. Also, the state and time boundary conditions in ψ were solved. The remaining 6 M 1 + 1 algebraic equations in Eqs.(91) – (94) were solved using a Newton-Raphson method and a FORTRAN code written on a SUN 3/260. The sparse, linearized equations were solved using subroutine MA28 from the Harwell subroutine library . This subroutine takes advantage of sparsity which leads to great computational savings.
Table II shows the convergence rate ofx^M 1+1=x T, the elapsed computer time for the first 5 iterations, and the percentage of zeroes in the Jacobian (i.e., the sparsity) versus the number of elements. The x(T) column shows that the 32-element case has almost converged on the exact solution. (Recall that the exact solution is x(T) = –1.) Note further that the approximate x(T) is not an upper or lower bound of the exact value, which is common in mixed formulations. The third column of Table II gives the elapsed computer time for five iterations. It is easily seen that there is a modest increase in computer time with an increase in the number of elements. Note that in some cases acceptable convergence criteria were satisfied in fewer than five iterations. This is because the answers obtained from a small number of elements (say 2 or 4) may be interpolated to generate initial guesses for a higher number of elements. Finally, the extremely sparse structure of the Jacobian is demonstrated in the last column. This strongly encourages the use of a smart sparse matrix solver such as MA28. This subroutine leads to quicker solutions and tremendous savings in memory allocation since only the nonzero elements of the Jacobian need be stored.
Table II. x(T), elapsed computer time and percent sparsity of Jacobian versus the number of elements M 1
| M 1 | x(T) | Time (sec) | Sparsity (%) |
--- --- |
| 1 | –7.5465 | 0.42 | 65.3 |
| 2 | –0.4439 | 0.44 | 80.5 |
| 4 | –0.8385 | 0.66 | 89.6 |
| 8 | –1.0685 | 0.76 | 94.6 |
| 16 | –1.0141 | 1.03 | 97.3 |
| 32 | –1.0034 | 1.52 | 98.6 |
| Exact | – 1 | — | — |
Results for the state and control are shown in Figures 20 and 21 for 4 and 8 elements and the exact solution. Note from Figure 21 that the 4-element case does not define the constraint boundaries very accurately; however, it is accurate enough to generate guesses for the 8-element case. Thus, in a problem with many constrained and unconstrained arcs, a small number of elements could still be used to generate guesses for a higher number of elements.
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Figure 20. State history versus time
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Figure 21. Control history versus time
One of the constraint multipliers, μ 1, is shown in Figure 22. Note that μ 1 = 0 when u< 1, so only half the multiplier history is shown. It can be seen that both the 4- and 8-element cases yield very poor approximations for the multiplier, indicating the need for more elements. However, from the control history plot, the 8-element answer seemed very accurate. This highlights the need to check all results and verify that the multiplier histories seem reasonable. Note also that, due to the symmetry in this problem, the nonzero segment of μ 2 is identical to μ 1.
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Figure 22. Half of the multiplier history versus time
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26 European Symposium on Computer Aided Process Engineering
2016, Computer Aided Chemical EngineeringTobias Keßler, ... Michael Mangold
2.2 Scalarization of constrained problems:
The continuation algorithms to be used in this work are able to compute parameter dependent solutions of algebraic equation systems. Because originally they were not intended to solve optimization problems, their application to optimization requires to convert the optimization problem into an algebraic problem.
Necessary and, for convex optimization problems, sufficient optimality conditions are the well known Karush-Kuhn-Tucker (KKT) conditions. Equality constraints do not pose a problem to the continuation algorithm, because they are always active and, hence, do not have different solution branches. Therefore, to simplify the notation, an optimization problem with a single inequality constraint:
(2)min x J x w,s.t.g x≤0,
is discussed in the following. Applying the KKT-conditions and replacing the inequality constraint with an equality constraint by introducing a slack-variable s we get the following set of equations:
(3a)ℒ=J x w+μ⋅g x+s 2,
(3b)∂ℒ∂x=∂J∂x+μ⋅∂g∂x=0,
(3c)∂ℒ∂s=2⋅μ⋅s=0,
(3d)∂ℒ∂μ=g x+s 2=0.
There are two cases we have to take into account, active and inactive inequality constraints. From equation(3c) it follows that s=0 for active inequality constraints and μ=0 for inactive inequality constraints. Now, assume without loss of generality that:
(4)μ=0,f o r w=w,μ<0,f o r w0,f o r w>w.
The examination of the sign of μ yields, that, though there exist solutions for all w, the only relevant solutions for s=0 are the solutions where μ>0 and, hence, w>w (active inequality constraints).
For inactive inequality constraints (w<w) we get:
(5)s 2=−g x.
Because the constraint is inactive and g(x)<0 holds, there exist two real solutions, namely:
(6)s=±−g x.
Figure 1 illustrates the feasible solutions of the optimization problem. In summary the inequality constraints require a switching strategy between solution branches at points where an inequality constraint becomes active or inactive.
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Figure 1. Solution branches for s and μ. The solid black line shows the valid solution branch and the gray line marked with circles shows the invalid solution branch for w<w, the dashed line marked with stars shows the valid solution branch for w>w
In the following we will explain the switching strategy used to switch branches at the critical points.
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Chapter
Resolving Spectral Mixtures
2016, Data Handling in Science and TechnologyA. Malik, R. Tauler
4.2 Nonnegativity Constraints
Nonnegativity constraints are probably the more commonly used constraints in curve resolution from the initial works of Lawton and Sylvestre and afterward [34,35]. Physical concentrations can be only positive or zero (C≥0) and in many spectroscopies, spectra values can be also only positive or zero (ST≥0). Both constraints can be better implemented as inequality constraints:
(15)g cneg,i,n T=-c i,n T≤0
(16)g sneg,j,n T=-s j,n T≤0
i=1,…,N R,j=1,…,N C,n=1,…,N
According to Eq. (15) for nonnegative concentrations, for each component or species there will be NR inequality constraints, and for N species this will give a total of N×NR inequality constraints. According to Eq. (16) for nonnegative spectra, for each species there will be NC inequality constraints, and for N species this will give a total of N×NC inequality constraints. Nonnegativity constraints together with the normalization/closure constraints are the more frequently applied constraints in MCR problems. However, they do not constrain the problem sufficiently to give unique solutions in general [25,32].
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3.4 Constraint handling
In the above implementations, we have only considered the simple limits or bounds because the examples we used are mainly unconstrained optimization problems. For constrained optimization problems with complex equality and inequality constraints, proper constraint handling techniques should be used to ensure the solutions obtained are feasible, satisfying all the constraints.
There are many constraint handling techniques, including penalty methods, dynamic penalty, evolutionary methods, multiobjective approaches, ϵ-constraint approaches, and others (Boyd and Vandenberghe, 2004; Yang, 2014). Interested readers can refer to more specialized literature.
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4 MCR Constraints and Their Implementation
Constraints implementation is a critical step in the optimization of the function in Eqs. (10) and (11). The solutions of Eqs. (10) and (11) are unbounded if no constraints are considered; i.e., there is an infinite number of possible solutions, and it is not possible to find any boundaries for them. However, in most of MCR problems, it is still possible to apply adequate constrains to decrease the extension of rotation ambiguities [6,31,32]. Commonly used constraints are schematically presented in Fig. 5. Constraints are implemented as mathematical conditions and drive the MCR optimization process to the final solutions. There are constraints linked to natural properties of the measurement of the data set and others that express mathematical conditions or models [6,9,33].
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Fig. 5. Examples of common constraints used in MCR. (A) nonnegativity, (B) unimodality, (C) closure, (D) hard modeling. Bold black profiles on the left plots are unconstrained. Red (gray in the print version) profiles on the right are profiles that have been constrained.
Source: Tauler R, de Juan A. Multivariate Curve resolution for quantitative analysis. Fundamentals and Analytical Applications of Multiway Calibration. In: Muñoz de la peña A, Goicoichea H, Escandar G, Olivieri A, editors. Data Handling in Science and Technology. Elsevier, 2015.
4.1 Normalization and/or Closure Constraints
It is possible to avoid the intensity or scale ambiguity in the cn or sn T profiles by using appropriate normalization and/or closure constraints. Intensity ambiguities arise because for any scalar k and species or component n profiles cn and sn T,
(12)c n k k-1 s n T=c n s n T
For a particular component n the concentration profile cn can be arbitrarily increased in an unlimited way by multiplying it by an arbitrary scalar number k if at the same time its spectrum sn T is decreased by the same amount by dividing it by the same number k. It is possible to limit the size of cn or sn T profiles using appropriate normalization and closure constraints. A spectral normalization constraint may be implemented using the norm of the species spectra, for instance forcing it to be equal to one, i.e., ||(sn(T))||=1. This is an equality constraint and is implemented in Eq. (10) by the equality constraint function,
(13)g norm,n T=1-||s n T||=0,n=1,…,N
If N species are present, this will give N equality constraints. Alternatively, other possible normalization constraints can be applied by constraining the signal height or maximum intensity of a profile to be equal to a constant value. Obviously this will also give N equality constraints implemented in a very similar way to the one given previously in Eq. (13).
A completely different constraint is the closure constraint, which is usually implemented on the rows of matrix C. By this constraint the sum of the elements of each row of matrix C is equal to a known constant. This is the case for instance in reaction-based systems, where a mass balance equation is obeyed by the concentration profiles of the species present in the system, i.e., ∑n=1 N c i,n T=TOT i. This closure constraint is implemented by the equation:
(14)g clos,i T=TOT i-∑n=1 N c i,n T=0,i=1,…,N R
where TOT i is the known total concentration of the species in the considered spectra (row of matrix D). This will give NR equality constraints, one for each row of matrix C. The number of equality constraints increases with this constraint, decreasing the degrees of freedom of the optimization.
Either normalization or a closure constraint or other normalization constraint applied during the optimization of the function defined by Eq. (11) avoid scaling problems in the variables during the optimization. If neither or both constraints are applied, the optimization will not work because of the absence of scale boundaries. On the other hand, only one of the two constraints, spectral normalization or concentration closure, is usually applied, otherwise the system becomes overly constrained. Closure constraint can only be applied if this is the case for the data, otherwise the solutions of Eq. (11) will not be correct. Conversely, when there is no closure in the concentrations, normalization of the species spectra ST is highly recommended to achieve a successful optimization and recover the shape of the species profiles, even if the true solutions are obviously not fulfilling this normalization constraint.
4.2 Nonnegativity Constraints
Nonnegativity constraints are probably the more commonly used constraints in curve resolution from the initial works of Lawton and Sylvestre and afterward [34,35]. Physical concentrations can be only positive or zero (C≥0) and in many spectroscopies, spectra values can be also only positive or zero (ST≥0). Both constraints can be better implemented as inequality constraints:
(15)g cneg,i,n T=-c i,n T≤0
(16)g sneg,j,n T=-s j,n T≤0
i=1,…,N R,j=1,…,N C,n=1,…,N
According to Eq. (15) for nonnegative concentrations, for each component or species there will be NR inequality constraints, and for N species this will give a total of N×NR inequality constraints. According to Eq. (16) for nonnegative spectra, for each species there will be NC inequality constraints, and for N species this will give a total of N×NC inequality constraints. Nonnegativity constraints together with the normalization/closure constraints are the more frequently applied constraints in MCR problems. However, they do not constrain the problem sufficiently to give unique solutions in general [25,32].
4.3 Selectivity and Local Rank Constraints
Selectivity and local rank constraints are often used constraints in curve resolution because they allow in many cases the partial or even the total resolution of the system and elimination of rotational ambiguities [5,6,15]. Selectivity and local rank constraints have a tremendous effect narrowing considerably the bands of feasible solutions, collapsing them into a unique solution. Selectivity and local rank constraints refer to the fact that in certain windows or regions of the data matrix, a particular species is known to exist while others are known not to exist. For instance, in some chemical evolving processes (like spectroscopic titrations or chromatography), it is a common situation that some of the species are not present at the beginning or at the end of the evolving process. Also, for spectroscopic signals, it can happen that a particular species does not have any signal in a particular spectral range. In these cases it is extremely helpful to use this information as a constraint defining for each species the concentration window where it exists and the concentration window where it does not (or where it does exist only at very negligible concentrations), the spectral window where it contributes to the measured signal and the spectral window where it does not. As it has been demonstrated , when the resolution conditions are appropriate, the use of the local rank information as a constraint allows the recovery of the true species profiles without ambiguities. In practice, however, these resolution conditions are often not completely achieved, as for instance in the case of embedded peaks in chromatography, or when the species profiles are extremely overlapped like in kinetic reaction-based systems.
Equality constraints are sometimes too strong during the optimization and, in fact, in many circumstances we cannot be completely sure that the values are exactly equal to zero or to another value in a certain data window. Usually it is known that a particular species does not exist at appreciable concentrations or that it does not contribute to the signal in an appreciable way. However, this does not mean that the concentration or the signal should be exactly zero in that window, but, below a threshold value. For this reason it is usually better just to select a threshold value and to force the values of the profile in the window to be lower than this threshold value. If that threshold value is equal to ɛ then the selectivity/local rank constraints can be described as,
(17)g csel,i,n T=c s e l,i,n T-ɛ≤0
(18)g ssel,j,n T=s s e l,j,n T-ɛ≤0
i=1,…,ncsel n,j=1,…,nssel n,n=1,…,N
The number of inequality constraints will depend on the number of values included in the selected windows, ncsel n and nssel n, for each species n considered.
Another type of possible constraints would be to fix some or all values in C or ST to a value exactly known. It may happen for instance, that the spectrum of one of the species be known in advance. Known spectra essentially define certain elements of the transformation matrix. In that case such information can be used to simplify the optimization problem through inequality constraints like
(19)g cknown,i,n T=c i,n T-c known,i,n≤thresh
(20)g sknown,j,n T=s j,n T-s known,j,n≤thresh
i=1,…,N R,j=1,…,N C,n=1,…,N
for a preselected low threshold value, thresh. If the whole concentration profile of one species is known, this gives NR equality constraints. If the whole spectrum of one species is known, this gives NC equality constraints.
4.4 Unimodality
In a wide variety of situations, concentration profiles have unimodal peak shapes. This is the case for instance for chromatographic elution profiles or also for concentration profiles in reaction-based systems (excluding oscillating reactions). On the contrary, spectra profiles are usually not unimodal in the general case. Some other types of instrumental signals may also be unimodal, as in electrochemistry. Whenever the shape of the profiles is unimodal, an additional constraint which may be useful is unimodality. In that case, what is intended is to avoid the formation of secondary maxima. The way used to implement unimodality constraints has been algorithmic. First the highest maximum is detected and then all the departures from the unimodal condition are constrained, discarding left and right maxima: at the left of the peak maximum,
(21)g unimod,i,n=c i-1,n T-c i,n T≤thresh
and at the right of the peak maximum,
(22)g unimod,i,n=c i+1,n T-c i,n T≤thresh
for a preselected low threshold value, thresh. This constraint can also include a small local departure from unimodality tolerance because of noise, which gives the amount of local increasing concentration tolerated at the right or at the left of the maximum concentration. See Refs. [7,37] for more detailed descriptions of the unimodality constraint. The number of inequality constraints from Eqs. (21) and (22) will change during the optimization depending on the shape adopted by the concentration profile c(T).
4.5 Model or Multilinearity Constraints
These are constraints applicable to multiset data using the extension of MCR bilinear model to the analysis of augmented data matrices. Although MCR gives by default a bilinear model, the conditions of trilinearity [38,39], multilinearity , or factor interaction (like in Tucker models) can be implemented as a constraint in a component-wise way. Therefore, completely bilinear, completely trilinear, or hybrid models can be set to be obtained in the final MCR results. Implementation of multilinear model constraint enforces decompositions to give unique solutions for the multifactor matrices (apart from scale and trivial permutation rotation ambiguities) and avoids the presence of rotational ambiguities associated with lower structured bilinear models . The fulfillment of this type of constraints implies for instance that a particular component (species) has the same concentration profiles in the different data sets (matrices) simultaneously analyzed, apart than the spectra profile already enforced to be the same by the default bilinear mode. An easy way to implement this in practice is checking for the common shape of the concentration profiles of this component in the different data sets by the singular value decomposition method . If the shape of the considered component concentration profile is the same in the different data sets, (fulfillment of the trilinear model), only one singular value should be larger than the noise level. A constraint can be implemented using this idea using the following equation:
(23)g tril,n=s v d 2/s v d 1≤thresh
where svd(2) and svd(1) refer respectively to the second and first singular value of the matrix having the k concentration profiles of the same component (n) in the k simultaneously analyzed data sets (k data matrices). thresh is a threshold value sufficiently small to constrain the considered profile of this component n to be practically invariant among different k data sets.
As shown in previous equations, final implementation of all these constraints may be performed using vectors of equality or inequality constraints. For instance spectra normalization constraints are implemented as equality constraints and nonnegativity constraints are instead implemented as inequality constraints (concentration and spectra should not be negative, but they can be zero or positive). All equality constraints are grouped together in a vector of equality constraints, g eq, and all inequality constraints are grouped together in a vector of inequality constraints, g ineq. See Ref. for a detailed description of the implementation of these and other type of constraints.
4.6 Hard Modeling
This constraint implies performing a model curve fitting task during the iterative optimization process [38,42]. When this constraint is applied for instance as a mass action law or as a kinetic rate law, the related concentration profiles do not present ambiguity. It enforces the concentration profiles to be fitted by a parametric physicochemical model, and the parameters of the model are obtained as an additional output. It can also be applied to pure analytical responses (in ST in Eq. 1), when the shape of the pure signal can be defined by a parametric equation. It is relevant to note that the hard modeling constraint and, particularly, some of the fitted parameters, e.g., total concentration in equilibrium or enzymatic models have been used for quantitative analysis purposes .
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12 International Symposium on Process Systems Engineering and 25 European Symposium on Computer Aided Process Engineering
2015, Computer Aided Chemical EngineeringA.M. Acevedo P., ... Galo A C. Le Roux
2.2 Pricing Interprocess Stream Using Slack Auctions
In this technique, the complicating constraints are rewritten as resource constraints (inequality constraints) to be used as a criterion for the algorithm convergence. A negative term is added to the maximization economic objective function of each subproblem, which is composed of a weight parameter (p) multiplied by the absolute value of the complicating constraints, enforcing their difference to decrease at each iteration. For economic purposes, p represents the price of the intermediate streams that is updated at each iteration (Rinaldo and Ungar, 2000).
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Condyloma Acuminatum
Editors: Jorma Paavonen MD; Elliot M. Levine MD, FACOG; Paritosh Prasad MD; Zbigniew Fedorowicz PhD, MSc, DPH, BDS, LDSRCS All Editors & Disclosures
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Condyloma acuminatum is a proliferation of anogenital skin and mucosa in response to infection by human papillomavirus, commonly referred to as anogenital warts.1 ,2 ,3
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The highest prevalence of genital warts is in sexually active female adults 20-24 years old and male adults 25-29 years old.3
An estimated 500,000 to 1 million new cases of genital warts are reported in the United States annually.2
Clinically apparent warts are reported in about 1%-2% of the sexually active population.2 ,3
The reported lifetime prevalence in the sexually active population in the United States is about 10%.3
SYSTEMATIC REVIEW BMC Infect Dis 2013 Jan 25;13(1):39 global prevalence of anogenital warts estimated to be 0.13%-5.1%
Study Details
based on systematic review of observational studies
systematic review of 32 studies evaluating prevalence of human papillomavirus (HPV) in adults
overall reported annual incidence of anogenital warts (new and recurrent) 160-289 per 100,000 persons
overall prevalence of anogenital warts 0.13%-5.1%
annual incidence rates of new genital warts in
men 103-168 per 100,000
women 76-191 per 100,000
incidence of recurrent genital warts in
men 163 per 100,000
women 110 per 100,000
peak incidence in
men aged 25-29 years
women < 24 years old
Reference - BMC Infect Dis 2013 Jan 25;13(1):39
SYSTEMATIC REVIEW Lancet 2019 Aug 10;394(10197):497 incidence of anogenital warts decreasing in females 15-29 years old and males 15-24 years old compared to prevaccination periods in high-income countries
Study Details
Obstetric_and_Gynecologic_Conditions incidence of anogenital warts decreasing in girls and women aged 20-29 years and boys and men aged 20-24 years compared to prevaccination periods in high-income countries (Lancet 2019 Aug 10)10/01/2020 05:03:33 PM
based on systematic review
systematic review of 65 studies comparing incidence or prevalence of human papillomavirus (HPV)-related endpoints (genital HPV infections, anogenital wart diagnoses, or histologically confirmed cervical intraepithelial neoplasia 2+ [CIN2+]) in pre-and postvaccination periods of general population in 14 high-income countries
29 studies assessed frequency of anogenital warts
after 5-8 years of vaccination anogenital warts diagnosis significantly decreased in
females aged 15-19 years (relative risk [RR] 0.33, 95% CI 0.24-0.36)
females aged 20-24 years (RR 0.46, 95% CI 0.36–0.6)
females aged 25-29 years (RR 0.69, 95% CI 0.53–0.89)
males aged 15-19 years (RR 0.52, 95% CI 0.37-0.75)
males aged 20-24 (RR 0.68, 95% CI, 0.47-0.98)
no significant differences in anogenital warts diagnosis in females 30-39 years old or males 25-39 years old
Reference - Lancet 2019 Aug 10;394(10197):497
Risk Factors
Risk factors for condyloma acuminatum include:
Specific sexual behaviors, such as:1 ,2
Unprotected sexual intercourse
Multiple sexual partners
Early age of initial sexual contact
History of other sexually transmitted infections
Immunosuppression, especially in patients with HIV infection:
CROSS-SECTIONAL STUDY Sex Transm Dis 2012 Nov;39(11):902 positive HIV status associated with increased risk of genital warts in female sex workers in Kenya
Study Details
based on cross-sectional study
1,182 female sex workers in Kenya with or without HIV-1 had pelvic exam
613 women (51.4%) were HIV-1-positive
women with HIV-1 had increased risk of genital warts (odds ratio 7.69, 95% CI 2.3-25.6)
Reference - Sex Transm Dis 2012 Nov;39(11):902
COHORT STUDY Lancet 2002 Jan 12;359(9301):108 women with HIV infection may have increased risk for condyloma acuminata
Study Details
based on prospective cohort study
925 women (52% HIV-1-positive) had gynecologic exam including colposcopy and human papillomavirus (HPV) DNA testing of cervicovaginal lavage twice yearly for median 3.2 years
vulvovaginal and perianal condyloma acuminata or intraepithelial neoplasia in 6% of women with HIV-1 vs. 1% of women without HIV-1 (p < 0.0001)
comparing women with vs. without HIV-1 in women without lesions at baseline
incident vulvovaginal or perianal lesions in 9% vs. 1% (p < 0.0001)
condyloma acuminata in 7% vs. 1% (p < 0.0001)
risk factors for incident HPV lesions included
HIV-1 infection (p = 0.013)
HPV infection (p = 0.0013)
decreased CD4 T lymphocyte count (p = 0.04)
history of frequent IV drug use (p = 0.02)
Reference - Lancet 2002 Jan 12;359(9301):108
COHORT STUDY Int J STD AIDS 2014 Jan;25(1):29 smoking and alcohol use may be associated with increased risk of external genital wart infection
Study Details
based on retrospective cohort study
22,802 patients (mean age 27 years) attending 1 of 2 genitourinary medicine clinics in Ireland from 1999 to 2009 screened for sexually transmitted diseases were included
101 patients diagnosed with HIV infection excluded from analysis
external genital wart infection in 23.6%
risk factors associated with external genital wart infection
smoking (odds ratio [OR] 1.41, 95% CI 1.29-1.54)
alcohol use (OR 1.35, 95% CI 1.18-1.55)
Reference - Int J STD AIDS 2014 Jan;25(1):29
Associated Conditions
Conditions that may be associated with condyloma acuminatum include:
Other sexually transmitted diseases
Other human papillomavirus (HPV)-related diseases, such as vulvar intraepithelial neoplasia (VIN), squamous cell carcinoma of the vagina, and penile carcinoma 2
Verrucous carcinoma may be associated with HPV. It is spread by local invasion and rarely metastasizes. It can vary histologically from pseudoepitheliomatous hyperplasia-like lesions to invasive squamous cell carcinoma. There are different clinicopathologic types based on the anatomic area.2
Giant condyloma of Buschke and Löwenstein occurs in the anogenital area.
Oral florid papillomatosis occurs in the oral cavity.
Carcinoma cuniculatum occurs on palmoplantar surfaces.
COHORT STUDY Eur J Obstet Gynecol Reprod Biol 2013 May;168(1):80 condyloma acuminata may be associated with cervical squamous intraepithelial lesions
Study Details
based on prospective cohort study
846 sexually active female adolescents in Brazil (mean age 15.8 years) with normal Pap smear and no genital warts followed for 2 years
63 adolescents (7.4%) presented with condyloma acuminata during follow-up period
abnormal Pap smear in 20.5% by end of follow-up
cervical squamous intraepithelial lesions in 70% of adolescents with genital warts
presence of genital warts associated with increased risk of cervical squamous intraepithelial lesions (relative risk 4.2, 95% CI 3.3-5.3)
Reference - Eur J Obstet Gynecol Reprod Biol 2013 May;168(1):80
Etiology and Pathogenesis
Pathogen
Human papillomavirus (HPV) is the causative agent of condyloma acuminatum.1 ,2
HPV is group of nonenveloped, double-stranded DNA viruses in the Papovaviridae family.
The types of HPV that cause genital warts are different from the types that cause anogenital cancers.
There are > 120 subtypes of HPV.
HPV-6/11 are low malignant risk subtypes and cause 90% of genital warts.
HPV types 31, 33, 35, 45, 51, 52, 56, 58, 59, and 68 are:
Associated with an intermediate malignant risk
Common causes of squamous intraepithelial neoplasia
Occasionally identified in visible genital warts, usually with co-infection with HPV-6/11
HPV-16/18 are high malignant risk subtypes associated with cervical cancer and anogenital cancers, usually with co-infection with HPV-6/11.
There has been an increasing prevalence of HPV infection over past 35 years in United States likely due to:2
Earlier age of first sexual contact
Increases in the total number of lifetime sexual partners
Transmission
Human papillomavirus (HPV) infection is typically transmitted through oral, anal, and genital sexual contact with an infected person.
Cases of autoinoculation have been reported.
Vertical transmission (from mother to child, typically during pregnancy and at the time of delivery) can also occur.
Reference - J Obstet Gynaecol Res 2017 Mar;43(3):429
It is often difficult to determine how or when a patient becomes infected with HPV since transmission can be asymptomatic.1
The incubation period ranges from 3 weeks to 8 months (mean 2-3 months after initial contact). The infection can remain dormant for prolonged periods with reported subclinical infection in up to 40%.2
Pathogenesis
Human papillomavirus (HPV) infection leads to progression of epithelial change.2
Viral particles penetrate both cutaneous and mucosal epithelium.
The virus invades and infects basal keratinocytes of the epidermis.
Mucosa can be infected anywhere along the genital tract.
Viral replication is restricted to the basal cell layer, and the life cycle is linked to progressive differentiation of epithelial cells.
Infected regions are marked by proliferation of viral DNA and formation of a warty papule or plaque.
The viral genome is composed of 6 early-open reading frames (E1, E2, E4, E5, E6, E7) and 2 late-open reading frames (L1, L2).2
Early-open E genes are important for regulatory function and encode proteins involved in viral replication and cell transformation.
Late-open L genes encode viral capsid proteins.
Differences in L1 define HPV subtypes.
Low-risk HPV subtypes remain separate from the host cell DNA and undergo replication independently.
High-risk HPV subtypes incorporate their DNA directly into the host cell’s genetic material.
Integration of viral and host cell DNA results in dysregulation and uncontrolled activation of E6 and E7 genes and promotes oncoprotein transcription that binds and inactivates tumor suppressor genes p53 and retinoblastoma.
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Related Topics
Human Papillomavirus (HPV) VaccineVulvar Intraepithelial Neoplasia (VIN)Vulvar Squamous Cell Carcinoma (VSCC)Cervical Cancer Screening
Produced in collaboration with American College of Physicians
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Related Topics
Human Papillomavirus (HPV) VaccineVulvar Intraepithelial Neoplasia (VIN)Vulvar Squamous Cell Carcinoma (VSCC)Cervical Cancer Screening
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1094 | https://fiveable.me/ap-physics-2-revised/unit-15/5-the-photoelectric-effect/study-guide/OLYoFyn8nutvrlZD | printables
🧲AP Physics 2 (2025) Unit 15 Review
15.5 The Photoelectric Effect
🧲AP Physics 2 (2025) Unit 15 Review
15.5 The Photoelectric Effect
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
🧲AP Physics 2 (2025)
Unit & Topic Study Guides
15.1 Quantum Theory and Wave-Particle Duality
15.2 The Bohr Model of Atomic Structure
15.3 Emission and Absorption Spectra
15.4 Blackbody Radiation
15.5 The Photoelectric Effect
15.6 Compton Scattering
15.7 Fission, Fusion, and Nuclear Decay
15.8 Types of Radioactive Decay
Photoelectric Effect Interactions
more resources to help you study
practice questionscheatsheetscore calculator
Electron Emission Fundamentals
The photoelectric effect occurs when electromagnetic radiation strikes a photoactive material, causing electrons to be emitted from the material. This phenomenon provides compelling evidence for the particle-like nature of light, as the energy of incident photons is directly transferred to electrons in the material.
Photoactive materials include certain metals like sodium and potassium, which have relatively low work functions
Semiconductors such as silicon and germanium also exhibit the photoelectric effect
The effect forms the basis for many modern technologies including solar cells, light meters, and night vision devices
Threshold Frequency
For electron emission to occur, the incident light must have a minimum frequency known as the threshold frequency. This critical concept helps explain why light behaves as discrete packets of energy rather than continuous waves.
When light frequency equals or exceeds the threshold frequency, electrons will be emitted regardless of light intensity
If light frequency is below the threshold, no electrons will be emitted even if the light is extremely intense
This observation contradicted classical wave theory, which predicted that sufficient light intensity of any frequency should eventually cause electron emission
For example, if a metal has a threshold frequency of 5×1014 Hz, light with a frequency of 6×1014 Hz will cause electron emission, while light with a frequency of 4×1014 Hz will not produce any photoelectrons, no matter how bright the light source.
Maximum Kinetic Energy and Frequency
The maximum kinetic energy of emitted electrons depends on both the frequency of the incident light and the material's work function. This relationship reveals important insights about energy conservation in quantum interactions.
The work function (ϕ) represents the minimum energy needed to remove an electron from a material
The relationship between maximum kinetic energy, light frequency, and work function is given by: Kmax=hf−ϕ where h is Planck's constant (6.63×10−34 J\cdots) and f is the light frequency
In a typical photoelectric effect experiment:
Two metal plates are placed in a vacuum chamber and connected to a variable voltage source
One plate is illuminated with monochromatic light, causing electrons to be ejected
The potential difference between plates is adjusted until the current drops to zero
This stopping potential directly relates to the maximum kinetic energy of the emitted electrons
Work Function of Materials
The work function is a characteristic property that represents the minimum energy required to remove an electron from a material's surface. It varies significantly between different materials and affects how they respond to light.
Materials with lower work functions emit electrons more easily when exposed to light
The work function determines the threshold frequency through the relationship: fthreshold=ϕ/h
Some approximate work function values:
Sodium: 2.3 eV
Potassium: 2.3 eV
Copper: 4.7 eV
Zinc: 4.3 eV
🚫 Boundary Statement
Work functions for materials will be provided on the exam when needed; students are not required to memorize work function values or factors that influence the work function of a material.
Practice Problem 1: Threshold Frequency
A metal has a work function of 2.0 eV. Calculate the threshold frequency for this metal. (Planck's constant h = 6.63 × 10^-34 J·s, and 1 eV = 1.60 × 10^-19 J)
Solution
To find the threshold frequency, we need to use the relationship between work function and threshold frequency:
fthreshold=ϕ/h
First, let's convert the work function from eV to joules: ϕ=2.0 eV×1.60×10−19 J/eV=3.20×10−19 J
Now we can calculate the threshold frequency: fthreshold=6.63×10−34 J\cdotps3.20×10−19 J=4.83×1014 Hz
Therefore, light with a frequency below 4.83 × 10^14 Hz will not cause electron emission from this metal.
Practice Problem 2: Maximum Kinetic Energy
Light with a frequency of 7.0 × 10^14 Hz strikes a metal surface with a work function of 1.8 eV. Calculate the maximum kinetic energy of the emitted electrons in eV. (Planck's constant h = 6.63 × 10^-34 J·s, and 1 eV = 1.60 × 10^-19 J)
Solution
We can use the photoelectric effect equation to find the maximum kinetic energy: Kmax=hf−ϕ
First, let's calculate the energy of the incident photons: hf=6.63×10−34 J\cdots×7.0×1014 Hz=4.64×10−19 J
Converting to eV: hf=4.64×10−19 J÷1.60×10−19 J/eV=2.9 eV
Now we can calculate the maximum kinetic energy: Kmax=2.9 eV−1.8 eV=1.1 eV
Therefore, the maximum kinetic energy of the emitted electrons is 1.1 eV.
Frequently Asked Questions
What is the photoelectric effect and how does it work?
The photoelectric effect is when photons hit a photoactive material and eject electrons (photoelectrons). Each photon has energy hf; to remove an electron the photon must have at least the material’s work function φ (so there’s a threshold frequency f0 = φ/h). If f ≥ f0 electrons are emitted, and the maximum kinetic energy of an ejected electron is Kmax = hf − φ. Kmax depends on the light’s frequency, not on how many photons hit the surface—that’s evidence light is quantized into photons. In lab setups you shine monochromatic light on a metal cathode in a vacuum cell, measure current vs. applied potential, and find the stopping potential where current → 0; eVstop = Kmax lets you determine φ. This is directly tested by AP (Topic 15.5: threshold frequency, work function, Einstein photoelectric equation). For a clear AP-aligned review, see the Topic 15.5 study guide ( More unit review and tons of practice problems are at the unit page ( and practice set (
Why do only certain frequencies of light cause electrons to be emitted from metals?
Only certain light frequencies eject electrons because light comes in photons with energy E = hf. Each electron needs at least the material’s work function φ to escape. If hf < φ (frequency below the threshold frequency), a single photon doesn’t have enough energy, so no electrons are emitted no matter how bright the light is. Once f ≥ f_threshold, electrons can be emitted and the maximum kinetic energy is K_max = hf − φ—so higher frequency (not higher intensity) makes faster photoelectrons. This behavior (threshold frequency, quantized photons, K_max = hf − φ) is exactly what the CED asks you to know for Topic 15.5. For a quick review and examples, check the Photoelectric Effect study guide on Fiveable ( For broader unit review and lots of practice problems, use the Unit 15 page and Fiveable practice questions (
I'm confused about threshold frequency - can someone explain what this means?
Threshold frequency is the minimum light frequency f_threshold needed to free an electron from a material by the photoelectric effect. If f < f_threshold, no electrons are emitted no matter how bright (intense) the light is—intensity only changes the number of photons, not their individual energy. Each photon has energy E = hf. To eject an electron you need hf ≥ φ, where φ is the work function (minimum energy to remove an electron). So f_threshold = φ/h. If hf > φ, emitted electrons can have excess kinetic energy: K_max = hf − φ (that’s the Einstein photoelectric equation on the CED). On the AP exam you won’t need actual work-function values (they’re given if needed), but you should be able to use K_max = hf − φ and relate it to stopping potential in experiments. For a quick review, check the Topic 15.5 study guide ( and try practice problems (
What's the difference between the number of photons hitting a material and the energy of each photon?
Number of photons vs. energy per photon: they’re two different things. The energy of each photon is E = hf (so set by frequency), and that determines whether an individual photon can overcome the material’s work function φ and how much kinetic energy an emitted electron can have (Kmax = hf − φ). The number of photons hitting the surface is basically the light’s intensity (photon flux). Increasing the photon number increases how many electrons are ejected per second (photoelectric current), but it does not change Kmax. If the photon energy is below the threshold frequency (so hf < φ) no electrons are emitted no matter how many photons hit. This distinction is exactly what the CED highlights for Topic 15.5 and is commonly tested (use Kmax = hf − φ on FRQs). For a quick review, see the Topic 15.5 study guide ( and grab practice problems at (
How do I use the equation Kmax = hf - φ to solve photoelectric effect problems?
Use Kmax = hf − φ exactly like an energy budget: each photon gives energy hf; φ (the work function) is the minimum energy to free an electron; the leftover is the maximum kinetic energy of the ejected electron. Steps to solve problems 1. Identify what’s given/asked (f, λ, φ, Kmax, or stopping potential Vstop). 2. If wavelength is given, convert to frequency: f = c/λ. 3. Plug into Kmax = hf − φ. If Kmax comes out negative → no electrons are emitted (f < threshold f0 = φ/h). 4. If they ask for stopping potential, use Kmax = eVstop (so Vstop = Kmax/e). 5. Watch units: h = 6.63×10^−34 J·s, c = 3.00×10^8 m/s, e = 1.60×10^−19 C. Example: given λ and φ in eV, convert φ to joules (φ[eV]×1.60×10^−19), find f = c/λ, compute Kmax, then Vstop = Kmax/e. This matches the CED essential knowledge (15.5.A.2–3). For more worked examples and practice problems, see the Topic 15.5 study guide ( and the AP Physics 2 practice bank (
What exactly is work function and why is it different for different materials?
The work function φ is the minimum energy needed to free an electron from a material—basically the energy barrier an electron must overcome to escape the metal (CED 15.5.A.3.i). In the photoelectric equation Kmax = hf − φ (CED 15.5.A.3.ii), φ sets the threshold frequency f₀: φ = h f₀. If hf < φ no electrons are emitted. Different materials have different φ because their electrons are bound differently: atomic/solid electronic structure, Fermi level, lattice arrangement, and surface chemistry (oxide layers, contamination) change how tightly electrons sit in the material. Metals with higher electron binding energy have larger φ and thus need higher-frequency light to emit electrons. On the AP exam you won’t need to memorize φ values (they’ll be given if needed). For a quick AP-aligned refresher, see the Topic 15.5 study guide ( For more review and practice, check the unit page ( and practice problems (
Why doesn't increasing the brightness of light below threshold frequency cause any electrons to be emitted?
Because light comes in photons, each with energy E = hf, emission needs every electron’s single-photon energy to meet or exceed the work function φ. The threshold frequency f₀ is defined by hf₀ = φ. If your light has f < f₀, each photon has hf < φ, so no single photon can free an electron—increasing brightness just gives you more low-energy photons, not more energy per photon, so no electrons are emitted. This is exactly what the CED stresses: emission requires a minimum frequency regardless of photon number (15.5.A.2.i and 15.5.A.2.ii), and Kmax = hf − φ only applies when hf > φ (15.5.A.3.ii). Note: very intense lasers can produce multi-photon effects, but that’s beyond AP scope and the typical photoelectric setup. For a quick review, see the Topic 15.5 study guide ( and practice problems (
Can someone explain the photoelectric effect experiment setup with the two metal plates?
Think of the classic photoelectric setup as a vacuum tube with two metal plates (a photoactive cathode and a collecting anode) connected to a variable power supply and an ammeter. Monochromatic light shines on the cathode; if each photon has energy hf ≥ φ (the work function), electrons are ejected (photoelectrons) into the vacuum. Some of those electrons reach the anode and produce a measurable current (photoelectric current). You vary the potential difference (make the anode more negative relative to the cathode) until the current drops to zero—that voltage is the stopping (or cutoff) potential. At that point eVstop = Kmax = hf − φ, so measuring Vstop vs. frequency lets you determine φ and confirm photons are quantized. The whole experiment is done in vacuum to prevent collisions and uses monochromatic light so hf is well defined (CED 15.5.A, 15.5.A.2–3). For a quick review, see Fiveable’s photoelectric study guide ( and practice problems (
I don't understand why the kinetic energy of emitted electrons doesn't depend on how many photons hit the material.
Good question—think of photons like individual bullets: each electron needs one photon with enough energy (hf ≥ φ) to escape. The maximum kinetic energy of an emitted electron comes from a single photon's energy minus the work function: Kmax = hf − φ (CED 15.5.A.3.ii). So if you increase the number of photons (light intensity) you increase how many electrons get hit per second → larger photoelectric current, but each electron still only gets energy from the one photon that hit it. That’s why Kmax (or the stopping potential in an experiment) doesn’t change with intensity, only with frequency; raise f and hf rises, so Kmax rises. Also remember the threshold frequency: below it no electrons are emitted regardless of how many photons hit (CED 15.5.A.2 and 15.5.A.2.i). For a short review, check the Topic 15.5 study guide ( and try practice problems ( to see current vs. stopping-potential graphs.
What happens when light frequency is exactly at the threshold frequency?
If the light’s frequency is exactly the threshold frequency f0, each photon has just enough energy to free an electron (hf0 = φ) but no extra to give it kinetic energy. So electrons can be emitted, but their maximum kinetic energy Kmax = hf − φ = 0. Practically that means emitted electrons barely escape the surface (zero KE), the stopping potential is essentially 0 V, and you’d measure a very small or easily interrupted photoelectric current (rate still depends on photon flux). This is exactly what the CED emphasizes: emission requires f ≥ f0, and Kmax = hf − φ (Topic 15.5). For extra practice or to review how this shows light’s particle nature, check the Topic 15.5 study guide on Fiveable ( and try related practice problems (
How does the photoelectric effect prove that light acts like particles instead of waves?
The photoelectric effect shows light acts like particles (photons) because experiments contradict a pure-wave picture. Key observations from the CED: (1) Electrons are only emitted if incident light has a minimum frequency f₀ (threshold frequency), independent of intensity. (2) Increasing intensity (more light waves per second) raises the number of emitted electrons but not their maximum kinetic energy. These match Einstein’s photon idea: each photon has energy E = hf, and one photon transfers its energy to one electron. If hf < φ (the work function), no electron is emitted regardless of brightness; if hf > φ, the maximum KE = hf − φ (Kmax = hf − φ). The stopping-potential experiment (adjust potential until current stops) measures Kmax and confirms the linear dependence on frequency, not intensity—strong evidence for quantized photons. For the AP exam, be ready to use Kmax = hf − φ and explain threshold frequency (Topic 15.5; see the Fiveable study guide for this topic: For extra practice, check unit resources and practice problems ( and
Why do we need to adjust the potential difference between plates until no current flows in the photoelectric effect experiment?
You adjust the potential until no current flows to find the stopping potential—the voltage that just prevents the fastest (maximum-energy) photoelectrons from reaching the anode. At that point the electric potential energy change eVstop equals the electron’s maximum kinetic energy, so Kmax = eVstop. Using Einstein’s photoelectric equation Kmax = hf − φ, you can therefore get eVstop = hf − φ and solve for φ or check the linear relationship between Vstop and frequency. That’s why the CED’s typical setup varies the potential until current = 0: it isolates the maximum kinetic energy of emitted electrons (e is the elementary charge) and gives direct experimental evidence for photons and the work function. For more AP-style review and practice on this topic, see the Photoelectric Effect study guide ( and unit overview (
What's the relationship between photon energy and electron kinetic energy in the photoelectric effect?
Photon energy (Eph) = hf. In the photoelectric effect a single photon gives its energy to one electron. If Eph is less than the work function φ (minimum energy to remove an electron) no electrons are emitted. If Eph ≥ φ, electrons are emitted and the maximum kinetic energy of an emitted electron is Kmax = hf − φ. So increasing the light’s frequency raises Kmax linearly; increasing intensity (more photons/sec) only raises the number of emitted electrons, not their individual energies. You can measure Kmax by the stopping potential Vstop: eVstop = Kmax. These are the exact AP ideas in CED 15.5.A (threshold frequency, work function, Einstein photoelectric equation). For a quick review, see the Topic 15.5 study guide (Fiveable) ( For more practice problems across the unit, check Fiveable practice (
I missed the lab on photoelectric effect - what were we supposed to observe and measure?
In the lab you would watch a photoactive metal emit electrons when hit with monochromatic light and measure two main things: the photoelectric current and the stopping (retarding) potential. Using a vacuum photoelectric cell, you’d vary light frequency (and intensity) and adjust the potential difference between cathode and anode until the current drops to zero—that stopping potential Vstop gives Kmax = eVstop. Key observations: (1) below a threshold frequency no electrons are emitted, (2) above threshold the maximum kinetic energy of electrons increases linearly with frequency (Kmax = hf − φ), and (3) changing intensity changes current (number of electrons) but not Kmax. From a plot of eVstop vs. f you can get Planck’s constant (slope = h) and the work function φ (intercept). This matches the AP CED goals (15.5.A, stopping potential, work function)—review the Topic 15.5 study guide ( and practice more problems at (
How do I know if a given frequency of light will cause photoelectric emission from a specific metal?
Check the metal’s work function φ (minimum energy to free an electron). Convert the given light frequency f to photon energy using E = hf (h = Planck’s constant). If hf ≥ φ (equivalently f ≥ f_threshold where φ = h f_threshold), electrons can be emitted; if hf < φ, no photoelectric emission occurs no matter the intensity. The maximum kinetic energy of emitted electrons is K_max = hf − φ, and you can measure K_max experimentally from the stopping potential V_s via eV_s = K_max. On the AP exam you may be given φ or f_threshold; use K_max = hf − φ and remember emission depends on frequency (not photon number), which is key evidence for photons (CED 15.5.A.2–3). For a quick review and practice problems, see the topic study guide ( and unit overview ( |
1095 | https://polisci.berkeley.edu/sites/default/files/people/u3827/CPS-2015-Paine-Less%20Distinctive%20than%20Claimed.pdf | Comparative Political Studies 1 –39 © The Author(s) 2015 Reprints and permissions: sagepub.com/journalsPermissions.nav DOI: 10.1177/0010414014564851 cps.sagepub.com Article Set-Theoretic Comparative Methods: Less Distinctive Than Claimed Jack Paine1 Abstract Proponents of set-theoretic comparative methods (STCM) sharply differentiate their approach from quantitative analysis—unlike many researchers who focus on integrating qualitative and quantitative methods. This article engages these opposing views by demonstrating shared foundations between STCM and quantitative techniques. First, it shows how the quantitative practice of analyzing cases that exhibit variation on both the explanatory conditions and the outcome—for example, all four cells of a 2 × 2 table—guards against misleading conclusions about necessary/sufficient conditions. Hence, conventional statistical ideas about association are relevant for STCM. Second, STCM’s tools for analyzing causal complexity share important features with regression interaction terms. Third, scrutinizing these shared foundations suggests how stronger theoretical and empirical standards for causal inference with deterministic hypotheses can be established. Focusing on shared foundations and recognizing that STCM does not genuinely break new inferential ground facilitate new opportunities for strengthening comparative research tools, rather than unproductively overemphasizing differences from mainstream methods.
Keywords causal inference, comparative method, determinism, necessary and sufficient conditions, QCA, qualitative comparative analysis, qualitative methods, regression, set theory 1University of California, Berkeley, CA, USA Corresponding Author: Jack Paine, Department of Political Science, University of California, 210 Barrows Hall, Berkeley, CA 94720, USA. Email: jackpaine@berkeley.edu 564851 CPSXXX10.1177/0010414014564851Comparative Political StudiesPaine research-article2015 at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 2 Comparative Political Studies The relationship between qualitative and quantitative methods—similari-ties, contrasts, and the feasibility of integration—is the focus of a major current debate in comparative politics. On one hand, many works stress the distinctive strengths of qualitative methods while also emphasizing that qualitative and quantitative traditions share many inferential goals. Brady and Collier (2010) advocate “shared standards” for different research tradi-tions, Lieberman (2005) presents an iterative method for combining statis-tical and case study findings, Dunning (2012) demonstrates the crucial contribution of qualitative work to evaluating natural experimental designs, and Seawright (2014a) advocates an integrative approach that bridges these two approaches.
On the other hand, several recent and influential books emphasize funda-mental differences between how qualitative and quantitative scholars approach social phenomena. Ragin (2008), Goertz and Mahoney (2012), and Schneider and Wagemann (2012) argue the core goal of qualitative research is to evaluate “complex” combinations of necessary and/or sufficient condi-tions. They propose a set-theoretic approach for studying these relationships. Given the focus of these three books and related work on set theory and cross-case comparisons, I refer to the techniques as set-theoretic comparative methods (STCM).
STCM scholars argue that there are sharp differences between their tools and quantitative methods. Mahoney, Goertz, and Ragin (2013) suggest that “in the social sciences, statistical and set-theoretic scholars adopt different approaches to causal analysis” (p. 75). Goertz and Mahoney (2012) maintain that “overcoming the quantitative-qualitative division in the social sciences is significantly a matter of better understanding the methodological differences between these two traditions along with the reasons why those differences exist” (p. 5).
Ragin (2008) begins his book by emphasizing sharp differences between STCM and quantitative methods. He underscores the contrast between “set-theoretic versus correlational connections” (pp. 6-10). A key example is found in Ragin’s argument that scholars, when studying necessary and suffi-cient conditions, should not incorporate cases that exhibit variation on both the posited causal conditions and the outcome—and that one of the cells of a 2 × 2 table is never needed to analyze either necessity or sufficiency (Ragin, 2000, p. 96; 2008, pp. 20-23).1 This approach is presented as sharply differ-entiating STCM from standard ideas of correlation and association in quanti-tative methods, as well as with quantitative case selection practices. Furthermore, in discussing “configurations of conditions versus ‘indepen-dent’ variables,” Ragin calls for a focus on alternative constellations of causal conditions. He claims that regression, by contrast, is limited to examining the at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 3 net effect of individual variables—a shortcoming Ragin sees as inherent even in regression with interaction terms.
STCM has commendably brought core issues in comparative analysis to the forefront of methodological attention. Its proponents routinely emphasize many foundational points about data analysis that should be heeded in all work, qualitative or quantitative. These considerations include careful case selection, close attention to measurement, and the inherent conditionality of most causal relationships. For these reasons, this approach has become an influential part of the broader reinvigoration of qualitative methods research in response to, and in part in reaction against, King, Keohane, and Verba (“KKV,” 1994).
However, by emphasizing the differences between their methods and quantitative research, STCM scholars have also posed an important challenge to “post-KKV” qualitative methods that seek integration with quantitative techniques—as with the studies just noted of Brady and Collier, Lieberman, Dunning, and Seawright.
This article attempts to advance this important debate by demonstrating shared foundations between STCM and quantitative methods—which, con-sequentially, counters STCM’s emphasis on differences. The core STCM procedures calculate associational measures for complex combinations of posited necessary/sufficient conditions. Contrary to STCM arguments, these procedures share core similarities with conventional quantitative techniques. Understanding these similarities not only helps scholars from different tradi-tions better understand each other’s research but also shows that STCM do not genuinely break new inferential ground. By strongly emphasizing differ-ences, STCM has unproductively isolated itself from mainstream methods. Furthermore, the focus in this article on common foundations also highlights a central shared challenge in much social scientific research, including STCM: moving from association patterns to infer causal relationships, on the basis of observational data.
Notably, the specific comparison with conventional quantitative methods employed below is vital not because the present analysis advocates using conventional quantitative tools. Rather, the fact STCM scholars frequently evoke this contrast to justify their own approach necessitates scrutinizing how STCM’s focus on differences may obscure shared foundations.
Following the overview section that summarizes key features of STCM, detailing the virtues of analyzing cross-case variance provides the first example of a crucial similarity between a core analytic procedure of STCM and quanti-tative techniques. Evaluating necessary condition hypotheses requires analyz-ing cases that exhibit variation on both the causal conditions and the outcome—as in quantitative research but contrary to the consensus of STCM at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 4 Comparative Political Studies scholars. To understand the importance of analyzing all four cells (in a 2 × 2 setting), suppose the hypothesis states X = 1 is a necessary condition for Y = 1. To evaluate this hypothesis, STCM calculates the percentage of Y = 1 cases with X = 1, hence only using two cells.2 Using the STCM metric, the data sup-port the necessary condition hypothesis if there are few Y = 1 cases with X = 0.
The problem with only using two cells, however, is that there may be few Y = 1 cases that are also X = 0 cases, for two reasons: (1) The data actually support the necessary condition hypothesis or (2) there are simply few X = 0 cases relative to X = 1 cases—which makes it essentially impossible to falsify the hypothesis using the STCM metric. Using information from all four cells ensures that consideration (2) will not create false positives about necessary relationships—in particular, by incorporating information from a cell that STCM scholars claim is never relevant for assessing necessity/sufficiency. An identical argument applies to evaluating the “relevance” of a sufficient condition hypothesis.
The next argument demonstrates inherent similarities between STCM and quantitative approaches to studying complex relationships, a central focus of STCM research.3 This argument compares STCM analysis of a truth table with regression analysis that employs multiplicative interaction terms. The analysis demonstrates how regression can be used to convey the same information about necessary/sufficient condition hypotheses as a truth table—in contrast to strong STCM skepticism about multiplicative regression interaction terms.
Researchers who acknowledge these core similarities will not be surprised that STCM and quantitative methods also share a crucial common limitation. This involves the shared challenge in social science of inferring causal con-clusions from associational relations. Examples of associational measures include STCM metrics for measuring the consistency and relevance of com-plex conditions as necessary or sufficient. Whereas current best-practice in quantitative textbooks and research focuses centrally on this shared chal-lenge, it has received less attention from STCM scholars. The final major argument therefore posits a best-case scenario in which a researcher has data perfectly consistent with a necessary/sufficient condition hypothesis (possi-bly involving complex conditions). It focuses on two issues that must be addressed before a compelling causal inference can be achieved: Freedman’s (2010) standard for comparing a deterministic hypothesis to a probabilistic benchmark and Waldner’s (2005) standard for evaluating hypothetical counterexamples.
The concluding section argues that these common foundations—including both core similarities and shared challenges—between STCM and quantita-tive methods raise a series of pressing issues. The bridging attempt offered by this article will hopefully clarify future methodological debates in at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 5 comparative politics. These not only include the relationship between STCM and quantitative methods but also between STCM and traditional qualitative techniques.
STCM Techniques for Analyzing Necessary/ Sufficient Conditions This section briefly outlines key elements of STCM, which provides building blocks for the subsequent discussion. Following the focus of STCM texts on 2 × 2 tables to argue for differences, I restrict attention to binary conditions throughout the article. The concluding section discusses possible extensions to multi-valued fuzzy sets.
STCM scholars view natural language, qualitative theory in social sci-ence, and qualitative research in general as inherently set-theoretic in struc-ture. Stemming from this view, STCM researchers focus on studying the necessary and sufficient conditions for set membership in the outcome of interest. If a condition X = 1 is necessary for a particular outcome Y = 1, then every Y = 1 case is also an X = 1 case. Therefore, X = 1 is a superset of Y = 1. If condition X = 1 is sufficient for Y = 1, then every X = 1 case is also a Y = 1 case. Therefore, X = 1 is a subset of Y = 1. Framing necessary and sufficient conditions in terms of sets has led STCM scholars to adopt the broader math-ematical language of set theory.
These scholars argue that it is reasonable to consider necessary and suffi-cient condition hypotheses even when empirical counterexamples to the hypothesis exist. For example, Braumoeller and Goertz (2000, 2002) under-score the importance of taking measurement error into account when evaluat-ing such hypotheses. STCM scholars have developed two novel measures for studying necessary/sufficient relations in noisy social science data. These measures can be used for either a single or for multiple conditions.
Consistency Scores and Subsets/Supersets A consistency score measures the extent to which the data support a claim of either necessity or sufficiency. The higher the percentage of X = 1 cases in the data that also achieve Y = 1—that is, the extent to which X = 1 is a subset of Y = 1—the more consistent the data are with a claim that X = 1 is sufficient for Y = 1. The higher the percentage of Y = 1 cases with the condition X = 1—that is, the extent to which X = 1 is a superset of Y = 1—the more consis-tent the data are with a claim that X = 1 is necessary for Y = 1 according to conventional STCM metrics. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 6 Comparative Political Studies Coverage Scores and Triviality The coverage score assesses what STCM scholars call the “triviality,” or “rele-vance,” of a condition (Schneider & Wagemann, 2012, p. 144). As an example of a trivial sufficient condition, suppose that every X = 1 case achieves Y = 1, meaning the data are perfectly consistent with the hypothesis. However, suppose that every X = 0 case also achieves Y = 1. Despite the high sufficiency consis-tency score for X = 1, we would intuitively think of X = 1 as a trivial sufficient condition for Y = 1. Because Y = 1 will always occur in the data set, regardless of the value of X, the absence of the trivial sufficient condition does not change the outcome. The coverage score for a sufficient condition is identical to the consistency score for a necessary condition. In this example, if we assume that there are an equal number of X = 0 and X = 1 cases, the coverage score is 0.5 because cases with X = 1 account for only half of the Y = 1 outcomes.
A similar concept of triviality, or relevance, applies to necessary condi-tion hypotheses. Assume that no X = 0 cases achieve Y = 1, implying the data are perfectly consistent with the hypothesis. But suppose that Y = 1 almost never occurs even when X = 1. For example, oxygen is a necessary condition for social revolution. The necessary condition coverage score—which is identical to the sufficient condition consistency score—detects that X = 1 is a trivial necessary condition because it will show that only a trivial percent-age of X = 1 cases have Y = 1.
Multiple Conditions In addition to the consistency and coverage scores for individual conditions, STCM also provides tools for analyzing clusters of multiple conditions. For example, a condition A = 1 may not be individually sufficient for Y = 1, but the conjunctural condition (A, C) = (1, 1) may be sufficient. This is an exam-ple of a complex sufficient condition. Furthermore, (D, E) = (1, 0) may also be a sufficient condition for Y = 1. This causal process therefore exhibits equifinality because there are multiple paths to the outcome.
Thus, the basic ideas of consistency, coverage, and multiple conditions pro-vide key building blocks for STCM analysis of necessity and sufficiency.
Core Similarity 1: The Virtues of Analyzing Cross-Case Variance Empirically evaluating necessary condition hypotheses requires analyzing cases that exhibit variation on both the causal conditions and the outcome4— which demonstrates a crucial similarity to quantitative research. This claim at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 7 runs against STCM arguments that necessary condition hypotheses can only be meaningfully tested using designs that lack variation on the outcome. Problematically, this accepted STCM procedure may lead one to conclude data are highly consistent with the necessary condition hypothesis simply because few cases lack the posited necessary condition, which produces arti-ficial support for the claim. In a 2 × 2 setting, guarding against this problem requires incorporating all four cells into the analysis. Appendix A demon-strates the STCM procedure for assessing the triviality of a sufficient condi-tion hypothesis faces a similar shortcoming that can also be fixed by incorporating information from all four cells.
There is wide agreement in the STCM literature that scholars should only focus on two cells at a time. Ragin (2008) argues the quantitative tradition of combining all four cells of a 2 × 2 table will “conflate different kinds of causal assessment” (pp. 7, 22). Braumoeller and Goertz (2002) argue that sampling cases from all four cells to test necessary condition hypotheses is “pointless” because “the theories that [they] examine don’t imply anything about the number or proportion of cases that should be found in [certain] cell[s]” (pp. 199, 200). Goertz and Mahoney (2012) state directly that “selec-tion on the dependent variable when testing necessary conditions follows directly from the definition of a necessary condition” (p. 179). Correspondingly, the standard STCM calculations of consistency and coverage for both neces-sary and sufficient conditions incorporate only two cells at a time—in con-trast to the quantitative approach of analyzing all four cells, such as with a regression coefficient or any standard measure of statistical association.
However, analyzing only two cells at a time can create misleading find-ings whenever the number of X = 1 and X = 0 cases differs. For example, consider a claim that X = 1 is necessary for Y = 1, implying that X = 0 cases with Y = 1 provide evidence against the hypothesis. If the number of X = 1 cases is large relative to the number of X = 0 cases, the data may appear to be highly consistent with the necessary condition hypothesis simply because there are relatively few X = 0 cases that could provide evidence against the hypothesis. This occurs because it is not possible for there to be many (X, Y) = (0, 1) cases if there are few X = 0 cases. The easiest way to guard against this problem is to compare the percentage of X = 1 cases with Y = 1 to the percent-age of X = 0 cases with Y = 1. This revised procedure incorporates all four cells.
The core argument complements and extends existing arguments that researchers should always incorporate all four cells when the data are avail-able. Seawright (2002) advanced this argument using a Bayesian model. The present arguments address two major concerns from responses published in the same issue of Political Analysis. First, Clarke (2002) argues Seawright at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 8 Comparative Political Studies used an implausible weighting procedure in his Bayesian likelihood function for introducing new cases into the analysis. In the three cross-tabulations presented below (Tables 2-4), both the total number of cases and the distribu-tion of X = 1 and X = 0 cells are fixed for each table. Therefore Clarke’s cri-tique does not hold here.5 Second, Braumoeller and Goertz (2002) critique Seawright for dismissing necessary condition hypotheses when deviant cases exist, whereas this section allows for the possibility of cases that are incon-sistent with a deterministic relationship. Separately, Freedman (2008) dis-putes Goertz’s (2008) advice to ignore certain cells in a 2 × 2 table. In contrast to the focus of this section, Freedman (2008) grounds his critique by appeal-ing to process tracing: At least in my experience, it is often hard to see where the cases go until you study them . . . Great work can be done with one cell, or even one case. Isn’t de Tocqueville’s Democracy in America a classic example of within-case analysis? (pp. 15-16) A Revised Approach to Calculating Necessary Condition Consistency Scores: The Importance of All Four Cells The exposition follows notation from the generic 2 × 2 table depicted as Table 1. To use substantively interesting terminology, suppose condition F refers to “fuel wealth” and A refers to the outcome “authoritarian rule.” The consistency score for fuel wealth as a sufficient condition for authori-tarian rule calculates the percentage of F = 1 cases with A = 1. Similarly, the consistency score for non-fuel wealth as a sufficient condition for authoritar-ian rule calculates the percentage of F = 0 cases with A = 1. Using the nota-tion from Table 1, these two terms can be expressed as follows: SC F A A F n n n FA FA Fa = = ( ) = = = ( ) = + 1 1 1 1 , % | , (1) Table 1. Notation for a 2 × 2 Table.
F = 0 F = 1 A = 1 nfA nFA A = 0 nfa nFa This table presents notation for each cell of a 2 × 2 table. In the subscript, uppercase refers to the presence of a condition and lowercase to its absence. F = fuel wealth; A = authoritarian rule. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 9 SC F A A F n n n fA fA fa = = ( ) = = = ( ) = + 0 1 1 0 , % | . (2) Examined individually, each term incorporates only two cells at a time.
In contrast, a bivariate regression coefficient combines information from all four cells. The regression coefficient for the association between fuel wealth and authoritarianism is defined as the covariance of F and A divided by the variance of F. In a 2 × 2 setting, this is simply the difference in aver-age outcome value between fuel and non-fuel cases, which is expressed by βfuel = + − + n n n n n n FA FA Fa fA fA fa . (3) Comparing Equations 1 through 3 shows the regression coefficient com-bines the two sufficient condition consistency scores, and Equation 3 can be re-expressed as βfuel SC SC = = = ( ) − = = ( ) F A F A 1 1 0 1 , , . (4) As proposed by STCM scholars, the consistency score for necessary con-ditions only incorporates information from two cells. The consistency score for fuel wealth as a necessary condition for authoritarian rule conveys the percentage of A = 1 cases with F = 1: NC F A n n n FA FA fA = = ( ) = + 1 1 , . (5) A crucial problem emerges here. The metric from Equation 5 can be large for two different reasons. First, nfA may be small because the data strongly support the hypothesis. Second, nfA may be small because there simply are not many cases that lack fuel wealth among the included cases, which implies there cannot be many (F, A) = (0, 1) cases. In other words, without accounting for differences in the frequency of non-fuel cases relative to fuel cases, the STCM necessary condition consistency score can create misleading findings.
Tables 2 through 4 demonstrate the potential for misleading findings by illustrating the importance of how cases are distributed across F = 0 and F = 1. The tables present three hypothetical 2 × 2 cross-tabulations. Because the num-ber of cases in both F = 1 cells is constant across the three tables, the STCM necessary condition coverage score is also the same. Therefore, none of the following concerns relate to this commonly used measure for triviality.6 With regard to whether fuel wealth is necessary for authoritarianism, according to Equation 5, Tables 2 and 3 have high consistency scores of 20/22 = 91% and 20/23 = 87%, respectively. In contrast, the consistency at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 10 Comparative Political Studies score in Table 4 is low, 20/40 = 50%. To understand why these consistency scores are problematic, consider the differences among the three tables. Tables 3 and 4 each have 38 more cases with fuel wealth than Table 2. The difference between Tables 3 and 4 consists of how the F = 0 cases are distrib-uted between the outcomes A = 0 and A = 1. In Table 3, almost all the addi-tional cases are in the (0, 0) cell, whereas in Table 4, the two F = 0 cells have an identical number of cases.
It is puzzling that the consistency score for Table 2 is high—especially when compared with Table 3. In Table 2, every fuel-poor case goes against the necessary condition hypothesis, whereas only 7.5% do so in Table 3. In Table 3, there are an equal number of fuel-poor and fuel-rich cases. Therefore, the rarity of fuel-poor cases with authoritarian rule indicates support for the hypothesis, a reasonable conclusion.
Table 3. F = 0 F = 1 A = 1 3 20 A = 0 37 20 This table is a hypothetical 2 × 2 table. It represents a fixed group of cases that are distinct from those in Tables 2 and 4. F = fuel wealth; A = authoritarian rule.
Table 2. Hypothetical Example of Misleading Necessary Condition Consistency Scores.
F = 0 F = 1 A = 1 2 20 A = 0 0 20 This table is a hypothetical 2 × 2 table. It represents a fixed group of cases that are distinct from those in Tables 3 and 4. F = fuel wealth; A = authoritarian rule.
Table 4. F = 0 F = 1 A = 1 20 20 A = 0 20 20 This table is a hypothetical 2 × 2 table. It represents a fixed group of cases that are distinct from those in Tables 2 and 3. F = fuel wealth; A = authoritarian rule. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 11 In contrast, Table 2 simply does not contain many F = 0 cases—which explains the rarity of (0, 1) cases. In fact, in Table 2, there is evidence that fuel-poor cases are more likely than fuel-rich cases to experience authoritar-ian governance. Nonetheless, the STCM calculation that incorporates only two cells suggests the data in Table 2 are highly consistent with fuel wealth being necessary for authoritarianism.
In another puzzling finding, the consistency score is considerably higher in Table 2 than in Table 4, even though a higher percentage of fuel-poor cases contradict the hypothesis in Table 2. The main difference between Tables 2 and 4 is that the number of fuel-poor cases is greater in Table 4.
To provide more systematic evidence for why overlooking this crucial simi-larity with quantitative case selection practices causes problems, Figure 1 sum-marizes 10,000 randomly generated 2 × 2 tables. In every simulated table, 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Percentage of F=1 Cases Necessary Condition Consistency Score Figure 1. Simulated relationship between the percentage of F = 1 cases and STCM necessary condition consistency scores.
This figure demonstrates the systematic tendency for necessary condition consistency scores to be higher in data sets with a higher percentage of F = 1 cases. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 12 Comparative Political Studies there is an equal chance for each of the four cells to be assigned anywhere between 0 and 20 cases. The horizontal axis presents the percentage of total cases assigned to F = 1. The vertical axis presents the STCM necessary con-dition consistency score for each of the 10,000 hypothetical tables.
An approach that incorporates only two cells at a time implies the distribu-tion of cases across F = 1 and F = 0 is unimportant. Crucially, this distribution can only be calculated by incorporating information from all four cells. Therefore, if STCM scholars are correct that one does not lose valuable infor-mation by discarding particular cells, there should be no systematic tenden-cies in this randomly generated data and the best-fit line should be flat. Instead, the black best-fit line shows a sharply positive association between the percentage of cases assigned to F = 1 and the necessary condition consis-tency score. That is, one is more likely to find support for their hypothesis using the STCM metric simply because their data set has a high percentage of F = 1 cases. Thus, neglecting to study all four cells can easily engender mis-leading conclusions.
A necessary condition consistency score calculation that incorporates all four cells would overcome these problems. For example, consider the follow-ing metric: NC cells SC F A n n n n n n n n n F A FA FA Fa FA FA Fa fA fA fa = = ( ) = + + + + = = 1 1 4 1 , ; , = ( ) = = ( ) + = = ( ) 1 1 1 0 1 SC SC F A F A , , . (6) Equation 6 incorporates all four cells.7 It adjusts for the frequency of fuel-poor cases relative to fuel-rich cases and does not report a high consistency score simply because there are few fuel-poor cases. Specifically, Equation 6 compares the percentage of fuel-rich cases with authoritarian governments to the percentage of fuel-poor cases with authoritarian rule. This contrasts with Equation 5, which compares the number of fuel-rich cases with authoritarian governments to the number of fuel-poor cases with authoritarian rule. By design, the revised measure only equals the STCM calculation when the numbers of fuel-poor and fuel-rich cases are equal. Comparing Equations 4 and 6 show that both a regression coefficient and the revised necessary condi-tion consistency metric incorporate the sufficient condition consistency mea-sures for both F = 0 and F = 1. The differences arise from the aggregation procedure, not from the number of cells used. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 13 Using Equation 6 to calculate necessary condition consistency scores for Tables 2 through 4, respectively, yields 33%, 87%, and 50%. Because there are equal numbers of F = 0 and F = 1 cases in Tables 3 and 4, the estimated consistency scores from Equations 5 and 6 are identical. Thus, the revised measure does not yield misleading conclusions when applied to non-skewed data. But the revised measure estimates a sharply different consistency score for Table 2 (33% vs. STCM’s 91%) by accounting for the unbalanced number of F = 0 and F = 1 cases, therefore guarding against misleading conclusions.
All Four Cells: Consistency or Triviality?
A potential counterargument to the position advocated here is that there are no inherent flaws with the existing STCM necessary condition consistency measure, and instead the concerns raised here apply only to the triviality of a necessary condition. Schneider and Wagemann (2012, pp. 235-237), for example, also discuss how differential numbers of X = 1 and X = 0 cases can create problems, but relate the concern to triviality rather than to consistency. This is not a compelling argument.
To use an illustrative example to support why skewed cases raise issues about consistency rather than triviality, consider every at-bat in Major League Baseball history as the set of cases. STCM procedures would conclude that not being Babe Ruth (X = 0) is a necessary condition for hitting a home run (Y = 1). After all, the overwhelming majority of home runs in baseball history have not been hit by Babe Ruth, that is, Y = 1 is nearly a perfect subset of X = 0.
The problem with this unwarranted conclusion does not stem from trivial-ity. Home runs occur somewhat frequently when the posited necessary condi-tion is present, that is, when players not named Babe Ruth come to bat.8 Instead, the problem with the conclusion is that the data are highly incon-sistent with the hypothesis. Babe Ruth at-bats (X = 1) do not imply the nega-tion of a home run (Y = 0), violating the contrapositive of the necessary condition hypothesis. In fact, the absence of the purported necessary condi-tion predicts a home run with considerably higher probability than the pres-ence of the condition. The STCM consistency metric misses this because the number of X = 1 cases is very small relative to the number of X = 0 cases. In contrast, a consistency measure that focuses on percentages rather than the number of cases easily avoids this problematic conclusion.
Furthermore, even those who reject the position that a skewed distribution of X cases is more closely related to consistency rather than to triviality still implic-itly embrace the conclusion that one needs to scrutinize all four cells. The stan-dard STCM consistency score calculation and Schneider and Wagemann’s (2012, p. 236) alternate necessary condition triviality measure jointly use all at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 14 Comparative Political Studies four cells.9 This reinforces the conclusion that the only way to account for a skewed X distribution is to incorporate information from all four cells.
In sum, to assess a necessary condition hypothesis, analysts should always study cases that exhibit variation on both the posited causal condition and on the outcome. This practice, which is central to quantitative research, guards against misleading findings rather than leads analysts to “conflate” (Ragin 2000, p. 96) necessity conditions and sufficient conditions. STCM and quan-titative methods share this crucial feature.
Core Similarity 2: Configurations of Conditions and Regression Interaction Terms Regression models with multiplicative interaction terms share a core similar-ity with STCM analyses of complex necessary and sufficient conditions. Because the quantities used in STCM to study complex conditions can be derived from quantities estimated by an interactive regression model, the regression results necessarily contain all the same information as the STCM analysis. This finding opposes pointed STCM arguments that they offer a distinct—and superior—alternative for studying complex causal patterns.
There is wide agreement in the STCM literature that STCM provides a bet-ter alternative than regression for studying complex relationships. STCM studies configurations of conditions using a “truth table,” which displays all logically possible combinations of the explanatory conditions under analysis by presenting one combination per row. A truth table provides the information needed to calculate the percentage of cases with each combination of condi-tions that achieves Y = 1, which is in turn used to compute consistency and coverage scores. In contrast, regression is purported to be unable to study complex relationships. Ragin (2008) characterizes regression as a method use-ful only for studying the net effects of individual conditions—in fact, one of his “four oppositions” between STCM and quantitative methods is “the analy-sis of causal complexity versus the analysis of net effects” (p. 9)—and he dismisses outright the ability of regression to analyze complex relationships. Ragin (2008) states that questions involving conditional relationships lie “out-side the scope of conventional net-effects analyses, for those approaches are centered on the task of estimating context-independent net effects” (p. 181). Many other scholars, such as Hall (2003, pp. 382-383), Becker (2002, p. 250), and Brown (2009, p. 415) express similar concerns that regression analysis is unable to adequately analyze complex relationships.
These arguments overlook how interactive regression terms serve as a cru-cial bridge between quantitative and STCM approaches. When the interac-tion terms are properly interpreted, it becomes clear that a regression model at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 15 provides all the information needed to calculate necessary and sufficient con-dition consistency and coverage scores for any complex combination of con-ditions—despite the fact that regression analysis is best interpreted as studying the effects of particular interventions.10 The analysis below demon-strates this point by analyzing a 2 × 2 table that involves a complex condi-tion.11 Because both regression and STCM can provide all the information needed to compute the number of cases in each cell of a 2 × 2 table, it follows that both approaches can be used to compute consistency and coverage scores for necessary and sufficient conditions—and therefore provide the same information for studying complex causal relationships.
To make the argument that truth tables and regression tables yield the same information, it is valuable to directly compare how a truth table and a regression table will display results from the same data set. The discussion focuses first on the truth table.
Truth Table: Computing the Number of Cases in Each Cell of a Complex 2 × 2 Table Suppose there are two conditions of interest, S and F, and an outcome A. Continuing the substantive example from above, F and A refer to fuel wealth and authoritarianism, respectively. S represents strong bureaucratic institu-tions. Table 5, in which the first two columns correspond to a truth table, summarizes a hypothetical set of data. It presents the four logically possible combinations of strong institutions and fuel wealth, the number of cases with each combination, and the percentage of cases with each combination of con-ditions that experience authoritarian rule.
Table 5. Notation for a Truth Table that Incorporates the Percentage of Cases per Row With A = 1.
S F Notation for number of cases Notation for % with A = 1 0 0 nsf psf 0 1 nsF psF 1 0 nSf pSf 1 1 nSF pSF For each logically possible combination of S and F, this table presents notation for the number of cases in the row and the percentage of those cases with A = 1. Denoting a generic condition as “X,” the subscript for either the number (n) or percentage (p) contains an uppercase X if X = 1, and a lowercase x if X = 0. Some STCM scholars include percentages in their truth tables (e.g., Basedau & Richter, 2014, p. 561), whereas others do not (e.g., Schneider & Wagemann, 2012, p. 106). A = authoritarian rule; S = strong bureaucratic institutions; F = fuel wealth. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 16 Comparative Political Studies Given this information, it would be natural for a STCM scholar to use QCA software to reduce subset redundancy and to calculate consistency and coverage scores. However, for the purposes of comparing STCM and regres-sion, it is useful to perform the calculations by hand. Suppose we wanted to learn about the necessary/sufficient consistency and coverage scores for hav-ing weak institutions and fuel wealth, that is, (S, F) = (0, 1).
Based on the discussion from the previous section, we know a 2 × 2 table will provide all the relevant information. Table 6 incorporates the notation from Table 5 and shows how to use the information to compute the number of cases in each cell. A complex condition can be summarized in binary form because we are only interested in its presence and its absence for computing consistency and coverage scores. If a case has strong institutions and/or lacks fuel wealth, then it does not have (S, F) = (0, 1). Therefore, [S = 1 OR F = 0] negates [S = 0 AND F = 1], and composes the other column in the table.
To understand why the symbols inside each cell represent how many cases would be observed in it, it is useful to begin with cell 1. The number of cases with weak institutions and fuel wealth that also experience authoritarian rule is simply the number of [S = 0 AND F = 1] cases multiplied by the percentage of cases with this combination of conditions that have A = 1, which is expressed by psF · nsF. This is shown in cell 1.
The number of cases in cell 2 is calculated using an identical procedure. If a case with weak institutions and fuel wealth does not have authoritarian rule, then it must have the negation of authoritarian rule. We know that 1 − psF is the percentage of cases with [S = 0 AND F = 1] that achieve A = 0. Multiplying this percentage by the total number of weak institutions/fuel wealth cases yields (1 − psF) · nsF, as shown in cell 2.
The terms in cells 3 and 4 are somewhat more cumbersome because there are three ways to have strong institutions and/or lack fuel wealth. The number of [S = 1 OR F = 0] cases is the sum of the number of cases that have either (a) weak institutions and no fuel, (b) strong institutions and no fuel, or (c) strong institutions and fuel. For cell 3, following an identical procedure as for the single condition in cell 1 yields psf · nsf + pSf · nSf + pSF · nSF. For cell 4, following an identical procedure as for the single condition in cell 2 leads to (1 − psf) · nsf + (1 − pSf) · nSf + (1 − pSF) · nSF.
Clearly, a truth table contains the information needed to fill in a 2 × 2 table for any combination of conditions the researcher desires to study.
Regression: Computing the Number of Cases in Each Cell of a Complex 2 × 2 Table Regression can also yield identical information for computing cells in a 2 × 2 table—and therefore for computing necessary/sufficient condition consistency at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 17 and coverage scores—although regression presents the information in a differ-ent form. Continuing the example from above, consider the coefficient esti-mates for the following model: A S F S F i S i F i SF i i i = + ⋅ + ⋅ + ⋅ ⋅ + β β β β ε 0 . (7) This model is called a fully saturated regression because it contains a mul-tiplicative interaction term for each combination of variables and includes all lower order terms. Showing how ordinary least squares (OLS) estimates each of β0, βS, βF, and βSF reveals how a fully saturated regression model provides the same information as a truth table for computing a complex 2 × 2 table, because the four β estimates collectively include all four percentages: psf, pSf, psF, and pSF. Throughout the discussion, I assume causal homogeneity. This implies that the β coefficients (which capture average effects) carry the same implications for the effect of particular interventions for each case.
The estimated constant term for the statistical model is β0. This term tells us the average outcome when both conditions equal 0. In other words, β0 is the percentage of cases with weak institutions and without fuel wealth that experience authoritarian rule—which by assumption is psf—and can be expressed as follows: % | , ( , ) .
A S F psf = ( ) = = 1 0 0 (8) Table 6. Summarizing the Truth Table as a 2 × 2 Table With a Complex Condition.
S = 1 OR F = 0 S = 0 AND F = 1 A = 1 (3) (1) psf · nsf+ psF · nsF pSf · nSf+ pSF · nSF A = 0 (4) (2) (1 − psf) · nsf+ (1−psF) · nsF (1 − pSf) · nSf+ (1 − pSf) · nSF This table takes information from Table 5 to calculate a 2 × 2 table in which the posited causal condition is complex, with each cell numbered by a term in parentheses. The second column provides the number of cases with weak institutions and fuel wealth. The first column provides the number of cases with strong institutions and/or lack fuel wealth. See Note 3 for the definition of a “complex” condition. S = strong bureaucratic institutions; F = fuel wealth; A = authoritarian rule; p = percentage; n = number. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 18 Comparative Political Studies This is the first of the four percentages we need from the regression results to calculate consistency and coverage scores.
To understand the other three regression coefficients, it is useful to think of regression as estimating the effects of particular interventions. The first hypo-thetical intervention of interest is what happens if a case with weak institutions and that lacks fuel wealth is changed to have strong institutions (while con-tinuing to be fuel-poor). The expected effect of the intervention is the average outcome when (S, F) = (1, 0) minus the average outcome when (S, F) = (0, 0): βS Sf sf A S F A S F p p = = ( ) = ( ) − = ( ) = ( ) = − % | , , % | , , .
1 1 0 1 0 0 (9) Because β0 tells us psf, we can add β0 to βS to calculate pSf from the regres-sion results. Thus, estimating βS provides the second of the four percentage terms that we need to calculate consistency and coverage scores.
The second hypothetical intervention of interest is what happens if a case with weak institutions and that lacks fuel wealth is changed to become fuel-rich (while continuing to have weak institutions). The expected effect of the intervention is the average outcome when (S, F) = (0, 1) minus the average outcome when (S, F) = (0, 0): βF sF sf A S F A S F p p = = ( ) = ( ) − = ( ) = ( ) = − % | , , % | , , .
1 0 1 1 0 0 (10) We can add β0 to βF to calculate psF from the regression results. Therefore, estimating βF provides the third of the four percentage terms that we need to calculate consistency and coverage scores.
The third hypothetical intervention of interest is what happens if a case with weak institutions and that lacks fuel wealth is changed to have both strong institutions and fuel wealth. The expected effect of this intervention can be expressed by the sum of three terms. The first component is the effect of only changing S, which is βS. The second component is the effect of only changing F, which is βF. The third component expresses the extent to which simultaneously changing both S and F differs from the sum of the effects of each individual intervention and is captured by βSF. Using the notation, this means that we can calculate the effect of the third intervention as % | , , % | , , .
A S F A S F S F SF = ( ) = ( ) − = ( ) = ( ) = + + 1 1 1 1 0 0 β β β (11) Because %[A = 1 | (S, F) = (1, 1)] = pSF and %[A = 1 | (S, F) = (0, 0)] = β0, we can solve Equation 11 to get pSF S F SF = + + + β β β β 0 . (12) at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 19 Thus, regression also allows us to calculate all four terms needed to com-pute consistency and coverage scores for necessary/sufficient conditions. Table 7 re-expresses Table 6, replacing the percentage terms from the truth table with the OLS coefficient estimates from the regression model in Equation 7.
A closer examination of the βSF coefficient also refutes the core STCM claim that conventional quantitative methods can only be used to estimate “context-independent net effects” (Ragin 2008, p. 181). Suppose fuel wealth promotes authoritarianism when institutions are weak, but exerts no effect on regime type when institutions are strong. In addition, suppose institutional quality does not exert an unconditional effect on authoritarianism and instead only modifies the effect of fuel wealth. Regression would reveal this relation-ship because βF would be positive, βS would be 0, and βSF would equal −βF. The negative βSF captures the fact that fuel wealth positively affects authori-tarianism when institutions are weak, but does not affect the baseline proba-bility of authoritarian rule when institutions are strong. Thus, for a case that originally has weak institutions and no fuel wealth, only one of the three possible combinations of changing either or both the two conditions will alter the baseline probability of authoritarianism: changing fuel wealth but not institutions, because βF > 0. Changing institutions but not fuel wealth has no effect because βS = 0. Changing both conditions also has no effect: βF + βS + βSF = βF − βF = 0. These findings are possible precisely because estimating a fully saturated regression assesses whether the relationships are conditional rather than constant.
In sum, multiplicative interaction terms from regression analysis can be used to compute consistency and coverage scores for complex necessary and Table 7. Summarizing the Regression Table as a 2 × 2 Table With a Complex Condition.
S = 1 OR F = 0 S = 0 AND F = 1 A = 1 β0 · nsf+ (β0 + βF) · nsF (β0 + βS) · nSf+ (β0 + βS + βF + βSF) · nSF A = 0 (1 − β0) · nsf+ [1−(β0 + βF)] · nsF [1 − (β0 + βS)] · nSf+ [1−(β0 + βS + βF + βSF)] · nSF This table replaces the percentage terms from the truth table with the coefficient estimates from the regression model in Equation 7. S = strong bureaucratic institutions; F = fuel wealth; A = authoritarian rule; n = number. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 20 Comparative Political Studies sufficient condition relationships, and regression is not forced to assume con-stant effects—belying the strong claims from the STCM literature that regres-sion is inherently unable to study causal complexity. Appendix B provides an additional example that features multiple paths to the outcome. The fact that the regression table presents the information differently than does a truth table should not obscure their inherent similarities. Furthermore, multiplica-tive interaction terms represent only one of many quantitative techniques for analyzing complex causality, with additional possibilities discussed in the conclusion.12 Shared Challenges: From Association to Causation Researchers who acknowledge these core similarities will not be surprised that STCM and quantitative methods also share common limitations. Drawing causal conclusions from associational relations poses great difficulties— regardless of how many cells of a 2 × 2 table are used, and regardless of whether STCM or regression is used to examine complex relationships. Whereas current best-practice quantitative textbooks and research focuses centrally on this challenge, it has received less attention from STCM schol-ars.13 Below I posit a best-case scenario in which the data are perfectly con-sistent with a necessary/sufficient hypothesis. This section focuses on two issues that must be addressed before a compelling causal inference can be achieved: Freedman’s (2010) empirical standard for comparing a determinis-tic hypothesis to a probabilistic benchmark and Waldner’s (2005) standard for evaluating hypothetical counterexamples.
Pointed critiques of conventional quantitative methods have led to a dra-matic rethinking of how cross-case comparisons can be translated into con-vincing causal claims. Quantitative methodologists have responded by producing best-practice advice that focuses centrally on causal inference, whether for analyzing field experiments (Gerber & Green, 2013), “natural” experiments (Dunning, 2012), or observational data (Morgan & Winship, 2007; Rosenbaum, 2002).
Although STCM scholars have also discussed causal inference issues (e.g., Mahoney et al., 2013), this section focuses on two issues specifically pertaining to deterministic inferences that require concerted attention. First, even if the data are perfectly consistent with a necessary or sufficient condi-tion hypothesis, it is useful to consider the likelihood the data could have been produced by an alternative probabilistic process. Applying this consid-eration—proposed by Freedman (2010)—demonstrates the generic difficulty of making convincing deterministic inferences using empirical evidence alone and the particular difficulties that arise with a small number of cases. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 21 Furthermore, the value-added of stating a hypothesis as necessary or suffi-cient—as opposed to expressing a parallel probabilistic claim—is question-able without scrutinizing why it is reasonable to believe that Y cannot occur if a posited necessary condition X is absent, or that Y must occur if a posited sufficient condition X is present. I therefore also consider Waldner’s (2005) standard for using hypothetical counterfactuals to evaluate deterministic hypotheses. Combining the empirical and theoretical approaches may sug-gest an avenue for creating stringent standards that more adequately assess deterministic hypotheses. Finally, because STCM scholars have raised important considerations regarding whether necessity/sufficiency claims are inherently deterministic, the end of the section engages this contentious topic.
Overall, scrutinizing the causal underpinnings of STCM hypotheses reveals the need for similar advancements as those that have improved quan-titative research in recent decades.
Data-Generating Process (DGP) Vis-à-Vis Data To distinguish associational patterns from causal inferences it is crucial to introduce the concept of a data-generating process (DGP) and to distinguish the DGP from the data actually observed. Suppose we have a best-case sce-nario for a deterministic hypothesis in which the cases are perfectly consis-tent with X = 1 being both necessary and sufficient for Y = 1. That is, every X = 1 case achieves Y = 1 whereas no X = 0 cases have Y = 1. If a scholar infers that X is in fact sufficient for Y from this associational pattern, they have implicitly made the following claim about the DGP: Pr Y X = = ( ) = 1 1 1 | . (13) In words, Y must occur if X is present. Therefore, any attempts to infer sufficiency must scrutinize how convincing the “must” component of their causal claim is.
Similarly, moving from association to inferring necessity implies: Pr Y X = = ( ) = 1 0 0 14 | . (14) In words, Y cannot occur if X is not present. Therefore, any attempts to infer necessity must scrutinize how convincing the “cannot” component of their causal claim is.
Crucially, we observe draws from the DGP but do not observe the DGP itself. Instead, we must impose assumptions to make inferences about the at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 22 Comparative Political Studies DGP. The remainder of the section discusses standards for inferring that a DGP is deterministic.
Freedman’s Standard for Assessing Probabilistic Alternatives The data may appear to strongly support a necessary/sufficient condition hypothesis when there are no empirical counterexamples. However, even in this ideal circumstance an empirical approach to evaluating deterministic hypotheses can produce misleading conclusions when scholars do not care-fully consider probabilistic alternatives that may be generating the observed data. It is never possible to fully disentangle deterministic from probabilistic alternatives solely on the basis of empirical observation15—an especially pressing concern when the number of cases is small. The following discus-sion elaborates upon Freedman’s (2010) standards for comparing determinis-tic hypotheses to probabilistic alternatives, using an example from applied STCM to substantiate the importance of this consideration.
Mahoney (2010) claims that, for Spanish America, the combination of (a) lack of a strong colonial legacy and (b) prolonged warfare during the 19th century without a major victory (jointly denoted as X = 1) was sufficient for low levels of economic development (Y = 0). Four countries in his study pos-sess this set of conditions, and all four have low levels of economic development.16 The pattern that Mahoney uncovers is fully consistent with a claim that the DGP involves Pr(Y = 0 | X = 1) = 1.17 However, suppose instead the DGP for X = 1 cases produced Y = 0 or Y = 1 with equal probability, that is, Pr(Y = 0 | X = 1) = Pr(Y = 1 | X = 1) = .5. We would still be fairly likely to observe four X = 1 cases achieve the same outcome (either all Y = 0 or all Y = 1), hence portraying a deterministic pattern. To create a best-case scenario for evaluat-ing the sufficient condition hypothesis, assume the same DGP governs all four cases. This can be conceptualized as a large urn in which 50% of the balls are yellow and 50% are red, that is, Pr(yellow ball | posited SC is pres-ent) = Pr(red ball | posited SC is present) = .5. If we take four independent draws from the large urn, there is a 2 × (0.5)4 = 12.5% chance of observing either all red or all yellow balls. Therefore, one out of every eight possible arrays of cases from the population of balls will be perfectly consistent with a deterministic process, even though this hypothetical DGP actually produces either type of ball with equal probability.
As an additional consideration, assuming the urn produces each type of ball with equal probability is particularly stringent for establishing evidence in favor of a probabilistic alternative. Suppose we instead assume that 90% of the balls are red and 10% are yellow: Pr(red ball | posited SC is present) = .9. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 23 If we take four independent draws, there is a 65.5% chance every ball will have the same color. This alternative possible DGP favors red over yellow balls, but also is not deterministic. Hence, observing that all four cases shar-ing a particular trait also achieve the same outcome provides weak evidence in favor of a deterministic hypothesis. One could reach an identical conclu-sion about necessity if they had four cases that lacked the posited necessary condition, and all four of these cases failed to achieve the outcome.
Probabilistic alternatives can never be empirically rejected unless we draw an infinite number of balls from the data-generating urn. For example, suppose the large urn produces 99% red balls and 1% yellow balls. If we take 298 independent draws from the urn, there is still a slightly higher than 5% chance that no yellow balls will be observed. While the 5% threshold is arbi-trary, it is notable because it is a commonly used statistical threshold for rejecting a null hypothesis of no effect. In the 99-1 urn example, different analysts could conceivably disagree on whether the 298 red draws provide strong evidence in favor of the deterministic hypothesis.18 For the present discussion, the main takeaway is that observing no deviant cases yields little support for a deterministic claim when there are few cases—regardless of how one states their probabilistic alternative. This is troubling because, at least in comparative politics, deterministic claims are usually made when analyzing few cases. In addition, because probabilistic alternatives can never be rejected solely on the basis of empirical observa-tions, we should always use additional standards to evaluate a deterministic hypothesis.
Waldner’s Standard for Evaluating Hypothetical Counterexamples Waldner (2005, p. 28) argues that scholars should not proclaim strong support for a necessary condition hypothesis—even if there are no empirical counter-examples—without scrutinizing why Y cannot occur when X is absent. Even if no alternate paths to Y are observed empirically in the set of cases, one still needs to evaluate hypothetical counterexamples to assess the plausibility of the cannot claim. A similar consideration calls into question claims of suffi-ciency absent a compelling argument that Y must occur when X is present. If a combination of conditions is sufficient for an outcome, then it should be difficult to posit a plausible hypothetical scenario in which the outcome could fail to occur when those conditions are present. Thus, scrutinizing hypotheti-cally possible scenarios composes an important component of evaluating a hypothesis that the DGP is deterministic. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 24 Comparative Political Studies Continuing the substantive example from above, the theory supporting the deterministic elements of Mahoney’s (2010) claims is likewise not very con-vincing. Importantly, Mahoney (2010) does provide compelling theoretical justifications for why the conditions he studies may have greatly increased the probability a case would achieve one outcome or another. However, his theoretical discussion does not convincingly ground the “must” claim inher-ent in his sufficient condition hypotheses nor the “cannot” claim inherent in his posited necessary conditions.
One of Mahoney’s (2010) complex sufficient condition hypotheses states that a weak tradition of mercantilist colonial institutions AND the presence of established liberal colonial institutions are jointly sufficient for high levels of development. Problematically for the claim, it is relatively easy to construct a hypothetical possibility that strongly suggests Pr (high development | pos-ited SC is present) < 1, in contrast to Mahoney’s implicit inference that Pr (high development | posited SC is present) = 1. Consider the out-of-region case Zimbabwe. This case illuminates how a country that possessed the con-ditions posited to be sufficient could have failed to achieve high develop-ment. Zimbabwe was imbued with British parliamentary institutions and a nascent industrial structure at independence but has been a developmental disaster—partly because the African majority government lashed out against the European settlers that originally established what are presumed to be “good” institutions.19 The point here is not to imply that Mahoney’s argument should be evalu-ated in terms of how it can be generalized. Rather, Zimbabwe functions as a counterfactual consideration for the Spanish American cases Mahoney does consider. Without a strong argument that a trajectory resembling Zimbabwe’s was not hypothetically possible for any Spanish American countries that pos-sessed the purported sufficient conditions for prosperity, we have serious grounds for questioning the deterministic “must” element of Mahoney’s (2010) hypothesis—which, again, is distinct from the issue of whether Mahoney analyzed conditions that strongly increased the likelihood of a par-ticular outcome.
In an example involving a necessary condition, Mahoney (2010) claims that possessing either of the following two complex conditions was necessary for achieving high economic development among the countries he analyzes: (NC #1) being neglected by the Habsburgs AND being either a colonial cen-ter or semiperiphery during the Bourbon era OR (NC #2) being neglected during both colonial eras AND victorious in warfare during the 19th century. As with the Zimbabwe example, examining out-of-region cases illuminates the possibilities that must be considered to strongly establish a combination of conditions as necessary for an outcome. It is useful to observe that Brunei, at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 25 Qatar, and the United Arab Emirates all scored in the highest development category in the United Nations’ Human Development Index in 2013, despite not inheriting strong institutions under British colonization nor achieving victory in warfare. Large oil reserves in these countries have provided an exceptional source of revenue that fall outside the standard “development” process. Thus, there exist conditions that could plausibly lead to development that do not require either Mahoney’s NC #1 or NC #2.
Crucially, the argument that an exceptional source of revenue can engen-der high levels of overall wealth—regardless of other development impedi-ments faced by the country—is plausible for the countries Mahoney (2010) analyzes. In fact, at least until the 1980s, Venezuela provides within-region evidence of this assertion. Propelled by oil revenues, between 1920 and 1980 Venezuela was one of the fastest growing countries in Latin America (Hausmann, 2003, p. 245). This conclusion suggests Pr(high development | ¬ NC #1 AND ¬ NC #2) > 0, violating the “cannot” element of the necessary condition claim.
One way of revising the necessary condition hypothesis in response to considerations raised by these out-of-region cases would be to posit an addi-tional qualification. The argument of necessity may hold, provided that the countries of concern do not have some exceptional source of revenue. However, then the worry arises that this qualification is simply another path to the outcome—which immediately changes the scope conditions for the original argument. Importantly, STCM scholars are very careful in thinking about scope conditions, as exemplified in Chapter 16 of Goertz and Mahoney (2012). However, claims of necessity and sufficiency require scrutinizing additional scope conditions about hypothetical possibilities that are not observed among the cases under inquiry, a topic STCM scholars have not addressed.
Toward a More Demanding Standard? Combining the Approaches A further example, which focuses on physical impediments, suggests a pos-sible avenue for creating viable standards. Here, Freedman’s (2010) and Waldner’s (2005) criteria for necessary/sufficient condition hypotheses are combined, resulting in a more demanding standard.
Mann’s (1993) discussion of the effects of the Industrial Revolution on economic changes in Europe posits a necessary condition hypothesis with more compelling deterministic underpinnings than the examples considered above. He states that between 1760 and 1914 there was a “truly exponential at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 26 Comparative Political Studies transformation in the logistics of collective power” (Mann, 1993, p. 12, emphasis added). The claim implicitly engages hypothetical counterexam-ples by suggesting that the outcome could not occur without this transforma-tion. Mann (1993) additionally argues that transportation infrastructure, economic growth, and military capabilities exceeded “all known historical rhythms” (p. 13, emphasis added), a reference to empirical observations.
The hypothesis that industrialization was necessary for the changes Mann describes is supported empirically. Over the course of thousands of years, countless civilizations had existed without a single counterexample to the hypothesis. But what distinguishes the necessary condition claim as a com-pelling deterministic hypothesis—as opposed to a parallel probabilistic claim—is that physical constraints on human production made it nearly impossible for any earlier society to reach the per capita income levels achieved by several 19th-century European countries. Therefore, in the absence of the posited necessary condition, it is difficult to conceive of alter-nate paths to the outcome for countries prior to the 19th century. The combi-nation of empirical evidence and hypothetical considerations suggests relatively strong evidence for the deterministic hypothesis.
Are Necessary/Sufficient Condition Hypotheses Deterministic?
Whereas the conclusions drawn from the two preceding sections apply regardless of the ontological status of necessary/sufficient condition hypoth-eses, the considerations levied in this section apply specifically to determin-istic hypotheses. This is an important distinction because STCM scholars reject the standard position that necessity and sufficiency are inherently deterministic propositions (e.g., Collier, Brady, & Seawright, 2010, p. 145), and instead prefer the term asymmetric. As examples, Goertz (2005) argues it does not matter whether necessary condition hypotheses are treated as deter-ministic or probabilistic, and Ragin frequently appeals to notions of “almost always” sufficiency (e.g., Ragin 2008, p. 49). Thus, it is important to address why STCM should be considered a deterministic method. Although STCM scholars have raised valuable points, they have yet to convincingly address critiques levied by scholars from diverse backgrounds.
Braumoeller and Goertz (2000) have articulated the clearest perspective on why it may still be useful to think in terms of necessity and sufficiency when the data are not perfectly consistent with a deterministic hypothesis, by appealing to measurement error. However, their defense does not invalidate the concerns presented in this section. Even if a data set contains empirical counterexamples and the researcher imposes the permissive assumption that all the deviant cases arose from measurement error, it is still useful to assess at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 27 the likelihood of observing the non-deviant observations under different probabilistic DGPs. Furthermore, the theoretical basis of the “must” or “can-not” claims should still be carefully scrutinized.
The standards presented here would need to be altered to accommodate determinism-with-exception positions that do not rely on measurement error. However, in their current form, alternative determinism-with-exception argu-ments face at least three main shortcomings. First, if we strip the “cannot” and “must” components of necessary/sufficient condition hypotheses, Waldner (2005) argues it is not clear what value-added these concepts retain. Any thresholds used for assessing “necessary enough” or “sufficient enough” will be arbitrary, and it is not entirely clear how to distinguish such claims from ones that would be made using quantitative techniques: If stating that P = .99 permits us to claim that the relationship is “extremely likely,” while P = .51 prompts us to claim that X is “necessary more often than not,” then there is no non-arbitrary reason to prohibit me from claiming that when P = .25, X is “hardly ever necessary” for Y and when P = .01, X is “virtually never necessary” for Y. All of these would count as necessary condition hypotheses. (Waldner, 2005, p. 28) Freedman (2010) makes a similar point in reference to whether outcomes can occur in the absence of almost necessary conditions: “‘Impossibility’ might just mean that the likelihood is below the cutpoint of 0.5 . . . Selecting cut-points is another famous problem” (p. 108).
Second, if STCM scholars want necessary/sufficient condition hypotheses to be evaluated as probabilistic, this choice requires explicitly modeling a stochastic component (Sekhon, 2005). None of the models presented in Ragin (2008), Goertz and Mahoney (2012), or Schneider and Wagemann (2012) contain a stochastic component. This implies the assumed causal structure is deterministic and renders as ambiguous the exact interpretation of “almost always” necessity or sufficiency.
Third, those who reject the standards presented in this section because they are only appropriate for deterministic hypotheses must replace them with viable alternatives. Without further steps, one cannot accept a claim that the DGP “almost always” produces Y = 1 when X = 1 is present—which entails an inference about the causal process—any more readily than one can accept that X increases the probability of Y simply by observing a positive regression coefficient linking the two. Expanding on the hypothetical urn examples from above, perhaps an almost-always sufficiency claim entails Pr(outcome | posited “almost always” SC is present) ≥ .9. What assumptions should we make to infer this process generated the data? Should we retain a at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 28 Comparative Political Studies weaker version of the “must” claim from a sufficiency hypothesis—which, again, is implied directly by the definition of a sufficient condition? Alternatively, a claim of “almost always” sufficiency could imply Pr(outcome | posited “almost always” SC is present) = 1 for 90% of the cases analyzed but not the other 10%. If this is indeed what the analyst means, then all the stan-dards proposed above apply fully for 90% of the cases—plus the additional requirement for the researcher to characterize which cases fit the scope con-dition for the sufficiency hypothesis.
Inferring any type of causal relationship is difficult with any method, and certainly none of this discussion presumes agreement with prominent quanti-tative scholars who argue that deterministic arguments are inherently uninter-esting or untestable (Clark, Gilligan, & Golder, 2006; Sekhon, 2004). However, a skeptical note is warranted. Without systematically employing demanding standards—perhaps similar to the ones discussed here or perhaps others—there is no reason to believe that associational patterns consistent with necessity or sufficiency allow us to infer the DGP is indeed determinis-tic. The stringent association-causation critiques levied on conventional quantitative methods cannot be dismissed in STCM. Expounding the simi-larities between STCM and quantitative methods clearly demonstrates the shared challenges for making compelling causal claims. This reveals vital considerations for future STCM research.
Conclusion Set-theoretic comparative methods (STCM) have crafted “a language for all those scholars that do not feel at ease with applying statistical principles and practices to their research” (Schneider & Wagemann, 2013a, p. 5). This arti-cle argues that while the language may well be distinctive, the research pro-cedures are more similar than has been recognized. In many regards, STCM share common foundations with quantitative research. Comprehending these areas of overlap not only helps scholars from different research traditions to better understand each other’s research but also highlights shared inferential challenges faced by both STCM and quantitative methods.
The arguments of shared foundations offered here raise three questions for future research. First, how strong are the common foundations in non-binary versions of STCM? Second, in addressing the shared goal of analyzing inter-actions and causal complexity, what can be learned by using tools employed in recent quantitative work? Third, to what extent is STCM an advance over conventional qualitative methods?
First, this article focuses on the binary version of STCM. Although it will be valuable for future work to evaluate possible extensions of shared at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 29 foundations to non-binary (fuzzy-set and multi-value) versions of STCM, it is important to point out why this discussion has focused on the binary ver-sion. This has been done (a) in the interest of expositional clarity; (b) because researchers outside the STCM tradition routinely conceive of necessary/suf-ficient conditions as inherently dichotomous concepts, implying the binary version commands wide interest; and (c) STCM scholars commonly focus on the 2 × 2 table to make their crucial argument that standard statistical ideas of correlation or association do not apply to STCM.
To the extent that divergences between STCM and quantitative methods emerge outside the binary setting, it will be important to (a) understand the circumstances in which STCM assumptions about taking the maximum or minimum of fuzzy sets should be preferred over alternative assumptions and (b) for STCM to address recent critiques of aggregating fuzzy sets to evaluate necessity and sufficiency (Braumoeller, 2013; Dunning, 2013).
Second, with regard to causal complexity, recent work on quantitative methods has devoted considerable attention to this topic. Kam and Franzese (2007) carefully review discussions that date back to the 1970s, underscoring obstacles to analyzing interactions and clarifying standards for applied work. Imai, Keele, Tingley, and Yamamoto (2011) and Glynn (2012) focus on causal mediation effects, and Hainmueller and Hazlett (2014) present a least squares method that relaxes linearity and additivity assumptions. Perhaps most directly related to STCM, Grimmer, Messing, and Westwood (2014) discuss machine learning techniques for analyzing complex relationships. Especially considering that many recent simulation studies demonstrate QCA algorithmic results are unreliable under various minor perturbations (Hug, 2013; Krogslund, Choi, & Poertner, 2014; Lucas & Szatrowski, 2014; Seawright, 2014b), it will be valuable for STCM scholars to address how their techniques relate to these recent quantitative advances.
Third, the argument that STCM is an improvement over traditional quali-tative methods is questionable. Work such as Goertz and Mahoney (2012) claims that because qualitative researchers routinely state hypotheses that either explicitly or implicitly reference necessary and sufficient conditions, most qualitative research should be recast in set-theoretic terms. But, as shown above, the ideas of subsets and supersets that are so important for STCM work can lead to misleading conclusions about associational relation-ships when the distribution of cases is skewed. Furthermore, STCM has not convincingly addressed the recent rethinking of causal inference that is so fundamental to the quantitative field. It is against the backdrop of this deficit that studies such as Mahoney (2010) stand out for meticulously using process tracing and structured case comparisons, which are staples of traditional qualitative methods. In this regard, new work on combining STCM with at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 30 Comparative Political Studies process tracing (Rohlfing & Schneider, 2013; Schneider & Rohlfing, 2013) also has great merit. Yet in these valuable studies, the distinctive contribution of the STCM component—as opposed to the contribution of traditional quali-tative tools—still needs to be demonstrated.
In sum, future research on comparative methods will be well-served by (a) exploring the nature and scope of potential shared foundations of the non-binary versions of STCM and conventional quantitative methods; (b) exam-ining the implications for STCM of new quantitative work on complex relationships; and (c) parsing out—in new work by STCM scholars that seeks to incorporate traditional qualitative tools—the actual value-added of the STCM component.
Appendix A Assessing Sufficient Condition Triviality With Skewed Cases The section on the first core similarity focused on problems with the STCM necessary condition consistency metric. The STCM procedure for evaluating the triviality of a sufficient condition hypothesis faces a similar limitation. Consider the hypothetical example in Table A1.
In this 2 × 2 table, the STCM consistency score for F = 1 as a sufficient condition for A = 1 is 1. This is a very sensible conclusion about an associa-tional pattern based on the data at hand—although, as discussed in the section on shared challenges, additional steps remain to achieve a compelling infer-ence about the fundamental “must” claim.
But the STCM sufficient condition coverage score (identical to the STCM necessary condition consistency score presented in Equation 5) will lead to the misleading conclusion that F = 1 is not a trivial sufficient condition for A = 1. As can be seen by examining the F = 0 cases, every F = 0 case also achieves A = 1. But because conventional STCM metrics only incorporate two cells at a time and therefore do not consider imbalance among F = 0 and Table A1. Hypothetical Example of Assessing Sufficient Condition Triviality With Skewed Cases.
F = 0 F = 1 A = 1 2 20 A = 0 0 0 F = fuel wealth; A = authoritarian rule. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 31 F = 1 cases, the STCM sufficient condition coverage score equals 20 / 22 = 91%. In contrast, Equation 6 corrects this problem and reports a coverage score of 1 / (1 + 1) = 50%. This revised calculation accurately reports that F = 1 is a trivial sufficient condition for A = 1.
Appendix B An Example of Studying Complex and Equifinal Relationships With Regression Extending the discussion from the section on the second core similarity, the following numerical example features equifinality to further clarify how regression can be used to study complex relationships. An equifinal causal process implies that for at least one outcome, there is more than one grouping of conditions that has a high sufficient condition consistency score, which I refer to below as “paths.” In other words, there are multiple paths to certain outcomes.
Levitsky and Way (2010) examine the causes of stable authoritarianism (A) among competitive authoritarian regimes. They focus on three condi-tions: the strength of Western linkage (W), the level of organizational power (O), and the strength of Western leverage (L). They claim there are two paths to stable authoritarianism: (W, O) = (0, 1) and (W, L) = (0, 0). They also claim there are two paths to the absence of stable authoritarianism: W = 1 and (O, L) = (0, 1). I represent their claims with hypothetical data, assuming that cases with conditions representing a particular path achieve that outcome 90% of the time. Table B1 presents the coefficient estimates that result from estimating a fully saturated regression model using ordinary least squares (OLS).
The following discussion explains how each of the coefficients are esti-mated and shows how using information from the regression table accu-rately captures a country’s pathway—that is, its probability of stable authoritarianism—after particular hypothetical interventions. As above, I assume causal homogeneity.
Constant. Because (W, L) = (0, 0) is a pathway to authoritarianism, low levels of all three conditions lead to A = 1 in 90% of cases. Thus, the constant term β0 = .9.
Linkage. The hypothetical intervention in which a country that originally lacked all three conditions gains high Western linkage changes its path from stable authoritarianism to unstable authoritarianism. High Western linkage at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 32 Comparative Political Studies (regardless of the value of other conditions) is a path to unstable authoritari-anism. Hence, this intervention lowers the probability of stable authoritarian-ism by .8 and β0 + βW = .1.
Organizational power. The hypothetical intervention in which a country that originally lacked all three conditions gains high organizational power does not change the expected outcome. Both with and without the intervention, the country is on a path to stable authoritarianism: β0 + βO = .9.
Leverage. The hypothetical intervention in which a country that originally lacked all three conditions gains high leverage changes its path from stable authoritarianism to unstable authoritarianism. High leverage is a path to unstable authoritarianism when organizational power is low: β0 + βL = .1.
Linkage × Organizational power. The hypothetical intervention in which a country that originally lacked all three conditions simultaneously gains both high Western linkage and high organizational power has the same effect as the hypothetical intervention in which a country that originally lacked all three conditions gains only high Western linkage. High Western linkage (regardless of the value of other conditions) is a path to non-stable authori-tarianism: β0 + βW + βO + βWO = .1.
Linkage × Leverage. The hypothetical intervention in which a country that originally lacked all three conditions simultaneously gains both high Western linkage and high leverage has the same effect as either individual interven-tion. As discussed above, either individual intervention switches a country Table B1. Regression Table for Hypothetical Representation of Levitsky and Way (2010).
Dependent variable is authoritarian stability Constant (β0) .9 Linkage (βW) −.8 Organizational power (βO) 0 Leverage (βL) −.8 Linkage × Organizational power (βWO) 0 Linkage × Leverage (βWL) .8 Organizational power × Leverage (βOL) .8 Linkage × Organizational power × Leverage (βWOL) −.8 at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 33 that originally lacked all three conditions from a stable authoritarianism path to an unstable authoritarianism path. Thus, βWL is positive to counteract that both βW and βL are negative: β0 + βW + βL + βWL = .1.
Organizational power × Leverage. The hypothetical intervention of high orga-nizational power counteracts the negative impact of the high leverage inter-vention. As discussed above, manipulating only leverage for a country that originally lacked all three conditions would switch its path from stable authoritarianism to unstable authoritarianism. However, additionally manip-ulating organizational power means the country remains on a stable authori-tarianism path. Hence, βOL is positive to counteract the negative βL term. β0 + βO + βL + βOL = .9.
Linkage × Organizational power × Leverage. The hypothetical intervention in which a country that originally lacked all three conditions gains all three con-ditions has the same effect as the hypothetical intervention in which a country that originally lacked all three conditions gains only high Western linkage. High Western linkage (regardless of the value of other conditions) is a path to non-stable authoritarianism: β0 + βW + βO + βL + βWO + βWL + βOL + βWOL = .1.
Acknowledgment The author thanks David Collier for tireless assistance and feedback. The article has also benefited greatly from discussions with and comments by Danny Choi, Kevin Clarke, Ruth Berins Collier, Thad Dunning, Zachary Elkins, Kenneth Greene, Simon Hug, Chris Krogslund, Marcus Kurtz, Sebastian Mazzuca, Katherine Michel, Gerardo Munck, Mathias Poertner, Roxanna Ramzipoor, Ingo Rohlfing, Jason Seawright, Sean Tanner, Guadalupe Tuñón, Kim Twist, Alison Varney, four anonymous review-ers, and editor Ben Ansell.
Author’s Note Any remaining mistakes are the sole responsibility of the author.
Declaration of Conflicting Interests The author declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.
Funding The author received no financial support for the research, authorship, and/or publica-tion of this article. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 34 Comparative Political Studies Notes 1. It is rare to find either STCM or quantitative articles that solely evaluate the relationship between a single binary condition and a binary outcome. However, STCM texts frequently use this setting to argue for fundamental differences. To directly compare arguments, I also examine a 2 × 2 setting in the analysis below. 2. The presence of deviant cases are not sufficient to eliminate the necessary condi-tion hypothesis if one accepts STCM arguments that measurement error must be taken into account when analyzing necessary and sufficient relationships among social scientific data. 3. A complex relationship is one in which multiple conditions must be combined to be either necessary or sufficient. Mahoney (2008) provides a useful discus-sion of INUS conditions (“insufficient but necessary part of a condition which is itself unnecessary but sufficient for the result,” p. 418) and SUIN conditions (“sufficient but unnecessary part of a factor that is insufficient but necessary for an outcome,” p. 419), the core concepts behind complex necessary/sufficient conditions. 4. This is not true if one accepts a single counterexample to invalidate a claim of necessity of sufficiency. In that case, only a single cell is relevant for the hypoth-esis. However, as stated above, STCM scholars frequently analyze data sets with one or more counterexamples, to which the following argument applies. 5. Of course, researchers rarely confront a “fixed” set of cases. There is consider-able scholarship in political science on how researchers should select cases to collect data for, which the present analysis does not contribute to. The argument here focuses on the next step in the research process and shows why—assum-ing a researcher already has data for all four cells—researchers should indeed examine all four cells from this fixed set of data. This contrasts Ragin’s (2008) and Goertz and Mahoney’s (2012) arguments about numerous examples of 2 × 2 tables they present: Scholars should ignore certain cells from their fixed set of cases. 6. The conventional necessary condition coverage score is identical to the sufficient condition consistency score presented in Equation 1. Schneider and Wagemann (2012, p. 236) present an alternate triviality measure that I discuss below. 7. This proposal for an alternative measure to compute necessary condition con-sistency scores should be primarily viewed as illustrative for demonstrating the importance of considering all four cells. Hopefully this equation will be useful as a starting point in future work for determining the best way to incorporate the concerns raised in this section. 8. Comparing the example presented above of oxygen as a necessary condition for social revolution—which exemplifies a trivial necessary condition—to the Babe Ruth example reveals important differences. The distribution of X = 0 and X = 1 cells is not particularly important in the former because there are no cases without oxygen that experience social revolution. Instead, the problem is that the presence of oxygen offers almost no predictive power for whether a social revolution will occur. In the Babe Ruth example, the skewed distribution of X = at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from Paine 35 0 and X = 1 cases completely drives the misleading conclusion. In this example, the absence of the necessary condition predicts a home run with considerably higher probability than the presence of the condition—in contrast to the absence of oxygen. 9. Using the notation from Table 1, Schneider and Wagemann’s proposed necessity relevance measure equals ( ) / ( ) n n n n n fa fA fa fA Fa + + + . Juxtaposing this term with Equation 5 demonstrates that all four cells are used when evaluating both the consistency and triviality of a necessary condition hypothesis.
10. Of course, this interpretation of regression results requires that the conditions of interest either have been manipulated by the researcher or are at least hypotheti-cally manipulable. It is difficult to attach a causal interpretation to conditions that do not possess this crucial property (Gerring, 2012, p. 207-212)—regardless of whether one thinks in terms of average effects or non-trivial necessary/sufficient conditions.
11. As an example of how a complex condition can be studied using a 2 × 2 table, suppose the posited necessary/sufficient condition is [S = 0 AND F = 1]. Cases with this combination of conditions would be listed in one column of the table (with one cell for Y = 1 and one for Y = 0), whereas cases possessing the negation of [S = 0 AND F = 1] would compose the other column in the table.
12. Importantly, the discussion above assumes that all logically possible combina-tions of conditions are empirically observed. In STCM terms, this means that the data do not exhibit “limited diversity.” In contrast, when limited diversity is present, regression cannot compute a coefficient for each interactive and lower order term. Both STCM and regression are forced to use strong and unverifi-able assumptions to make inferences about the data when logically possible combinations of conditions are not empirically observed. Possibly an important avenue for future research would be to compare the assumptions imposed by each method and to assess circumstances in which one should be preferred over another. Schneider and Wagemann (2013b) provide a recent contribution that focuses on the STCM approach to handling limited diversity.
13. Collier (2014) provides a similar argument.
14. Equation 14 states the contrapositive of a necessary condition hypothesis in probability terms. It is logically equivalent to state the direct definition in prob-ability terms: Pr(X = 1 | Y = 1) = 1. However, it does not make sense to condition on Y when modeling a causal process because X must occur temporally prior to Y for X to be a cause of Y.
15. A similar concern applies to quantitative results: Even strong correlations should not be accepted as causal without a plausible research design or other supporting evidence for the hypothesis.
16. The only issue raised here about Mahoney’s (2010) contribution relates to his claims that he has identified and found strong evidence that a particular set of conditions is necessary and sufficient for his outcomes. He presents a nuanced historical framework accompanied by careful process tracing. As shown here, however, these commendable attributes of his research do not strongly support his claims of necessity or sufficiency. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from 36 Comparative Political Studies 17. To avoid confusion, note that Equation 13 states the data-generating process (DGP) if X = 1 is sufficient for Y = 1, whereas the Mahoney example focuses on whether X = 1 is sufficient for Y = 0. In addition, the “test” proposed in this section is intended to illustrate the core point about the importance of probabi-listic alternatives, as opposed to suggesting that this exact test should be used for assessing probabilistic alternatives. There are many possible ways to model a probabilistic DGP, and additional work is needed to scrutinize the most appropri-ate tests for assessing probabilistic alternatives.
18. Freedman (2010) argues empirical data will never produce strong evidence for a deterministic hypothesis. Using the term “population” instead of “data-generat-ing process” and referring to observed draws from the population as a “sample,” he asks what we can conclude about observations with trait U if the sample does not include any observations with this trait. “If the fraction of U’s in the sample is small, that proves U is rare in the population (modulo the usual qualifications). However, unless we make further assumptions, it is impossible to demonstrate by sampling theory that there are no U’s in the population” (Freedman, 2010, pp. 110-111).
19. See Good (1976, p. 605) for evidence on colonial economic development, and Compagnon (2011, ch. 6) for evidence on post-independence institutional dismantling.
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Author Biography Jack Paine is a Ph.D. Candidate in the Department of Political Science at the University of California, Berkeley. His research focuses primarily on applying game theoretic models to study civil war and authoritarian politics. He is particularly inter-ested in the effects of oil wealth and of pre-colonial/colonial legacies. at UNIV CALIFORNIA BERKELEY LIB on January 19, 2015 cps.sagepub.com Downloaded from |
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Square root property
The square root property is a method for solving quadratic equations.
Method
First we must convert the quadratic to either the form or .
First format
take the square root of both sides
so the roots are and .
Second format
take the square root of both sides
isolate
and those are the roots.
Retrieved from "
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1098 | https://math.stackexchange.com/questions/1170255/every-subgroup-of-mathbb-z-n-is-closed-under-multiplication | abstract algebra - Every subgroup of $(\mathbb {Z_n},+)$ is closed under multiplication - Mathematics Stack Exchange
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Every subgroup of (Z n,+)(Z n,+) is closed under multiplication
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I am stuck in this proof that every subgroup of (Z n,+)(Z n,+) is also a subring. which requires me to prove it is closed under multiplication. I have to show if a,b∈G<Z n a,b∈G<Z n , then a b∈G<Z n a b∈G<Z n
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asked Mar 1, 2015 at 12:14
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I suggest starting with describing what the subgroups of Z n Z n are. This problem will be much easier if you describe what G G looks like.Michael Burr –Michael Burr 2015-03-01 12:18:23 +00:00 Commented Mar 1, 2015 at 12:18
2 Integer multiplication can be modeled as repeated addition, i.e. 5a = a+a+a+a+a.John McGee –John McGee 2015-03-01 12:18:24 +00:00 Commented Mar 1, 2015 at 12:18
Getting stuck is not too hard - what did you try? I would suggest induction on b b, and prove that G G is closed under multiplication by every natural number.Carl Mummert –Carl Mummert 2015-03-01 12:18:28 +00:00 Commented Mar 1, 2015 at 12:18
@JohnMcGee thanks. So a b∈G a b∈G because a b=b+b+...a t i m e s∈G a b=b+b+...a t i m e s∈G Bosnia –Bosnia 2015-03-01 12:22:29 +00:00 Commented Mar 1, 2015 at 12:22
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(Z n,+)(Z n,+) is a cyclic group, and any subgroup G G is likewise cyclic. This means, in particular, that if a,b∈G a,b∈G, that:
a=k g=g+g+⋯+g(k a=k g=g+g+⋯+g(k times)
b=m g=g+g+⋯+g(m b=m g=g+g+⋯+g(m times)
for some integers k,m k,m and a generator g g of G G.
Now compute a b a b (mod n n) -can you think of a way to somehow use the distributive law for integers, and then reduce mod n n?
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answered Mar 1, 2015 at 14:54
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a b=(k m)g a b=(k m)g, correct?Bosnia –Bosnia 2015-03-01 15:24:54 +00:00 Commented Mar 1, 2015 at 15:24
Yes, but you should be able to prove this (induction will work, but arguing from integers and reducing mod n n is "faster").David Wheeler –David Wheeler 2015-03-01 15:43:38 +00:00 Commented Mar 1, 2015 at 15:43
is g=g c d(a,b)?g=g c d(a,b)?abel –abel 2015-03-15 15:49:08 +00:00 Commented Mar 15, 2015 at 15:49
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If you know about rings, then that should be easy to prove:
Every additive subgroup of Z Z is in fact an ideal.
The additive subgroups of Z/n Z/n correspond to the additive subgroup of Z Z that contain n Z n Z.
So, every additive subgroup of Z/n Z/n is in fact an ideal, and hence closed under multiplication.
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answered Mar 1, 2015 at 12:24
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1099 | https://math.stackexchange.com/questions/502855/evaluate-the-limits-that-exist-without-lhopitals-rule | calculus - Evaluate the Limits that Exist without L'Hopital's Rule - Mathematics Stack Exchange
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Evaluate the Limits that Exist without L'Hopital's Rule
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I have four limit problems, for homework, but I don't quite understand them. I will only ask for help on one of them because that should be all I need to understand these. According to the Calculus professor, 'I do not know L'Hopital's Rule, yet.' Therefore, I may not use it L'Hopital's Rule. We have went as far as to understand lim x→0 lim x→0 sin x x=1 sinx x=1.
The problem is:
lim x→0 x 2−x sin 3 x lim x→0 x 2−x sin3 x
Thank you, for your help.
-Rux
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edited Sep 23, 2013 at 22:00
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What else do you know as far as the class is concerned? Are you familiar with the sum expansion of sin x sinx? How many terms would be required to solve this problem?abiessu –abiessu 2013-09-23 21:35:12 +00:00 Commented Sep 23, 2013 at 21:35
take out x x-factor common from numerator, and use this to show that the limit is −1/3−1/3 S L –S L 2013-09-23 21:37:38 +00:00 Commented Sep 23, 2013 at 21:37
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Hint:
x 2−x sin 3 x=x(x−1)sin 3 x=3 x sin 3 x⋅x−1 3 x 2−x sin3 x=x(x−1)sin3 x=3 x sin3 x⋅x−1 3
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answered Sep 23, 2013 at 21:38
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Okay, I understand this. Thank you, for the help. I can get it from here.JRuxDev –JRuxDev 2013-09-23 21:51:02 +00:00 Commented Sep 23, 2013 at 21:51
You're welcome.mrf –mrf 2013-09-23 21:55:19 +00:00 Commented Sep 23, 2013 at 21:55
Just to continue with this answer, you factor our the x x from the top in the first step. Then, you can multiply the top and bottom by 3 and split that up into the final equation you have listed. I see that 3 x sin 3 x 3 x sin3 x is equivalent to sin x x sinx x which is 1. So the next step would be x x-1 3 1 3? And, so x=0 x=0, since the limit is lim x→0 lim x→0. 0−1 3 0−1 3 = −1 3−1 3?JRuxDev –JRuxDev 2013-09-23 22:18:23 +00:00 Commented Sep 23, 2013 at 22:18
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Hint: Divide the numerator and denominator by x x
lim x→0 x 2−x x sin 3 x x=lim x→0 x−1 sin 3 x x lim x→0 x 2−x x sin3 x x=lim x→0 x−1 sin3 x x
Can you use any formula for limits in the denominator now?
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edited Sep 23, 2013 at 21:46
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answered Sep 23, 2013 at 21:38
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Hint:
lim x→0 sin x x=…lim x→0 sinx x=…
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edited Sep 23, 2013 at 21:46
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answered Sep 23, 2013 at 21:39
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563How to prove that lim x→0 sin x x=1 lim x→0 sinx x=1?
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